unique showdown probabilities for two preflop hand matchups

eastbay · #1 06-22-2005, 11:37 PM

There are 1326 possible preflop hands. There are roughly 1326^2 total preflop matchups (some of which are invalid because they share a card).

There are then roughly 1.7M entries in a full win/lose/tie matchup matrix.

Clearly some of these are equivalent. For example, AhTs vs. KhTs has exactly the same w/l/t breakdown as AcTh vs. KcTh, because both the rank and suit relationships are the same.

Q1: How many unique showdown probabilities are there for every possible two-hand matchup?

eastbay

GFunk911 · #2 06-23-2005, 01:34 AM

[ QUOTE ]
There are 1326 possible preflop hands. There are roughly 1326^2 total preflop matchups (some of which are invalid because they share a card).

There are then roughly 1.7M entries in a full win/lose/tie matchup matrix.

Clearly some of these are equivalent. For example, AhTs vs. KhTs has exactly the same w/l/t breakdown as AcTh vs. KcTh, because both the rank and suit relationships are the same.

Q1: How many unique showdown probabilities are there for every possible two-hand matchup?

eastbay

[/ QUOTE ]

Is this what you're looking for. Here it is only for when both sides are suited

There are 78 different ABs hands you can make.
If Player A has ABs, Player B can only make 55 different suited hands of the same suit.

78*78 (ABs v CDs where suits are different) + 78*55 (ABs v CDs where suits are same). = 10374 different probabilities for suited v suited.

Obviously there are many 50%'s (KTs v KTs, KJs v KJs, etc) bt they are really different probabilities. I assume you want them seperate.

Suited v Unsuited+Unpaired
78(#suited)*78(#off and not of suited's suit) + 78*66(#off where Card A is of suited's suit) + 78*66 (#off where Card B is off suited's suit) = 16380. It's highly possible there a divide or multiply by 2 error in there, although I suspect I possibly got it right by accident.

Suited v Pairs

78*13(#pairs with none of suited's suit)+78*11 (#pairs with one of suited's suit) = 1872

eastbay · #3 06-23-2005, 02:00 AM

[ QUOTE ]
[ QUOTE ]
There are 1326 possible preflop hands. There are roughly 1326^2 total preflop matchups (some of which are invalid because they share a card).

There are then roughly 1.7M entries in a full win/lose/tie matchup matrix. (It's symmetric, but ignore that for the moment.)

Clearly some of these are equivalent. For example, AhTs vs. KhTs has exactly the same w/l/t breakdown as AcTh vs. KcTh, because both the rank and suit relationships are the same.

Q1: How many unique showdown probabilities are there for every possible two-hand matchup?

eastbay

[/ QUOTE ]

Is this what you're looking for. Here it is only for when both sides are suited

There are 78 different ABs hands you can make.
If Player A has ABs, Player B can only make 55 different suited hands of the same suit.

78 (ABs v CDs where suits are different) + 55 (ABs v CDs where suits are same). = 133 different probabilities for suited v suited.

Obviously there are many 50%'s (KTs v KTs, KJs v KJs, etc) bt they are really different probabilities. I assume you want them seperate.

[/ QUOTE ]

Right, I'm interested in full w/l/t values, not just "pot equity". So KhTh vs. KsTs is not quite the same as KhJh vs. KsJs, for example. It is possible that there are "coincidental" non-unique probabilities in there somewhere. I'm not interested in figuring out what those might be.

My answer is over 100k, btw, just to give an idea.

eastbay

GFunk911 · #4 06-23-2005, 02:02 AM

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
There are 1326 possible preflop hands. There are roughly 1326^2 total preflop matchups (some of which are invalid because they share a card).

There are then roughly 1.7M entries in a full win/lose/tie matchup matrix.

Clearly some of these are equivalent. For example, AhTs vs. KhTs has exactly the same w/l/t breakdown as AcTh vs. KcTh, because both the rank and suit relationships are the same.

Q1: How many unique showdown probabilities are there for every possible two-hand matchup?

eastbay

[/ QUOTE ]

Is this what you're looking for. Here it is only for when both sides are suited

There are 78 different ABs hands you can make.
If Player A has ABs, Player B can only make 55 different suited hands of the same suit.

