**23 people**. In a room of just 23 people there’s a 50-50 chance of at least two people having the same birthday. In a room of 75 there’s a 99.9% chance of at least two people matching.

Put down the calculator and pitchfork, I don’t speak heresy. The birthday paradox is strange, counter-intuitive, and **completely true**. It’s only a “paradox” because our brains can’t handle the compounding power of exponents. We expect probabilities to be linear and only consider the scenarios we’re involved in (both faulty assumptions, by the way).

Let’s see why the paradox happens and how it works.

## Problem 1: Exponents aren’t intuitive

We’ve taught ourselves mathematics and statistics, but let’s not kid ourselves: it’s not natural.

Here’s an example: What’s the chance of getting 10 heads in a row when flipping coins? The untrained brain might think like this:

“Well, getting one head is a 50% chance. Getting two heads is twice as hard, so a 25% chance. Getting **ten** heads is probably 10 times harder… so about 50%/10 or a 5% chance.”

And there we sit, smug as a bug on a rug. No dice bub.

**After pounding your head with statistics**, you know not to divide, but use **exponents**. The chance of 10 heads is not .5/10 but .5^{10}, or about .001.

But even after training, we get caught again. At 5% interest we’ll double our money in 14 years, rather than the “expected” 20. Did you naturally infer the Rule of 72 when learning about interest rates? Probably not. Understanding compound exponential growth with our linear brains is hard.

## Problem 2: Humans are a tad bit selfish

Take a look at the news. Notice how much of the negative news is the result of acting without considering others. I’m an optimist and *do* have hope for mankind, but that’s a separate discussion :).

In a room of 23, do you think of the 22 comparisons where **your** birthday is being compared against someone else’s? Probably.

Do you think of the **231** comparisons where someone who is not you is being checked against someone else who is not you? Do you realize there are so many? Probably not.

The fact that we neglect the **10 times as many** comparisons that don’t include us helps us see why the “paradox” can happen.

## Ok, fine, humans are awful: Show me the math!

The question: What are the chances that two people share a birthday in a group of 23?

Sure, we could list the pairs and count all the ways they could match. But that’s hard: there could be 1, 2, 3 or even 23 matches!

It’s like asking “What’s the chance of getting one or more heads in 23 coin flips?” There are so many possibilities: heads on the first throw, or the 3rd, or the last, or the 1st and 3rd, the 2nd and 21st, and so on.

How do we solve the coin problem? Flip it around (Get it? Get it?). Rather than counting every way to get heads, **find the chance of getting all tails, our “problem scenario”**.

If there’s a 1% chance of getting all tails (more like .5^23 but work with me here), there’s a 99% chance of having **at least one head**. I don’t know if it’s 1 head, or 2, or 15 or 23: we got heads, and that’s what matters. If we subtract the chance of a problem scenario from 1 we are left with the probability of a good scenario.

The same principle applies for birthdays. Instead of finding all the ways we match, **find the chance that everyone is different, the “problem scenario”**. We then take the opposite probability and get the chance of a match. It may be 1 match, or 2, or 20, but somebody matched, which is what we need to find.

## Explanation: Counting Pairs (Approximate Formula)

With 23 people we have 253 pairs:

(Brush up on combinations and permutations if you like).

The chance of 2 people having different birthdays is:

Makes sense, right? When comparing one person's birthday to another, in 364 out of 365 scenarios they won't match. Fine.

But making **253 comparisons** and having them *all* be different is like getting heads 253 times in a row -- you had to dodge "tails" each time. Let's get an approximate solution by pretending birthday comparisons are like coin flips. (See Appendix A for the exact calculation.)

We use exponents to find the probability:

Our chance of getting a single miss is pretty high (99.7260%), but when you take that chance hundreds of times, the odds of keeping up that streak drop. Fast.

The chance we find a match is: 1 – 49.95% = 50.05%, or just over half! If you want to find the probability of a match for any number of people n the formula is:

## Interactive Example

I didn’t believe we needed only 23 people. The math works out, but is it real?

You bet. Try the example below: Pick a number of items (365), a number of people (23) and run a few trials. You’ll see the theoretical match and your actual match as you run your trials. Go ahead, click the button (or see the full page).

As you run more and more trials (keep clicking!) the actual probability should approach the theoretical one.

## Examples and Takeaways

Here are a few lessons from the birthday paradox:

**$\sqrt{n}$**is roughly the number you need to have a 50% chance of a match with n items. $\sqrt{365}$ is about 20. This comes into play in cryptography for the birthday attack.- Even though there are 2
^{128}(1e38) GUIDs, we only have 2^{64}(1e19) to use up before a 50% chance of collision. And 50% is really, really high. - You only need 13 people picking letters of the alphabet to have 95% chance of a match. Try it above (people = 13, items = 26).
- Exponential growth rapidly decreases the chance of picking unique items (aka it increases the chances of a match). Remember: exponents are non-intuitive and humans are selfish!

After thinking about it a lot, the birthday paradox finally clicks with me. But I still check out the interactive example just to make sure.

## Appendix A: Repeated Multiplication Explanation (Exact Formula)

Remember how we assumed birthdays are independent? Well, they aren’t.

If Person A and Person B match, and Person B and C match, we know that A and C must match also. The outcome of matching A and C depends on their results with B, so the probabilities aren’t independent. (If truly independent, A and C would have a 1/365 chance of matching, but we know it's a 100% guaranteed match.)

When counting pairs, we treated birthday matches like coin flips, multiplying the same probability over and over. This assumption isn’t strictly true but it’s good enough for a small number of people (23) compared to the sample size (365). It’s unlikely to have multiple people match and screw up the independence, so it’s a good approximation.

It’s unlikely, but it can happen. Let’s figure out the real chances of each person picking a different number:

- The first person has a 100% chance of a unique number (of course)
- The second has a (1 – 1/365) chance (all but 1 number from the 365)
- The third has a (1 – 2/365) chance (all but 2 numbers)
- The 23rd has a (1 – 22/365) (all but 22 numbers)

The multiplication looks pretty ugly:

But there’s a shortcut we can take. When x is close to 0, a coarse first-order Taylor approximation for $e^x$ is:

so

Using our handy shortcut we can rewrite the big equation to:

But we remember that adding the numbers 1 to n = n(n + 1)/2. Don’t confuse this with n(n-1)/2, which is C(n,2) or the number of pairs of n items. They look almost the same!

Adding 1 to 22 is (22 * 23)/2 so we get:

Phew. This approximation is very close, plug in your own numbers below:

Good enough for government work, as they say. If you simplify the formula a bit and swap in *n* for 23 you get:

and

With the exact formula, 366 people has a guaranteed collision: we multiply by $1 - 365/365 = 0$, which eliminates $p(\text{different})$ and makes $p(\text{match}) = 1$. With the approximation formula, 366 has a near-guarantee, but is not exactly 1: $1 - e^{-365^2 / (2 \cdot 365)} \approx 1$ .

## Appendix B: The General Birthday Formula

Let’s generalize the formula to picking *n* people from *T* total items (instead of 365):

If we choose a probability (like 50% chance of a match) and solve for *n*:

Voila! If you take $\sqrt{T}$ items (17% more if you want to be picky) then you have about a 50-50 chance of getting a match. If you plug in other numbers you can solve for other probabilities:

Remember that m is the *desired chance of a match* (it’s easy to get confused, I did it myself). If you want a 90% chance of matching birthdays, plug m=90% and T=365 into the equation and see that you need 41 people.

Wikipedia has even more details to satisfy your inner nerd. Go forth and enjoy.

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Herman HiddemaApril 26, 2007 at 2:44 amThe math here is actually wrong. The chances of individual pairs are not independent. You math would work if you take each pair and have them name a random number between 1 and 365.

With this math, taking a group of 365 people still results in a non-zero chance that they all have different birthdays.

Aleksandar ProkopecJuly 16, 2017 at 7:43 am+1 exactly what I thought when I was reading this. And another +1 for the pidgeonhole counterexample ;)

Aleksandar ProkopecJuly 16, 2017 at 7:45 am(however, there’s an appendix talking about that)

AnonNovember 4, 2017 at 7:12 amIf there is a group of 365 people there would be a non-zero chance they have different birthdays. They could each have a birthday on each different day of the year. You would need 366 people (ignoring feb29) for it to be guaranteed that there’s a pair with the same birthday

KalidApril 26, 2007 at 4:15 amThanks for the info, you’re right. I did some more digging (good paper here) and birthdays aren’t mutually independent.

If Person 1 = Person 3, and Person 3 = Person 5, there isn’t an independent event that Person 1 = Person 5. The probability of 1 matching 5 has already been determined by the other statements.

From what I was able to gather, this is only a problem if there are existing overlapping pairs. For a small n relative to the number of outcomes (365), it’s unlikely to have multiple matches that affect the probability, so assuming independence may be ok for computing approximations.

Too many topics, too little time. » Understanding the Birthday Paradox | BetterExplainedApril 26, 2007 at 4:56 am[…] Understanding the Birthday Paradox | BetterExplained: Understanding the Birthday Paradox […]

Carnival of Mathematics Edition #6 at nOnoscienceMay 4, 2007 at 10:56 am[…] Regarding your birthday, whether you are savvy with Hamming’s error correcting code or not, listen to Kalid Azad when he presents Understanding the Birthday Paradox posted at BetterExplained in which he explains the Birthday Paradox from statistics. […]

Techniques for adding the numbers 1 to 100 | BetterExplainedMay 7, 2007 at 4:30 pm[…] Understanding the Birthday ParadoxUnderstanding the Pareto Principle (The 80/20 Rule)Law of Unintended ConsequencesSpeed Up Your Javascript, Part 2: Downloadable Examples!Number Systems and Bases Like it? Share on: […]

AnonymousMay 9, 2007 at 2:09 pmThe last formula is incorrect, it should be:

n ~ sqt(-2 ln(1-p)) sqt(T)

^^^

or else you are finding the probability to miss.

KalidMay 9, 2007 at 2:55 pmThanks for the tip! I fixed up the article to use p(different) and p(match), which is much more clear.

PseudonymMay 13, 2007 at 9:43 pmThe “take-away lesson” about GUIDs is wrong. GUIDs are (theoretically) guaranteed to be globally unique, because they include such things as the MAC address of your network card (something which is globally unique until some cheap NIC manufacturer starts recycling them) and the current time.

The catch is that because of the time factor, the current GUID algorithm won’t last forever. We will run out in a couple of centuries.

KalidMay 14, 2007 at 9:08 pmHi, that’s a good point about MAC addresses. However, if you consider GUIDs as just a giant random number (for the purposes of the exercise), you are looking for how many “items” out of a pool of 2^128 you can distribute before having a 50% chance of collision.

For the birthday paradox, it’s about 23 items (of a pool of 365) before a 50% chance of collision. For GUIDs, it will be roughly 2^64 items before a 50% chance of collision.

