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The odds of being dealt A,A are 220 to 1. So the ~ odds of one of 5 opponents having A,A is 44 to 1. (220/5).
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That's incorrect. First of all, you neglected to take into account your own cards. The odds of being dealt AA is indeed 220-to-1, but the probability that one
particular opponent gets dealt AA when you don't hold an A is 6/C(50,2) = 6/1225 or about 203-to-1. Then as an approximation, you can multiply this by 5 to compute the odds that one of your 5 opponents holds AA as 5*6/1225 =~ 39.8-to-1. This is a slight approximation because it double counts the cases where 2 opponents hold AA, which has probability C(5,2)/C(50,4), and this must be subtracted off. So the exact answer is 5*6/1225 - C(5,2)/C(50,4) =~39.9-to-1.
You asked about this
here and were given this answer.
For more details, see my other posts in this thread which link to the explanation of the
inclusion-exclusion principle.