You have a standard pack of 52 playing cards. You then shuffle them and begin to draw out cards until you have three of a kind. What is the most likely number of cards drawn when this happens?

You then shuffle another pack of 52 playing cards into the pile. What happens to the expected number of cards now? (i.e. does it double / halve / stay the same?)

A binomial random variable X represents the number of successes that occur in k trials for a given experiment that can result in either a "success" or a "failure". The expected number of successes in k trials, with success having probablility p, is given by E[X]=kp.

In our case, a "success" occurs in the event that, having chosen n cards, at least one and no more than one of the "n choose 3" ( =n!/[(n-3)!*3!] ) subsets of three cards contains a three of a kind. The probablity of success for any given 3 card subset is given by (3/51)*(2/50)=3/1275, since, given the first card, there is a 3/51 chance that the second card is the same rank, and given the first two cards, there is a 2/50 chance that the third card is the same rank as the first two.

So, for this problem, the number of trials k is given by "n choose three", where n is the number of cards we have chosen so far. p is as already given above, 3/1275. We wish to equate this such that kp=1, that is, the expected number of 3 of a kinds just equals one.

So we have

n!/[(n-3)!*3!] * 3/1275 = 1

n(n-1)(n-2)/2550=1

n^3-3n^2+2n-2550=0.

Solving the above for n yields a number between 14 and 15, 14 yielding an expected number of three of a kinds less than 1, and 15 yielding a number greater than 1. So we will say that 15 cards is the expected number for at least one three of a kind.

If we increase the number of decks to two decks, then p, the chance of a success, will change. It will now be given by 7/103*6/102 = 7/1751.

now we have

n!/[(n-3)!*3!] * 7/1751 = 1

This results in a value between 12 and 13. So now 13 is the expected answer.