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Both presented solutions so far have basically been the probability that any one person will get a royal times the number of players at the table (and accounting for the few cases where the royal is on the board).
But this can't be correct, becasue all opponents share the community cards. In order for the presented solutions to be correct, we would have to effectively be dealing 7 cards to one player, scooping them up and dealing 7 more cards to the next player, etc.
I'll post what I think is the correct answer later, but I'd like additional input, please.
Anyone?
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I gave you the
exact solution. I guarantee you that it is correct, and the method is fundamental. It should be completely understandable for anyone with a mastery of basic probability theory, in particular, the probability of a union of events. It is actually an example of the
inclusion-exclusion principle.
To make the solution easier to understand, ignore the case of a royal on the board for the time being. Only consider the cases where a player makes a royal by using 1 or both cards from his hand. It should be clear that no more than 1 player can make a royal flush this way. That is, the event of the ith player making a royal flush using 1 or both hole cards is mutually exclusive of every other player making a royal. The probability that one of the 7 players holds a royal this way is the probability of a union of 7 mutually exclusive events, which is simply the sum of the probabilities for each individual player, and since these probabilities are the same, it is 7 times the probability for 1 player. The fact that there are community cards does not change this fact. If you don't understand this point, then this is where your efforts need to be directed.
We are computing the probability of a union of 7 events, where each event is the probability that the ith player has a royal. The probability of a union is always the sum of the probabilities for each individual player, minus the probabilities of the pairwise intersections, plus the probabilities of the 3-way intersections, etc. However, in this case, the only intersection is the 7-way intersection when all 7 players have a royal due to a royal on the board. Either 1 player has a royal, or all 7 have a royal, but it is impossible for 2 to 6 players to have a royal. Therefore, the first term of inclusion-exclusion multiplies the probability of a single player having a royal by 7. Since this will count each case where all 7 players have a royal 7 times, the final term of inclusion-exclusion subtracts off 6 times the probability of all 7 players having a royal, so that these are counted only once. This completes the calculation.
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In order for the presented solutions to be correct, we would have to effectively be dealing 7 cards to one player, scooping them up and dealing 7 more cards to the next player, etc.
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This is false. The calculation would be different in the case you describe because it would be possible for 1 to 7 players to have a royal flush, and they could even have the same royal flush, so there would be 7 terms in the inclusion-exclusion solution. If the probability of 1 player getting a royal in 7 cards were P, then the probability for the case you describe would not be 7P, but would be 1 - (1-P)^7 =~ 1 - (1 - 0.003232%)^7 =~ 0.0226% or 1 in 4420. Note that in this case the hands are independent, while in the original problem they are not.
Also, the calculation would be different still for 7-card stud, which you mentioned. In that case, 1 to 4 players could have royals, so there would be 4 terms in the inclusion-exclusion solution. However, in all of these cases, the first term will be 7 times the probability of making a royal in 7 cards, and this term will dominate, with the other terms being relatively small, so that the numerical answers to all these cases will be fairly close.