Last Weeks Question:
On the opening night of The Engineers 150 year anniversary bash at the Monte Carlo Royale (please publisher! Ad!), the manager announces that they will give a free ticket and £100 of chips to the first person in line whose birthday is the same as someone who has already bought a ticket.
You have the option of getting in line at any time. Assume you do not know anyone else’s birthday, and birthdays are distributed randomly throughout the year, etc., what position in line gives you the greatest chance of being the first duplicate birthday?
Last Week’s Answer
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
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/D
P(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 = 0
x = (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 then 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.