Probability that someone in a 10 handed game is dealt AA

[King Yao]

Probability that someone in a 10-handed game is dealt AA...is this correct or close?

1 - [(48/52) + (4/52 x 48/51)]^10 = 4.4%

[jwvdcw]

I'm no math wiz, but I think you should just take the odds of getting aces and multiply it by 10.

[PairTheBoard]

I believe it is much much more complicated than that. I have yet to see a simple solution.

Also I'm afraid I don't see the logic behind your formula.

PairTheBoard

[BruceZ]

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I'm no math wiz, but I think you should just take the odds of getting aces and multiply it by 10.

[/ QUOTE ]

That would be 10/221, and this is close enough for all practical purposes, but it double counts the times 2 players have AA, so to be exact, we must subtract the probability that 2 players have AA. The probability that 2 specific players have AA is 1/C(52,4) = 1/270,725 since there is just 1 way to choose all 4 aces out of 270,725 possible 4 card combinations. To get the probability that 2 players have AA out of 10 players, we multiply this by the number of ways to choose 2 players out of 10, which is C(10,2) = 45. This is now exact since no more than 2 players can have AA, that is, these are mutually exclusive. So all together the exact probability that someone has AA is 10/221 - 45/270,725 or about 4.5%.

[PairTheBoard]

There's problems with that. For one thing the events of each specific player being dealt Aces are not independent- which is required for adding probabliliteis. That's the worst problem. For another it double counts the rare case when AA are dealt twice.

PairTheBoard

[BruceZ]

[ QUOTE ]
I believe it is much much more complicated than that. I have yet to see a simple solution.


[/ QUOTE ]

See below!

6/C(52,2) * 10 - 1/C(52,4) * C(10,2) =~ 4.5%.

[BruceZ]

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There's problems with that. For one thing the events of each specific player being dealt Aces are not independent- which is required for adding probabliliteis.

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Independence is not required to add probabilities. Adding probabilities requires that the events are mutually exclusive. Events are mutually exclusive if they cannot both happen at the same time. Since two players can have AA, these are not mutually exclusive. Independence is required to multiply probabilities, such as if we did 1 - (220/221)^10. This is also a close approximation. Events are independent if the occurence of one event does not affect the probability of the other event.

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For another it double counts the rare case when AA are dealt twice.

[/ QUOTE ]

This is the only problem, and it is minor since the probability of two players having AA is so small.

[King Yao]

I am not confident I am right, that is why I ask...but here's my logic.

1 - [(48/52) + (4/52 x 48/51)]^10 = 4.4%

(48/52) = probability that the first card for the first player is not an Ace
(4/52 x 48/51) = probability that if the first card is an Ace, that the second card is not an ace

[(48/52) + (4/52 x 48/51) = probablity that both cards aren't an ace. This equals 220-1.

then I take it to the 10th power. each player has a 220/221 chance of not have AA. if the cards are independent (ah, now that I am writing this, I realize this is probably where the problem is), then raising them to the 10th power gets the probability that no one has AA.

Subtracting that term from 100% gets the chance that at least one player has AA.

Is the problem with my formula where I noted above in ()?

[King Yao]

thanks...it looks like jwv's much simpler approximation is much closer than mine. thanks.

[BruceZ]

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thanks...it looks like jwv's much simpler approximation is much closer than mine. thanks.

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In this particular case it is a little closer. Your approximation assumes independence, and his assumes mutually exclusive events. Yours gives a lower bound, and his gives an upper bound. That is, the exact answer will lie above yours and below his. They are both very close, and they are usually both very usable approximations for card problems. The exact answer when required is obtained from the inclusion-exclusion principle.