unique showdown probabilities for two preflop hand matchups

[eastbay]

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]

[ 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]

[ 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]

[ 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]

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

eastbay

[GFunk911]

[ 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]

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]

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]

[ 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]

[ 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 /images/graemlins/spade.gif K /images/graemlins/heart.gif. When B has no shared rank, you can still distinguish Q /images/graemlins/spade.gif J /images/graemlins/heart.gif from Q /images/graemlins/heart.gif J /images/graemlins/spade.gif. I think you undercounted by 55*78.

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


http://twodimes.net/h/?z=138760
cards win %win lose %lose tie %tie EV
A/images/graemlins/spade.gif K/images/graemlins/heart.gif 1114667 65.10 588268 34.36 9369 0.55 0.654
Q/images/graemlins/spade.gif J/images/graemlins/heart.gif 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/images/graemlins/spade.gif K/images/graemlins/heart.gif 1114667 65.10 588268 34.36 9369 0.55 0.654
J/images/graemlins/spade.gif Q/images/graemlins/heart.gif 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/images/graemlins/spade.gif 7/images/graemlins/heart.gif 951868 55.59 752043 43.92 8393 0.49 0.558
J/images/graemlins/spade.gif Q/images/graemlins/heart.gif 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/images/graemlins/spade.gif 7/images/graemlins/heart.gif 951910 55.59 752000 43.92 8394 0.49 0.558
Q/images/graemlins/spade.gif J/images/graemlins/heart.gif 752000 43.92 951910 55.59 8394 0.49 0.442

[pzhon]

I decided to try to use Burnside's Lemma. Sometimes this makes hard problems easy, and sometimes it makes easy problems hard.

Burnside's Lemma (Cauchy):

Let G be a group acting a set S. The number of orbits of the group action equals

Sum |x fixed by g| |G|^-1 .
g in G

Here, the relevant set is unordered pairs of hands, and the relevant group is S4 permuting the suits. (For generalizations, it might be more convenient to use ordered n-tuples of distinct hands, and S4 (+) Sn.) The orbits are the equivalence classes of pairs of hands, precisely what we want to count.

There are 24 elements of S4, so there are 24 terms to add, but many of these are equivalent. Ther are only 5 conjugacy classes of permutations of 4 elements: The identity, switching two suits, switching two pairs of suits, cycling 3 suits, and cycling 4 suits.

/images/graemlins/spade.gif The identity. There is only one permutation of this type. It fixes all (52C2)(50C2)/2 = 812,175 unordered pairs of hands, which contributes 812,175/24 = 33,840.625 to the total. (Yes, it does seem strange that this isn't an integer. The total will be an integer. If it weren't, it would indicate there is an error in the calculations.)

/images/graemlins/spade.gif Six permutations switch a pair of suits. Suppose we are switching all clubs and diamonds. Which ordered pairs of hands are fixed by this? There must be either 0, 1, or 2 clubs.

Subcase 0: With no clubs or diamonds, all of the suits in the two hands must be spades or hearts. There are (26C2)(24C2)/2 ordered pairs of hands fixed because they have no clubs or diamonds.

Subcase 1: There is one club and one diamond. For the switch not to change the hands, the club and the diamond must be in the same hand. That means we have a pair with suits /images/graemlins/club.gif /images/graemlins/diamond.gif and a hand with neither club nor diamond. There are 13*(26C2) possibilities.

Subcase 2: There are two clubs and two diamonds. Either switching the suits switches two offsuit hands, X /images/graemlins/club.gif Y /images/graemlins/diamond.gif <-> X /images/graemlins/diamond.gif Y /images/graemlins/club.gif, or it switches two suited hands, X /images/graemlins/diamond.gif Y /images/graemlins/diamond.gif <-> X /images/graemlins/club.gif Y /images/graemlins/club.gif, or it fixes the hands, X /images/graemlins/club.gif X /images/graemlins/diamond.gif, Y /images/graemlins/club.gif Y /images/graemlins/diamond.gif. There are (13C2) possibilities of each type, or 3(13C2) total for this subcase.

In total, switching clubs and diamonds fixes (26C2 24C2 / 2) + 13*(26C2) + 3*(13C2) = 49,231 pairs of hands. Since there are 6 permutations of this type, this contributes 49,231*6/24 = 12,327.25.

