Another Tack on the Royal Flush

[BruceZ]

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I think you are basically saying that the chance that anyone at a 7-handed table will get a royal is just 7 times thae chance that one specific person will get a royal. Am I correct?

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Minus 6 times the probability of the board being a royal. See my solution.

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I'm a little confused.

Let's say there are 2 games running, each with 7 players. One is stud, the other HE. By your logic, could we not conclude that the chances that someone at the stud game will make a royal are the same as at the HE table?

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No, you cannot conclude that by my logic, nor by any correct logic, as it is false. The calculation for stud is different. In stud, it is possible for 1 to 4 players to have royals, while in Hold'em, only 1 player can have a royal unless all 7 players have royals. However, in both cases, the numerical answers will be similar as they will be dominated by the same term 7*4*C(47,2)/C(52,7).

[GrunchCan]

I appreciate the much more in-depth explanation you have provided. I do have what I would consider to be a basic understanding of probabilities, but I am unfamiliar with the inclusion-exclusion principle to which you refer.

In my further research on this problem, I saw elsewhere a reference to this, and I was in fact starting to search on this principle to understand it when I took a brief detour to see if any other replies had been made to my OP.

Anyway, while I don't understand everything you've said (becasue of a lack of education), I certianly have the capacity to understand it. And I am following your pointers now.

Thanks again, and I'm processing what you said...

[BruceZ]

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I appreciate the much more in-depth explanation you have provided. I do have what I would consider to be a basic understanding of probabilities, but I am unfamiliar with the inclusion-exclusion principle to which you refer.

In my further research on this problem, I saw elsewhere a reference to this, and I was in fact starting to search on this principle to understand it when I took a brief detour to see if any other replies had been made to my OP.

Anyway, while I don't understand everything you've said (becasue of a lack of education), I certianly have the capacity to understand it. And I am following your pointers now.

Thanks again, and I'm processing what you said...

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The inclusion-exclusion principle is only needed to take into account the royal on the board, and this only makes a small contribution to the final result. If we look only at cases where a player makes a royal with 1 or both hole cards, then we just have the probability of a union of mutually exclusive events, and this is the sum of the probabilities of each event, or in this case, 7 times the probability for 1 player.

[GrunchCan]

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he probability of a union of mutually exclusive events is the sum of the probabilities of each event, or in this case, 7 times the probability for 1 player.

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(Emphasis mine)

This then is what I need to consider, since it is this which I'm having a hard time accepting. I'm not suggesting you're wrong any more; just that I can't comprehend how that could possibly be right. I'm probably just tired & dumb.

[BruceZ]

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he probability of a union of mutually exclusive events is the sum of the probabilities of each event, or in this case, 7 times the probability for 1 player.

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(Emphasis mine)

This then is what I need to consider, since it is this which I'm having a hard time accepting. I'm not suggesting you're wrong any more; just that I can't comprehend how that could possibly be right. I'm probably just tired & dumb.

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Note that "the probability for 1 player" here means the probability that 1 specific player makes a royal using 1 or both hole cards. That is what makes it mutually exclusive. It is not the total probability of making a royal in 7 cards which is 4*C(47,2)/C(52,7) which includes royals on the board. The fraction of royals which are totally on the board is 1/C(7,5), since there are C(7,5) ways to choose the royal cards, so the fraction for which the player uses 1 or both hole cards is [C(7,5)-1]/C(7,5) = 20/21. So the probability of a player making a royal using 1 or both hole cards is 20/21 * 4*C(47,2)/C(52,7).

Now only some flops will allow a royal, so assuming that we have one of those flops, only 1 player can have a hand that makes a royal, which could require 1 specific card or 2 specific cards. If the probability of holding this hand is P, then a player will hold this hand a fraction P of the time, and since each player has the same probability P of holding the required hand, and since these fractions P do not overlap, then it should be clear that one of the 7 players will hold this hand a fraction 7P of the time.

Do you see it now?

This gives us an alternative way of computing the full answer (for 7 players including royals on the board) as a check. This is 7 times the probability that a player makes a royal using 1 or both hole cards, plus the probability of a royal on the board which is 4/C(52,5). This is:

7 * 20/21 * 4*C(47,2)/C(52,7) + 4/C(52,5)

= 1 in 4608, same as before, so that checks.

[GrunchCan]

I do understand what you're saying now. Thanks very much for your explanations & patience.