Another Tack on the Royal Flush

GrunchCan · #1 11-23-2005, 12:32 PM

You are observing a 7-handed game of texas hold'em. What is the probability that by the river someone (or everyone) would have a royal flush, if they stayed in the hand?

In other words, if nobody folds and all seven players show down, what is the probability that at least one player will have a royal flush?

I think I know the answer, but I don't want to taint the replies, and I want to check my answer (a friend of mine came up with the answer when I couldn't figure it out & asked him).

#2 11-23-2005, 12:52 PM

I'll give this a shot.

The chance of getting a royal flush in diamonds when you take 7 cards is 0.000808016%. So the chance of getting a royal flush in any suit is 4 times that or 0.003232062%.

The chance of that happening at least once when you do it 7 times is 1-(1-0.00003232062)**7 or 0.022622241%.

Or once in 4,420 hands.

GrunchCan · #3 11-23-2005, 01:11 PM

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?

But that can't be right, becasue of the community cards. If this were stud, it might be right. But becasue it's hold'em, you need to account for all seven players having 5 cards in common.

BruceZ · #4 11-23-2005, 01:16 PM

[ QUOTE ]
You are observing a 7-handed game of texas hold'em. What is the probability that by the river someone (or everyone) would have a royal flush, if they stayed in the hand?

In other words, if nobody folds and all seven players show down, what is the probability that at least one player will have a royal flush?

[/ QUOTE ]

For both 3 and 4 parts on the board, only 1 player can have it, so the only way that more than 1 player can make a royal is if the board is a royal, and then all 7 players have it. The probability of making a royal with any 7 cards is 4*C(47,2)/C(52,7). If we multiply this by 7, it will count the cases where there is a royal on the board 7 times, so we then subtract off 6 times the probability that the board is a royal, which is 6*4/C(52,5).

7*4*C(47,2)/C(52,7) - 6*4/(52,5)

=~ 0.0217% =~ 1 in 4608.

BruceZ · #5 11-23-2005, 01:18 PM

[ QUOTE ]
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?

[/ QUOTE ]

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

GrunchCan · #6 11-23-2005, 01:24 PM

[ QUOTE ]
[ QUOTE ]
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?

[/ QUOTE ]

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

[/ QUOTE ]

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?

#7 11-23-2005, 01:35 PM

I'd say that if the board in HE is 'royal friendly' that increases the chance that all 7 will get a royal. And if the board is 'royal unfriendly' it decreases the chance of all players. I'd think (without doing the math) that that exactly outweighs each other, so you can do the math for 7 independent hands and it gives the same results.

But that's just an assumption I have no facts or math to base on.

BruceZ · #8 11-23-2005, 01:51 PM

[ QUOTE ]
I'd say that if the board in HE is 'royal friendly' that increases the chance that all 7 will get a royal. And if the board is 'royal unfriendly' it decreases the chance of all players. I'd think (without doing the math) that that exactly outweighs each other, so you can do the math for 7 independent hands and it gives the same results.

But that's just an assumption I have no facts or math to base on.

[/ QUOTE ]

In order for you to multiply by 7, you need the 7 hands to be mutually exclusive, not independent. Mutually exclusive means that only 1 player can have a royal. This is not quite true because when there is a royal on the board, all 7 players have a royal. By multiplying by 7, you are counting the cases where there is a royal on the board 7 times. You must subtract off 6 of these as I did in my solution so that you only count these once. These are the only cases where more than 1 player can have a royal.

GrunchCan · #9 11-23-2005, 03:02 PM

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?

BruceZ · #10 11-23-2005, 06:19 PM

[ QUOTE ]
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?

[/ QUOTE ]

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.

[ QUOTE ]
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.

[/ QUOTE ]

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.