Probability Question

[BruceZ]

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You shuffle a deck of 52 cards and start dealing off the top.

a. How many cards should you expect to turn over before you get to the second ace?

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Consider the 4 aces plus 1 non-ace. In order for the non-ace to occur before the 2nd ace, the non-ace must occur in positions 1 or 2 of these 5 cards, and this has a probability of 2/5. This is true for each of the 48 non-aces, so adding these 48 probabilities of 2/5 gives the expected value of the number of non-aces before the 2nd ace, which is 48*2/5. Adding the 2 aces, the expected value of the number of cards occurring up to and including the 2nd ace is 48*2/5 + 2 = 21.2.


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b. The 7th club?

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Consider the 13 clubs plus 1 non-club. In order for the non-club to occur before the 7th club, the non-club must occur in positions 1-7 of these 14 cards, and this has a probability of 7/14 or 1/2. This is true for each of the 39 non-clubs, so adding these 39 probabilities of 1/2 gives the expected value of the number of non-clubs before the 7th club, which is 39*1/2. Adding the 7 clubs, the expected value of the number of cards occurring up to and including the 7th club is 39*1/2 + 7 = 26.5.