Poker math problem, not strategically relevant.

[Saddlepoint]

In a 10-handed hold'em game, if everyone sees the river for free, what are the odds of somebody having a Royal Flush at the end? And, would it be equal to (10)*(the odds of a Royal Flush being possible from 7 cards)? No, right?

I just am not completely sure how to attack it.

[Darryl_P]

It's not because there is some chance that everyone has a royal, ie. if the board comes up royal. Your way would incorrectly count that 10 times, when we only need to count it once. I suppose we could just adjust for that difference and go with your calculation, or:

Focus on the board and divide it into 3 cases:

1) 3 parts on board
2) 4 parts on board
3) 5 parts on board

Find the probability of each and multiply it by the pr. that someone has it, given a board like that. That gives 3 numbers that you just add up.

[BruceZ]

[ QUOTE ]
It's not because there is some chance that everyone has a royal, ie. if the board comes up royal. Your way would incorrectly count that 10 times, when we only need to count it once. I suppose we could just adjust for that difference and go with your calculation, or:

Focus on the board and divide it into 3 cases:

1) 3 parts on board
2) 4 parts on board
3) 5 parts on board

Find the probability of each and multiply it by the pr. that someone has it, given a board like that. That gives 3 numbers that you just add up.

[/ QUOTE ]

For both 3 and 4 parts on the board, only 1 player can have it, so we can just multiply the probability that 1 player makes it (with 1 or 2 in his hand) by 10 and add the probabilty that all 5 board cards make a royal. Or multiply the probability of making it with any 7 cards by 10, and subtract off 9 times the probabilty of making it with 5 cards.

10*4*C(47,2)/C(52,7) - 9*4/(52,5)
=~ 1 in 3233.