/images/graemlins/spade.gif Three permutations switch two pairs of suits, e.g., /images/graemlins/spade.gif <-> /images/graemlins/heart.gif, /images/graemlins/diamond.gif <-> /images/graemlins/club.gif. How many pairs of hands are fixed by this?

Subcase 0: 0 spades. Then there are two diamonds and two clubs, just like subcase 2 of the switch /images/graemlins/diamond.gif <-> /images/graemlins/club.gif. There are 3*(13C2) pairs of hands fixed this way.

Subcase 2: 2 spades. There are two spades and two hearts, just like subcase 2 of the switch /images/graemlins/diamond.gif <-> /images/graemlins/club.gif. There are 3*(13C2) pairs of hands fixed this way.

Subcase 1: There is one card of each suit. If the switch fixes the hands, they must look like X /images/graemlins/spade.gif X /images/graemlins/heart.gif, Y /images/graemlins/diamond.gif Y /images/graemlins/club.gif, and there are 13^2 such possibilities. If the switch exchanges the hands, they either look like X /images/graemlins/spade.gif Y /images/graemlins/diamond.gif, X /images/graemlins/heart.gif Y /images/graemlins/club.gif, or X /images/graemlins/spade.gif Y /images/graemlins/club.gif, X /images/graemlins/heart.gif Y /images/graemlins/diamond.gif. There are 2*13^2 more possibilities of this type, for a total of 3*13^2 in this subcase.

In total, /images/graemlins/spade.gif<-> /images/graemlins/heart.gif, /images/graemlins/diamond.gif<-> /images/graemlins/club.gif fixes 3(13C2) + 3(13C2) + 3*13^2 = 819 pairs of hands. Since there are 3 permutations of this type, the total contribution is 819*3/24 = 121.875.

/images/graemlins/spade.gif Eight permutations rotate 3 suits, e.g., /images/graemlins/spade.gif -> /images/graemlins/heart.gif -> /images/graemlins/diamond.gif -> /images/graemlins/spade.gif. The 3-cycle must fix the hands, and no hand can contain 3 suits, so for this to fix a pair of hands, the hands consist of all clubs. There are 13C2 * 11C2/2 = 2145 unordered pairs of hands suited in clubs. The contribution of this case is 2145*8/24 = 715.

/images/graemlins/spade.gif Six permutations rotate 4 suits, e.g., /images/graemlins/spade.gif -> /images/graemlins/heart.gif -> /images/graemlins/club.gif -> /images/graemlins/diamond.gif -> /images/graemlins/spade.gif. Since no hand can contain all 4 suits, for a 4-cycle to fix a pair of hands, it must switch the hands. All cards have the same rank, so the hands look like X /images/graemlins/spade.gif X /images/graemlins/club.gif, X /images/graemlins/heart.gif X /images/graemlins/diamond.gif. There are 13 such pairs for each 4-cycle. The total contribution is 13*6/24 = 3.25.

So, the total number of equivalence classes of unordered pairs of hands is 33,840.625 + 12,327.25 + 121.875 + 715 + 3.25 = 47,008.

To get the count of ordered pairs, it is not quite as simple as multiplying by 2. Some classes of unordered pairs correspond to only one equivalence class of ordered pair, namely, the ties. The number of equivalence classes of ordered pairs is 2*(47008)-ties. The 13 pairs can be tied in 1 way, the 78 suited hands can be tied in 1 way, and the 78 offsuit hands can be tied in two ways (disjoint suits, or shared suits), for a total of 247 ties. The number of equivalence classes of ordered pairs is 93,769.


As a check, here is my attempt at the straightforward method. I'll count equivalence classes of ordered pairs of hands (A,B).

Case (pair,pair): There are 13 possible pairs for A. For each, B can be of the same rank in one way. If B is of a different rank, there are 12 possible ranks, and B can share 0, 1, or 2 suits, for a total of 13*(1+12*3)= 481.

Case (pair, suited): If a rank is shared, there are 12 possibilities for the other rank. If no rank is shared, either a suit is shared or not, for 2*(12C2) possibilities. Total: 13*(12+2*(12C2))= 1872.

Case (pair, offsuit): If a rank is shared, there are 12 choices for the other rank, and 2 choices for whether the suit is shared or not. If no rank is shared, there are 12C2 choices for the ranks, and 4 choices for which suits are shared. Total: 13(12*2 + (12C2)4) = 3744.

