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Frisbee's answer is correct. But it only answers for eight players (although the other numbers are hidden in that table). Since the OP asked for 8-10, I will add the answers for nine and ten players. The odds of an ace not occurring in the next x cards dealt (after you're dealt two aces) is ( 48!/(48-x)! )/( 50!/(50-x)! ). That's the probability that no one ...


These are the chances (assuming you have no ace): These are valid only preflop assuming that there are 50 cards left in the deck (you are holding 2) and you are one of the players (2 players = 1 opponent with 2 cards; 3 players = 2 opponents with 4 cards and so on)


Pick is it 8 or 10 At 8 I think it is 0.4114 Or 1.4306 : 1 This is for exactly one ace out - (not two) Using combination (2/1) * (48/13) / (50/14) (2 aces need 1) (48 non ace need 13) / (50 cards need 14) Both other aces out would be 0.0743 Add them for 1 or 2 aces out = 0.4857 You can get that same number with 1 - 48 / 50 * = ...


Squeeze play is a specific move that works under the following circumstances: A very aggressive and loose player raises preflop (wide range) A loose player after him calls, since he knows the raiser will have a wide range You currently have a relatively tight table image What you do with a squeeze play is re-raising the pot big, such that the original ...

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