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Old 09-17-2005, 07:26 PM
AaronBrown AaronBrown is offline
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Join Date: May 2005
Location: New York
Posts: 505
Default Re: chances of flush vs flush

With 47 cards out, there are 47*46/2 = 1,081 possible sets of pocket cards. There are four spades out higher than your 8, plus four other spades. These can be arranged into pocket cards in 24 ways. 24/1,081 = 2.2%. With four other players, there's about an 8.8% chance that one of them was dealt two spades that give a higher flush than yours. Many of these hands would have folded preflop in most games.

One of the four spades higher than 9 could be paired with 39 non-spades in the deck, that's 4*39 = 156, or a 14.4% chance than a single other player holds a higher spade. If exactly one other player holds a spade, there are 7 cards that can beat you. The chance of one of them showing up on the turn or river is 1 - (38/45)*(37/44) = 30.0%. Multiplying by the 14.4% chance that gives a 4.2% chance that one player will beat you.

It gets a bit complicated to figure four other players, but it's not a bad approximation to multiply by 4 again and get 16.8%. Add to the 8.8 and you get 25.6%, or about one chance in four that your flush will lose if any player holding a spade stays in to showdown. The actual probability of loss is closer to your 8% estimate, given how many of the hands that beat you are unlikely to have been played.
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