Case (suited, suited): If the suit is shared, there are (13C2)(11C2) possibilities. If the suits are not shared, every collection of ranks produces a separate case, so there are (13C2)^2 possibilities. Total: (13C2)(11C2)+(13C2)^2 = 10,374.

Case (suited, offsuit): There are 13C2 choices for the suited hand. If a suit is shared, there are 11 choices for the rank of the third card, and 12 choices for the last card. If no suit is shared, all 13C2 pairs of ranks give different cases. Total: (13C2)(11*12)+(13C2)^2 = 16,380.

Case (offsuit, offsuit): There are 13C2 choices for the ranks of the first hand.

Subcase (0 suits in common): There are 13C2 choices for the ranks of the second hand. (13C2)(13C2)

Subcase (1 suit in common): There are 2 choices for which card is shared, 12 choices for the rank of the second hand's card of the shared suit, and 12 choices for the rank of the last card. (13C2) (2*12*12)

Subcase (2 suits in common): For convenience, let the first hand be A /images/graemlins/spade.gif K /images/graemlins/heart.gif. The second hand has one spade and one heart. If the spade is a king, there are 12 choices for the heart. If the spade is not a king, there are only 11 choices for the heart, since the hand must not be a pair. (13C2)(12+11*11).

Total: (13C2)^2 + (13C2)(2*12*12) + (13C2)(12+11*11) = 38,922.

The other cases are counted by symmetry:

Case (offsuit, suited): Same as (suited, offsuit), 16,380.
Case (offsuit, pair): Same as (pair, offsuit), 3744.
Case (suited, pair): Same as (pair, suited), 1872.

Total: 481 + 10374 + 38922 + 2(16380 + 3744 + 1872) = 93,769.

[tworooks]

Wouldn't you be pissed if you wrote all that and you were wrong?

[pzhon]

[ QUOTE ]
Wouldn't you be pissed if you wrote all that and you were wrong?

[/ QUOTE ]
First, that I used two methods and came up with the same number makes me trust my answer.

Second, if it turns out that I am wrong, I wouldn't really be upset. I knew Burnside's Lemma, but hadn't applied it to anything nontrivial. This problem gave me an excuse to see how it works in practice.

[eastbay]

[ QUOTE ]

There are 24 elements of S4, so there are 24 terms to add, but many of these are equivalent. Ther are only 5 conjugacy classes of permutations of 4 elements: The identity, switching two suits, switching two pairs of suits, cycling 3 suits, and cycling 4 suits.


[/ QUOTE ]

I know nothing of group theory terminology so I didn't try to follow your Burnside's lemma method.

Let me give my general approach which I think is somehow related to your "conjugacy class" idea, which was designed to have some hope of generalizing to the N-way case (where enumeration by inspection is awful.)

Step 1: Construct the suit permutation operators which span the space of the equivalence class of any given matchup.

Step 2: Find a method for applying said operators such that the space is indeed spanned for every possible matchup. Redundancy is not important.

After some rumination I came up with two logically distinct operators:

f1: cycle all suits
f2,k: cycle suit k through all remaining unused suits, k=1..4

Q1a: Does this span the space of all equivalence classes?

My method for generation of all members of the space is:

H1 H1k=f2,k(H1)
H2=f1(H1) H2k=f2,k(H2)
H3=f1(H1) H3k=f2,k(H3)
H4=f1(H1) H4k=f2,k(H4)

With the equivalence class being hopefully fully enumerated in the generated set of matchups.

Q1b: Does this really span the space of every possible matchup's equivalence class?

While this may seem a bit of a strange approach, what I am looking for here is really an automated equivalence class generation method that generalizes to the N-way problem.

eastbay

[pzhon]

[ QUOTE ]

Step 1: Construct the suit permutation operators which span the space of the equivalence class of any given matchup.

[/ QUOTE ]
You mean you want a set of operations to perform on any matchup to find all equivalent matchups? The equivalent matchups are all found by permuting the suits, and it is possible that all 24 permutations give you distinct hands, e.g., A/images/graemlins/spade.gif K/images/graemlins/heart.gif vs. T/images/graemlins/heart.gif 8/images/graemlins/diamond.gif, or A/images/graemlins/spade.gif K/images/graemlins/heart.gif vs. Q/images/graemlins/diamond.gif J/images/graemlins/club.gif. Do you want a set of generators for the permutations of the 4 suits?

