Assume someone is drawing cards continuously from a deck of cards (without replacement) and stops until he/she gets the 3 or Hearts.

What is the expected minimum rank (Ace = 1, J = 11, Q = 12, K = 13, and so on) among the cards he/she draws?

I think there are three cases to consider: 1) ace before 3 of Hearts 2) ace after 3 of Hearts, but deuce before 3 of Hearts 3) ace and deuce after 3 of Hearts

For case 1) probability is 4/5, 2) is 4/25, 3) 1/25.

Thus, the expected value is 4/5*1 + 4/25*2 + 1/25*3 = 31/25

Is what I am thinking wrong?


You’ve calculated two of the probabilities incorrectly.

The probability of getting the $\heartsuit 3$ before getting any of the aces and deuces is $\frac19$: it must be the first of those $9$ cards. Let $p$ be the probability of getting a deuce but no ace before the $\heartsuit 3$. Either you get an ace before the $\heartsuit 3$, or you get a deuce but no ace before the $\heartsuit 3$, or you get the $\heartsuit 3$ before any of the aces and deuces, and these three events are mutually exclusive, so $\frac45+p+\frac19=1$, and $p=\frac4{45}$.

  • $\begingroup$ Your answer is right. $\endgroup$ – CaptainObvious Oct 4 '12 at 2:39
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    $\begingroup$ @hdj, if you are satisfied with Brian's answer, you can "accept" it by clicking in the check mark next to it. $\endgroup$ – Gerry Myerson Oct 4 '12 at 7:21
  • $\begingroup$ Wait a minute. @Brian, can you give out the way to calculate the probability (4/45) by other means instead of 1 minus others? $\endgroup$ – CaptainObvious Oct 4 '12 at 16:58

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