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## Main Question or Discussion Point

Let's say we have a very simple game of two-card poker, where the deck consists of only 4 aces and 4 kings. There are 28 possible combinations: Six pairs of aces, six pairs of kings, and 16 ace-king hands. In play against a single opponent, both players are dealt two cards, face down. The probability of either player being dealt a pair (aces or kings) is 6/28, while the probability of an ace-king combination is 16/28.

Let's say you are dealt a pair of kings and want to calculate the probability of winning. Before the deal, there was a 6/28 probability that your opponent would be dealt a pair of aces. However, you know you got two kings, so you know your opponent could have been dealt only 1 of the 6 possible pairs of kings and that they could have been dealt only 8 of the 16 possible ace-king combinations. Given your pair of kings, your opponent could have only one of 15 possible hands (6 pairs of aces, 1 pair of kings, and 8 ace-king combinations). To calculate the probability that they are holding a pair of aces, do you use 6/15 instead of the original 6/28?

My second question has to do with combining probabilities. What if, in the scenario above, there were three players instead of two? Each of your opponents has the same probability of having been dealt a pair of aces. If you are trying to determine the probability of beating both opponents, the probabiliies should be added together, shouldn't they? 1 - (6/15 + 6/15) would make your probability of winning only 0.20, which seems awfully low, given that there are six hands that will beat you, one that will tie, and 8 that will lose to you.

What am I missing?

Let's say you are dealt a pair of kings and want to calculate the probability of winning. Before the deal, there was a 6/28 probability that your opponent would be dealt a pair of aces. However, you know you got two kings, so you know your opponent could have been dealt only 1 of the 6 possible pairs of kings and that they could have been dealt only 8 of the 16 possible ace-king combinations. Given your pair of kings, your opponent could have only one of 15 possible hands (6 pairs of aces, 1 pair of kings, and 8 ace-king combinations). To calculate the probability that they are holding a pair of aces, do you use 6/15 instead of the original 6/28?

My second question has to do with combining probabilities. What if, in the scenario above, there were three players instead of two? Each of your opponents has the same probability of having been dealt a pair of aces. If you are trying to determine the probability of beating both opponents, the probabiliies should be added together, shouldn't they? 1 - (6/15 + 6/15) would make your probability of winning only 0.20, which seems awfully low, given that there are six hands that will beat you, one that will tie, and 8 that will lose to you.

What am I missing?