There’s a bit more information here:

http://en.wikipedia.org/wiki/UUID

Hope this helps,

-Kalid

JalynJanuary 29, 2018 at 6:37 amhow are you going to site a wikipedia article… smh

AllanJune 18, 2007 at 6:58 amcan the math in the birthday paradox applicable to pick3 lottery?

KalidJune 22, 2007 at 12:10 amHi Allan, I’m not too familiar with the rules of Pick3, but I’ll take a shot.

The birthday paradox helps find the chance that any two random numbers will “collide” in a set.

In Pick3, you don’t really care if two guesses collide… you want the guess to collide with the winning number. In this case, two losing tickets that both guessed 123 (when the real answer was 999) isn’t helpful.

I may be missing something though!

n(t)July 22, 2007 at 2:52 pmHey, great blog.

“” A coarse first-order Taylor approximation for e^x is: \displaystyle{e^x \approx 1 + x}”

that’s just valid if x

n(t)July 22, 2007 at 2:53 pm[…] if x

n(t)July 22, 2007 at 2:54 pm[..] if x is far less than 1

Le blog d'Alex ChauvinJuly 24, 2007 at 2:50 amLe paradoxe des anniversaires…Je suis tombé par hasard (c’est souvent le cas sur Internet de nos jours) sur ce paradoxe des anniversaires qui stipule que dans une réunion regroupant 23 personnes, la probabilité que deux d’entre elles soient nées le même jour est……

Ashton CarrOctober 16, 2007 at 12:02 pmI am doin a science fair experiment on this i need help–and i need to know if the math is over my head??!!

KalidOctober 16, 2007 at 1:27 pm@nt: Thanks for the tip, I updated the article to make that more clear.

@Ashton: Hi Ashton, you might want to ask your math teacher to see if you’ve covered the necessary topics in class. You’ll probably need statistics and combinatorics.

zhaoNovember 27, 2007 at 4:27 pmhello kalid,

i read a few of your articles and think they are freaking awesome.

thanks and keep up the good work.

KalidNovember 27, 2007 at 7:30 pmHi Zhao, thanks for the comment! I’ll try to keep cranking out the posts :).

Pigeon Birthdays | If Chaos Were OrganizedDecember 10, 2007 at 8:59 am[…] After reading another math explanation on why that’s true, I know that I understand it now. Sure, I might not be able to repeat (or fully understand) the math equations which generate the percentage, but I can identify the bottom line of understanding — when written in POE (plain ol’ English): […]

demiDecember 11, 2007 at 7:23 amHeyy ;; i have no clue how to do this!

abcDecember 19, 2007 at 7:50 amI think that the math behind this birthday paradox is wrong..

The chance of two people having same birthdays is 1/365 = 0.0027397

therefore p(n)= 0.0027397 ^C(n,2)

if we take an example of 23 people

we get p(23)= 0.0027397 ^ 253 ~=0

so how is it possible??

KalidDecember 19, 2007 at 8:03 amHi, you’re correct 1/365 is the chance of 2 people having the same birthday. However, (1/365)^253 would be the chance of 253 people having the *same* birthday! (Which, as you see, is pretty close to zero).

For this problem, it’s important not to mix up 1/365 (the chance of 1 collision) and 364/365 (the chance of no collision). We first find the chance that somehow, everyone manages to be different:

p(23 people have different birthdays) = (364/365)^253

If there is a 40% chance that everyone is different, there is 1-40% = 60% chance that there was an overlap somewhere. Hope this helps. (Technically, we are assuming independent events but that subtlety is not important for the main point).

abcDecember 23, 2007 at 7:26 amhi,

(364/365)^253 means that 253 people have different birthdays

when you check this for 366 people , there is a >=100% chance for the birthday paradox.

but when you use this fomula we get the answer as 1 – 2.6 * 10^-80 which is less than 1

why is it so??

AND I have never seen two people having the same birthday in my group which has a greater strength than 23.this cannot be a coincidence!!!

I still doubt that there is a 50% chance of people having the same birthday

KalidDecember 23, 2007 at 7:35 amHi, when you make the probability like (364/365)^253, you are assuming independent events. What this means is that each comparison is “fresh”, with no memory of the past. It would be like having 2 people pick the same number out of 365, and choosing a different number each time.

This approximation makes the math easier, and is ok for small values. If you want the actual %, take a look at Appendix A.

Yep, the paradox seems strange, doesn’t it? Take a look at this page and run some experiments on your own to see:

http://betterexplained.com/examples/birthday/birthday.html

As you click “run trial”, you will see the actual match percentage for 23 people approach 50%, which is the predicted one. Hope this helps.

How to Develop a Sense of Scale | BetterExplainedJanuary 30, 2008 at 8:16 pm[…] A GUID, or large ID number used in programming, is at no risk of running out. How many are there? Well, we could give everyone a copy of the internet, every second, for a billion years… and still have enough GUIDs to identify each page. See how much bigger that is than “2^128″? (For the geeks: yes, the birthday paradox makes the chance of collision much higher). […]

AnonymousMarch 27, 2008 at 5:06 pmthe math for the birthday paradox is in fact quite simple, the “problem scenario” probability is in fact

364/365 times 363/365 times … times (364-22)/365

you should think like this.

-person one chose a day of the year as a birthday

-person two chose a day of the year as its birthday, BUT DIFFERENT than person one’s choice.

-so on

-person 23 does the same BUT DIFFERENT than previous people choices.

This is exactly what I wrote above in probabilities

AnonymousMarch 27, 2008 at 5:07 pmOh yeah,… sorry the last fraction is (365-22)/365

bye

KalidMarch 27, 2008 at 7:00 pmYep, that’s right. Sometimes that multiplication can be long to do out — see Appendix A for a shortcut.

BrittanyApril 15, 2008 at 3:59 pmThanks for this…im gonna use this as an idea for science fair!

Testing to see if the Birthday Paradox holds true.

23 in a room, 50% chance two will match!

Can’t wait!

KalidApril 15, 2008 at 6:36 pmSounds great Brittany! And if you have 75 people at your fair, you’re almost guaranteed to have a match :).

JonMay 30, 2008 at 6:17 amIt’s funny. There are actually two birthday paradoxes. The other comes from logic and is actually, actually, according to Quine, a veridical paradox, where it appears to be paradoxical, yet is proven true anyway, the fact that someone turns 7 when they are twenty-eight years old (born feb. 29), much like this birthday paradox.

What is interesting is that the two overlap. So to properly treat the birthday paradox (your version) you would have to take this into account.

So a very interesting treatment would be: what happens to the probability of sharing a birthday when you take into account feb 29, twins, triplets, etc, the fact (i believe) that there are higher frequencies of babies born during certain times of the year than others.

I might work this out, if asked, but I don’t think it would work out to 50% out of 23. It would be interesting to see how close it was though.

Carnival of Mathematics Edition #7 « Unruled NotebookNovember 1, 2008 at 10:41 pm[…] Regarding your birthday, whether you are savvy with Hamming’s error correcting code or not, listen to Kalid Azad when he presents Understanding the Birthday Paradox posted at BetterExplained in which he explains the Birthday Paradox from statistics. […]

PaulaJanuary 2, 2009 at 10:10 amthx 4 the info it was confusing but really good, im going 2 use this 4 my science fair project

Finance Blog » Blog Archive » The Birthday ParadoxJanuary 18, 2009 at 8:09 am[…] In class I used the example of "The Birthday Paradox" whilst discussing behavioural finance, and was reminded that I also raised the issue – without fully explaining it – during the Quantitative Methods course. To compensate, here's a list of explanations. The paradox comes from the unintuitive finding that the probability that any of 23 random people sharing a birthday is a whopping 50%. The problem is that when I select 23 people in class, there's only a 50-50 chance it comes off. In future I might select 30 students, since then there;s a 70% chance of a match, and I assume this would still be sufficiently surprising to demonstrate the point. The explanation that makes most sense to me is as follows: […]

Links of the Day - February 12, 2009 « stuffthought: Adi’s BlogFebruary 12, 2009 at 11:22 am[…] Better Explained is a website where this guy, Kalid Azad, explains things really well. Like, really well. So far, the three main catagories of explanations are “Math and Numbers,” “Programming and Web Development,” and “Business, Writing & Communication.” But, for example, check out his explanation of one of the more fascinating things that I learned back in Game Theory at CTY (oh good times…): the Birthday Paradox. Yes, if you’re in a room with 22 other people, the likelyhood of two people having the same birthday is just over 50%. Crazy, huh. […]

The Birthday Paradox « Math BlogMarch 14, 2009 at 8:19 pm[…] The Birthday Paradox The Birthday Paradox explained and a interactive example included on the page. Explore posts in the same categories: Paradox […]

AlbertApril 1, 2009 at 10:53 amDoes the dependency matter really at all?? I have just read it once, so maybe I don’t get it yet, but it seems you are just looking for at least 1 match?

50/50 chance of at least one match? If that is the case why would the dependency matter?

AlbertApril 1, 2009 at 12:02 pmIt seems since you are looking at each individual group at a time, that each event would be independent from the rest. Therefore looking at each group separately each group has a 1/365th possibility of matching?

hmm

I don’t know

LynneApril 1, 2009 at 9:37 pmI have a question: 6 people, one movie being advertized…3 people having the same birthday…and same birthday show on advertisement at the same time. What would the ‘chances’ be? This was an actual event.

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LynneMay 3, 2009 at 2:01 pmI asked a question on 4-1-09 about the ‘birthday paradox’ and an actual event. The reason I would like to know the ‘odds’ of that happening, is because it was one event out of four similar events. Any suggestions on where I can find some answer to the ‘odds’ other than here?

Better Explained « Xavier Seton’s BlogMay 7, 2009 at 12:36 am[…] Statistics: Combinations & permutations, Birthday Paradox, Bayes’ Theorem, […]

Birthdays and paradoxes | Polymath ProgrammerJune 22, 2009 at 2:01 am[…] Then I thought I should at least leave you with something. So I’ll introduce you to the concept of the birthday paradox. Basically, given a group of 23 people, what’s the percentage chance that any 2 persons in that group share the same birthday? (hint: it’s higher than you think. Here’s a less clinical explanation of the paradox than the one from Wikipedia.). […]

SomaJuly 2, 2009 at 7:52 pmHi Ashton, you might want to ask your math teacher to see if you’ve covered the necessary topics in class. You’ll probably need statistics and combinatorics.

tamekaJuly 30, 2009 at 2:25 pmall my brothers and sisters(not by both same parents)have the birthday of 2or 16

The Birthday Paradox | by Jeremia FroylandAugust 1, 2009 at 1:13 pm[…] The math gets somewhat complicated, but you can check it out in more detail: Understanding the Birthday Paradox and Wikipedia’s page. […]

SeanSeptember 11, 2009 at 1:06 pmThe explanations given are all approximations, in order to get an exact result you follow the start to Appendix A, but don’t attempt to simplify with e^x. The solution is actually fairly simple, for n possibilities (days in the year) and k events (people at the party) we get a probability of:

1 – (P(n-1,k-1)/(n^(k-1))).