[ QUOTE ]

f1: cycle all suits
f2,k: cycle suit k through all remaining unused suits, k=1..4


[/ QUOTE ]
I'm not sure what you mean by these. Is f1 the operation /images/graemlins/spade.gif -> /images/graemlins/heart.gif -> /images/graemlins/diamond.gif -> /images/graemlins/club.gif -> /images/graemlins/spade.gif? If so, what is f2,1? f2,2? Are you making these change, based on the matchup? You don't need to do that.

Any 4-cycle and any 3-cycle will generate all 24 permutations. (The only subgroups of S4 containing a 4-cycle are the powers of the 4-cycle, a group of order 8 that looks like the symmetries of a square, and all of S4. The first two don't include an element of order 3, so adding this generates all of S4.) However, you need to apply the 4-cycle and 3-cycle multiple times, and not just as f^a t^b. You have to be able to get 24 distinct results.

For n hands, here is a simple Monte Carlo technique: For any a random set of hands, you can compute x, the number of equivalent n-tuples of hands. Empirically find A, the average value of 1/x. The total number of nonequivalent hands is A*(total # of n-tuples of hands). This may converge rapidly enough for your purposes, but it depends on having a good way to generate hands uniformly randomly. Was this your plan?

[eastbay]

[ QUOTE ]

You mean you want a set of operations to perform on any matchup to find all equivalent matchups?


[/ QUOTE ]

Yes.

[ QUOTE ]

The equivalent matchups are all found by permuting the suits


[/ QUOTE ]

Right.

[ QUOTE ]

, and it is possible that all 24 permutations give you distinct hands, e.g., A/images/graemlins/spade.gif K/images/graemlins/heart.gif vs. T/images/graemlins/heart.gif 8/images/graemlins/diamond.gif, or A/images/graemlins/spade.gif K/images/graemlins/heart.gif vs. Q/images/graemlins/diamond.gif J/images/graemlins/club.gif. Do you want a set of generators for the permutations of the 4 suits?


[/ QUOTE ]

Not sure what you mean. I want permutations that remain within the equivalence class.

[ QUOTE ]

f1: cycle all suits
f2,k: cycle suit k through all remaining unused suits, k=1..4


[/ QUOTE ]
I'm not sure what you mean by these.



[/ QUOTE ]

It's very simple.
A1K2 vs Q2J3 under f1 becomes A2K3 vs. Q3J4.

A1K2 vs. Q2J3 under f1,1 becomes A4K2 vs. Q2J3 (suit k=1 is cycled through all suits that don't otherwise appear in the matchup.)

[ QUOTE ]

Are you making these change, based on the matchup?


[/ QUOTE ]

No.

[ QUOTE ]

Any 4-cycle and any 3-cycle will generate all 24 permutations.


[/ QUOTE ]

I don't know what you mean by 4-cycle or 3-cycle, and I'm no t sure what you mean by "all 24" permutations. Certainly some equivalence classes do not contain 24 matchups.

[ QUOTE ]

(The only subgroups of S4 containing a 4-cycle are the powers of the 4-cycle, a group of order 8


[/ QUOTE ]

I don't know what a group is, or what a subgroup is.

eastbay

[pzhon]

[ QUOTE ]
[ QUOTE ]

You mean you want a set of operations to perform on any matchup to find all equivalent matchups?


[/ QUOTE ]

Yes.

[ QUOTE ]

The equivalent matchups are all found by permuting the suits


[/ QUOTE ]

Right.

[/ QUOTE ]
Ok, there are 24 ways to permute the 4 suits.

1 permutation does nothing.
6 switch 2 suits, e.g., /images/graemlins/spade.gif <-> /images/graemlins/heart.gif.
3 switch 2 pairs of suits, e.g., /images/graemlins/spade.gif <-> /images/graemlins/heart.gif, /images/graemlins/diamond.gif <-> /images/graemlins/club.gif.
8 cycle 3 suits, leaving 1 fixed, e.g., /images/graemlins/spade.gif -> /images/graemlins/heart.gif -> /images/graemlins/diamond.gif -> /images/graemlins/spade.gif.
6 cycle 4 suits, e.g., /images/graemlins/spade.gif -> /images/graemlins/club.gif -> /images/graemlins/diamond.gif -> /images/graemlins/heart.gif -> /images/graemlins/spade.gif.

