Odds of AA running into KK at full table? - Page 3

tylerdurden · #21 07-20-2004, 05:32 PM

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you need to use the inclusion exclusion principle, or else you will double count the times that two players get dealt AA.

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If I had double-counted some events, my number would have been *higher* than yours, but mine is lower. [img]/images/graemlins/confused.gif[/img]

tylerdurden · #22 07-20-2004, 05:33 PM

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Nope, this doesn't work. If someone has AA, then the chance that the next person does NOT have KK is .95677. But if they don't have KK, then the chance that the next person does goes up. Do you see why?

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No, I don't see it. If we were looking at each players hand as it came out, then dealing the next one, yeah, this would be right. But I'm assuming we're dealing all ten hands simultaneously, like at a normal table.

aloiz · #23 07-20-2004, 05:57 PM

Yea the number you got was lower because you over approximated odds of not getting AA. To figure out the exact odds that all the people at the table do not get AA you need to take P(p1 not have AA) * P(p2 not have AA|p1 not have AA) * P(p3 not have AA|p2 and p1 not have AA) etc...

As you continue given the fact that x amount of people don't have AA decreases the odds that the current player does not have AA. Thus your number was slightly high, making the probability that at least one player having AA slightly low.

Note that you could also use inclusion/exclusion instead of conditional probability to figure out the exact odds that no one has AA. P(no one has AA) = 10 * P(a single person doesn't have AA) - P(two people don't have AA) + P(3 people don't have AA) - P(4 people don't have AA) ....

aloiz

tylerdurden · #24 07-20-2004, 06:34 PM

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As you continue given the fact that x amount of people don't have AA decreases the odds that the current player does not have AA.

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I still think this is the wrong way to figure this. All hands are dealt (effectively) simultaneously. I don't know what the guy in front of me has before I pick up my cards, so I can't calculate based on what he has or doesn't have, so for computational purpses, his hole cards are effectively the same as the ones still in the deck (i.e. unknown).

A given player's chances of getting AA in the hole on any give hand are always 6/((52*51)/2), no matter how many people are playing. It doesn't matter what position he's in, or how many people get their cards before or after him.

aloiz · #25 07-20-2004, 08:44 PM

You're right in the sense that if you take any specific player at a ten person table the odds that that person is dealt AA is 1/221. However if you now multiply that number by 10 to calculate the odds that at least one player at a ten person table gets AA you are double counting the times that 2 people get AA.

Here's a simplified version. Say you have a four card deck. As Ah 2s 2h, and you deal 1 card to two players. The odds that any given player gets an ace would be 1/2, but the odds that at least one player gets an ace is not 2 * 1/2 =1, because both two's can be dealt. To figure out the odds that at least one player gets dealt AA we take 2 * odds that a given player is dealt AA and subtract the odds that both players get AA. The odds that both players get AA is 1/6. so 1-1/6 = 5/6. Now we can check this because we know the only time that we don't deal an ace is if both two's are dealt. The odds that both two's are dealt would be 1/2 * 1/3 = 1/6.

hope that makes sense.

aloiz

Cosimo · #26 07-20-2004, 09:02 PM

The chance that you have an ace is 14%. The chance that the next guy has an ace is 14%. The chance that someone at the table has an ace: is it 10x14%? What does it mean for there to be a 140% chance that someone has an ace? That's meaningless. That means that the chance that no-one has an ace is -40%, which is again absurd. You can't multiply probabilities for dependent events; only for independent events.

Once you've given a card to one person you can't give it to anyone else. The cards that you hold are not independent of what the next guy holds, or else it would be possible for the entire table to have aces. That's 20 aces. It doesn't matter if you simultaneously distribute the cards, because you still can't give the same card to two people.

One of the tricks of statistics is to take a bunch of simultaneous but dependent events and serialize them. This is not an approximation--it's in fact the recommended way of breaking down complex problems. So whether we say that these events are simultaneous or not isn't really the point. More appropriately, if the dealer gave a pair of kings to one player, it is less likely that he "at the same time" gave a pair of kings to another player.

Take all the bazillions of combinations of hands that can be dealt to ten people. What fraction contains at least one pair of kings? That's X.

Take the subset of those bazillions that includes at least one pair of kings. What fraction contains a second pair of kings? This fraction will be smaller than X--there's only two more kings left, yet every other card is still possible.

tylerdurden · #27 07-20-2004, 10:50 PM

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However if you now multiply that number by 10 to calculate the odds that at least one player at a ten person table gets AA you are double counting the times that 2 people get AA.

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Oh, you're talking about the first calculation I did? I already admitted that was wrong and submitted a different one. And even then, in my first calculation, I didn't count anything double, I just plain did it wrong (see Cosimo's explanation of why it was wrong).

tylerdurden · #28 07-20-2004, 10:50 PM

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The chance that you have an ace is 14%. The chance that the next guy has an ace is 14%. The chance that someone at the table has an ace: is it 10x14%? What does it mean for there to be a 140% chance that someone has an ace? That's meaningless.

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Uh, yeah. I got that part, that's why I re-did my answer. Didn't you read my 2nd calculation?? At least you have the right reason why my 1st attempt was wrong.

aloiz · #29 07-21-2004, 12:52 AM

I thought I answered your latest calculation. However the example in my previous point should explain why both approximation methods, whether you calculate the odds that someone gets AA and multiply by 10, or you calculate the odds that no one gets AA and raise it to the tenth power are both approximations and not accurate. If you use 10 * 1/221 as the odds at least one person gets AA preflop you're double counting the times two people get AA. If you use (1320/1326)^10 as a way to calculate the odds that no one gets AA preflop this is also wrong, as you do not have 10 independent events.

If you go back to the four card deck example. Your way of calculate the odds that no one is dealt an ace preflop would be (1/2) ^ 2 = 1/4. However the actual odds that no one gets an ace is 1/2 * 1/3 = 1/6.

If I'm still missing your calculation then I apologize, but none of your calculations that I’ve seen are exact.

aloiz

DMBFan23 · #30 07-21-2004, 01:12 AM

I would think that the probability of being dealt (not AA) IS independent, since the cards are being dealt simultaneously and assessed "instantaneously" if that makes sense.

IMO, it's the same principle as not counting the expected value of outs that are dead in other people's hands when calculating pot odds...you just ignore their cards, since we cannot know what they have.

So returning to our example, we can only know, given the two cards we are considering, the odds that they are not AA. only when we assume something as a stipulation does conditional probability apply (ie given that x, y happens so often...). from there finding p', p' = odds no one has AA, 1-p gives us the odds of someone having AA. this also takes care of all double counting, since we are finding the following probability: "it is not the case that no one has AA." this could be one or two players with AA.