Where P(n,k) is the number of ways to pick k elements from a set of n, or n!/(n-k)!.

This will give an exact solution, the probability of finding two people with the same birthday from a crowd of 23 is more accurately: 50.7297234%

I hope that this makes sense, if it doesn’t, look at the page on combinatorics and/or think about the fact that (1-(j/n)) = ((n-j)/n) with reference to Appendix A.

RoySeptember 13, 2009 at 12:20 pmAren’t there 366 possible birthdays? (feb 29)

angelinaFebruary 3, 2010 at 5:55 pmthis is really interesting!! im doing a math project on this !! nice topic!!

KalidFebruary 5, 2010 at 2:31 am@angelina: Awesome, glad you liked it!

keith hoyesFebruary 13, 2010 at 6:59 amthe way i intuitively see the 50% is like this,

imagine throwing 23 point blobs of paint at a calender on the wall. Then move all the dates that hit into a tidy ~5×5 square in the corner.

Now throw another 23 blobs of paint at the wall.

To me, it is almost inconceivable that no paint blobs will now touch the dates in the 23 blob square in the corner (50/50 maybe)

Carnival of Mathematics Edition #7 « Unruled NotebookFebruary 15, 2010 at 3:31 am[…] are savvy with Hamming’s error correcting code or not, listen to Kalid Azad when he presents Understanding the Birthday Paradox posted at BetterExplained in which he explains the Birthday Paradox from […]

AnonymousFebruary 16, 2010 at 11:55 amI got it after third heading….HAHAHAHAHAHA!!!

GeorgeMarch 17, 2010 at 8:32 amThank you.

Well detailed and nicely structured guide for a really misinterpreted problem.

Problemy z losowaniem | Moim subiektywnym okiemApril 1, 2010 at 2:11 pm[…] maksymalnej ilości kombinacji (bez dzielenia). Ta ilość prób wg ogólnego wzoru (wyprowadzenie tutaj w załączniku B) to tylko ok. 1,18 * sqrt(N) gdzie N oznacza ilość wszystkich możliwych […]

citizen428.blog()April 13, 2010 at 2:20 pmThe Birthday Paradox in Clojure and Incanter…The Birthday Paradox is an interesting little problem in probability theory. To quote Wikipedia:

“[…] the birthday problem, or birthday paradox pertains to the probability that in a set of randomly chosen people some pair of them will have t…

ArdyaApril 18, 2010 at 9:21 pmgreat explaination. its helping. i’m doing a math project about diehard randomness test…. can you help me to understand of the test that is called birthday spacing test?

MelissaMay 5, 2010 at 10:10 pmYour formula P(different) = e to the power of – (22*23/2*365) is incorrect. If you punc that in to a TI-83 plus, you’ll get zero as an answer. You need 22*23 in it’s own bracket and the same for the 2*365. The division sign would stay outside of the two brackets, in the middle. It should really look like this: P(different) = e to the power of – (22*23)/(2*365). It took my forever to understand what was wrong with the equations until I finally clicked ‘very close’ and say the other calculations. Haha. Hope this helps anyone who was as confused as I was. I thought my TI-83 was broken! (:

CaroMay 31, 2010 at 3:18 amUmm, I’m not a mathamatician… so please excuse me if this is a stupid comment. I understand the principle behind the calculations. I even agree that they are correct. However, one thing which seems to me to be incorrect is the assumption that birthdays are prefectly evenly distributed throughout the year. An equal weight (likelyhood) is being assigned to each day of the year. I think in reality that there are far more birthdays at certain times of the year, and therefore on certain days of the year (9 months after xmas, valentines day, etc.) I don’t see anywhere where this is being included into the calculations. Can you explain? Thanks.

KalidMay 31, 2010 at 10:32 pm@Caro: Great question! Yes, you are absolutely right — we currently assume that birthdays are distributed evenly. To simplify the problem, we ignore the possibility that birthdays could have a certain spread — realistically, it may be slightly more than 23 birthdays to account for this. But, I doubt the real distribution is very much different from the ideal one (certain holidays only celebrated in certain countries, etc.) so it might all average out reasonably. But that’s a great point to bring up.

hemant singh yadavJune 2, 2010 at 3:19 amfirst i got a bit confused

dat wat d helll is birthday paradox

but after reading dis

it is damn easy

thank U

very much

jOHHNJune 10, 2010 at 4:47 amIn our retirement village we have a birthday book, which contains about 80 names. The birthdays are read out each month to a gathering of about 25 people. If a certain person is there the same time as me,we have a match. Otherwise. NO

The correctorJune 18, 2010 at 3:43 amPlease stop confusing people. Let’s stop the confusion all over the world with this annoyingly wrong principal. I don’t mean computatively wrong. I mean, it is wrong to call it the birthday principal. It is a number principal with 365 set numbers principal.

This problem has been confusing people for the longest time, because no one will explain that it does not do what people think it is supposed to do. Which is calculate the odds that 365 people in a room will find someone else with their birthday.

The problem itself is actually very easy to understand. Even I can understand it and I never learned any advanced math. The equation is cheating. It has nothing to do with any applicable birthdays. There is no reason to delete each match after it is made.

The correctorJune 18, 2010 at 3:49 amThis is not a paradox. This is a simple math problem, and its title confuses people into thinking that something impossible is happening, when its not, they are just being confused by an incorrectly named title of a principal.

The Birthday Paradox in Clojure « citizen428.blog()August 13, 2010 at 8:44 am[…] don't really find this counter-intuitive, but as Kalid from BetterExplained accurately observed "exponents aren't intuitive" and "humans are a tad bit selfish". Here's a quote regarding that last […]

The Birthday Paradox | SlashmarksOctober 13, 2010 at 2:56 pm[…] math gets somewhat complicated, but you can check it out in more detail: Understanding the Birthday Paradox and Wikipedia’s […]

GavinDecember 14, 2010 at 3:01 pmOld thread, but still interesting. Here’s a simpler way of doing it – look at your Facebook birthdays, how many shared birthdays are there?

Indeed, export your friends birthdays, pick a sample of 23, and see if they match up – quite surprising!

KatMarch 2, 2011 at 12:33 pmLOL! At first I thought about my classroom, and instinct said how unlikely it was that two people had the same birthday, and then I realized we had a set of twins…

KalidMarch 2, 2011 at 2:37 pm@Kat: Hah, an even easier way to see it in action :)

Roosevelt Grier: The Life of a Gentle Giant | Eleventh StackMarch 3, 2011 at 10:29 am[…] a birthday with someone is more complex than might be imagined. The concept even has a name: the Birthday […]

Book Review: Here Comes Everybody by Clay Shirky « wolvesinthepianoMarch 25, 2011 at 11:18 am[…] people and too many possible interactions. Shirky explains this phenomena in Chapter 2 as ‘The Birthday Paradox.’ There are an infinite number of people then the number of interactions is even higher. […]

Interesting Science: The Birthday Paradox | SmithwareJune 19, 2011 at 4:18 pm[…] can read more on the Birthday Paradox on Wikipedia and this article by Kalid over at Better Explained. Also, special thanks to Kalid for proofreading this for me. So, […]

AndyJuly 13, 2011 at 7:55 pmYour equation right after you mention “the multiplication looks pretty ugly” looks like it could be computed using factorial(!) notation, which many scientific calculators have:

1*(1-1/365)*(1-2/365)*…*(1-22/365)=

1*(364/365)*(363/365)*…*(343/365)=

365!/{(365-22)!*365^23}

But 365! is likely too big for many calculators to handle.

KalidJuly 18, 2011 at 9:08 pm@Andy: Great point — and yep, probably much too large for normal calculators.

The Birthday Paradox « The New PrintAugust 3, 2011 at 8:20 am[…] mental math trick involved is known as “The Birthday Paradox.” You can skip directly the more technical explanation if you prefer. The question wasn’t about whether or not two people had one particular […]

James SmithAugust 9, 2011 at 12:44 pm@gavin I went ahead and created a Facebook app to show the Birthday Paradox with your friends: http://apps.facebook.com/thebirthdayparadox/

NathanNovember 1, 2011 at 3:56 pm…and then sometimes I put too much egg white into the batter….

oh dang it!! Wrong website!!

JoeDecember 5, 2011 at 6:14 amOn average, Facebook members have befriended 130 other Facebook members, which means there is nearly 100% chance that every average member of Facebook shares a birthday with a person on his/her friend list. If any staff member from Facebook is listening, I’d like to know if the outcome matches the theory!

kalidDecember 5, 2011 at 5:15 pm@Joe: Actually, in this case it means among your 130 friends it’s almost a 100% chance that two of them share a birthday (not necessarily with you though! Friend A and friend B could have a birthday in common).

sonnyDecember 20, 2011 at 9:09 pmWhy doesn’t the following work :

We start with the first person in the group. The probability of another person in the group with the same birthday is 22/365 (since the probability of any one person having the same birthday is 1/365 and these are independent probabilities). Then we go to the next person. There is 21/365 chance of finding another person in the group with the same birthday. The next one is 20/365 and so on. And since these are all independent if each other we can add the probabilities, which gives us 253/365. This is the probability of finding 2 people in the group with the same bday.

What am I missing here?

Thanks!

kalidDecember 20, 2011 at 9:19 pm@sonny: You can’t add the probabilities :). By that reasoning, if we had 30 people in the group, the chance would be (1 + 2 + … + 30) / 365 = 435 / 365, which is greater than 1. It is the right idea to consider each pair, though.

sonnyDecember 21, 2011 at 4:51 pmThanks Khalid. So is there a way to solve solve this without using the ‘negative’.. that is not by calculating the probability of someone else in the group not having the same bday? Do it directly instead?

kalidDecember 21, 2011 at 11:48 pm@sonny: Great question — I don’t think my probability knowledge is strong enough :). The issue is you need to enumerate every possible type of collision: 1 with 3, 1 and 2 with 3, 1 and 3 and 14… all of which are “problem scenarios”. It’s a bit like writing a spellcheck where you keep track of the possible typos vs. having the correct word and seeing if what you wrote is different from that :).

ChrisJanuary 5, 2012 at 5:13 pmTHis is great your the bomb man, how did you figure this our, science fair project here we go

kalidJanuary 9, 2012 at 1:30 pm@Chris: Glad you liked it.