These 5 essentially different classes of permutations are indexed by the cycle structures, the 5 partitions of 4: 1+1+1+1, 2+1+1, 2+2, 3+1, and 4. Those were the 5 cases I used in my application of Burnside's Lemma.

What more do you want than these 24 operations? Note that this set of 24 does not depend on the pairs of cards.

[ QUOTE ]
f1,1 becomes A4K2 vs. Q2J3 (suit k=1 is cycled through all suits that don't otherwise appear in the matchup.)

[ QUOTE ]

Are you making these change, based on the matchup?


[/ QUOTE ]

No.

[/ QUOTE ]
You seem to have contradicted yourself. You say the operations both don't depend on the matchup, and depend on the unused suits.

Also, you hadn't mentioned f1,1 before, only f1 and f2,k.

You still haven't been clear what you mean by "cycle through." What you have said is ambiguous, since you have not stated what is involved in the cycle, and when you have more than 2 elements in a cycle, there are multiple possible cycles.

I think you should adopt a standard notation for permutations (http://en.wikipedia.org/wiki/Permutation) (see the abstract algebra section of the linked page) rather than trying to invent your own notation.

[jason_t]

[ QUOTE ]
[ QUOTE ]
Wouldn't you be pissed if you wrote all that and you were wrong?

[/ QUOTE ]
First, that I used two methods and came up with the same number makes me trust my answer.

Second, if it turns out that I am wrong, I wouldn't really be upset. I knew Burnside's Lemma, but hadn't applied it to anything nontrivial. This problem gave me an excuse to see how it works in practice.

[/ QUOTE ]

You're trained in math, aren't you? Burnside's lemma owns.

[eastbay]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]

You mean you want a set of operations to perform on any matchup to find all equivalent matchups?


[/ QUOTE ]

Yes.

[ QUOTE ]

The equivalent matchups are all found by permuting the suits


[/ QUOTE ]

Right.

[/ QUOTE ]
Ok, there are 24 ways to permute the 4 suits.

1 permutation does nothing.
6 switch 2 suits, e.g., /images/graemlins/spade.gif <-> /images/graemlins/heart.gif.
3 switch 2 pairs of suits, e.g., /images/graemlins/spade.gif <-> /images/graemlins/heart.gif, /images/graemlins/diamond.gif <-> /images/graemlins/club.gif.
8 cycle 3 suits, leaving 1 fixed, e.g., /images/graemlins/spade.gif -> /images/graemlins/heart.gif -> /images/graemlins/diamond.gif -> /images/graemlins/spade.gif.
6 cycle 4 suits, e.g., /images/graemlins/spade.gif -> /images/graemlins/club.gif -> /images/graemlins/diamond.gif -> /images/graemlins/heart.gif -> /images/graemlins/spade.gif.

These 5 essentially different classes of permutations are indexed by the cycle structures, the 5 partitions of 4: 1+1+1+1, 2+1+1, 2+2, 3+1, and 4. Those were the 5 cases I used in my application of Burnside's Lemma.

What more do you want than these 24 operations? Note that this set of 24 does not depend on the pairs of cards.


[/ QUOTE ]

You must be using some specialized meaning for the word "cycle", especially since you are confused by my (what seems to me to be plain English) use of it.

For example, your "cycle 3 suits, leaving 1 fixed" doesn't seem to remain within the equivalence class, unless you take care not to use the 1 suit which has remained fixed. This was the essence of my f2 operators.

For example:

A1K2 vs. Q3J4

cycling the first 3 suits means to me:

A2K3 vs. Q4J4

except that those two matchups are not in the same equivalence class.

eastbay

[eastbay]

[ QUOTE ]

For example:

A1K2 vs. Q3J4

cycling the first 3 suits means to me:

A2K3 vs. Q4J4

except that those two matchups are not in the same equivalence class.

eastbay

[/ QUOTE ]

I see what must mean now. You mean:

A1K2 vs. Q3J4 -> A3K1 vs. Q2J4

Yes, that makes sense.

However, while your way of organizing the permutations is nice on paper, it looks problematic from an algorithmic perspective. For example, given the matchup:

A1K1 vs. Q1J1

the "switch two suits" permutation doesn't apply. There aren't two suits to switch. I don't see an efficient way to only apply those permutations which are applicable to any given matchup, whereas that was very natural given my forms.

