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I have a unique opportunity to present to a very large group of people ($2{,}000$ in a theatre hall) about how chance works and how human intuition can be way off to guess likeliness.

Rather than present the classic birthday problem to them (that is discussed many times very well in Math SE), I wanted to have some audience participation instead to illustrate the issue of chance and human intuition on the answer more directly with them, by asking a series of questions (see below) to get to the most common birthday in that entire audience and count how many hands are up for that.

If I were to do this, how many people are likely to share that most common birthday?

Obviously, the many permutations among $2{,}000$ people mean that every single person will share so many birthdays with so many other people, and some birthdays will be less likely than others, but what number will I see specifically for that most common one? You can take any kind of confidence criteria that would be reasonable, such as a minimum number to expect with $50\%$ certainty.

That way when I know the day-of exactly how many people are actually attending, I can update the final guess appropriately, and of course the answer can be more universally applicable to any crowd of $N$ people similarly.

Edit: Since the number isn't likely going to be very high, a follow-up suggestion that I liked was the question "Given that answer, how many birthdates should I look to ask for, such that $y$ people raise their hands?"

Then I may see that asking just $10$ dates gets over $100$ people, or $15$ gets $300$ etc. and I can get that impressive number I'm looking for.

  1. To get the final answer, the questions I'd ask would be: "Whose birthday is in the first half of the month?", then list the 6 months that seems to get more hands and pick the winner, then ask of those "Whose birthday is among days 1-10? 11-20? 21-28/30/31?", then ask whichever of those 3 groups has the most hands up "Is it odd or even?" and then list the 4-6 options and count each until we have a winner. I'll have assistants all over the theatre to help with the counting. I'd appreciate a comment if there's a more efficient way to do this that wouldn't be confusing. (Edit: See comments; this actually isn't as effective as I thought, so other suggestions welcome!)

  2. I think this approach for such a large crowd would be most effective, since intuition without any statistics that would lead someone thinking you'd need $183$ people to have a $50/50$ chance that two share a birthday and be shocked to hear it's $23$, could apply the opposite direction and I could suggest that since $2000/365 = 5{.}47$, maybe $5$ or $6$ people would share the most common birthday to add more of an impact when we see the actual answer. I could just pick a random date or my birthday and see the number of hands, but I think this "most common birthday" approach could be really effective.

Henry
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Benny Lewis
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    Especially if these people were mostly born in the same year (as you might typically get in a university class, for example), the assumption that all days are equally likely is actually pretty far from the truth. Besides some seasonal effects, relatively few people are born on weekends and holidays (elective caesarians and induced labour tend not to be scheduled for those days, because the obstetricians would rather not work then). – Robert Israel Jun 22 '17 at 03:01
  • Then there was the time I had a pair of identical twins in my class... – Robert Israel Jun 22 '17 at 03:01
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    The "Balls into Bins - A Simple and Tight Analysis" paper seems to give a formula for this where $n=365$ and $m=2000$. I am not good enough at math to understand or apply their formula ... Someone else please Google this paper, use their formula, and report on the expected number. – Bram28 Jun 22 '17 at 03:23
  • @RobertIsrael That's true but this group will be pretty varied by age, so the instance of weekends on particular birthdays is likely to get averaged out surely? But I can see how this approach would definitely be affected for an audience likely to be the same age. Good food for thought! – Benny Lewis Jun 22 '17 at 03:30
  • Do you know anything about the subject that you are going to present to 2000 people? – Carsten S Jun 22 '17 at 04:33
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    @CarstenS Wow, that's a little harsh, no? The subject is people's misperceptions about luck vs statistics. The audience won't be receptive to Mathematics, so I have to simplify my explanations when Math is involved. My background is in engineering, so I had a lot of statistics in my course and I'll understand answers people may give, and want to be correct if I allude to numbers in my talk, but it won't be about the actual Mathematics but our understanding of chance. But I'm rusty, so I'm not able to get to this answer myself... which is why I asked. – Benny Lewis Jun 22 '17 at 04:54
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    This problem is discussed in a Wikipedia article. The research on this question has been on finding asymptotic formulas for ranges the maximum will fall in with high probability. It's unlikely there will be a formula for the most likely maximum. The best you can do in this situation is probably to do repeated random simulations. – user49640 Jun 22 '17 at 05:07
  • Among other things wrong with your method, counting people in an audience is hard. We do this frequently for formal votes in my town. If the numbers are close it can take several minutes to get an answer. Now do this repeatedly to perform your procedure. – David K Jun 22 '17 at 05:56

2 Answers2

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The method to get to the most common shared birthday won't work as pointed out by Bram28. But it's easy to get to an estimate of the number, using the fact that on average on each day there are $\lambda = \dfrac{2000}{365}\approx 5.48$ birthdays. We then approximately have a Poisson distribution for the probability to have $n$ birthdays on some given day:

$$P(n) = \frac{\lambda^n}{n!}\exp(-\lambda)$$

This means that the expected number of days with $n$ birthdays will be $\mu(n) = 365 P(n)$. The number of days with $n$ birthdays will be approximately distributed according the the Poisson distribution with mean $\mu(n)$. This means that for $n$ such that $\mu(n) \geq \log(2)$, the probability of there being a day with $n$ birthdays will be larger or equal to 50%; this means that $n$ must chosen to be $13$ or smaller.