JamalJanuary 18, 2012 at 6:20 amI’m no mathematician but I am very intrigued by it . I have come up with a simple answer for this problem for thoughs who think in a way I do. I start off by assuming that there is on average 30 days in each month so imagine a calendar with 30 days so don’t imagine the specific days but when you think that each month has the same numbers it actually makes more since so instead of writing all the math think in normal ration terms. If you get 23 people in a room with the same birth number not month then you know you have about a 23 out of 30 chance not that’s pretty high well there’s only 12 months and thoughs chances slim down a bit but anyway that’s my quick thought on it

royJanuary 18, 2012 at 7:08 amHi, Kalid. I loved this post. Very interesting stuff!

I work with someone who was born on the same day in the same year in the same state only a few hours away (opposite coast from current residence) and only 2–3 hours apart. What are the chances?

kalidJanuary 20, 2012 at 1:23 am@Jamal: Interesting way to think about it — breaking it down by birth “day” and then birth “month” (might be easier to see how common it is).

@roy: Thanks, glad you liked it! Wow… there should be a name for that, virtual twin :).

KatJanuary 26, 2012 at 12:00 pmThank You for the awesome facts!!! I love it. Really helped with my algebra project xD. :)

kalidJanuary 27, 2012 at 10:37 am@Kat: Thanks!

KatJanuary 31, 2012 at 5:26 pmNo problem. I totally understand this and Im only in 7th grade! woo hoo. But yea. Im gonna reccomend to my friends some of them are doing the birthday paradox project too :) Thanks again Kalid!

kalidFebruary 3, 2012 at 12:44 am@Kat: Whoa, that’s awesome you’re getting this so early, you’ve got quite a head start! More than welcome!

How many of your Facebook friends have a birthday on Valentine’s Day ? | | Furious PurposeFurious PurposeFebruary 13, 2012 at 6:37 am[…] and it took only 36 or so to find them. How is this possible ? Well, it’s called the Birthday paradox, and it is based on the inability of our brains to think in exponentials, and to consider scenarios […]

KristinaMarch 22, 2012 at 11:02 amWhen I was in 7th grade my science teacher bet that there weren’t 2 people in our class of about 30 people who had the same birthday. We laughed our butts off at him because right away we had a set of twins in the class. Even once we removed them we all said our birthdays and we found the set of twins, me and a guy who all had the same birthday (Aug 3). There was also another pair of unrelated people who had the same birthday.

kalidMarch 25, 2012 at 3:39 pm@Kristina: Yep, with 30 people it starts to get pretty likely there’s be an overlap! Pretty amazing.

DonMarch 22, 2012 at 5:50 pmThe math is slightly flawed in the respect that there are actually 366 days/year during leap years. Very interesting though.. :-) -d

kalidMarch 25, 2012 at 3:38 pm@Don: True :). Might need to make slight adjustments or plug in 365.25 into the equation =)

NathanMarch 22, 2012 at 10:25 pmWent through entire grade school without anyone sharing my same birthday?

elementary K-6

secondary 7classes twice a year for another 6 years.

Some explain the chances of that happening?

kalidMarch 25, 2012 at 3:42 pm@Nathan: The trick to remember is the paradox is about everyone else not getting overlaps either (i.e. Billy and Joey could have an overlap, and it would count).

Hire Jim Essian - Friday Roundup: The “SHAME-Us Catuli” EditionMarch 30, 2012 at 9:29 am[…] birthday paradox checks out. Pre was born not only on the same day as I am, but in the exact same year. And, since […]

bamboletta: natural, handcrafted companions for little ones - » Blog Archive » About those Serendipity’s ….April 10, 2012 at 7:15 pm[…] have the same birthday. Seem counter-intuitive? Here’s a great description of the paradox: http://betterexplained.com/articles/understanding-the-birthday-paradox/ For Bamboletta drawings, let’s assume that there are 21 drawings, and 200 entrants each. In […]

AV-Media Trelleborg - Statistik, sannolikhet, fotbollsspelare och strumporApril 16, 2012 at 1:51 am[…] 57 så är det 99 % sannolikhet. En trevlig och enkel förklaring till hur det går till hittas på better explained. Vill du läsa mer om problemet och andra logiska kullerbyttor rekommenderas ett besök på David […]

MohammadApril 25, 2012 at 6:15 amHi

I need to calculate the probability of concurrency of 3 or more accident which are the same in the particular period. Is there any way to do this?

AToJuly 10, 2012 at 2:12 amIndeed “only consider the scenarios we’re involved in”. Thanks for the remanding.

ChrisAugust 2, 2012 at 10:15 amOK, I’m awful at math. I’m writing a piece right now about Olympians sharing the same birthday. Every day I log on to the official Olympic website and check the birthdays. There are roughly 10,960 athletes competing. From what my untrained eye can tell, it averages to about 30 birthdays per day. Can anyone help me out and explain the math a little better for me? Writing is my thing. Math makes my brain hurt. Thanks!

kalidAugust 7, 2012 at 1:55 pmHi Chris! Yep, for about 10,960 athletes you’d expect 10960 / 365 ~ 30 birthdays per day. In a room (or specific event), you can use the formula to figure out the chance of at least two people having a common birthday. If a track heat has 12 people, there’s a 16% chance of two people having the same birthday (see the formula at the bottom, but it’s 1 – e^(-12*11/(2 * 365)).

Happy Birthday from London: Breaking down Olympic birthdays - Bear Down and BlogAugust 2, 2012 at 1:30 pm[…] fact, all you need is a pool of 75 athletes for a 99.9 percent chance that two of them will end up running around the Olympic Village in their birthday […]

ChrisAugust 9, 2012 at 9:21 amThanks Kalid! That’s perfect.

Here’s what I came up with if anyone if interested. Just having some fun with numbers and the Olympics.

If I got something wrong, let me know.

http://tucsoncitizen.com/bear-down-and-blog/2012/08/02/happy-birthday-from-london-breaking-down-olympic-birthdays/

AbdulAugust 15, 2012 at 10:07 pmIn high school I recall my teacher explaining this paradox. She said theoretically if there were 23 students in our class, the probability of two or more students have the same birthday is 50 percent. So my question is, in a class everyone is born in the same year would this reduce the probablity?

kalidAugust 16, 2012 at 10:53 pm@Abdul: Great question. Offhand, I don’t think people being born in the same year should change things.

ShambhuAugust 17, 2012 at 10:10 amSolution below is much simpler. Just find probabilty of each one having diff B’day and then subtract from 1 to get 0.507297234 answer. Any thoughts?

364/365*363/365*362/365*361/365*360/365*359/365*358/365*357/365*356/365*355/365*354/365*353/365*352/365*351/365*350/365*349/365*348/365*347/365*346/365*345/365*344/365*343/365=0.492702766

1 -0.492702766 = 0.507297234

kalidAugust 21, 2012 at 12:44 am@Shambhu: Yep, that works! But it’s a pain to compute manually. The formula in Appendix A gives a shortcut vs. having to do all those 23 multiplications out.

Grey MattersNovember 5, 2012 at 9:58 amI think you discount the formula at the beginning of Appendix A (1-1*(1-1/365)*(1-2/365)*…) too much by jumping immediately into an approximate shortcut. Let’s see what happens if we simplify that equation first.

The denominator of the equation is simple to work out – it’s 365 multiplied by itself as many times as there are people. For x people, the denominator will be 365^x.

The numerator also has a familiar pattern. For x people, It will be 365*364*363*…*365-x.

So, we have a pattern something like a factorial, but that stops after x numbers. How do we handle that? Yep, it’s our old friend the permutation formula!

So, the short form of the formula would be P(365,x)/365^x. Writing the long form of the formula, we end up with: x!/(((365-x)!)365^x)

Prefer to think of it with combinations instead of permutations? Permutations are just combinations with redundancies taken into account to focus on particular orders of events, or mathematically: P(365,x)=(C(365,x))x!

This makes the full formula: ((C(365,x))x!)/365^x

Yes the formula you write out at the start of Appendix A looks bad, but it simplifies quickly to a clear and understandable form. Examining it several ways in terms of combinations and permutations helps make it clearer.

Grey MattersNovember 5, 2012 at 10:06 amAs I wrote the formula above, that is, of course, the formula for no 2 people sharing a birthday.

To find the probability of at least 2 people sharing a birthday, as mentioned, we still need to subtract all that from 1.

Five Statistics Problems That Will Change The Way You See The World | Delaware ReasonNovember 14, 2012 at 7:10 am[…] Source: Better Explained […]

evelynDecember 14, 2012 at 11:36 ami’m doing the project too

It *could* just be coincidence | Education|Education Details and Additional Information|Details about education resourcesDecember 28, 2012 at 8:15 am[…] http://betterexplained.com/articles/understanding-the-birthday-paradox/ […]

KrisDecember 29, 2012 at 2:40 pmHey.

I was wanting to take leap day into account and so I figured I should use 365.25

And by the way my sample size is 50 people

Would this work…

50*49=2450

2450/2=1225

so 1225 combinations

364.5/365.5=99.7264022%

so 99.7264022 is the chance of a combination not matching

.997264022^1225=3.48%

so 3.48 is the chance that all 1225 combinations don’t match

1-3.48%=96.5131327%

so 96.5131327 is the cance of at least one of them matching

That is how I worked it out and I’m not sure if it is correct so help me out please.

KrisDecember 29, 2012 at 2:47 pmHey.

I just realized that in my math I used 364.5 and 365.5 instead of 364.25 and 365.25 and that messed it up, but if I change that, did I have the right idea?

Haha. Oops.

Mark GoldmanJanuary 2, 2013 at 3:39 pmThanks so much the explanation. I apologize–I’m confused on one point which indicates: “…we could list the pairs and count all the ways they could match. But that’s hard: there could be 1, 2, 3 or even 23 matches!”

I don’t understand why wouldn’t the limit on matches be 22? The “target” can’t match with himself, (or can he?) Sorry, I’m sure I’m missing something obvious.

“…we could list the pairs and count all the ways they could match. But that’s hard: there could be 1, 2, 3 or even 23 matches!

SharkFebruary 1, 2013 at 10:27 pmSomething doesn’t add up here. The first calculator shows that the birthday example with 365 persons would result in a 100% match, meaning at least 2 persons should have the same birthday. But it’s possible that all 365 persons have different birthdays (the first person born on January 1, the second on January 2 and the last on December 31).

GeorgeDecember 7, 2017 at 7:34 pmIf you ever find that, you owe them all a steak dinner.

KalidFebruary 2, 2013 at 12:31 amHi Mark, great catch. Yes, that should be 22 matches, appreciate the correction.

Hi Shark — the equation is a probabilistic argument. In fact, you hit “100%” (i.e., the limit of the javascript programming language) at around 90 people. At that point, the difference between 99.9999999… and 100.0 is too small to represent on the computer!

So it is theoretically possible to have 365 random people with 365 different birthdays. Practically, at around 100 randomly chosen people, you are virtually guaranteed to have a match (i.e., the probability of not having one is tiny, too small to be shown on a computer :)).

Mark GoldmanFebruary 2, 2013 at 12:50 amKhalid-thanks for the clarification and the website–it’s a remarkable service and resource.

