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.

The fewest draws is 4 and the greatest number possible is 40. Therefore the solution for expected outcome is the finite series given by the following:

4*P(4 cards) + 5*P(5 cards) + ... 40*P(40 cards)

Case 1: 4*(1 * 39/51 * 26/50 * 13/49)

In this case, each sucessive draw reveals a new suit. For all other cases, let 'N' imply 'new suit' and 'O' imply 'old suit'. Each successive case will include 39, 26, and 13 in its numerator for each combination of Ns and Os. For example with case 2:

Case 2: 5*[(P(NONNN) + P(NNONN) + P(NNNON)]

Of course the first N refers to the first card which is always a 'new suit' and the probability of its draw is identically 1 which I will hereafter disregard.

Case 2: 5[(12/51 * 39/50 * 26/49 * 13/48) + (39/51 * 24/50 * 26/49 * 13/49) + (39/51 * 26/50 * 36/49 * 13/48)]

Factoring out common elements shows that the expression is:
(5*39*26*13*47!/51!)*(12+24+36)

And we see that each term in the series will be similarly constructed given the proper indexing for 'old suits'. I am beginning to see something with i-2 choose i-4 where i is the number of cards to make our hand., but it is still fuzzy.

Posted by Eric on 2006-06-06 12:08:06