78 (ABs v CDs where suits are different) + 55 (ABs v CDs where suits are same). = 133 different probabilities for suited v suited.

Obviously there are many 50%'s (KTs v KTs, KJs v KJs, etc) bt they are really different probabilities. I assume you want them seperate.

[/ QUOTE ]

Right, I'm interested in full w/l/t values, not just "pot equity". So KhTh vs. KsTs is not quite the same as KhJh vs. KsJs, for example. It is possible that there are "coincidental" non-unique probabilities in there somewhere. I'm not interested in figuring out what those might be.

My answer is over 100k, btw, just to give an idea.

eastbay

[/ QUOTE ]

Yes, i had a total brain fart and said 78+55, instead of (78+55)*78. I already FMP.

eastbay · #5 06-23-2005, 02:05 AM

The matchup matrix is symmetric, but ignore that for the time being. We can divide by two at the end.

eastbay

GFunk911 · #6 06-23-2005, 02:10 AM

[ QUOTE ]
The matchup matrix is symmetric, but ignore that for the time being. We can divide by two at the end.

eastbay

[/ QUOTE ]

right, it can clearly be done in a consistent way and then divide by 2 at the end. my muddled brain is just f'ing it up, inconsistently dividing by 2 sometiems and not other times .

GFunk911 · #7 06-23-2005, 02:32 AM

For 2 offsuit hands that are both unpaired, I am getting 38,922 possibilities. Does this match your answer? It seems low to me.

(6084+5616+5616+5616+5616+5187+5187)

I'm fairly sure I made at least two errors, but sleep beckons. If you're actually interesting in some assistance or confirmation, as opposed to simply posing a question, PM me.

Math+Night=Errors

SumZero · #8 06-23-2005, 04:55 AM

My quick SWAG:

Part 1:
First consider hand A is suited. There are 13*12/2 such hands.
Part 1a: Consider hand B suited in the same suit. There are 11*10/2 such hands.
Part 1b: Consider hand B suited in some other suit. There are 13*12/2 such hands.
Part 1c: Consider hand B unsuited with 1 card in the hand A suit. There are 11*13 such hands.
Part 1d: Consider hand B unsuited with 0 cards in the hand A suit. There are 13 + 13*12/2 such hands.
Conclusion of part 1: A * B = 78 * 367 = 28,626.

Part 2:
Now consider hand A is a pair. There are 13 such hands.
Part 2a: Consider hand B suited in the same suit as one of the pair cards. There are 12*11/2 such hands.
Part 2b: Consider hand B suited in a different suit than the pair cards. There are 13*12/2 such hands.
Part 2c: Consider hand B unsuited with 1 card in a paired card's suit. There are 12[pairs] + 12*12[unpaired] such hands.
Part 2d: Consider hand B unsuited with both cards in each of the paired card's suit. There are 12[pairs] + 12*11/2[unpaired] such hands.
Part 2e: Consider hand B unsuited with neither card in each of the paired card's suit There are 13[pairs] + 13*12/2[unpaired] such hands.
Conclusion of part 2: A * B = 13 * 469 = 6,097.

Part 3:
Now consider hand A is an unsuited unpaired hand. There are 13*12/2 such hands.
Part 3a: Consider hand B suited in the higher card suit. There are 12*11/2 such hands.
Part 3b: Consider hand B suited in the lower card suit. There are 12*11/2 such hands.
Part 3c: Consider hand B suited in an unused suit. There are 13*12/2 such hands.
Part 3d: Consider hand B paired in the suits of hand A. There are 11 such hands.
Part 3e: Consider hand B paired with 1 card in the suit of the higher card of A and one card in an unused suit. There are 12 such hands.
Part 3f: Consider hand B paired with 1 card in the suit of the lower card of A and one card in an unused suit. There are 12 such hands.
Part 3g: Consider hand B paired in the unused suits. There are 13 such hands.
Part 3h: Consider hand B unpaired unsuited in the two suits of A. There are 11*10/2[no shared ranks]+12[share rank with lower card of A in high card suit]+11[share rank with higher card of A in low card suit but no double count of shared lower card of A in high card suit] such hands.
Part 3i: Consider hand B unpaired and unsuited with one card in same suit as the higher card of A and the other card in an unused suit. There are 12*12 such hands.
Part 3j: Consider hand B unpaired and unsuited with one card in the same suit as the lower card of A and the other card in an unused suit. There are 12*12 such hands.
Part 3k: Consider hand B unpaired unsuited and sharing no suits with A. There are 13*12/2 such hands.
Conclusion of part 3: A * B = 78 * 702 = 54,756.

