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]

[ QUOTE ]
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]

[ QUOTE ]
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.

[/ QUOTE ]

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.

[ QUOTE ]
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]

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

[/ QUOTE ]

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 (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Board=probability&Number=417 383&Forum=probability&Words=inclusion-exclusion&Match=Entire% 20Phrase&Searchpage=0&Limit=25&Old=1year&Main=4169 81&Search=true#Post417383).

[PairTheBoard]

Yes of course. I seem to be making a lot of mistakes these days.

PairTheBoard

[PairTheBoard]

Yes, I see it now. In the term (4/52 x 48/51), I thought it said (4/52 + 48/51) which made no sense to me. Did you change it or did I misread it originally?

And as Bruce says this is a lower aproximation because the term (4/52 x 48/51)^10 should be:

P(1st player no AA) * P(2nd player no AA Given 1st player no AA) * ... * P(10th player No AA Given 1st 9 players No AA).

Conditioned on the previous players not having AA the deck should becoming slightly more rich in Aces, making succeeding probabilities of not have AA less and the product less. Thus making 1-Product slightly Greater than your answer.

PairTheBoard

[PairTheBoard]

Is that an exact solution Bruce? If so it must be the slickest I've ever seen. Could you explain it please? I'm afraid I'm not seeing it.

PairTheBoard

[BruceZ]

[ QUOTE ]
Is that an exact solution Bruce? If so it must be the slickest I've ever seen. Could you explain it please? I'm afraid I'm not seeing it.

[/ QUOTE ]

Yes it's exact. The explanation was here (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Number=750250&page=0&view=ex panded&sb=5&o=14&vc=1). It's a simple application of the inclusion-exclusion principle. We start by adding the probabilities even though the events are not mutually exclusive. This is the first term 10*6/C(52,2). This has the effect of double counting the deals where 2 players have AA, so we subtract that off in the second term to get the exact answer. I used a little trick for this second probability by noting that 2 players must have all 4 aces, and there is just 1 combination of 4 cards out of C(52,4) that is 4 aces. Then since only 1 pair of players can have AA, the pairs of players are mutually exclusive, so multiply 1/C(52,4) by the number of pairs C(10,2) to get the probability of 2 players having AA, and we're done. So the final exact answer is 10*6/C(52,2) - C(10,2)/C(52,4).

Let me know if you see it OK now. I have quite a few similar examples all over this forum and in the archives. Search for inclusion-exclusion.

[PairTheBoard]

Wow. That is nice. I would have never thought of that
C(10,2)/C(54,4)
and I don't think I've ever seen it before. Identify the two players that are to get the AA,AA. Then the probability that they get dealt AA,AA is the same as the probabilty that when 4 cards are dealt they are all Aces. 1/C(52,4). Then C(10,2) ways of idendifying the two players. As you say, they are mutually exculsive.Amazing.

That's what I hate about these kinds of problems. If you see the trick they can be easy. If you don't they can get a lot more complicated.

Thanks,

PairTheBoard

[scottjack]

The probability of any single person getting pocket aces is 6/1326 (6 different pairs of aces, divided by 1326 possible 2 card hands) Given that, the probability that a single person is NOT dealt pocket aces is 1320/1326

1-(1320/1326)^10= .044338542

Looks like we agree, pretty darn close anyway

[Ed Miller]

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

Here is the rule I use for approximations like this...

If the probabilities involved are very SMALL (1/221 qualifies as small) then the probability of the event occurring in one of X trials is approximately X/221 (again, as long as X isn't too large). So I'd approximate the chance of someone getting aces to be 10/221, but I wouldn't approximate the chance of someone getting a pocket pair to be 10/17. 1/17 is too large.

Similarly, you can approximate the chance to hit a two outter on either the turn or river as 2/47 * 2, but not a ten outter as 10/47 * 2.

[bigbull266]

Using combinations there are 1326 different 2 card combinations in a 52 card deck. 6 of those combinations can be pocket aces or about .4%. I just read an excellent book on how to calculate all these kinds of scenarios by Mike Petriv. The book is called Hold'em's odd(s) book.