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.
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!)
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.
Answer
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.
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