Chances someone has an ace (or any other holding)

[Joost1982]

You get dealt AA once in every 221 hands (statistically). So the chance is ~ 0,45%. Now, when there are 6 people on your table, can you multiply this chance times six to get the chance that SOMEONE has AA on your table? (so 0,45*6=2,7%) ?

I'm kind of a math n00b and I think this is not the correct way to calculate it, is it?

[pzhon]

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when there are 6 people on your table, can you multiply this chance times six to get the chance that SOMEONE has AA on your table? (so 0,45*6=2,7%) ?


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It's close, but not quite right. You have computed the average number of AA hands, 6/221 ~ 0.027149. If there could be only one AA hand at a time, this would be the probability. However, it is also possible (though quite rare) for two players to have AA. You've counted these hands twice. To count these only once, you should subtract the probability that two players get AA.

The probability that two particular players both have AA is 1/52C4 = 1/270,725. There are 6C2 = 15 ways to choose two players out of 6, so the probability that two players get AA is 15/270,725.

So, the probability that at least one player gets AA is 6/221 - 15/270,725 ~ 0.027094. So, you were not far off.

An approximation people commonly use is to assume that the hands are independent, as though they are dealt from different decks. The probability that 6 hands from different decks are all not AA is (220/221)^6, so the probability that at least one has AA is 1-(220/221)^6 ~ 0.026844. That's less accurate in this case, since the events that the players get AA are closer to exclusive than independent.

It's a bit odd to ask about the probability while ignoring what you have. If you have one ace, you block half of the AA combinations, which roughly cuts the probability of running into AA in half. If you don't have an ace, that raises the probability that an opponent has AA.

[Joost1982]

Thanks for the calculations!

[AngusThermopyle]

When the probabilities are small, adding them together (which is what multiplying by 6 is doing) does not introduce significant errors, so it suffices for ballpark computations.

When the probabilities are large, you can see where there are problems.

Six handed game:
Odds a specific player is suited: 12/51 , or around 24%.
Using your method to find odds that at least one player is suited would give around 6*24 = 144%, which is clearly false.