Suppose you hold 97652 after the second draw.

An opponent bets, and stands pat. Let's assume he has either 97642 or 98432 and the play of the hand does not make one hand any more likely

How much more likely is it that he has 98432 (only two duplicated cards) vs the 97642 (four duplicated cards)?

Or maybe that is not the best way to phrase the question, as I want to look at many combos better or worse than our 97652.

When compared to 97652, what adjustment factor should I possibly apply to 97542 or 98432? And then other hands based upon the common cards? Thanks...

Basically I want to sum up the Xs above and below our hand the 97652 accounting for card removal


Possible hands starting with
Strong 2 Card range (234-267; 238-268)
75432 x
76432 x
76532 x
76542 x
85432 x
86432 x
86532 x
86542 x
86543
87432 x
87532 x
87542 x
87543 x
87632 x
87642 x
87643
87652 x
87653
95432 x
96432 x
96532 x
96542 x
96543
97432 x
97532 x
97542 x
97543
97632 x
97642 x
97643
97652 This is your hand
97653
97654
98432 x
98532 x
98542 x
98543
98632 x
98642 x
98643
98652 x
98653
98654
98732 x
98742 x
98743
98752 x
98753
99754
98762 x
98763
98764

If you are just interested in the raw counts, leaving everything else to the side (including possibility of a flush), then, unless I am missing something, it is a matter of simply tallying the various cases.

Of course, if you hold a card of a particular rank, that means that there are only three remaining cards of that rank for Villain to hold (rather than four).

This means that, generally speaking, each "duplicated" rank multiplies Villain's possible holdings of a specific five-card hand by 3/4.

Assume you have five cards of five different ranks. For specificity, assume you hold 97652. Further assume you know that Villain also holds five cards of five different ranks. Then:

- the number of specific possible hands Villain can hold with 0 overlapping cards with yours (say KQJT3) is simply (4^5) = 1024

- the number of specific possible hands Villain can hold with 1 overlapping card with yours (say T9843) is simply (4^4)*(3^1) = 768

- the number of specific possible hands Villain can hold with 2 overlapping cards with yours (say 98432) is simply (4^3)*(3^2) = 576

- the number of specific possible hands Villain can hold with 3 overlapping cards with yours (say 87642) is simply (4^2)*(3^3) = 432

- the number of specific possible hands Villain can hold with 4 overlapping cards with yours (say 97642) is simply (4^1)*(3^4) = 324

- the number of specific possible hands Villain can hold with 5 overlapping cards with yours (97652) is simply (3^5) = 243.

Then what remains is to tally how many hands you are interested in determining for Villain fall into each category.

Of course, there are always complications (flushes, pairs, etc.) that may need to be considered, but I think these are the basics.

Hope this helps.

Thanks, I think this is probably helpful. Flushes are negligible and cancel out anyway.

I'm just interested in examining the relativity of how often villain has hands that are better and worse than mine, not the absolute number.

So since two duplicates is the least possible in our universe of 9 high or better hands the 98432 should be assigned a factor of 1, while the 97642 has four duplicates and thus should have a factor of (324/576) = .5625.

I know I'm the one trying to use this, but does this make sense to you?

Yes, that is a good way to think about it.