Odds of your oppenent holding AA against your KK (10 handed)
 

Can anyone tell me what the odds are of running into AA against your KK in a ten handed hold 'em game? Thanks.

Re: Odds of your oppenent holding AA against your KK (10 handed)
 

There are 4 Aces in the deck, so there are 6 possible 2-card combinations of pocket Aces. (4*3 / 2*1)

Taking out your own hand leaves 50 other cards in the deck, which can be combined in 2-card hands some 1,225 different ways (50*49 / 2*1).

Since you're not holding any Ace, that leaves those 6 combinations of pocket Aces among the total possible 1,225 combinations. So, the probability of your Heads-up opponent picking up (being dealt) one such pocket-Aces combination is 6/1225 = 0.00489795 or 0.489795%, which is also the probability of your Heads-up opponent picking up a pocket pair from a specific rank when you're holding no card from that rank.

You are playing now against 9 opponents, but you should not multiply that by 9 to find the answer for a 10-handed game : it is impossible for your opponents to simultaneously "pick" cards from those 1,225 combinations. (When one of them picks up for instance A[img]/forums/images/icons/diamond.gif[/img]5[img]/forums/images/icons/club.gif[/img] , all the combinations containing the A[img]/forums/images/icons/diamond.gif[/img] "disappear".)

This is a problem that only a qualified engineer can solve.

Re: Odds of your oppenent holding AA against your KK (10 handed)
 

For these types of problems, the following approximation is always within a few tenths. The probability of a single opponent having AA is 6/(50*49/2) = 6/1225. The probability that someone has AA with 9 opponents is 1 minus the probability of no one having it, and this is approximately 1-(1-6/1225)^9 = 4.3% or 1 in 23.1. This is approximate since we are treating the hands as independent, and they are not, but the approximation is excellent.

In this case, we could also just multiply the probability of one person having KK by 9 to get 9*(6/1225) = 4.4% or 1 in 22.7. As we shall see, this is even closer to the exact answer which turns out to be 1 in 22.77. The only reason this is approximate is that it double counts the cases of two people having KK, which would have to be subtracted from this to get the exact answer, but that probability is very small. It has nothing to do with the fact that the opponents cannot "simultaneously pick", and someone could get just an A, as someone incorrectly stated. That is an issue for the first approximation where we are assuming independence. For the second approximation, we are summing probabilities which only assume mutually exclusive events. Note for example that if we had AA, the probability that a particular other player has AA is 1/1225, and the probability that one of 9 players has AA is EXACTLY 9/1225 because it is not possible for more than one other player to have AA. The probability of each player having AA is 1/1225, the same as if they "simultaneously picked".

Whether this second approximation is always closer than the first approximation is an area I would have to research further. I have always used the first approximation, and it is always sufficiently close.

A short while ago, I worked out the exact answer for many of these types of problems for the WPT TV show. These used 9 or 4 total players. You can see these results with an explanation of how they are derived at:

http://www.twoplustwo.com/forums/sho...rue#Post212797

The post "more exact answers" makes some corrections to the formulas explained earlier in the thread. The AA vs. KK problem was actually solved exactly several different ways. It is probably never necessary to be this precise (except for TV).

Here is the exact formula for AA vs. KK 10 handed:

[ 6*9*P(48,16)/2^8-6*C(9,2)*P(46,14)/2^7 ] / [ (P(50,18)/2^9 ) ] = 1 in 22.77.

typo
 

In this case, we could also just multiply the probability of one person having KK by 9 to get 9*(6/1225) = 4.4% or 1 in 22.7. As we shall see, this is even closer to the exact answer which turns out to be 1 in 22.77. The only reason this is approximate is that it double counts the cases of two people having KK

Replace the KKs with AA.

Feedback
 

This is in response to some questions I have received about this calculation.

1. There was no disagreement between my results and those of J'Aoudbe in the WPT thread. Our results were identical. His calculation used a different method than mine, and he computed KK vs. AA for the 10-handed case, which is what was asked in this thread, whereas I used my method to compute the 9-handed case, which was needed for the TV show. The results my method produces for the 10-handed case, given above in this thread, are identical to his results for the 10-handed case using his method. Similarly, his method produces identical results to mine for the 9-handed case. I did not recognize he was doing the 10-handed case at first, so I initially claimed there was a problem with his method, but in fact there was not, and this was retracted. Everything is in perfect agreement and God is in his heaven. My results were generalized to many other hands, which I believe would be much more cumbersome with the other method. On the other hand, his method introduces a useful trick which can no doubt be of benefit elsewhere. Also, all my results were independently verified by another poster.

2. As for why we can just multiply 1/1225 by 9 or 6/1225 by 9, this all we are doing:

P(A u B) = P(A) + P(B) - P(AB)

Where u means union of events. P(A u B) is the probability of A OR B occurring. P(AB) means probability of A AND B occurring. Note that when we add P(A) + P(B) we double count the probability that A and B occurred, which is why we subtract off P(AB).

generalizing:

P(A u B u C u ...) = P(A) + P(B) + P(C) - P(AB) - P(AC) - P(BC) + P(ABC) + ...

In other words, first we add the probabilities of each event, then we subtract off all pairwise combinations of the events occuring twice, then we add back all 3-way combinations of the event occuring 3 times which got SUBTRACTED twice in the previous step, etc... The adds and subtracts alternate until all combinations are exhausted. The pairwise probabilities are usually computed all in one term as the probability of the event occurring twice, regardless of how it occurs twice. Same for the other terms. In practice, these terms usually become small quickly, so it is not necessary to compute more than the first 2 or 3. My calculation for the TV show required computing 3 terms for all cases. Obviously, this gives the exact result for the 4-handed case. This method is called the inclusion-exclusion principle, because we count by including too many things, then excluding some of them, etc.

