I have a probability problem with cards and the expected value of a card rank.
I have a deck of 52 cards. I draw cards without replacement. While drawing cards from the deck, a first ace is drawn at rank $k$ (that is the $k^{th}$ card drawn is an ace, all previous were not). We want to find the expected number of additional draws until we get an ace.
My idea is to follow this route:
If I call $X$ the random variable of the rank of the second ace, $N=52$ the total number of cards, $p = N -k$ the remaining number of cards after the first ace is drawn, the idea is to plug the expectation value of $X$ $$E[X] = \sum_{i=1}^{p-3}i P(X=i)$$
But the formula doesn't seem to simplify. What would be your take at this?
Best Answer
The five strings of non-aces can be permuted amongst each other, conserving probability. Note that some of the strings may be empty and where there are multiple empty strings, the 120 possible "abstract" permutations will produce fewer than 120 physical permutations of the cards.
The symmetry implies that the expected length each string of non-aces is the same, or 48/5 for a standard deck of cards.
The expected waiting time to the first ace, or from the first to the second ace, or ace $k$ to ace $k+1$, is the preceding number $+1$. That is the expected length of a full interval of non-aces followed by an ace. So 53/5 is the answer for a standard deck.
The general answer is (cards+1)/(aces+1), by the same argument.