finding the optimal strategy 
You have a deck of 32 playing cards. Somebody draws one card after another and shows them to you. At any point of time you may bet that the next card is black. If it is indeed black you earn $10, otherwise nothing. If you don't do anything you earn nothing as well. Find the optimal strategy.

In other words you should find the point of time where the quota of remaining red cards in the deck is maximal. 
Any hints how to to this?
 A: I take as additionnal assumption that if you do nothing, you bet on the last card rather than automatically losing.
There is no optimal strategy. Imagine the game this way. Instead of betting on the next card, you bet on the last card. Note that you still interrupt the game at any time you want you just bet on a different card.
The last card has the same probability of being black as the first one on the remaining card since you know nothing of how they are ordered. (i.e. every permutations is possible). The expected number of times you win this game is thus the same as in yours.
When do you win my game? Well you win if your last card is black, which ultimately happens half of the time.
This simple (symmetry) argument shows that the game is fair and there is no optimal stategy
A: You are looking for an hint or the solution?
Hint: wrote an explicit formula for $E[n,k]$ as a function of $E[n-1,k]$, $E[n-1,k-1]$, $n$ and $k$. Try to find a closed form that satisfy the above equation.
edit: $n$ are the number of cards, and $k$ the number of black ones. The solution works for any number of cards, so there is no need to worry for 32 or 52 decks.
A: You may make a bet at any time; in particular, you may either make a bet when there are an odd number of remaining cards or an even number of remaining cards.
With this in mind, we will prove the claim that no strategy exists which can make the chances of winning greater than $50\%$, by considering $2$ cases. For each, we will look at $P(r,b)$ where $r$ and $b$ denote the number of red and black cards remaining, so that $P(r,b)$ is the probability that $r$ red cards remain and $b$ black cards. Obviously, the chances of winning are greater than $50\%$ if $P(r>b) > P(b>r)$.
Case $1$: The number of remaining cards is odd (say, $2n+1$).
Observe that $P(0,2n+1) = P(2n+1,0)$. [It's just as likely to have no black cards as it is to have no red cards]. Also $P(1,2n) = P(2n,1)$, and so on until (and including) $P(n,n+1) = P(n+1,n)$. In total, $P(r>b) = P(b>r)$, so the odds of winning are not greater than $50\%$.
Case $2$: The number of remaining cards is even (say, $2n+2$).
Using reasoning similar to Case $1$, we have $P(0,2n+2) = P(2n+2,0); P(1,2n+1) = P(2n+1, 1)$; and so on until (and including) $P(n,n+2) = P(n+2,n)$. For all these possible outcomes, $P(r>b) = P(b>r)$, so the chance of winning is not greater than $50\%$.
The only distinction in this case is the outcome where $(r,b) = (n+1,n+1)$. Sadly this does not help our chances. Since the number of cards remaining is equal for each suit, our probability of winning is still $50\%$.
