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#10
 irchans Senior Member Join Date: Sep 2002 Posts: 157 Re: A 7 Card Stud probability question

I will take a shot at this one. I will assume that the 3 face down cards are random. In a real game, this assumption is very wrong so the final probability is merely a guide. We will also have to ignore the possibility of straightflushes. Computations like this one often have many cases so they are frequently wrong. Don't believe the result until someone has confirmed it.

Suppose the opponents up cards are p p x1 x2. There are 3 down cards. There are C(52-4, 3) = 17296 possible sets of downcards. If he has a full house or four of a kind, then the downcards would have to be px-, xx-, pp-, ppx, pxx, or xxx. The size of each of these sets is

px- 2*6*40 = 480 ways
xx- 2*3*40 = 240 ways
pp- 1*40 = 40 ways
ppx 1*6 = 6 ways
pxx 2*2*3 = 12 ways
px1x2 2*3*3 = 18 ways
xxx 2 ways.
x1x1x2 3*3 = 9 ways
x1x2x2 3*3 = 9 ways

The resulting probability is (480+240+40+6+12+18+2+9+9)/17296 ~= 4.7 %.

Another way to derive the probability is p1 + p2 - p3 where

p1 is the probability that exactly 2 of the down cards have values of p, x1, or x2,
p2 is the probability that all of the down cards have values of p, x1, or x2, and
p3 is the probability that the opponent has 3 pair.

p1 = 3*(40/48*8/47*7/46)
p2 = 8*7*6/(48*47*46)
p3 = 6* (40/48*3/47*3/46.)

p1+p2-p3 ~= 4.7 %.

Dynasty might point out that in a real poker game, the opponent is more likely to have a pocket pair or paired door card. Also, the betting and cards exposed by the other players add a lot of information.

Hope that helps.