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3

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 ...

-4

Jack is definitely right, Frisbee, thus you're very wrong. The odds of getting dealt pocket aces are 1/221 (4/52*3/51). Odds that another ace has been dealt out preflop is 2/50, so roughly 4%.

2

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 * = ...

0

I have seen it on TV a few times Based on hand independent it is 0.000256 (quad) * 0.0017 (boat) = 0.00000044 But since they share a pair on the board less than that I think the real calc is 0.0017 * 0.00101 = 0.00000172 From the boat there are 45 cards out and need to match the pair on the board So the odds of getting quadded are: (2/2) / (45/2) = ...

0

A squeeze play is two players in the pot and you are in late position and you want to blast them both off the pot. If you have a player that will open light and based on their image another player that will call with a wide range based on that and you are in late position then you are getting good pot equity to take down the pot pre flop. What is ...

1

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 ...

2

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)

-1

I don't know why no one has answered this for you yet. It's a reasonable thing to want to know, but as others may have mentioned, should be more for novelty than for using to make your in-game decisions. Anyways, here's a breakdown I put together. A lot of figures are fractions of percents, so I simplified it by rounding everything up to the nearest whole ...

-1

You shouldn't have played it so aggressively. Someone with even a King is likely to call you if they are the only person left to act. At that point, having such low pairs, there a lot of different ways you can lose over the course of the next two cards. They could make a higher two pair. The board could pair, counterfeiting your two pair. Or like what really ...

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