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Imagine I am drawing cards (with replacement) from a standard deck (52 cards, 13 of each suit). I am trying to compute the expected number of cards I will need to draw in order to have at least one card of each suit.

My computer simulation tells me that the answer is 8.3, but I have no idea how to solve the problem using math.

Any ideas?

Thanks!

Solution: Thanks to Byron Schmuland who pointed me in the direction of the correct answer.

By linearity of expectation,

E(number of cards to get all suits) = E(number of cards to get 1 suit) + E(number to get a different suit) + E(number to get a third different suit) + E(number to get a 4th different suit)

E(number of cards to get 1 suit) = 1

E(number to get a different suit) = 1/(3/4) = 4/3 (by the geometric distribution)

E(number to get a 3rd different suit) = 1/(2/4) = 4/2

E(number to get a 4th different suit) = 1/(1/4) = 4

1 + 4/3 + 4/2 + 4 = 8.3333

Angband Addict
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    This is the coupon collector's problem. The theoretical answer is 8.3333 which suggests that your simulation is way off. Take a look at http://en.wikipedia.org/wiki/Coupon_collector%27s_problem –  Sep 02 '14 at 20:12
  • Ok, I will check my code :) Meanwhile, would someone provide the solution? – Angband Addict Sep 02 '14 at 20:14
  • Found the bug. It is indeed 8.3 :) Thanks for the tip about the coupon collector's problem. I will write up a solution once I have fully understood it. – Angband Addict Sep 02 '14 at 20:16
  • It's only the coupon collector problem if the card deck is infinite. For a 52-card deck the answer will be different. – MJD Sep 02 '14 at 20:20
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    I think it is the coupon collector problem, because the cards are drawn with replacement. So in a sense, the deck is infinite. – Angband Addict Sep 02 '14 at 20:23
  • Yes, you're quite right. I didn't realize that you were drawing with replacement. – MJD Sep 02 '14 at 20:38

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