Question

In his spare time, Henry shuffles a standard deck of 52 playing cards. He then turns the cards up one by one from the top of the deck until the third ace appears. What is the expected number of cards Henry needs to turn up to get the third ace

1. The expected number of cards Henry needs to turn up to get the third ace is 31.8
Given: We need to find what is the expected number of cards Harry needs to turn up to get the third ace and Harry has a standard deck of 52 cards.
Now, we know there are 4 ace cards. Let’s consider the ace cards and non-ace cards separately.
There are (52-4) = 48 non-ace cards.
Let us consider that the non-ace cards will be cut by an ace card.
So, there are 5 possible ways in which the ace cards can cut the division, provided
We are having an equal number of non-ace cards in between the appearance of each ace card which means the setup is symmetric among the non-ace cards.
So let us name the spaces between ace cards as s1, s2, s3, s4, and s5.
Therefore, the position of the third ace card is equal to s1 + s2 + s3 + 3.
The expected value of this position is E[s1 + s2 + s3 + 3].
By linearity of expectation, E[s1 + s2 + s3 + s4] is E[s1] + E[s2] + E[s3] + 3.
As the setup is symmetric between the five places, E[s1] = E[s2] = E[s3] = E[s4] = E[s5]
And since E[s1 + s2 + s3 + s4 + s5] = E[s1] + E[s2] + E[s3] + E[s4] + E[s5] = 48
Therefore, E[s1] = E[s2] = E[s3] = E[s4] = E[s5] = 48 / 5
The expected value of position is E[s1] + E[s2] + E[s3] + 3
= 48 / 5 + 48 / 5 + 48 / 5 + 3
= 3 * 48 / 5 + 3
= 31.8
Hence, the expected number of cards Henry needs to turn up to get the third ace is 31.8
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