52-pickup is the practical joke "game" where the mark finds out that the game consists of the proposer spraying all 52 cards of a deck into the air, letting them fall onto the floor, and then the mark has to pick them up.

Suppose all 52 cards are now on the floor, each card randomly and independently face up or down with probability 1/2.

What is the probability that the sum of the face-up cards (counting A, J, Q, K as 1, 11, 12, 13 respectively) is a multiple of 13?

Probability for each possibility of face-up cards:(1/2)^52.
Required Probability: {(1/2)^52}*(ΣS(n)); n is multiple of 13 upto 364.

S(n): coefficient of x^n in (1+x+x²+...+x^13)^4
S(n): coefficient of x^n in (1-x^14)^4*(1-x)^(-4)
Coefficient of x^n in (1-4x^14+6x^28-4x^42+x^56)*(1-x)^(-4)
There is a formula for computing coefficient of x^n in
(1-x)^(-4), but I dont remember it.

So, sum all such S(n) and substitute in above eqn to get the result :)

Posted by Praneeth on 2008-01-16 04:06:35