Odds on poket pairs question

[Poker21]

the odds of being dealt a pair of aces before the flop is .45% or so correct? so if im at a table with ten players does that mean that someone will have pocket aces 4.5% (.45 * 10) of the time???

[EverettKings]

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so if im at a table with ten players does that mean that someone will have pocket aces 4.5% (.45 * 10) of the time???

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No. They'll be close, but lower.
The odds of any of them having aces is:
The odds of the first player having them (.0045)
+ The odds the first player doesnt have them (.9955) but player 2 does (.9955)*(.0045) = .00448
+ The odds neither of the first 2 have them, but player 3 does = .9955*.9955*.0045 = .00446
etc etc

So it's actually .0045*(.9955)^0 + .0045*(.9955)^1 + .0045*(.9955)^2 + ... +.0045*(.9955)^n-1 where n is the number of players you have. I dont really want to plug it out for you, but because the individual odds are so slim (.0045 or .45%) your original math won't be too far off. It'll probably be close to 4% though.

And the 0.45% odds of getting Aces is correct. On the first card you must be dealt one of the 4 aces, the odds of which are 4/52 (4 aces in a deck of 52), and on the second card there are 3 aces to catch out of 51 deck cards, so 3/51. Both of these must work, so the odds are (4/52)*(3/51) = .0045 or .45%.

[fnord_too]

no, it means the probability someone will have AA is 1 - ((1-.0045)^10) which comes out to a hair over 4.4%. This is a common mistake in figuring probability. In this case the difference is very small, but let me give you another:

Odds of rolling a 6 on a six sided die are 1/6. If you roll a die 6 times you do not have a probability of 1 of rolling a six, there are plenty of times you will roll a die six times in a row and never see a 6. The best way to figure out these independant event probabilities, where you are looking for the likelihood that something will happen at least once in so many tries, is to calculate the odds of it not happening then take this to the power of trials you have. This is the likelihood that the event will not occur, which when subtracted from 1 will give you the liklihood that it will occur.

For the dice example, 1-((5/6)^6) = ~2/3.

If you are looking at the probability of not rolling N on an N sided die in N rolls, the likelyhood approaches 1/e as N approaches infinity. But I digress.

[BruceZ]

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no, it means the probability someone will have AA is 1 - ((1-.0045)^10) which comes out to a hair over 4.4%. This is a common mistake in figuring probability.

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It's a common mistake, and so is yours. You cannot get the exact answer by multiplying the probabilities of each player not having AA because, unlike your dice example, the hands are not independent. His method is a starting point to the exact answer, but it double counts the times 2 players have AA. To get the exact answer, we must subtract off the probability of 2 players having AA which was double counted. This is 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) which I have explained about a million times now. Note that the probability of two specific players having AA is 1/C(52,4), and we can multiply this by C(10,2) to get the exact probability of 2 players having AA since these are mutually exclusive (no more than 1 pair of players can have AA). So the exact probability of at least 1 player having AA out of 10 players is (1/221)*10 - C(10,2) / C(52,4) =~ 4.36%.

Note also that when you hold AA, the probability that another specific player holds AA with you is 1/1225, so the probability that one of 9 players holds AA with you is exactly 9/1225. This is exact since no more than one other player can have AA.

[BruceZ]

See my above post in this thread for the simple way to get the exact answer.

[EverettKings]

Bruce, tell me why I'm wrong here:
The odds of NOBODY having AA is (1-.0045)^10, correct?
Therefore, the odds of NOT NOBODY (hence, somebody, but it could be one or two players) is 1-(1-.0045)^10 as the poster before you stated. Why is that calculation wrong?

[BruceZ]

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The odds of NOBODY having AA is (1-.0045)^10, correct?

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NO! The probability of a single player NOT having AA is 1-.0045. The probability of 2 players NOT having AA is NOT (1-.0045)^2 because the cards removed by the first player affect the odds of the second player, so they are not independent. The correct probability of 2 players NOT having AA would be the probability of the first not having AA (1-.0045) multiplied by the conditional probability that the second player does not have AA given that the first player does not have AA. This conditional probability is different from 1-.0045 and slightly lower since AA is now slightly more likely.

P(A and B) = P(A)*P(B) only if A and B are independent.
P(A and B) = P(A)*P(B|A) always, even if A and B are not independent.

[MushashiAce]

So if the odds of another player having pocket Aces Pre-flop whe you have Aces is 9/1225, then how would you figure another player having pocket Aces or Kings when you have AK?

[EverettKings]

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The correct probability of 2 players NOT having AA would be the probability of the first not having AA (1-.0045) multiplied by the conditional probability that the second player does not have AA given that the first player does not have AA. This conditional probability is different from 1-.0045 and slightly lower since AA is now slightly more likely.


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I considered this, but in this case we MUST consider the events independent. If the condition is true that player 1 does not have AA, he either has one or zero aces. If he has no aces, then as you said it makes player 2's AA a little more likely, but if he has an ace, it makes player 2's AA significantly less likely. Because we do not know exactly how player 1's lack of AA is going to affect the deck, we must treat player 2's AA as an independent probability.

You seem pretty confident in your answer, so I really do believe that you're correct, but still I'm curious if and where my logic is failing here.

[aloiz]

You need to calculate both cases.

case 1 P(player 1 has ace) * P(player 2 has AA|player 1 has A)

case 2 P(player 1 has no ace) * P(player 2 has AA|player 1 has no ace)

add the two cases together to get the probability that player two has an AA. Obviously if you tried to figure this out for all ten players it would be nearly impossible. Using inclusion exclusion is much easier.

aloiz

[BruceZ]

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I considered this, but in this case we MUST consider the events independent.

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That's absolutely false. They are NOT independent.

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If the condition is true that player 1 does not have AA, he either has one or zero aces. If he has no aces, then as you said it makes player 2's AA a little more likely, but if he has an ace, it makes player 2's AA significantly less likely.

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Correct.

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Because we do not know exactly how player 1's lack of AA is going to affect the deck, we must treat player 2's AA as an independent probability.

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Not so; we CAN compute the effect. Aloiz has the right idea, except we want player 2 to NOT have an A:

P(player 1 has ace) * P(player 2 does not have AA | player 1 has A) +
P(player 1 has no ace) * P(player 2 does not have AA | player 1 has no ace)

Evaluating these terms gives:

4*48 / C(52,2) * [C(50,2) -3 ] / C(50,2) +
C(48,2) / C(52,2) * [C(50,2) - 6] / C(50,2)

= 0.99095392

This is slightly different from (220/221)^2 = 0.9909707 which you are suggesting, and the difference gets larger with more players.

Note that the easiest way to compute this is by inclusion-exclusion:

2/221 - 1 / C(52,4) = 0.00904608 that at least 1 ace is out.

1 minus this is 0.99095392, in exact agreement with the above.

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You seem pretty confident in your answer

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I'm 100% confident. I've done this kind of problem so many times I can literally do it in my sleep. It is well known that poker hands are not independent. I've watched countless people try to use independence as you have done, or mutual exclusivity (multiplying by 10). Both often give excellent approximations. Neither are exact except for the times that mutual exclusivity applies such as AA vs. AA. To be exact you need inclusion-exclusion.

[BruceZ]

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So if the odds of another player having pocket Aces Pre-flop whe you have Aces is 9/1225, then how would you figure another player having pocket Aces or Kings when you have AK?

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9*6/1225 - C(9,2)*6/C(50,2)*3/C(48,2) = 1 in 22.9.

I computed several of these last year for the producer of the WPT TV show (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Number=213933&page=&view=&sb =5&o=&vc=1). Note that the method I used for those was more complicated than what I am using now.