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

[BruceZ]

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.

It is an approximation, but it differs from the exact answer only by the probability of two people having AA, and this is a very small amount. Even when we compute the exact answer, we start by computing this number 9*6/1225, and in this case that gets us over 99% of the way to the final answer. Then we compute another term which gets us all the way to the final answer.

Imagine a group of gamblers, each of whom plays either poker or blackjack. One way to try to count the gamblers would be to count the ones that play poker + the ones that play blackjack, but that that would count twice the ones who play both. So we would end up with a number that was too big, and it would be too big by exactly the number who play both. Then if we knew how many played both, we could subtract this off to get the correct number of gamblers.

When we compute probabilities, we are really just counting things. We are counting the number of ways certain things can happen, and then we divide by the total number of things that can happen. In this case, we are trying to count things, namely the number of ways another AA can be out, similar to the way we were counting the gamblers. We are counting the ways each player can have AA, and then summing these together to get 9*6/1225, but in so doing we counted twice all the times two players had AA.

Remember Venn diagrams from school? You had big circles that represented sets. Where they overlapped was the intersection of the sets. The union of all the sets was the set of all the members that were in any of the circles. Suppose you wanted to count the number of members in the union of all the sets. If you counted the elements in each of the circles and added them together, then the members that were in two of the circles would get counted twice, the ones that were in 3 of the circles would get counted 3 times, etc. You can picture a Venn diagram for the poker case. Each player has a big circle that represents the times he has AA, 9 circles in all. Each circle will overlap each other player's circle, but there will be no places where 3 or more circles overlap because it's impossible for 3 or more players to have AA. Our approximation is counting the members in each circle and adding them all together. This is P(1) + P(2) + P(3) + ... + P(9) = 6*9/1225. In so doing, we count twice the members in the places where the circles overlap. This overlap is small, so the result is close to the answer we want. For the exact result, we compute the number of members in the overlapping sections, and subtract this off. For more complicated problems, there may be places where 3 or more circles overlap. In that case we proceed the same way, except now when we subtract off all the places where two circles overlap, we have subtracted off twice all the places that 3 circles overlap. So then we have to compute the places that 3 circles overlap, and add this back on. We keep doing this iteratively until we either get the exact answer, or until these areas of overlap become so small that we can ignore the rest of them and say we are close enough. This is the inclusion-exclusion principle described above, and used for the answers in the WPT thread.

[BruceZ]

I realized that there is a much simpler way to do these problems than the way I did them on the WPT thread. You don't need all the permutations and combinations. For example, for KK vs. AA you can just do this for the EXACT answer:

6*9/C(50,2) - C(9,2)/C(50,4) = 1 in 22.77.

The second term is the probability of 2 people having AA. There are C(9,2) ways to pick the two players, and 1 way they can have all 4 aces out of the C(50,4) ways they can have 4 cards.

All the problems can be done almost this simply, except they will have 3 terms instead of 2 for the desired level of accuracy.

-Bruce

[irchans]