The answer is 20.

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You should try to be the 20th person in line.

Suppose you are the Kth person in line. Then you win if and only if the K-1 people ahead all have distinct birthdays AND your birthday matches one of theirs. Let A = event that your birthday matches one of the K-1 people ahead B = event that those K-1 people all have different birthdays Then Prob(you win) = Prob(B) * Prob(A | B)

(Prob(A | B) is the conditional probability of A given that B occurred.)

Now let P(K) be the probability that the K-th person in line wins, Q(K) the probability that the first K people all have distinct birthdays (which occurs exactly when none of them wins). Then

P(1) + P(2) + ... + P(K-1) + P(K) = 1 - Q(K) P(1) + P(2) + ... + P(K-1) = 1 - Q(K-1)

P(K) = Q(K-1) - Q(K) <--- this is what we want to maximize.

Now if the first K-1 all have distinct birthdays, then assuming uniform distribution of birthdays among D days of the year, the K-th person has K-1 chances out of D to match, and D-K+1 chances not to match (which would produce K distinct birthdays). So Q(K) = Q(K-1)

(D-K+1)/D = Q(K-1) - Q(K-1)(K-1)/D Q(K-1) - Q(K) = Q(K-1)(K-1)/D = Q(K)(K-1)/(D-K+1)Now we want to maximize P(K), which means we need the greatest K such that P(K) - P(K-1) > 0. (Actually, as just given, this only guarantees a local maximum, but in fact if we investigate a bit farther we'll find that P(K) has only one maximum.) For convenience in calculation let's set K = I + 1. Then Q(I-1) - Q(I) = Q(I)

(I-1)/(D-I+1) Q(I) - Q(I+1) = Q(I)I/DP(K) - P(K-1) = P(I+1) - P(I) = (Q(I) - Q(I+1)) - (Q(K-2) - Q(K-1)) = Q(I)*(I/D - (I-1)/(D-I+1))

To find out where this is last positive (and next goes negative), solve x/D - (x-1)/(D-x+1) = 0

Multiply by D

(D+1-x) both sides: (D+1-x)x - D*(x-1) = 0 Dx + x - x^2 - Dx + D = 0 x^2 - x - D = 0x = (1 +/- sqrt(1 - 4*(-D)))/2 ... take the positive square root = 0.5 + sqrt(D + 0.25)

Setting D=365 (finally deciding how many days in a year!), desired I = x = 0.5 + sqrt(365.25) = 19.612 (approx).

The last integer I for which the new probability is greater than the old is therefore I=19, and so K = I+1 = 20.

Computing your chances of actually winning is slightly harder, unless you do it numerically by computer. The recursions you need have already been given.