The idea here is to quickly discover all members of each equivalence class for the N-way case. Counting the classes was just a sanity check on my method.

eastbay

[pzhon]

[ QUOTE ]
However, while your way of organizing the permutations is nice on paper, it looks problematic from an algorithmic perspective.

[/ QUOTE ]
I still don't see what difficulties there are and what alternative you offer.

[ QUOTE ]
For example, given the matchup:

A1K1 vs. Q1J1

the "switch two suits" permutation doesn't apply. There aren't two suits to switch.

[/ QUOTE ]
There are always 4 suits. You can always apply all 24 permutations of the 4 suits.

/images/graemlins/spade.gif <-> /images/graemlins/heart.gif applied to A/images/graemlins/spade.gif K/images/graemlins/spade.gif vs. Q/images/graemlins/spade.gif J/images/graemlins/spade.gif results in A/images/graemlins/heart.gif K/images/graemlins/heart.gif vs. Q/images/graemlins/heart.gif J/images/graemlins/heart.gif.

/images/graemlins/diamond.gif <-> /images/graemlins/club.gif applied to A/images/graemlins/spade.gif K/images/graemlins/spade.gif vs. Q/images/graemlins/spade.gif J/images/graemlins/spade.gif results in A/images/graemlins/spade.gif K/images/graemlins/spade.gif vs. Q/images/graemlins/spade.gif J/images/graemlins/spade.gif.

[ QUOTE ]
The idea here is to quickly discover all members of each equivalence class for the N-way case. Counting the classes was just a sanity check on my method.

[/ QUOTE ]
You can always apply the 24 permutations of the suits to get a complete list of equivalent match-ups. Sometimes there will be only 3 distinct match-ups repeated 8 times. Sometimes there will be 24 distinct collections of hands.

[eastbay]

[ QUOTE ]
[ QUOTE ]
However, while your way of organizing the permutations is nice on paper, it looks problematic from an algorithmic perspective.

[/ QUOTE ]
I still don't see what difficulties there are and what alternative you offer.

[ QUOTE ]
For example, given the matchup:

A1K1 vs. Q1J1

the "switch two suits" permutation doesn't apply. There aren't two suits to switch.

[/ QUOTE ]
There are always 4 suits. You can always apply all 24 permutations of the 4 suits.

/images/graemlins/spade.gif <-> /images/graemlins/heart.gif applied to A/images/graemlins/spade.gif K/images/graemlins/spade.gif vs. Q/images/graemlins/spade.gif J/images/graemlins/spade.gif results in A/images/graemlins/heart.gif K/images/graemlins/heart.gif vs. Q/images/graemlins/heart.gif J/images/graemlins/heart.gif.

/images/graemlins/diamond.gif <-> /images/graemlins/club.gif applied to A/images/graemlins/spade.gif K/images/graemlins/spade.gif vs. Q/images/graemlins/spade.gif J/images/graemlins/spade.gif results in A/images/graemlins/spade.gif K/images/graemlins/spade.gif vs. Q/images/graemlins/spade.gif J/images/graemlins/spade.gif.


[/ QUOTE ]

Ok, I thought "switch two suits" meant A1K2 vs Q1J2 -> A2K1 vs.Q2J1, which clearly wouldn't make sense if there was only 1 suit in the hand. The language here is ambiguous.

Now I wonder if I know what you meant by "cycle 3 suits." Is it:

A1K2 vs Q3J4 -> A3K1 vs. Q2J4, as in 1,2,3 are "cycled" to 3,2,1?

eastbay

[pzhon]

[ QUOTE ]
Now I wonder if I know what you meant by "cycle 3 suits."

[/ QUOTE ]
There are 8 ways to cycle 3 suits. One of them is described by /images/graemlins/spade.gif -> /images/graemlins/heart.gif -> /images/graemlins/diamond.gif -> /images/graemlins/spade.gif. That means the new matchup has a /images/graemlins/heart.gif where the original matchup had a /images/graemlins/spade.gif, a /images/graemlins/diamond.gif where the original had a /images/graemlins/heart.gif, and a /images/graemlins/spade.gif in place of every /images/graemlins/diamond.gif. Every /images/graemlins/club.gif stays the same.

This is how the permutations of {/images/graemlins/spade.gif, /images/graemlins/heart.gif, /images/graemlins/diamond.gif, /images/graemlins/club.gif} act on a deck of cards or the set of pairs of hands. The 24 permutations of the suits have the structure of a group. The action on the set of pairs of hands has the structure of a group action.