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 value of distinct suits (Spades, Hearts, Diamonds, Clubs) among all of cards he/she draws?

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With or without replacement? – joriki Oct 4 '12 at 17:15
without replacement. – CaptainObvious Oct 4 '12 at 17:19
Since the position of the 3 of hearts is uniform across the deck, this is $$\frac{1}{52}\sum_{i=1}^{52}E(S_i)$$ where $S_i$ is the number of suits represented in $i$ random cards without replacement, including the 3 of hearts. My guess is it sums up to close to four. – Arthur Oct 4 '12 at 17:53

We have $1$ for the suit of Hearts that's always present. By linearity of expectation, the expected number of the other three suits that we see is three times the probability of seeing a particular one of them. The $3$ of Hearts and the $13$ cards of the suit are in uniformly random order, so the probability that the $3$ of Hearts is the first of the $14$ is $1/14$, whereas the probability that at least one card from the suit comes before it is $1-1/14$. Thus the total expected number of suits is $1+3(1-1/14)=4-3/14=53/14\approx3.8$, which, as Arthur expected, is quite close to $4$.