[Robk]

One way of doing it is using the following formula for drawing without replacement.

Say you have G "good" and B "bad" marbles in a jar making N total marbles (ie G + B = N), and you draw out n of them ( you don't put one back after drawing it). The probability of having g "good" and b "bad" ones (ie g + b = n) is given by P = ((G c g)*(B c b))/(N c n). (where x c y = x!/(y!*(x-y)!). Now we can apply this to the card situation. For example when there are no other spades out we can find the probability of making the flush by finding the probability of missing and subtracting this from one. You miss if you draw either 4 or 3 "bad" cards, ie non spades. So N = 42 (the number of unseen cards), n = 4 (the number you will get), G = 10 (number of spades left), B = 32 (the number of nonspades). Now when b = 4, g = 0, we have P = (10 c 0)*(32 c 4)/(42 c 4) = .3213. And when b = 3, g = 1, P = (10 c 1)*(32 c 3)/(42 c 4) = .4431. Now if you get any more than one "good" card you will have a flush. So the chances of missing are .3213 + .4431 = .7644, ie you'll make your flush with P = 1 - .7644 = .2356. You can find the other probabilities in the same way. There might be a neater way of doing it but this is all I could think of right now.