Steve WilliamsFebruary 9, 2013 at 4:18 pmThis is the best webpage on the Birthday Paradox that I’ve found!

We are doing an elementary school “science” project. We picked the Birthday Paradox and did 40 trials (using mostly the internet) and came out with the expected (though counter-intuitive) result of about 50% pairs.

Now we’ve gotten to writing the “conclusion” of the report and realize that the answer involves apparently college-level math! Question: is there a simplified way to explain the paradox, at least to hint at why it works, that a smart elementary school student could understand?

Thanks!

kalidFebruary 11, 2013 at 11:34 amHi Steve, great question. I might try this: put the kids in groups of 10 and have them guess (before they start) how many handshakes they need so everyone in the group shakes hands. They might guess 10 (or 9, since that’s how many THEY need to do), but you’ll see it’s quite a large number (10*9/2 = 45). In the same way, the number of “birthdays to check” is not you against everyone else (22), but everyone by everyone (a much larger number). In rough terms, that’s why the odds are much closer to 50-50 instead of 22/365.

Karim Mohammed IB STUDENTFebruary 22, 2013 at 10:55 pmI used the Birthday Paradox concept in a math project of mine, and they told me some professor objected to it cause it’s his idea. It was my research and they were my results and raw data. What do you think i should do, submit it or change it?

JimFebruary 27, 2013 at 5:56 pmSo if I ask 23 people to pick a number between 1 and 365, there’s a 50% chance that at least two of them would choose the same number. This is the scariest thing I’ve ever heard of.

DreamBox Learning® : October 5th is the Most Popular Birthday! The Birthday Paradox ExplainedMarch 21, 2013 at 10:37 am[…] you’d like a further explanation of the birthday paradox, BetterExplained has step-by-step instructions, as well as a birthday paradox calculator. And if you have an […]

Shawn KeenanApril 14, 2013 at 12:19 pmMy son used the birthday paradox for his science fair project. I do have one question. Is it possible to calculate the odds of 3 people in a group of 25 having the same birthday (month/day)? Could I do this using Appendix C, by simply changing the /2 to /3 in the R3 line {pairs = (people * (people -1)) / 2}?

KGWMay 4, 2013 at 8:36 amMy 3rd grade daughter is testing the birthday paradox for a science fair project. The math itself is a little difficult, but testing the paradox is easier to understand. We have tested 14 samples and 12 samples produced a birthday match. Wouldn’t you expect it to be closer to half the samples? Is there an optimal sample size? I saw someone mention they did 40 samples.

kalidMay 4, 2013 at 1:29 pmHi KGW, not sure what you mean about samples. I.e., you tested 14 samples (of 23 people), and of those 14 samples, 12 had a match? Yep, in theory it should be about half, but with a relatively small population, it’s easy to skew. Also, in the real world, birthdays probably aren’t perfectly evenly distributed, and the “clumpiness” may make matches easier.

For the purposes of the paradox though, it’s still startling that such small groups have so many matches! (So I think the experiment still makes that point :)).

RonMay 29, 2013 at 11:41 amI just completed the following survey: I asked people in the office to pick a number between 1 and 365 inclusive. I moved around office so nobody could hear any other person’s response.

It took 20 tries before one person matched an earlier response! Interestingly, the 18th person and the 20th person I asked picked “6.” So, if I had chosen the respondents in a different order, I could have gotten the match earlier.

Just Because Life is Random, Doesn’t Mean I Have to Like It | AIMS Education FoundationJune 13, 2013 at 4:05 am[…] Wil Reimer, author of the Historical Connections books, did the Birthday Paradox problem [http://betterexplained.com/articles/understanding-the-birthday-paradox/] in our class of 30+ students and darn it, there was a pair of students with the same birthdate. […]

LenoxusJune 14, 2013 at 7:42 pmIf you’re interested in “triples”… a room with 88 or more people has an over-50% chance of at least one birthday being held by at least three people. For a room with 733 or more people, it’s guarenteed! (Per the pigeonhole principle. Hypothetically, a room of 732 people could consist of 366 pairs of people each sharing a birthday unique to that pair. The next person to walk into the room must have a birthday belonging to one of the pairs.)

I’ve been trying and failing to solve the problem for uneven distributions, such as if we assume all birthdays except Leap Day are equally likely and Leap Day is 25% as likely as the rest. There doesn’t seem to exist a simple formula for it on the Net…

paulAugust 13, 2013 at 5:30 pmI always explain it with a dartboard example. Put up a dartboard with 365 squares on it, put on a blindfold, and start throwing. You can start to see that randomly hitting it will reduce the space that you can hit that has not been hut before. Humans can “see” that analogy pretty well.

TrinaAugust 30, 2013 at 2:08 pmThis is a wonderful website i learn alot from this

SAMAKSHISeptember 12, 2013 at 6:15 amSomebody explain i still dont get it

Crazy Facts » In a room with 23 people, there’s a 50/ …September 12, 2013 at 9:46 am[…] a room with 23 people, there’s a 50/50 chance that two people share the same […]

LenoxusSeptember 12, 2013 at 3:48 pmSAMAKSHI:

The more people there are, the more opportunities for two of those people to share a birthday. When there are 23 people in a room, the number of opportunities is SO big that the chances of a match between two people is better than half.

So about half of all 23-person rooms should be able to say “Yes, two of the people in this room share a birthday”.

We’re not interested in a specific birthday, such as January 15. We’re asking about the situation where any two people have a birthday in common.

It can help to remember that if there were 367 people in the room, then the chances we have a hit are not just very high, but are actually 100%. EVERY room with 367 or more people has two or more people sharing a birthday. That’s because the only other possibility is that there are 367 unique birthdays in that room, and there aren’t that many birthdays to go around.

What if there were 366 people? Then it’s possible they each have their own birthday (including a Leap Day birthday!), but that’s EXTREMELY unlikely. So the chances of a match are very close to 100%.

What if there were 365 people? Just a bit less. 364 people? Slightly less than that.

… and so on, down to 23. At 23 people, the chances are just over 50%. At 22, they are below 50%. At 2, they are about 1/365, or well below 1%.

The whole thing is just a graph that curves differently than you might expect.

ClarkeyDecember 11, 2013 at 8:49 pmGreat explanation.

For those who continue to doubt Mathematical equations (for some reason). I encourage you go to a random number generator website such as random.org and choose the field between 1 and 365. Write the numbers down and see how many you write down before you get a match/repeated number….not long! :)

Where a lot of people seem to get confused and doubt the equation is that they are stuck on themselves (such as comment 73)…It is not the odds of YOU having the same number as someone else in the room, its the odds of anyone having the same number as anyone, which you explained well. Great site.

Day 769: Happy Birthday To Possibly Both Of You - 1000 Words, 1000 DaysFebruary 7, 2014 at 5:46 am[…] best just to take my word for it. Or try it out here – this page lets you run the a sample set of any size and keeps track of the matches for you. I […]

Question #4 What is your Birthday? | Jontherose351February 9, 2014 at 8:36 pm[…] intrigue in this paradox (further explained here) is that the chance of having matching birthdays in a group of people seems extremely high for the […]

The Quick Guide to GUIDs | BetterExplainedApril 20, 2014 at 7:13 pm[…] the birthday paradox shows us the chance of a collision as GUIDs are used. It’s very, very unlikely that GUIDs will […]

Birthday Paradox and Forensic Science | The Nightly BrewMay 19, 2014 at 10:15 pm[…] (The reasoning leading to this conclusion is presented in the paper, and there are many other more detailed explanations of the paradox.) This is a surprising result and highlights how unintuitive probability can […]

Rickyvalle21May 22, 2014 at 4:39 amGetting 10 heads in a row is actually .5^9 Because the .5 is accounting after you flipped a coin once.

The Birthday Paradox - MRI Software BlogMay 22, 2014 at 9:21 am[…] The real answer is a measly 23. Once your community grows to 57 people, there is a 99% chance of two residents sharing a birthday. Welcome to the birthday paradox. […]

The Birthday Paradox | WeedBoxMay 22, 2014 at 10:01 am[…] this and you want to know how this all works, then you should check out the original post by Kalid Azad. We must warn you though, it’s […]

VinceMay 29, 2014 at 12:03 amExact and easy formula is:

P(at least one shared birthday) = 1 – 364!/((365^n-1)*(365-n)!)

= 1 – 365!/((365^n)*(365-n)!)

Where n is the number of people in the room, and 365 is used as the number of days, thus not taking into account leap years.

The formula was found by simplifying:

P = 1- 365/365 * 364/365 * … * (365-(n-2))/365 * (365-(n-1))

Can someone help me?June 3, 2014 at 4:41 pm[…] 2014 05:28 pm @cicerone imposter, Here's another good article about the birthday paradox. http://betterexplained.com/articles/understanding-the-birthday-paradox/ 0 Replies GA_googleFillSlot("a2kTopicLeaderboardEnd"); […]

spopeJuly 8, 2014 at 6:37 amI don’t see how the interactive example is correct. If you put 1000 items and only 142 people it says there is a 100% chance of a match. Surely this isn’t correct, for example if person 1 picked 1, person 2 picked 2, 3 picked 3 and so on up to 142 there wouldn’t be a match. Not very likely I know but possible (in fact is it not just as possible as every other selection?) , so there can’t be 100% likelihood of a match. Plus there are many other combination that wouldn’t create a matching pair.

If you keep with the birthday example all you need is 86 people for 100% chance of a matching pair, but it is “possible” for 86 people all to have different birthdays so how can it be 100%.

Sorry if I’m way off the mark, I just don’t get it.

What is it about the laws of averages and statistics that makes it so that if you have as few as 30 people in a room, it is likely that two of them have the same birthday? - QuoraJuly 23, 2014 at 9:51 am[…] with just 23 people.There is a breakdown of the maths to calculate the exact percentages here: http://betterexplained.com/artic…Embed QuoteUpvote • Comment Loading… • Written just now Add your answer, or answer […]

The Birthday Problem: Or, We’re Terrible at Estimating Probability | You're Doing It WrongJuly 27, 2014 at 11:32 am[…] There’s great explanation of the math here: Understanding the Birthday Problem. […]

meAugust 4, 2014 at 1:58 pm@spope – I think it probably rounds.. so it would be 99.999102% (or something like that) so it rounds to 100%.. just a guess..

BoAugust 8, 2014 at 5:27 amThe formula can’t be exact. Using this formula, you would calculate a (though small) chance that in a room of 366 people there would be not a single mutual birthday. This can’t be correct as there are in this case more people than days in a year. Therefore the possibility to have not a single same birthday should be as zero as it could ever be.

