Find the expected number of 5's in a 6 card hand. I am having trouble with finding the expected value. I feel like my answer is wrong and I cannot figure out what it is that I am doing incorrectly. The problem is a cribbage hand consists of 6 cards dealt from a standard 52 card deck. Each player is required to discard two of these cards into a separate hand called 'the crib.' Find the expected number of 5's in a 6 card hand. I defined $X$ to be the number of 5's in a 6 card Cribbage hand from a standard 52-card deck. So, $X$ is hypergeometric. Using the formula , $$E[X] = n \frac{r}{N}$$ I defined $N = 52$ since it is the sample size, $n = 6$ since it is the subgroup size, and $r = 4$ since there are only 4 fives in a deck (sample size). When I plug everything in I get $$E[X]=0.4615$$ I do not know if I am defining my variables wrong or if there is something I have to do beforehand.

Your answer is correct. Each card contributes an expected $\frac{4}{52}=\frac{1}{13}$ to the expected number of 5s and there are 6 cards so $$E(\text{number of 5s})=\frac{6}{13}\approx 0.4615$$ Any reasons why this answer seems wrong to you? A couple of simple point that might help to clear possible confusion:

The expected number of 5s is different from the probability of at least one five (because the expectation counts hands with 2 5s in them twice).

$E(X+Y)$ is _always_ equal to $E(X)+E(Y)$, it doesn't matter if $X$ and $Y$ are correlated, which is why it is often easier to calculate expected values than probabilities.