[ThinkQuick]

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If you want to stick with your assumptions about what's in the 18 hole cards that you don't see, then do the rest of your math with fewer cards left in the deck. (i.e. 6/32 * 5/31 . . . )

You'll end up with something very close to Sklansky's numbers. His are more accurate because you really don't know anything about other people's hole cards. There could be zero spades there or 10 spades there. They are just as unknown to you as the part of the deck that hasn't been dealt yet.

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The first paragraph of this response is great but the second part isn't correct.

To the OP:
You can get the exact same probability if you properly account for the number of spades in the other players hands. You said there would be 3 spades in these 18 cards.
But you need to apply basic probabilities. You are sequestering 4 spades for your hand and the flop, as well as 1 non spade for the flop).

so their hands will contain (9spades/47remaining cards)*18 cards dealt = 3.446 spades within their 18 cards.

Now there are 13spades - 2 in your hand - 3.446 in other hands - 2 on the flop = 5.554 spades remaining

there are 52 - 20 - 3 = 29 cards to come.


5.554 spades in 29 cards = .1915
this is the same as
9 spades in 47 cards = 0.1915

therefore p of getting a flush will be the same in both cases

note that the 9/47 even showed up our calculations.

anyways the point is you can do it 'your' way with the proper math, and it comes out to the same answer. However, the 'D.Sklansky' way is MUCH simpler.