chances of flush vs flush

[MickeyHoldem]

None of the following calculations take into consideration cards that may or may not be played or folded, and assume that all hands are played to the river and showndown.

On the flop, the prob. of any 1 player having 2 spades...
p(1) = (8c2)/(45c2)
the prob. of any 2 players having 2 spades...
p(2) = (8c4)/(45c4)
similarly for 3 and 4 players...
p(3) = (8c6)/(45c6)
p(4) = (8c8)/(45c8)

Now there are 4 spades higher (and 4 lower) than yours, so p(1) & p(2) must be adjusted for the times they don't beat you...
q(1) = p(1) * (1 - (4c2)/(8c2))
q(2) = p(2) * (1 - (4c4)/(8c4))

Now by inclusion/exclusion we see that the prob. that no one is ahead of you on the flop (with 4 oppenents) is...
P(f) = 1 - 4*q(1) + 6*q(2) - 4*p(3) + p(4) = .913876
so the prob. that you are losing is 8.6124%


After the turn/river, there are 3 possible situations...
(a) no additional spades.... occurs (37c2)/(45c2)
(b) 1 additional spade.... occurs (8*37)/(45c2)
(c) 2 additional spades... occurs (8c2)/(45c2)

In (a), the situation is the same as the flop calculations above, except for there being 43 cards unknown cards instead of 45. You are ahead...
P(a) = .905883

For (b) the spade that fell could be higher or lower than your 7 (equal chance). If it's higher, there are 3 cards that beat you; lower, 4. So you're ahead here...
P(b) = .5 * (40c8)/(43c8) + .5 * (39c8)/(43c8) = .477311

For (c), things are more complicated again...
(c1) both spades are higher.... occurs (4c2)/(8c2)**
(c2) one higher, one lower.... occurs 4*4/(8c2)
(c3) both spades are lower.... occurs (4c2)/(8c2)
so similar to (b) you're ahead...
P(c) = (4c2)/(8c2) * (41c8)/(43c8) + 16/(8c2) * (40c8)/(43c8) + (4c2)/(8c2) * (39c8)/(43c8) = 0.535167

So finally we can combine these, and find out that you're ahead...
P = (37c2)/(45c2) * P(a) + (8*37)/(45c2) * P(b) + (8c2)/(45c2) * P(c)
= .767260
so the prob. that you will lose is .232740


**note when 2 higher spades fall here, you will only chop when no other player holds one of the remaining 2 higher spades.... this occurs .399342% of the time [img]/images/graemlins/wink.gif[/img]

[MickeyHoldem]

[ QUOTE ]
Yes, with 8 spades, there are 8*7/2 = 28 ways to combine two of them. But with only 4 higher spades, there are 4*4/2 = 8 ways for someone two have two higher spades plus 4*4 = 16 ways to have a higher one and a lower one. 8 + 16 = 24, so that's how many ways someone can have two spades that result in a higher flush. That's what I was trying to compute; and that's why I can multiply my 2.2% by 4.

[/ QUOTE ]
Should be...

4c2 = 6 ways to have 2 higher spades
4*4 = 16 ways to have 1 higher and 1 lower

6 + 16 = 22 ways for a higher flush

[AaronBrown]

Thank you for the correction.