Probability question...

[ajohnson16]

Was thinking about this last night, and have no clue how to solve this, if it even can be solved.

In a standard table, how many board combinations are there to make top set the nuts?

What about five-handed, or heads up?

To anyone who can answer it, thanks in advance for satisfying my curiosity.

i've never seen how many different board combinations there can be... must be a huge #.

i don't think you're thinking about it correctly. you have your set (or trips, i get confused - one is pocket pair and the other pair on the board). then you really have to look at the board and see if there's danger.

i thought your question is way too broad.

[KJL]

I am not sure exactly what you are asking here. If you are asking how many boards in Texas Hold'em make top set the nuts, then it does not matter what size table it is at.
In order for top set to be the nuts, than the board cannot have any pairs, flushes, or straight combinations. I have no idea on how to do this, but maybe I'll try it later.

[ajohnson16]

After thinking about it all day, I also realized that the size of the table would not matter.

I figured this is a very difficult question to answer, and if no one can get it, it's okay with me.

By set, I mean PP with the third card on the board.

[WordWhiz]

I just brute forced it out. I may have missed something, but I did double check. I get 65 total. For QQQ, the board ranges from QJ762 to Q8732. For AAA, from AK984 to AJ762. I get 10 nut boards for a set of queens, 27 for kings, and 28 for aces, if anyone wants to compare answers. Because aces go high and low, none of the ace combos have a 5 or a 10 in them. It's impossible for a set of aces to be the nuts if there's a five or ten on the board--a straight's always possible.

[Bubu]

damn ! the site ate my answer. Anyway here it is - again.
More briefly....

If anyone knows of a better way of counting the solutions
please let me know. The trick is to note that the smallest
set making a nut hand is a set of queens. We can then
enumerate the possible boards. First ignore suits

23456789TJQ
23 78 Q
23 7 9 Q
23 7 T Q
23 7 JQ
2 4 7 9 Q
2 4 7 T Q
2 4 7 JQ
2 5 7 T Q
2 5 7 JQ
2 67 JQ

That's ten for queen set nut. The counting for a king set
and an ace set are similarly tedious. I have not done.
->N_queen_nut=10

Now to count ways to avoid flushes.

case 1.
rainbow suits which needs the one board card to
carry a repeated suit. There are 4 ways of choosing
what suit to repeat. Then there are 5 ways of choosing
what board card to carry that repeated suit. The remaing
4 distinct suits can be distributed in 4! ways amongst the
remaingin board cards ->4*5*4!

case 2.
3 suits on board so 2 cards carry distinct repeated suits.
Choose the first repeated suit in 4 ways. Choose the boards
cards carrying this suit in 5 choose 2 ways (10 ways). The second repeated suit can be chosen in 3 ways. The board cards carrying this suit can then be chosen in 3 choose 2 ways (3 ways). The remaining card can be of any of the two remaing suits.->4*10*3*3*2

other cases
2 sutis on board leads to possible flushes. so that's it !

=>non-flush_combos=4*5*4!+4*10*3*2

The answer is :

(N_queen_nut+N_king_nut+N_ace_nut)*non-flush combos


Hope this is right. Please if anyone has a better way
of counting let me know.


edit:
What the hell ! How do align the columns this ? I put a nice grid to help with the counting...