How Many Distinct Birthdays in a Room of k People?

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SUMMARY

The discussion focuses on calculating the expected number of distinct birthdays, j, in a room of k people on a planet with n days in a year. The formula derived is j = n - n * (1 - 1/n)^k, which accounts for the probability that a given day is not a birthday for anyone. The conversation also touches on combinatorial aspects, specifically using binomial coefficients to express the number of ways to arrange birthdays. The final expected number of distinct birthdays is given as nk/(n+k-1).

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This is a restatement of the vocabulary problem which I introduced in https://www.physicsforums.com/showthread.php?t=293553. Perhaps these terms will be more familiar/less ambiguous.

Suppose we are on a planet where each year has n days, and in a room with k people. If birthdays are uniformly distributed throughout the year, how many distinct birthdays, j, do we expect to find in the room? (Alternatively, how many birthdays n-j are NOT represented in the room, on average?)
 
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Here's the answer (someone else's idea): j = n - n * (1-1/n)^k. The probability that a given day is nobody's birthday is (1-1/n)^k, so the expected number of days that are nobody's birthday is n * (1-1/n)^k.
 
I think that, since its distinct birthdays (from what I understand, distinct bdays are only the ones that don't coincide on the same day), you will have nCk bdays total, thus the expected number on any given day would be (nCk)/n since they're uniformly distributed.

There will also be nC(n-k) bdays that don't happen, once again the expected number would be nC(n-k)/n since they're uniform. I'm not sure about this, but I think its intuitive.
 
Let X be the number of distinct birthdays, and for 1 \leq i \leq n define Xi to be 1 if there's at least one person whose birthday is on the ith day, and 0 otherwise. Then X = X1 + X2 + ... + Xn, so:

E(X) = \sum _{i=1} ^n E(X_i) = nE(X_1)

E(X1)
= Prob(at least one person has their birthday on day 1)
= 1 - Prob(no one has their birthday on day 1)
= 1 - (# of ways to arrange k birthdays amongst n-1 days, allowing repetition)/(# of ways to arrange k birthdays amongst n days, allowing repetition)
= 1 - \binom{n+k-2}{k} / \binom{n+k-1}{k}

So the final answer is:

\frac{nk}{n+k-1}
 
AKG said:
define Xi to be 1 if there's at least one person whose birthday is on the ith day, and 0 otherwise.
But doesn't the question as for distinct birthdays, ie Xi would be 1 if there is only one birthday on day i and 0 if there is not only one birthday on day i?
 

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