Hard | Probability |

What is the expected number of cards that need to be turned over in a regular 52-card deck in order to see the first ace?

Hint

Hint: the locations of the four aces in the deck divide it into the five groups X1, ...,X5.

Answer

53/5

Solution

Define X1,X2…X48, such that Xi = 1 if ith card turns over before 4 aces, 0 otherwise. Thus, total number of cards turned to see first ace = 1 + sum (Xi)

using linearity of expectation, E(X) = 1 + sum ( E(Xi))

Now consider the ith card and the four aces, all the orders are equally likely: X,A,A,A,A Or A,X,A,A,A, or 3 others. They are equally likely when we have no knowledge about position of these Aces with respect of the deck. Hence we have 5 equally likely slots for ith card: 1 A 2 A 3 A 4 A 5. We are interested in 1st slot.

Hence E(Xi) = 1/5 for all i = 1 to 48

Thus E(X) = 1 + 48/5 = 53/5 = 10.6

This can be done by recursive equation f(n) = (4/n)*1 + ((n-4)/n)*(1+f(n-1), where f(n)=expected cards to flip in a deck of n cards, but I definitely can't solve this without a computer.

A shorter explanation is to consider the 52 cards uniformly distributed over (0,1), so on average they're at k/53 for k=1,2,3,…,52. The four aces are on average at 1/5, 2/5, 3/5, 4/5. So 0.2=k/53 implies k=10.6, done!

using linearity of expectation, E(X) = 1 + sum ( E(Xi))

Now consider the ith card and the four aces, all the orders are equally likely: X,A,A,A,A Or A,X,A,A,A, or 3 others. They are equally likely when we have no knowledge about position of these Aces with respect of the deck. Hence we have 5 equally likely slots for ith card: 1 A 2 A 3 A 4 A 5. We are interested in 1st slot.

Hence E(Xi) = 1/5 for all i = 1 to 48

Thus E(X) = 1 + 48/5 = 53/5 = 10.6

This can be done by recursive equation f(n) = (4/n)*1 + ((n-4)/n)*(1+f(n-1), where f(n)=expected cards to flip in a deck of n cards, but I definitely can't solve this without a computer.

A shorter explanation is to consider the 52 cards uniformly distributed over (0,1), so on average they're at k/53 for k=1,2,3,…,52. The four aces are on average at 1/5, 2/5, 3/5, 4/5. So 0.2=k/53 implies k=10.6, done!

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