Flush Probability
1. Given a full ring table in NLHE (9 players), what is the maximum number of players who can flush?
2. In a heads up game, the probability both players flush is approximately
A. 0% B. 0.5% C. 1.0% D. > 1.0%
Detailed math analysis to provide your answer (estimate) not allowed 😀
See below for my right or wrong answers . I didn't see a HIDE option
[CODE]1. 9 Board flushes
2. B Not so easy for me to analyze but i got 0.402%[/CODE]
6 Replies
1. 9 - Four hearts on the board, each hand with one heart.
or 5 hearts on the board and 0 to 8 hearts to players
By feel, I would guess B is the closest; my math just barely agrees, saying
Spoiler
0.72%
Having slept and done it a few ways now, I'm less worried that I made a brainfart, so I'll share my calculation.
Spoiler
The answer will be 4x the probability of flush+flush in a specific suit, so let's talk about clubs. If >6 clubs are dealt total, both players have a flush no matter the configuration of the 9 dealt cards. When 5 or 6 clubs are dealt, the configuration matters.
So, we can split the calculation into two: P(>6 clubs) and P(5 or 6 in the right config).
(We don't need to, but I'll show it this way first.)
P(>6 clubs) = [C(13,9) + 39*C(13,8) + C(39,2)*C(13,7)] / C(52,9)
For the probability involving configurations, I find it helpful to use a denominator of C(52,13), which is all possible configurations of clubs within the 52 cards (ignoring ranks). Choosing a convenient denominator for the problem at hand has been the theme of recent threads.
P(5 or 6 well-configured clubs) = [C(43,8) + (5*2*2+4)*C(43,7)] / C(52,13)
There are 9 dealt cards and 43 undealt. If only 5 clubs are dealt, they must all be on the board, with the remaining 8 being among the 43 undealt. If 6 clubs are dealt, there can be four on the board (5 ways) and one in each player's hole cards (2 possible spots for each club), or there can be five on the board with the 6th being one of the 4 hole cards. There are again 43 spots for the 7 undealt clubs.
All told: P(both players flush) =
4{
[C(13,9) + 39*C(13,8) + C(39,2)*C(13,7)]/C(52,9) + [C(43,8) + (5*2*2+4)*C(43,7)]/C(52,13)
}
But since C(52,13) is the denominator for some of it, we might as well use that denominator for all of it:
4[C(43,4) + 9*C(43,5) + C(9,7)*C(43,6) + (5*2*2+4)*C(43,7) + C(43,8)] / C(52,13)
I did the following analysis
Prob(2 flush) =Sum(x=3 to 5)[ Pr(x cards of 1 suit on board) *Pr(Both players have needed flush cards given x board)]
Example: Board has 3 of one suit
Prob(both flush) = [4*C(13,3)*C(39,2)/C(52,5)]*[C(10,4)/C(47,4)]
If board has 3 of one suit,then 4 cards of the remaing 10 of the suit have to be dealt to the two players
I reviewed the calcs using probabilites instead of combos and made the novice mistake of not accounting for order (permutations). When I made the correction I got heehaw's result.
Oh, BTW, technically, you should eliminate straight flushes