Qué es la paradoja del cumpleaños y por qué me hace especial | Traduquímica et al.October 10, 2014 at 2:49 am[…] Kalid. Understanding the Birthday Paradox. Better Explained. [fecha de consulta: junio […]

What is the birthday paradox and why does it make me special? | Either a linguist or a chemist beOctober 10, 2014 at 5:48 am[…] Kalid. Understanding the Birthday Paradox. Better Explained. [viewed June […]

SleatOctober 28, 2014 at 2:58 pmBo, good example. When you have 366 people the probability of everyone having different birthdays has a term which is zero. When you multiply by this zero term you get zero, meaning a probability of one that two people share the same birthday.

Jenny GilmoreNovember 5, 2014 at 11:35 amHow come towards the end you minus it from 1?

ClarkeyNovember 5, 2014 at 4:34 pmHi Jenny,

When you are calculating, you are working out how many persons DON’T have the same birthday as another. The ‘minus 1’ gives you your chances of another person having the same birthday.

1) 23 individuals x 22 partners = 506 couples. Divided by 2 for unique couples (i.e…Tom and Jane are the same couple of Jane and Tom)

2) 364/365 (Days of the year someone might NOT have the same birthday as you) Yx 253 couples. = 0.4995, being the chances of someone NOT having the same birthday as another person…. -1 to view -0.5005…which is only a quick way of seeing the answer when the real equation is 1 – 0.4995 to give you your answer being 0.5005 (50.05% likely of someone having the same birthday as another.

The Paradox Of Choice: Why More Is Less | The Book Review HubNovember 11, 2014 at 8:44 am[…] Understanding the Birthday Paradox | BetterExplained – As you run more and more trials (keep clicking!) the actual probability should approach the theoretical one. Examples and Takeaways. Here are a few lessons from the …… […]

danNovember 29, 2014 at 12:35 pmI approached this by figuring that there are 253 pairs and each pair has a 1/365 chance of having the same birthday. The probability that at least one pair would have the same birthday is 253*(1/365). Where is the flaw in my reasoning?

Thanks

anonymousNovember 30, 2014 at 6:52 amIn a room of 32 people a teacher pickedv a pupil at random. What is the probability of him picking someone with the same birthday as him?

LenoxusNovember 30, 2014 at 11:38 amdan: There are two problems with your approach.

One is that you are treating the probabilities as independent when they aren’t. Consider just three people: A, B, and C. There are three pairs out of these people (AB, AC, BC). By itself, the probability that a given pair has the same birthday is 1/365. But if AB shares a birthday, and BC shares a birthday, then the probability that AC shares a birthday is not 1/365 but simply 1. However, for the sake of a simple estimate this fact doesn’t make much difference, because with only 23 people, triple-birthdays are very rare.

The second problem is that independent probabilities don’t work that way — you can’t just add them up. The probability of rolling a one on a 6-sided die is 1/6. But the probability of rolling at least 1 one out of two dice is

not2/6. If it worked that way, then rolling 6 dice would give us a probability of 6/6, or 1 — not only guaranteeing at least 1 one, but at least 1 of every number, since all of the numbers are equivalent here. Of course it doesn’t work that way — if I roll six dice, getting a “straight” of all the numbers is extremelyunlikely, not guaranteed.Instead, you work out the dice probabilities by asking the opposite question: What is the probability of rolling anything but a one? With two dice, there are five-times-five equally-likely ways to do this (as if the dice only had 5 sides): a two and a two, a two and a three, etc, all the way to a six and a six. And here are 36 total equally-likely possible combinations: one-one, one-two, and so on. So our answer to

thisquestion is 25/36. That means the answer to our original question is just 1-25/36, or 11/36.Getting back to the birthdays, the way to determine the probability with 23 people is not to add the independent probabilities, because then you get a false guarantee of a pair at 28 people (there are 378 pairs, so you would find a probability of 378/365 for a pair, and that number can’t even be right in itself — probabilities should never exceed 1). Instead, we do multiplication, and what we multiply is the inverse scenario: 364/365, or the probability that two people don’t share a birthday.

We multiply this number by itself 253 times (in other words, raise it to the 253rd power). The result is a number slightly

less than1/2: 0.4995. This means that the probability for the reverse question is slightly more than 1/2, about 0.5005.However, this is slightly off because of the non-independence factor. The more correct way to calculate it is to build the sets of possible rooms that lack matches, and this isn’t as hard as you might think. (It’s simpler to understand than the methods shown in the blog post.)

The first person who enters the room is permitted to have any of 365 birthdays. The second person we put into it is only allowed one of 364 birthdays, because they can’t be a match. The third person is allowed one of 363 birthdays, because she can’t match either of the first two. All these possibilites multiply out, for the twenty-three people: 365 x 364 x 363x 362 x 361… x 345 x 344 x 343.

That’s our numerator: The total set of equally-likely combinations that fulfill a principle of lacking any matched pairs. For our denominator, we need the total set of possible combinations in general, and that would be 365 x 365 x 365 x 365… twenty-three times.

The resultant number is about 0.4927 — which, as before, is just under one-half. Hence, although we have a more precise probability, we still find that the same number of people (23) is the threshold here.

ClarkeyNovember 30, 2014 at 6:33 pmAnonymous (Ref – Comment 118)

Your example is different from the birthday paradox. Your example asks what the probability is of someone having the same birthday as a specific person…as opposed to someone having the same birthday as anyone in the room.

To answer your question…you could have 10,000 people in a room it doesn’t make a difference….the odds of picking a singular pupil at random and them having the same birthday as you is 1/365 (not halved again as some may think, as you are trying to match a pre-existing specific number)

Hope that makes sense?

-Clarkey-

Quantity Has A Quality All Its Own | CraptardDecember 22, 2014 at 12:46 am[…] The number 421 is not just a 42 tacked onto the front of a 1, although that’s nearly sufficient by itself, with its doubling down progression so common to discrete mathematics, computers, and losing all your money gambling. As Castañeda might have put it in Tails Of Power, “Cuatro, dos, uno, *nada*! It simply ends … vanishes without a tres.” 421 is prime and is indeed the only number of that pattern that is prime, up through the 23-digit number 10,245,122,561,286,432,168,421. It happens to be the number of days between Douglas Adams’ birth date and mine, a fact which amuses me no end and why I came to consider it. Call it Zeno’s Birthday Paradox. […]

Shubham KumarJanuary 31, 2015 at 7:22 amSir, I read it before the heading “Interactive Example” and hats off to you. You have explained it so nicely, that it can actually feel what is happening in this “paradox”!

Thanks a lot sir!

Human Intuition Is Usually Wrong Facing Uncertainty | hewenjingFebruary 5, 2015 at 8:29 am[…] Birthday Problem: In a room of 23 people, what’s the probability of two people having the same birthday? What about there are 75 people? The answer is 50% and 99.9%! Explanation: http://betterexplained.com/articles/understanding-the-birthday-paradox/ […]

JoshFebruary 5, 2015 at 2:06 pmIt is actually much easier than what is explained here.

The first person could have any birthday during the year.

This would be 365/365

The second could only have one birthday that matched (regardless of the number of people in the class.

This would be 1/365

Using the multiplication rule of probabilities that one event AND another will occur:

365/365 x 1/365

1 x 0.00274

0.00274

ClarkeyFebruary 5, 2015 at 5:38 pmSorry Josh, but it isn’t that simple and you are actually wrong. How do you use that equation to get 50.05% probability?

Certainly not having a go here, as this is what the website is about. However:

1) the example you use doesn’t give an answer to the actual question involving 23 people.

2) using your calculations is not the calculation of probably as you have the potential to get an answer greater than ‘1’, which in this circumstance you should never have an answer greater than ‘1’, that just isn’t probability… you need to look at the chances of someone NOT having the same birthday as someone else and work from there….

JoshFebruary 5, 2015 at 5:47 pmStraight from my Statistics class and WebAssign:

Same Birthday: Suppose two people are randomly selected from a class of 35 students. What is the probability that they have the same birthday? Round your answer to 3 significant digits*.

WebAssign will check your answer for the correct number of significant figures.0.00274

Correct: Your answer is correct.

seenKey 0.00274

………………………………………..

*Significant Digits: Here are some probabilities expressed to 3 significant digits.

You start counting digits from left to right starting with the first non-zero digit.

0.123 0.0123 0.00123 0.102 0.350 0.300

Solution or Explanation

This is tricky because it doesn’t matter how large the class is. Also, it doesn’t matter what the first person’s birthday is. The probability that the second person has the same birthday is

1

365

= 0.0027397 ≈ 0.00274

to three significant digits.

ClarkeyFebruary 5, 2015 at 6:00 pmJosh….it matter immensely how big the classroom is…

Under your example of 35 people you have 1190 pairs (35x 34)….factor in that is pair is twice (1:you & I, 2: myself and you) and you have 595 unique pairs.

lets go PROBABILITY calculations now….

Chances of someone NOT having the same birthday as the other

364/365 = 0.997260274. Ans ‘y’ 595 = 0.1954649. 1 – 0.1954649

Chances of two people having the same birthday is 80.4536%

Use the same maths as above for 25 people and indeed you get you 50.05%.

In your calculation you haven’t factored in the most important part of the calculation….the amount of people in the equation :)

ChezniFebruary 6, 2015 at 11:25 amIn two of my class (one a class of 20 and the other class of 21), there are two other people that share the same birthday as me, so 3 total. I find that really weird.

AnonymousFebruary 25, 2015 at 3:55 pmwhy is 23 the number of people required for a probability of 50 % of two people having the same birthday? its my homework question^

AnonymousFebruary 26, 2015 at 11:41 amDon’t make sense

Two math anthologies.March 13, 2015 at 11:39 am[…] have the same birthday is thus a product of all of these individual probabilities (the formula is here); the 23rd person added to the group drops the probability that there is no birthday match under […]

TrigtfApril 15, 2015 at 3:19 amAll depents on how frequently those births happen. Look at the below diagram, http://www.todaysparent.com/blogs/on-our-minds/birthdays-most-common/, the actual probability for someone to have a birth in a certain date is not equal for all 365 days. There are factors, we dont know (probably weather , moon etc), that pushes a lot of people to have birth in Jun, Jul, Aug, Sep, Oct. So that explains the probabilty to have 2 people in a class with same birthdate since most people are between those months, like 80% are in that range of 150 days.