Count Iblis
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  • Hey! So being 'surprised' when the number is over 15, and 'really surprised' if the number is over 20 wasn't too far off, it seems like! – Bram28 Jun 22 '17 at 03:50
  • @Brian28 Hey, your intuition was close! Good job! – Benny Lewis Jun 22 '17 at 03:52
  • @Brian28 I guess even with 13 people for the most likely birthday, if I just list the 10 birthdays at the end and get around 10 common birthdays per date, that gets a bunch of people putting their hand up. So it wouldn't be about finding the one day, but showing how a particular sample of birthdays have over 100 people sharing birthdays in various ways. That's not quite the same, but still its own version of impressive, no? – Benny Lewis Jun 22 '17 at 03:55
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    @BennyLewis Oh ... That's not bad ... Yeah, so the question may be: how many days of the year do we need to pick so that at least (10%? 20%? What would be a good number here?) of the audience has their birthday on one of those days? But still, the problem is how to find those days and people interactively ... Sakerh Malyala suggested maybe putting a web form so people with their phone can pick their birthday ... And the computer does the rest .... – Bram28 Jun 22 '17 at 04:00
  • @BennyLewis I am guessing (there's me guessing again ...) that the number of days you need in order for at least half of the audience to have their birthday in one of those days is a good bit less than 100 ... In fact, I would not be (too) surprised if this was below 50. So yeah, that could be counterintuitive: that you only need less than 20% days in the year int order to get half of the birthdays covered ... but as I said, I am guessing again: we should ask the real mathematicians if they can give a prety good estimation as to the number of days needed ... I bet they can .... – Bram28 Jun 22 '17 at 04:14
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I don't have any answer to your main question, but I do want to point out that your method may not find the date with the most shared birthdays. For example, suppose there are 10 people that all share April 10 as their birthday, and that no other date has that many shared birthdays. However, now also suppose that 900 people have their birthday in the first 6 months, and 1100 in the second six months ....

Bram28
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  • its an intuitive, but flawed method. +1 agreed. – Saketh Malyala Jun 22 '17 at 03:03
  • @SakethMalyala I wonder what the chances are that this method would result in the most shared birthday date ... I bet they would be fairly small, actually... an interesting mathematical question, maybe. Or more to the point: is there a better method? What is the most efficient way to find the most shared borthday? Or at least a very highly shared birthday? – Bram28 Jun 22 '17 at 03:08
  • That's true! Drafts... Is there some way to get around it in a small number of questions? I guess any similar iteration of such questions is going to be biased towards the group having a higher number, even if a specific day doesn't :-/ Is there a series of questions that is at least /less/ biased towards doing that? – Benny Lewis Jun 22 '17 at 03:23
  • @BennyLewis Good questions! And an interesting problem in and of itself! If only I had answers .... :(. But I did find a paper that seem to provide a formula to your main question :). ... If only I could understand it ... :(. I mentioned the paper in the Comments to your Post, so I think someone will be able to give you an answer to that question at least fairly soon ... But yeah, you of course need to have an interactive demonstration that works! – Bram28 Jun 22 '17 at 03:26
  • @BennyLewis By the way ... my gut feeling is that you don't get nearly as 'shocking' of a result as with the original birthday problem. I'd be surprised if the number turns out to be over 15, and really surprised if it was over 20 ... And even with 20 people in the audience sharing the same birthday I think you'd get a lot of shrugs .... But hey, it's birthdays! Maybe my intuition is just as wrong on this one as it is with the original birthday problem! :) – Bram28 Jun 22 '17 at 03:37
  • make a forms, display the link on a projector, and write a simple program on sheets to compute the most shared birthday(s) – Saketh Malyala Jun 22 '17 at 03:44
  • @BennyLewis Indeed, with a group of 2000, maybe the thing to do is to have an interactive demonstration of the law of the big numbers. That is, your audience may expect more variation regarding something than there is ... thinking of something interesting ... – Bram28 Jun 22 '17 at 03:46
  • @Bram28 I know - that's the main reason I want to do the Math(s) ahead of time. It could turn out to be 100, which would be impressive if I introduce the idea of 5 or 6, but if it's 15-20 then I've to take another approach to make the same impact. If you think of something that could be impressive that I could get from that number of people, do share! Of course, people aren't super imaginative, so I could get them to answer a subjective question that could have lots of answers but that they actually pick just a few of them, but that's more of a psychological thing since they pick the answer. – Benny Lewis Jun 22 '17 at 03:50
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    @BennyLewis Well, Count Iblis seems to have found a fairly reliable answer ... the expected number is about 13 ... so yeah, that just gets you a bunch of shrugs. And the problem is that you're up against the Law of the big numbers. When you flip a coint 10 times, we can easily get a large deviation from the expected number (e.g you get 80% heads where the expectation is 50%), but with 2000 flips, the expected deviation will be much smaller. So again, I would recommend to work with the Law of Big numbers, i.e. to do something that demonstrates that ... in some interactive way. – Bram28 Jun 22 '17 at 03:58
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    Maybe you can use the audience to get a large sample of some other probability paradox--such as having them pair off in 1000 pairs and each pair simulates one game of the Monty Hall problem. – David K Jun 22 '17 at 06:03
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    @DavidK Yeah, good idea! Benny Lewis: maybe you should create another post, and ask that more general question: giiven this opportunity with 2000 people in the audience, what fun things can you do with the audience to demonstrate some things about chance, probability and statistics? I bet you'll get a nice variety of answers. – Bram28 Jun 22 '17 at 11:55
  • Good idea indeed! OK, I just wrote up another question that focuses more on the answerer giving a better suggestion: https://math.stackexchange.com/questions/2332412/interactive-problem-with-large-end-result-proof-for-expected-result-to-demon Thanks! – Benny Lewis Jun 22 '17 at 14:24