[BruceZ]

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case at first, so I initially claimed there was a problem with his method, but in fact there was not, and this was retracted.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.