Flush over flush

[Jman]

What is the probability of someone having a HIGHER flush if I have a:

T high flush with 3 other people at the river?

Q high flush with 2 other people at the river?

K high flush with 1 other person at the river?

Thanks!

[amerksmann22]

with a 4 flush on board?
or with a 3 flush on board?
be more specific, thanks.

[Jman]

Sorry I meant with 3 of one suit on the board. Thanks!

[aloiz]

I'll take a stab.

Probability that your 10 high flush is beaten against three other players = 3 * 4 * 7 / C(45,2) = .085

Q high flush against two players = 2 * 2 * 7 / C(45,2) = .028

K high flush against one player = 7 / C(45,2) = .007

Edit: this assumes that the T, Q, and K are in your hand

aloiz

[Jman]

Thanks aloiz. [img]/images/graemlins/laugh.gif[/img]

[BruceZ]

[ QUOTE ]
Probability that your 10 high flush is beaten against three other players = 3 * 4 * 7 / C(45,2) = .085

Q high flush against two players = 2 * 2 * 7 / C(45,2) = .028

K high flush against one player = 7 / C(45,2) = .007

Edit: this assumes that the T, Q, and K are in your hand

aloiz

[/ QUOTE ]

I hope you realize that only the last of these is exact. For the exact answer you cannot simply multiply by the number of players since this double counts the times that two players have you beat, and triple counts the times that 3 players have you beat. You must use the inclusion-exclusion principle for the exact answer.

Also, note that there are 4*7 + C(4,2) better hands in the first case, not just 4*7, because players can have 2 over cards. In the second case, there are 2*7 + 1 not just 2*7 for the same reason.

Refer to this post for JTs vs. 8 opponents. Note the correction also. That problem was on the flop, but you should be able to adapt this to the current problems. As you can see, the counting is very tricky, but fortunately only the first term makes significant difference in each case.

Q high flush against 2 players:

2*(2*7+1) / C(45,2) - 2*7*1*6 / C(45,2) / C(43,2)

= 3.0%

T high flush against 3 players:

3*[4*7 + C(4,2)] / C(45,2) -
C(3,2)*[4*7*3*6/4! + C(4,2)*(2*7 + 1) ] / C(45,2) / C(43,2)*2

= 10.2%

Convergence is obtained to 0.1% with just 2 terms.

[aloiz]

Thanks, but shouldn't the number of flushes that can be made in the the first case be 4 * 4 + 6 and the second case be 2 * 6 + 1?

And if I wanted to do the last term for the T high flush example would the following be correct?

(4*4 * 3*3 * 2*2/4! + C(4,2) * 4 * 3)/C(45,2)/C(43,2)/C(41,2) * 3!

aloiz