Suppose someone continuously draws one card each time from a deck of cards (without replacement), until he/she gets the 3 of Hearts.

What is the expected maximum value among all of cards he/she draws?

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What is the set of values that cards have? (And how many cards are there to begin with; I suppose $52$?). –  Marc van Leeuwen Oct 6 '12 at 16:52
The value for A is 1, J is 11, Q is 12, and K is 13. The number of cards to begin with is 52 (4*13). –  CaptainObvious Oct 6 '12 at 16:56
@MarcvanLeeuwen Thanks for the comments on my answer, which I've now deleted. You were quite right - my answer was nonsense, and frankly, I'm a bit embarrassed by it, now that I re-read it. –  user22805 Oct 7 '12 at 18:05

Clearly chances are pretty good that the maximum will be $13$, so I'll propose the following alternative scoring procedure to compute the maximum attached to a given permutation of the $52$ cards. It starts with a score of $13$, and subtracts zero or more penalties (each of one unit) under certain circumstances.

1. A penalty is attributed if among the five cards { "three of hearts" (which I shall abbreviate #) and the four kings}, it is # that comes first of the five

2. A penalty is attributed if among the nine cards { #, 4 Kings, 4 Queens}, it is # that comes first of the nine.

3. A penalty is attributed if among the $13$ cards { #, 4 Kings, 4 Queens, 4 Jacks}, it is # that comes first of the $13$.

...

10. A penalty is attributed if among the $41$ cards { #, 4 Kings, 4 Queens, ... 4 fours}, it is # that comes first of the $41$.

It is easy to see that the maximum of the numbers in the permutation up to the # inclusive equals $13$ minus the total of the penalties, in all cases.

These events are of course highly dependent, but the linearity of the expected value does not require independent variables. We can therefore compute the expected value of the maximum as $13$ minus the expected values of each of the forms of penalties. But the latter are quite easily seen to be respectively $\frac15$, $\frac19$, $\frac1{13}$, ... $\frac1{41}$. So the answer for the expected value of the maximum is $$13-\frac15-\frac19-\frac1{13}-\cdots-\frac1{41}\approx12.34932$$

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