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For question 1, could I take the number of hand combinations that make a higher pair, divide by the number of total hand combinations (1326) and then multiply by the number of other players at the table?
For example, I have KK. (6 / 1326) * 5 = 0.0226
I have QQ. (12 / 1326) * 5 = 0.0452
EDIT:
I just realized that the total hand combinations is no longer 1326 because I have two known cards in my hand. I guess that makes the total hand combinations (50*49)/2 = 1225
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Using 1225, this is the first term of inclusion-exclusion. Using this one term will be very accurate in cases like this one where the probability that 2 or more opponents hold one of the hands in question is small, which occurs when there are relatively few hands in question. Usually only 2 or 3 terms are required for an accuracy better than 0.1%. In the series of articles by
Brian Alspach on inclusion-exclusion titled "I'm In...No I'm Out", he gives tables with some of the numbers that you are interested in.
In cases where there are many hands in question, such as when you are computing the probability that an opponent holds one of many possible pairs, the approximation of independence is often accurate. That is, if the probability that a single opponent holds one of these hands is p, then the probability that he does not hold one is 1 - p, and the probability that one of 5 opponents does not hold one is approximately (1 - p)^5, so the probability that one of 5 opponents holds one is approximately 1 - (1 - p)^5.