When we have AA, and we want to compute the probability of someone else having AA, it is impossible for even 2 other people to have AA, so there are no pairwise terms, and we can just add the individual probabilities as:

P(someone else has AA) = P(player 1 has AA) + P(player 2 has AA) + P(player 3 has AA) + ...P(player 9 has AA)

The probability of each player getting AA before the deal given that you will get AA is 1/1225, so the above sum is 9/1225.

Suppose instead of cards, we had each of the 1225 hands written on 1225 pieces of paper, and these were distributed to the other 9 players. The above method would still apply. Each player still has a 1/1225 chance of getting the AA. Each player is symmetrical and has this same probability. The above sum is 9/1225. This is true even though in this case, there are combinations of hands that are not possible in a real card game. For example, someone could have AA, and someone else could have A5. For that matter, each of the 9 players could have an A!. The probability works out the same. There are C(1225,9) ways 9 players can choose 9 hands, ignoring order. There are C(1225,8) ways one of them can have AA. C(1225,8)/C(1225,9) = 9/1225, same as the other method.

Similar comments would apply to the case where you have KK, and you want to know probability that someone has AA. In that case, you would need to subtract the probability that 2 players have AA to get the exact answer. We can approximate this case by simply ignoring the small probability that 2 people have AA, and simply compute as before 9*(6/1225) = 1 in 22.7 for this probability.

3. Note again the typo from the first post. It is not another KK you are worried about, but 2 people having AA, as should be clear from the above.

Correction to slips of paper problem
 

The probability is actually C(1224,8)/C(1225,9) and this is indeed exactly 9/1225. In fact it is

(1224*1223*...*1217/8!)/(1225*1224*...*1217/9!) = 9/1225.

Notice that this is the same result as the card game because only 1 other player can have AA. It works for other hands as long as we are GIVEN that only one other player has that hand, such when we make this approximation when we have KK and we only consider one player having AA. As soon as we consider that more than 1 other player can have this hand, this won't be the same. For example, in this game 6 other players could have AA when we have KK. So this should not be considered a general model for a card game.

Re: Correction
 

Thanks for the input, Bruce. I asked you for clarifications through a PM, but if you prefer to take it back here, it is quite alright.

The original question was "Can anyone tell me what the odds are of running into AA against your KK in a ten handed hold 'em game?" I took that to mean the poster wanted to find out abt the odds of running into any AA, which means one or two opponents having pocket Aces. You may have understood the question more clearly, though.

This is why I objected that the example with the "1,225 pieces of paper" on which each of the C(50,2) combinations is written, cannot aply in the "real world". The probability is indeed appoximated with [9*(6/1225)] but it is not exact. If we accept that more than 1 opponent can hold AA, this is no longer "considered a general model".

Thanks for clarifying.

Re: Correction
 

The original question was "Can anyone tell me what the odds are of running into AA against your KK in a ten handed hold 'em game?" I took that to mean the poster wanted to find out abt the odds of running into any AA, which means one or two opponents having pocket Aces.

I AM answering the question of the probability of running into one or two AA. The approximation is not exact, but it's within a cat's whisker. It gives 4.4% which is the exact answer rounded off, and it gives 1 in 22.7 compared to 1 in 22.77 for the exact answer. This is always close enough for all practical poker purposes. This approximation does not assume that only one player gets AA, that would be a different number (though equally close in the other direction). By ignoring this possibility it actually double counts these cases.

This is why I objected that the example with the "1,225 pieces of paper" on which each of the C(50,2) combinations is written, cannot aply in the "real world". The probability is indeed appoximated with [9*(6/1225)] but it is not exact.

The slip of paper analogy really has nothing to do with this approximation, and I really have no use for it, because that would allow multiple players to get AA, and it would give the wrong answer in that case. I don't think in terms of that for this approximation. All that is required is to notice that the probability of any player getting AA is 6/1225, and it's the same for every player, so summing them gives the first order approximation. If you subtract the probability of two people getting AA from this, you get the remaining 0.01% to the exact answer (4.39% or 1 in 22.77). That is what is answered at the bottom of the original post.

trying again
 

Let's try this one more time. Forget the scraps of paper. Throw them in the trash, they're confusing you. You're sitting at the table, and everyone gets two cards. Nobody has looked at their cards yet. You peek at yours and see KK. Now each of those other players cards have a probability of exactly 6/1225 to be AA. If you were to just add these 9 6/1225's together to get 9*6/1225, that would be extremely close to what we are looking for, the probability of at least one more AA. The only thing that happened is that we double counted all the times that two people got AA. But those times are very rare, which is why this is so close. If we really wanted an exact answer, we would compute the probability of two people getting AA, and we would find this is 0.01%, and we would subtract this little bit off from 9/1225 to get the exact answer. At the bottom of my original post, there are combinatorics for both of these terms, and the first will evaluate to exactly 6*9/1225.

Thanks again
 

"Forget the scraps of paper. Throw them in the trash."

More PC to put them in the recycle bin.

"If you were to just add these 9 6/1225's together to get 9*6/1225, that would be extremely close to what we are looking for, the probability of at least one more AA."

I fully agree, of course. I was only counting on you for the exact answer which I got in the WPT thread.

"...to get the exact answer. At the bottom of my original post, there are combinatorics for both of these terms, and the first will evaluate to exactly 6*9/1225."

This is what I don't understand. How can [(probability of one player having AA)*(number of opponents)] be approximate and exact at the same time? As usual, I must be missing something.