A birthday puzzle

[irchans]

Variation 1 (fairly easy if you can do the orignial variation)

The manager of a movie theatre announces that a free ticket will be given to the first person in line whose birthday and birth hour is the same as someone else who has already purchased a ticket. You can get in line at any time. The birthdays and hours are distributed uniformly at random over 365 days and 24 hours. Which position in line should you take?


Variation 2
(Hard. I don't think there is a closed form expression for the answer, but maybe good approximations exist.)

The manager of a movie theatre announces that a free ticket will be given to the first person in line whose random number is the same as someone else who has already purchased a ticket. You can get in line at any time. The random numbers are uniformly distributed integers ranging from 1 to N. Which position in line should you take? (The answer is a formula which depends on N.)

[pzhon]

[ QUOTE ]

Variation 2

The manager of a movie theatre announces that a free ticket will be given to the first person in line whose random number is the same as someone else who has already purchased a ticket. You can get in line at any time. The random numbers are uniformly distributed integers ranging from 1 to N. Which position in line should you take? (The answer is a formula which depends on N.)

[/ QUOTE ]
Actually, I think this might have been asked in this forum before.

Let P(k) be the probability that the first duplicate is the kth person in line.

P(k) = (N-1)(N-2)...(N-(k-2)) (k-1) / N^(k-1)

We'd like to know when P(k+1) is smaller than P(k). When this happens, we want to be the kth person in line.

P(k+1)/P(k) = (N-(k-1))k / (N (k-1))

After a little algebra, we see that this is 1 when k(k-1)=N, or when k = 1/2 + sqrt(N+1/4).

When 1/2 + sqrt(N+1/4) is an integer, it is equally good to stand in this place or the next. These are the optimal places. For example, when N=2, it is equally good to stand in place 0.5+sqrt(2.25)=2 and 3.

When 1/2 + sqrt(N+1/4) is not an integer, the optimal place is given by rounding 1/2 + sqrt(N+1/4) up to the next integer. For example, when N=365, you want position ceiling(0.5+sqrt(365.25)) = ceiling (19.61) = 20.

[irchans]

Lovely solution. I didn't sit down to figure it out before reading your post. I should have!