If you keep drawing cards from a shuffled deck, what is the expected number of draws to complete at least one card from each suit?

The expected value is the average number of draws that it would take if you repeated the experiment many times.

(In reply to Beginning.. by Eric)

I have now calculated the expected outcome as the product of:

39*26*13*12!/51! and the triple sum:

Sum from i=2 to 14 of the sum from j=i+1 to 27 of the sum from k=j+1 to 40 of the expression k*(26-i)!*(39-j)!*(52-k)!/[(14-i)!*(27-j)!*(40-k)!]

This is so damn messy I am almost certian to be making an error. Nevertheless, the approach I am taking is to consider every possible combination of NOOOONONOONOOOOOOOOOOO... strings ('N' again means 'new suit', 'O' means 'old suit'. Where the first card is always a new suit and there are no more than 40 cards dealt. i,j,and k refer to where in the string the second, third and fourth 'new suits' appear. Then I calculate the expected outcome of each particular combination and sum across all. Because each term has 39*26*13*12!/51! in common I have factored it out.

I sure wish some programmer would come along and finish this off...

Edited on June 6, 2006, 10:01 pm

Posted by Eric on 2006-06-06 21:46:17