Re: Quick question: at least one player dealt pocket pair
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.
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