Will I get called? Help!

Let's say its folded to me in the cut-off with AKo in a no-limit cash game, and I make a raise of 5x the bb. Let's also say I know my three remaining opponents have a calling/raising range of 77+, AKs, AKo. What's the possibility of one of them playing with me? Does the math change if my hand was a pair of queens, or does the formula work no matter what?

Rick.

[AaronBrown]

You have listed 64 hands another player would call on. Before you pick up your hand, any other player could have 52*51/2 = 1,326 different hands, so the probability of a single opponent calling you is 64/1,326 = 4.8%.

Once you pick up your hand, there are only 50*49/2 = 1,225 different hands another player can have. Your AKo knocks out 3 AA, 3 KK, 2 AKs and 5 AKo, for 13 hands. So the chance of one particular other player calling you is 51/1,225 = 4.2%. It's approximately (but not exactly) correct that the chance of at least 1 out of three other opponents calling you is 1 - (1 - 51/1,225)^3 = 12.0%.

If you had QQ you would knock out 5 QQ, so there would be 59/1,225 remaining hands and the chance of at least 1 out of three other players calling would be approximately 13.8%.

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You have listed 64 hands another player would call on. Before you pick up your hand, any other player could have 52*51/2 = 1,326 different hands, so the probability of a single opponent calling you is 64/1,326 = 4.8%.

Once you pick up your hand, there are only 50*49/2 = 1,225 different hands another player can have. Your AKo knocks out 3 AA, 3 KK, 2 AKs and 5 AKo, for 13 hands. So the chance of one particular other player calling you is 51/1,225 = 4.2%. It's approximately (but not exactly) correct that the chance of at least 1 out of three other opponents calling you is 1 - (1 - 51/1,225)^3 = 12.0%.

If you had QQ you would knock out 5 QQ, so there would be 59/1,225 remaining hands and the chance of at least 1 out of three other players calling would be approximately 13.8%.

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Thanks for the reply. I got as far as you did in figuring out the number of combinations that were left, and then dividing this into the number of possible combinations. Its the "but not exactly" part that I'm confused about. How would one deduce the 'exactly' part of it? I'm unsure if it matters though, as I don't know how inaccurate the 'non-exactly' answer is.

I believe your answer to be (very?) slightly inflated, a miniscule amount over. There will be times, for instance, with both opponents have a playable hand, as well as times all three will. This is of course unlikely, so your answer can't be far from off. I wonder what the math is to get the 'exact' solution?

[AaronBrown]

It's not very inaccurate. It ignores the dependence among people's hands. For example, if one player doesn't call, he's somewhat less likely to hold an Ace than average, making the next player slightly more likely. This would make more difference if you assumed, say, that players called on any hand with an Ace. But you assume enough variety in the hands that it doesn't make much difference.

The full calculation would require enumerating many different cases. It's not hard to write a program to do it, but it's a tedious job to do by hand. Anyway, give the probable error in your assumptions, there's no point to doing a precise calculation.

[jtr]

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Anyway, give the probable error in your assumptions, there's no point to doing a precise calculation.

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Excellent point Aaron. I see this mistake often in poker-related calculations. (Usually mine.)