HyperThreadingJuly 2, 2015 at 4:01 amWhy you, for “x!”, do not use Stirling formula: x!=x^(x+0.5)*exp(-x+1/(12*x)-1/(360*x^3)+1/(1260*x^5)-1/(1680*x^7)+1/(1188*x^9) …), (“x” do not need to be whole number, but it have to be greater than, say, 10) so something like n!/((n-m)!*n^m) (n=365, m=23, for example), may look like (n/(n-m))^(n-m+0.5)*exp(-m+(1/n-1/(n-m))/12-(1/n^3-1/(n-m)^3)/360)+…), and this way it is not problematic to calculate “complementary probability”? After that you subtract that from one and … Of cause “IEEE float” format is not accurate enough for this formula, and it does not allow argument of exponential function to have absolute value greater than 88 or so, but that is a story for another day …

HyperThreadingJuly 2, 2015 at 4:16 am… so 1000!/(858! * 1000^142) is 2.69985824…e-5, and probability that among 142 “items” that share 1000 “unique item numbers” two or more have the same “item number” is 99.99730014176% (the last digit is “rounded up”), and it is not so close to 100%, even for “IEEE float” format, to be rounded to 100% …

HyperThreadingJuly 2, 2015 at 4:24 am… I am sory, but I forgot to mention that exp(-142) is too small for “IEEE float”. I swear: no more comments from me in a foreseeable future :-)

KatJuly 8, 2015 at 4:00 pmThat’s confusing.

AnonAugust 7, 2015 at 2:24 pmThe probability of at least two people sharing a birthday in a group of 22 people is about 50.72972343239854072, not 50.05%. Kalid’s method faultily assumes that the probability any pair sharing a birthday is independent of the probability of another pair sharing a birthday, which is not the case because the pairs contain some of the same people. The exact probability of at least 2 people sharing a birthday out of a group of x people can be calculated by the formula 365Px/365^x, or 365!/((365-x)! 365^x).

AnonAugust 7, 2015 at 2:26 pmCorrection: the above is the probability of at least two people sharing a birthday out of a group of 23, not 22, people.

JeffSeptember 20, 2015 at 9:07 pmYou really want to blow someone’s mind?

with a true random selection of 230 people, merely ten times the birthday paradox, there’s almost a 50% chance of not only having two people with the same birthday – but two people with the same birthDATE. (with a 100 year pool.)

Mathematically, it says that number is 191.11… (365.2425*100 = 36524.25 sqrt(36524.25) = 191.113186…)

However realistically, it doesn’t close in on 50% until you get above 220s

[365.2425 is the actual days per year to take leap years into consideration.]

And if the people are randomly selected from a certain pool – such as a college population – the chances increase greatly, obviously…

Isn’t math fun?

Robert walkerOctober 30, 2017 at 2:30 amHow come the last 4 digits in my ssn is the year I was born and the day I was born on??

MikeykNovember 29, 2017 at 7:59 pmOh great. My brain just splattered. Thanks.

Deborah karrOctober 23, 2015 at 8:56 amAnd if your population group was born born nine months after a natural disaster ?? ?? No electricity for the television, doesn’t everything get skewed ? ? ? ? ?

syedOctober 26, 2015 at 4:22 amI understand the math part of this question but I have to answer this question in sort of algorithm way because we are using computers so can someone help me.

maxiMarch 30, 2016 at 7:36 pmYou shouldn’t introduce multiplication without mentioning it is not accurate but a good enough approximation hack. People’s math feeling and instinct got fu*ked up by that. When I see this “answer”, my math alarm ring and tells me that no way. It turns out it is incorrect. Getting an approximation and getting a math theory / answer is different. Don’t put it like they are the same.

Jake SaundersApril 27, 2018 at 8:54 pmYou don’t sound really bright. Just sayin’. I’m curious as to what you mean by “multiplication… is not accurate”

abbyApril 20, 2016 at 6:56 pmCan you please tell me what are the chances of two people dating and both of our parents have the sabre birthdays as in my boyfriend’s mom has the same birthday as my dad and my boyfriend’s dad has the same birthday as my mom what are the chances of that? ???

Terry RuckerJune 11, 2016 at 12:51 pmI don’t know about the math, way beyond me, but my old algebra teacher would try this experiment in every class he taught. In the three classes I had with him each one had pairs. Go figure.

PeterJuly 4, 2016 at 9:25 amI was not familiar with this paradox. To make things worse, I have zero knowledge of statistics. So I wrote my own little program and tested for 23 people in a room, and also for 72.

With 23 people in the room, here are number of matched birthdays after running 1000 random tests each time:

476

487

518

527

480

510

492

520

527

499

Obviously, close to 50%.

And with 72 people in the room. here are results for 1000 tests each time:

999

1000

1000

997 (I actually ran more than this number of tests and this one really sticks out – got it only once!)

1000

1000

1000

1000

With that one exception of 997 matched birthdays, every time it was either 999 or 1000.

That shows that out of 1000 tests with 72 people in the room, almost every time there was at least one pair matched.

So the math explained in this article works, the explanation on the other hand – I think it needs to be a bit better :)

Here is my stab at a better explanation:

Lets say that you are under impression that even if we had 180 people in the room, they could all still have different birthdays. And they could (theoretically). Lets imagine they are children and that room has 360 slots (for simplicity I am ignoring five days that are left over, bear with me, I am trying to make this simple to understand). Now, we put those kids in every other slot, so 180 slots are taken. Kids being kids, one of them is bound to full around and move left or right. As soon as ONE kid does that, there is an overlap, and we have a match. So, it becomes 50% chance with only ONE child given an opportunity to “misbehave”.

Now, imagine allowing two kids to misbehave? Your chances grow quite a bit.

Now, we put only 72 kids in their slots. In a room with 360 slots, that means every fifth slot will be occupied. Any one kid needs to move by four slots left or right to overlap with another kid – and we have a match. That means that with only one kid being allowed to be random, we already have around 20% chance of a match (or is that 25%?), and that is if the kid only moves four spaces in one direction. Allow two kids to do that, and your chances of a match increase. Allow it for 72 kids, and you have a match for sure.

Hope that helps and hope I am at least in the ballpark with my interpretation.

ermmhh..August 10, 2016 at 9:02 pmCopy/pasting part of your article:

“Adding 1 to 22 is (22 * 23)/2 so we get:

\displaystyle{p(different) \approx e^{-((23 \cdot 22) /(2 \cdot 365))} = .499998}

Phew. This approximation is very close and good enough for government work, as they say. If you simplify the formula a bit and swap in n for 23 you get:

\displaystyle{p(different) \approx e^{-(n^2 / (2 \cdot 365))}} “

What is this sorcery?!

ermmhhAugust 10, 2016 at 9:11 pmI thought copy/pasting would work better.. ahh well..

I was refering to (23*22) becoming (n^2) with no other change in the formula.

SomeoneAugust 15, 2016 at 2:21 am” The chance of 10 heads is not .5/10 but .5^10, or about .001.” – WRONG! It’s .5^9, which equals .002

The first heads is free! You can flip coins as long as you like (you didn’t narrow down the problem) , it doesn’t matter how many times you hit tails first. It only counts *after* you hit your first “heads”! – that’s the set-up you gave us, that’s what you have to work with. It says “how many times can you hit this in a row” it did not read “what’s the chance of hitting heads each time, when flipping a coin exactly 10 times.” (for that you gave the [in that case: _*correct*_ – ] answer – but since that wasn’t the question, you failed.)

Understanding your own word-problems is important.

SomeoneAugust 15, 2016 at 2:23 am*which is about .002

Chris HallMay 5, 2018 at 9:07 pmNo, the first heads is not “free”. It is still a 0.5 chance, regardless of how many tails you’ve flipped beforehand. The chance of getting another nine in a row is 0.5^9, which, when multiplied by the first 0.5, gives the chance of getting ten in a row as 0.5^10.

The probability of getting the same *result* (either heads or tails) ten times in a row is 0.5^9, because it doesn’t matter what the first flip is (that one is “free”); it only matters that the next nine match it.

EvanAugust 20, 2016 at 8:47 pmUsing the birthday calculator, once you put 86 people in the room It gives a 100% chance of 2 of them having the same birthday. 100%? Wouldn’t that mean that there is no combination of 86 people where they all have different birthdays? But of course there are.

kalidAugust 21, 2016 at 11:28 pmThe calculator uses a statistical formula and is rounded. Statistically, there’s a near-guarantee that in a group of 86 random people, two people will share the same birthday. The actual percentage is 99.9955

BenSeptember 17, 2016 at 6:26 amYou need a group of 367 people to be certain at least two share the same birthday., since the 366th person might have been born on 2/29 during a leap year.

CHABOINovember 9, 2016 at 3:37 pmGonna be honest, didn’t understand 40% of this but it was still really fascinating. You obviously find this a lot of fun and I want to tell you to keep arguing and doing it!

The FPaxos “Even Nodes” Optimisation | slash dev slash nullMarch 22, 2017 at 4:28 pm[…] obtain a majority of three. The leader has to wait until the slowest of two response arrives. The birthday paradox effect means that even with a low probability of any single response being slow there is a […]

Születésnap paradoxon – lesz olyan, aki osztozik a tortán? | Comsci blogMarch 25, 2017 at 7:29 am[…] Understanding the Birthday paradox – szimulációval Scientific American – a borítókép is tőlük származik […]

17 Unbelievable Scientific Facts Bordering on the ImpossibleApril 7, 2017 at 3:31 am[…] Source: Betterexplained […]

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TimApril 10, 2017 at 5:00 pmI actually thought the multiplicative approach is the more intuitive one.

How did this happen? « Jim's Random NotesJune 9, 2017 at 10:55 pm[…] or slept through the part of his algorithms class where they studied random selection and the birthday paradox (also, birthday problem). The comment, “There is a tiny chance that a generated code may […]

Wing Forward: FileMaker in New MexicoJuly 3, 2017 at 10:02 am[…] UUID being duplicated; developers care if any duplicate occurs, which is an example of the Birthday Paradox. And then I’ll introduce real life variables to get a sense of the […]

15+1 Ακραία Επιστημονικά Γεγονότα που μοιάζουν βγαλμένα από Ταινία Φαντασίας. Το 13ο θα σας Τρομοκρατήσει! – MyMind.grJuly 5, 2017 at 3:09 pm[…] Σύμφωνα με μαθηματικούς υπολογισμούς, έστω ένα από τα μέλη της ομάδας έχει γενέθλια την ημέρα Χ, τότε υπάρχει μόνο 364 ημέρες για τον δεύτερο, 363 για τον τρίτο, κλπ. Για να υπολογίσετε την πιθανότητα αυτή, το μόνο που χρειάζεται είναι να πολλαπλασιαστούν αυτές οι αναλογίες: .. 365/365 * 364/365 * 363/365 κτλ. Ως αποτέλεσμα, λιγότερο από το 0.01% των μελών της ομάδας έχουν διαφορετικές ημερομηνίες γέννησης. (Πηγή) […]

15+1 Ακραία Επιστημονικά Γεγονότα που μοιάζουν βγαλμένα από Ταινία Φαντασίας. Το 13ο θα σας Τρομοκρατήσει! – Destora.comJuly 5, 2017 at 3:10 pm[…] Σύμφωνα με μαθηματικούς υπολογισμούς, έστω ένα από τα μέλη της ομάδας έχει γενέθλια την ημέρα Χ, τότε υπάρχει μόνο 364 ημέρες για τον δεύτερο, 363 για τον τρίτο, κλπ. Για να υπολογίσετε την πιθανότητα αυτή, το μόνο που χρειάζεται είναι να πολλαπλασιαστούν αυτές οι αναλογίες: .. 365/365 * 364/365 * 363/365 κτλ. Ως αποτέλεσμα, λιγότερο από το 0.01% των μελών της ομάδας έχουν διαφορετικές ημερομηνίες γέννησης. (Πηγή) […]

In an office of 57 people, there is a 99% chance that 2 of them share a birthday – FavRiverJuly 8, 2017 at 2:57 am[…] Links: Businessinsider.com – Betterexplained.com […]

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Sweet32 Nearly a Year On: Vulnerabilities People Stop Caring About – Just another WordPress siteJuly 20, 2017 at 7:33 pm[…] Sweet32 and attacks like it are very simple. They’re based off of the same concept as the Birthday Problem – the mathematical “paradox” where in a group of only 23 people there’s a […]

Sweet32 Nearly a Year On: Vulnerabilities People Stop Caring About – FULLACCESS.IOJuly 20, 2017 at 7:38 pm[…] Sweet32 and attacks like it are very simple. They’re based off of the same concept as the Birthday Problem – the mathematical “paradox” where in a group of only 23 people there’s a […]

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JhillOctober 4, 2017 at 6:58 pmP(different) = P(365,23)/(365^23)

Fred HOctober 29, 2017 at 8:25 amThis would only account for the ‘same birthday’ if that meant the same day and month. To actually have the ‘same birthday’ would require the same day, same month, and same year. That dramatically reduces the probability of this happening — unless everyone in the room happens to have been born in the same year.

