Probability Challenge

22 students is the max.

At 23 students the probability that the 23rd student will NOT share their birthday with any od the other 22 (and of course that htey are all exclusive) is:

364!/342!/365^22 = >50%.

With probability problems it is generally better to work on the probability that the event will not occur than the probability that it will.

Guest wrote: "...(and of course that they are all exclusive) is: 364!/342!/365^22 = >50%."

The exclusive prerequisite is important, because if they allow twins, the answer will be different!

Correct. The maximum students per class is 22. Nice job.

Ok lets look at the probability of a class with two students in it having two students in it having the same birthday is

P₁ = 1/365

Now if there were three students in the class the probability of identical birthdays would be

P₃ = 2/365 + 1/364

So if we had 19 students we would have

P₁₉ = 18/355 + 17/364 + 16/363 + ………. + 1/348 ≈ 47.6%

So the answer is 19 as 20 would put it well over the 50% mark.

I think that Masu has calculated the expected number of identical birthdays. This counts (for example) a class with two pairs of shared birthdays as two events, whereas it is only one in terms of the way the question was posed.

Hmmm...Two valid answers for the same question?

I believe our guest has correctly calculated the probability that the 'next' student, once you reach the 23rd added student, will have a duplicate birthday as someone in the room...but in an exponential (Ion Saliu) set where there may already be a duplicate...like the "pick three" lotteries.

Masu, on the other hand, may be giving us the correct permutation for a unique set, where we assume that each previous sutdent in the class is already known to have a unique birthday out of the possible 365, and the probability that the nth student will not duplicate an existing member...like the lottery-ball situation, or BINGO...which may better satisfy the "any of the students" criteria.