Quick question: at least one player dealt pocket pair

[slickpoppa]

I could probably calculate this myself, but I' probably screw it up. If I am in a 10 handed game and I don't have a pocket pair, what is the probability of at least one other player having a pair? What about in a 6 handed game?

[KJL]

I am not sure if this is right but I think there is a 53% chance that someone has a pocket pair at a 10 person game, 1/17*9. So a 6 handed game would be 29%, 1/17*5.

[ajv]

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I am not sure if this is right but I think there is a 53% chance that someone has a pocket pair at a 10 person game, 1/17*9. So a 6 handed game would be 29%, 1/17*5.

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This would mean that in a 19 person game the probablity that one has a pair would be 18*100%/17=105.8%

[BruceZ]

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I could probably calculate this myself, but I' probably screw it up. If I am in a 10 handed game and I don't have a pocket pair, what is the probability of at least one other player having a pair? What about in a 6 handed game?

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Here is the solution for 8-handed from spectator's point of view (easily adapted to 9 or 5 opponents).

Here is the solution when you hold 33 and want the probability of an opponent holding a pair > 33.

[gol4pro]

This gets into conditional probability if you're a player at the table, because someone else is MORE likely to be dealt a pair if you are, and they are less likely to be dealt one if you aren't.

Anyways, assuming you're a spectator.

Chance of at least no PP being dealt 1-(16/17)^n. where n=number of players at the table.

so 1 handed = 1/17 (duh, we all know that)
2 = 256/289 = about 1/8
3 = 4,196/4913

Then just multiply it on out. Keep in mind, this is the chance that one is NOT dealt.

The probability of one being dealt is difficult to calculate, because you have to use binomial theorem (unless someone knows another way?! Tell me), which although doable, is too much of a pain in the ass for me to want to do right now.

[BruceZ]

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This gets into conditional probability if you're a player at the table, because someone else is MORE likely to be dealt a pair if you are, and they are less likely to be dealt one if you aren't.

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The conditional part just manifests itself in a simple change to the numbers used in the inclusion-exclusion calculation I linked to.


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Anyways, assuming you're a spectator.

Chance of at least no PP being dealt 1-(16/17)^n. where n=number of players at the table.

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That's an approximation since it assumes that the players' hands are independent, though it is a very good approximation in this case. This is discussed in the first inclusion-exclusion thread I linked to. Use the inclusion-exclusion method to get the exact answer (to any desired accuracy).