Ok, so in a game of Texas Hold 'Em you have two hearts in your hand. The next two cards places are both hearts, but the third one is a spade. What are the odds that by the end of the round you will have a Flush (5 of same suit).
I've asked someone else and they said it would just be 1/4 + 1/4 = %50, but for some reason I think it's more complicated than that.
Ok, you have 5 cards showing, meaning 47 remain. You have 4 hearts showing, meaning 9 remain.
There are 47-choose-2 ways of picking the next two cards. = 47!/2!(47-2)! =1081 possible ways. This will be the denominator.
Now then, the number of ways of picking a heart and something else (including another heart) is 9*46 = 414. That is, there are 9 possible hearts. After that, there will be 46 cards remaining in the deck. This is the numerator.
So the probability of picking up the 5th heart is 38.30%