This ‘birthday paradox’ has been used (on NPR yesterday, for example) to try to explain away the fact that 3,000,000 U.S. voters in the 2016 elections had the ‘same last name,’ ‘same first name,’ and ‘same birthday.’ The commentator used the birthday paradox to explain away 2,700,000 of those matches — but that is incorrect unless you limit ‘same birthday’ to mean just the same day and month, and ignore the year of birth.

GregDecember 14, 2017 at 9:45 amVery well done and explained. My wife is a teacher with typically 20-26 students each year and she’s always claimed 1/2 the time there’s a match. I always thought it counterintuitive until your explanation. Thank you!

THE BORN IDENTITY – Grey is OkayJanuary 16, 2018 at 8:00 am[…] think we’re special snowflakes. In a room of only 23 people -yes only 23- there’s a 50% chance you share a birthday with someone else. […]

coincidence? – The Return Of MathJanuary 25, 2018 at 7:59 am[…] to calculate what can happen in life, it’s what makes it fun. Do you want to know with whom do you share birthdays? […]

Michael W LewisJanuary 28, 2018 at 10:34 amHow many people do you need in a room so 2 people were born on the same day and same year?

Jake SaundersApril 27, 2018 at 8:50 pmThat’s a much more complicated problem since the chance of someone being born on a single day is estimated at 1/365 (which I’m sure is statistically not true… in the southeast, more people are born nine months after football season). But the chance of having someone from the same year requires calculating the odds that any person was born in the same year as you and thus requires estimating the likelihood that someone will live for “x” number of years. If you’re 100 years old then the chance that someone in the room was born in the same year is very small.

B GradFebruary 26, 2018 at 6:12 pmI thought about it this way. There are 12 months and UP TO 31 days in a month. Since we would need to have two people have the same month, 13 people would guarantee a month match. Now if we had 31 MORE people, (for a total of 44) it would virtually guarantee that two people had both month and day match. Certainly we know the month will match. But these are not truly independent since the 13 people to catch the month also have a day associated with it. If those 13 days are different then we would have a month match but not a day match. If we added 31 more people, we would for sure have a day match as well.

I realize that the exact answer is not obtained but having 44 in a room would virtually lock it.

EliasMarch 25, 2018 at 3:04 amSure, if you would add 31 more people that you knew were born in the same month this technique would work. But since this is not what you mean (as i interpret it), means that these 31 persons’ birthdays could be distributed over several months without coinciding with any other date that is already “taken”. To exemplify, with one person already in every month, you could fill January with 30 people (without any match) and with the remaining 1 person, you could put in February on a date that the February-person would not be born on. Thus, it is not “locked” if you would have 44 people in a room. It is only completely locked when you have 366 people, (even though wit hfewer people you could have an almost certain match. For instance, 57 persons would give you a 99,0% chance of a match). ‘

I hope this was somewhat helpful in understanding the paradox.

Douglas NorbergMarch 3, 2018 at 3:25 pm(23 * 22)/2 =253; How is 253 half of 365?

I get (20*19)/2 = 190 or slightly more than 365/2 = 182.5

Jake SaundersApril 27, 2018 at 8:45 pm253 is just the number of possible pairs in a room of 23 people. 23 options for the first person, then take them out of the equation and there’s 22 people left. So combinations are 23 x 22 but you can’t count (a) paired with (b) and (b) paired with (a), so you divide by 2.

softturboMarch 21, 2018 at 10:54 pmThe correct, simple, and exact calculation (which was also the way I was taught in high school) is here:

http://mathforum.org/dr.math/faq/faq.birthdayprob.html

Taking the c(n,2) th exponent of (364/365) is not really the correct way to think about this problem. The main reason why the p(different) decreases so quickly is because the number of available (untaken / different) birthdays decreases as the number of people increase. And this formula does not really take this into account. Also with this formula, p(different) does not really go down to 0 with 366 people, at which point the probability of having two people with the same birthday should be nothing short of certainty (100%).

OP needs to go back to high school and learn the basics before showing off his fancy but ultimately incorrect math skills.

kalidMarch 22, 2018 at 11:08 amPlease read Appendix A. The exact formula is not simple to compute, try working out 365! / ((365-n)! * 365^n) with n=23 on a pocket calculator. (The shortcut is to rely on calculus and substitute the Taylor series for the exponential function, also not simple.)

softturboMarch 22, 2018 at 11:50 amOn the contrary, it’s actually not that hard to calculate this on a pocket calculator at all. If the pocket calculator supports calculating e^x, then it certainly would support x^y. And even without support of factorials, it’s easy to see that 365! / (365-n)! with n=23 would simply reduce down to multiplying 365*364*363…*(365-n+1). i.e. 22 numbers, which is perfectly manageable. I suggest you try working it out yourself, the smart way.

And might I mention that the exact formula in,its simplest form, as you stated above, was never shown in the article. And there was no real effort in trying to explain the PROPER way to solve this problem first before jumping into a shortcut method. And you chose the really rough approximation method that did not really follow the correct line of thinking. And in doing so, it strikes me that this article seems to be more of an effort to show off and confuse people than to educate.

Please read https://en.m.wikipedia.org/wiki/Birthday_problem

That is how it should be done if you were trying to compare and contrast various methods of arriving at the solution.

kalidMarch 22, 2018 at 12:14 pmMultiplying 22 3-digit numbers without error doesn’t seem simple. (If you tried it, you’d see calculators lose precision at some point, so you get a rough approximation anyway.)

The walkthrough of the exact calculation is already in the appendix. The very last paragraph references the wiki article. The speculations for the article’s existence are curious.

softturboMarch 22, 2018 at 12:34 pmThe only curious speculation I might have would be whether one has ever used or tried to use a pocket calculator.

Personally I have seen many of shopkeepers use a calculator to add up more than 23 non-consecutive decimal numbers without fail. And the difficulty of multiplying 23 numbers on a calculator is the same as adding them, in case you have never tried it as I have suggested.

I do appreciate your effort in trying to explain this. But sometimes even with one’s best efforts, it’s just would not suffice. Just like I would not expect to win the 100m sprint in the Olympics even with my best efforts. The bar is a lot lower here in this case. But that still doesn’t mean there’s not a standard to meet.

kalidMarch 22, 2018 at 3:29 pmIdeas for standards are your own. Thankfully I’m not beholden to them.

Out of respect for you, I won’t discuss the original version of your comment you edited, nor the difference in computational requirements for adding 23 3-digit numbers vs. multiplying them.

softturboMarch 22, 2018 at 3:58 pmThis blog is yours. You are free to hold whatever standard you want.

Reagan ColbertApril 4, 2018 at 1:30 pmI had a math assignment due today discussing this exact problem (the odds of 2 in 23 having the same birthday) so I really appreciate this article!! ;)

StatalyzerMay 9, 2018 at 11:39 amWouldn’t the math be 364/365 * 363/365 * 362/365…. etc

kalidSeptember 26, 2018 at 10:17 amYep, that’s the exact formula (see appendix A). It’s harder to compute by hand vs (364/365)^n.

Burner EmailMay 10, 2018 at 9:00 amPlaying with this tool helped me learn and teach my students

https://lukezirngibl.github.io/birthday-paradox/dist/

Peter CrowleyMay 21, 2018 at 11:43 pmProbability of 3 birthdays (or more) the same in a group of 350? A group of 7? both scenarios presented to me in the last 3 days.

i can see a way to “batter” the problem with repetitative spreadsheet calcs – does anyone know a more elegant way?

C MoonJune 7, 2018 at 12:51 amthank you for this. my maths teacher showed this to us but didnt explain it becuase “we wouldnt understand” and its “above GCSE level”. i finally get it.

Alf MullinsJune 10, 2018 at 9:23 amThe squads for the football World Cup have been announced containing 23

players per team. Satisfyingly, 16 of the 32 teams contain players with

shared birthdays.

Darren WoolcockJune 13, 2018 at 8:23 pmBut not all birthdates are created equal: http://thedailyviz.com/2016/09/17/how-common-is-your-birthday-dailyviz/

What impact would this have?

CyanhydeJuly 4, 2018 at 12:31 pmThe above formula for calculating the probability of matches doesn’t account for having more people than items.

I.E. If you have 4 people picking 3 items (let’s say primary colors) there’s guaranteed to be an item match. However, especially for small numbers, the calculator (which is using that formula) still displays a probability of less than 100%.

MarkAugust 12, 2018 at 5:45 amI was unsatisfied with the approximation at first because it says there is a chance that no people share a birthday if there are 367 people. But it’s really really close to zero, so I guess it’s good enough for any realistic purpose.

yuperOctober 26, 2018 at 4:41 amAfter reading through this article I have found a major inconsistency that creates confusion. The question is asked in a way that states “how many people do you need for a 50% chance that 2 share a birthday?”

The answers are given in terms of the question actually being “how many people do you need for a 50% change that at least 2 share a birthday?”.

The missing, but very important operative phrase being “at least”.

Manish ManishJanuary 3, 2019 at 12:29 amSatisfying.

How Wars Change Birth Dates – chitnotesMarch 24, 2019 at 5:40 pm[…] probability and statistics in undergraduate (found a link of an article with similar content here for the mathematically savvy), where in a room of 23 people, there’s a 50/50 chance that two […]