So I might have miscounted something but that gives me:

28,626 + 6,097 + 54,756 = 89,479 such hands.

But it is late so I may have miscounted.

eastbay · #9 06-23-2005, 10:28 AM

[ QUOTE ]

So I might have miscounted something but that gives me:

28,626 + 6,097 + 54,756 = 89,479 such hands.

But it is late so I may have miscounted.

[/ QUOTE ]

I got 83,473. I found my answer via an equivalence class generating algorithm (because I wanted my method to be general for the N-way problem), so it's hard to tell where the discrepancy might be. Pretty close, though.

eastbay

pzhon · #10 06-23-2005, 05:10 PM

[ QUOTE ]

Part 3h: Consider hand B unpaired unsuited in the two suits of A. There are 11*10/2[no shared ranks]+12[share rank with lower card of A in high card suit]+11[share rank with higher card of A in low card suit but no double count of shared lower card of A in high card suit] such hands.

[/ QUOTE ]
My count agrees with yours except in this step. Suppose hand A is A [img]/images/graemlins/spade.gif[/img] K [img]/images/graemlins/heart.gif[/img]. When B has no shared rank, you can still distinguish Q [img]/images/graemlins/spade.gif[/img] J [img]/images/graemlins/heart.gif[/img] from Q [img]/images/graemlins/heart.gif[/img] J [img]/images/graemlins/spade.gif[/img]. I think you undercounted by 55*78.

Actually, this example is flawed, since those matchups give the same w/l/t probabilities. However, A[img]/images/graemlins/spade.gif[/img] 7[img]/images/graemlins/heart.gif[/img] vs. Q[img]/images/graemlins/spade.gif[/img] J[img]/images/graemlins/heart.gif[/img] is different from A[img]/images/graemlins/spade.gif[/img] 7[img]/images/graemlins/heart.gif[/img] Q[img]/images/graemlins/heart.gif[/img] J[img]/images/graemlins/spade.gif[/img].

http://twodimes.net/h/?z=138760
cards win %win lose %lose tie %tie EV
A[img]/images/graemlins/spade.gif[/img] K[img]/images/graemlins/heart.gif[/img] 1114667 65.10 588268 34.36 9369 0.55 0.654
Q[img]/images/graemlins/spade.gif[/img] J[img]/images/graemlins/heart.gif[/img] 588268 34.36 1114667 65.10 9369 0.55 0.346

http://twodimes.net/h/?z=121452
cards win %win lose %lose tie %tie EV
A[img]/images/graemlins/spade.gif[/img] K[img]/images/graemlins/heart.gif[/img] 1114667 65.10 588268 34.36 9369 0.55 0.654
J[img]/images/graemlins/spade.gif[/img] Q[img]/images/graemlins/heart.gif[/img] 588268 34.36 1114667 65.10 9369 0.55 0.346

http://twodimes.net/h/?z=1047170
cards win %win lose %lose tie %tie EV
A[img]/images/graemlins/spade.gif[/img] 7[img]/images/graemlins/heart.gif[/img] 951868 55.59 752043 43.92 8393 0.49 0.558
J[img]/images/graemlins/spade.gif[/img] Q[img]/images/graemlins/heart.gif[/img] 752043 43.92 951868 55.59 8393 0.49 0.442

http://twodimes.net/h/?z=318619
cards win %win lose %lose tie %tie EV
A[img]/images/graemlins/spade.gif[/img] 7[img]/images/graemlins/heart.gif[/img] 951910 55.59 752000 43.92 8394 0.49 0.558
Q[img]/images/graemlins/spade.gif[/img] J[img]/images/graemlins/heart.gif[/img] 752000 43.92 951910 55.59 8394 0.49 0.442