unique showdown probabilities for two preflop hand matchups

[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.

[img]/images/graemlins/spade.gif[/img] 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.)

[img]/images/graemlins/spade.gif[/img] 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 [img]/images/graemlins/club.gif[/img] [img]/images/graemlins/diamond.gif[/img] 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 [img]/images/graemlins/club.gif[/img] Y [img]/images/graemlins/diamond.gif[/img] <-> X [img]/images/graemlins/diamond.gif[/img] Y [img]/images/graemlins/club.gif[/img], or it switches two suited hands, X [img]/images/graemlins/diamond.gif[/img] Y [img]/images/graemlins/diamond.gif[/img] <-> X [img]/images/graemlins/club.gif[/img] Y [img]/images/graemlins/club.gif[/img], or it fixes the hands, X [img]/images/graemlins/club.gif[/img] X [img]/images/graemlins/diamond.gif[/img], Y [img]/images/graemlins/club.gif[/img] Y [img]/images/graemlins/diamond.gif[/img]. 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.

[img]/images/graemlins/spade.gif[/img] Three permutations switch two pairs of suits, e.g., [img]/images/graemlins/spade.gif[/img] <-> [img]/images/graemlins/heart.gif[/img], [img]/images/graemlins/diamond.gif[/img] <-> [img]/images/graemlins/club.gif[/img]. 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 [img]/images/graemlins/diamond.gif[/img] <-> [img]/images/graemlins/club.gif[/img]. 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 [img]/images/graemlins/diamond.gif[/img] <-> [img]/images/graemlins/club.gif[/img]. 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 [img]/images/graemlins/spade.gif[/img] X [img]/images/graemlins/heart.gif[/img], Y [img]/images/graemlins/diamond.gif[/img] Y [img]/images/graemlins/club.gif[/img], and there are 13^2 such possibilities. If the switch exchanges the hands, they either look like X [img]/images/graemlins/spade.gif[/img] Y [img]/images/graemlins/diamond.gif[/img], X [img]/images/graemlins/heart.gif[/img] Y [img]/images/graemlins/club.gif[/img], or X [img]/images/graemlins/spade.gif[/img] Y [img]/images/graemlins/club.gif[/img], X [img]/images/graemlins/heart.gif[/img] Y [img]/images/graemlins/diamond.gif[/img]. There are 2*13^2 more possibilities of this type, for a total of 3*13^2 in this subcase.

In total, [img]/images/graemlins/spade.gif[/img]<-> [img]/images/graemlins/heart.gif[/img], [img]/images/graemlins/diamond.gif[/img]<-> [img]/images/graemlins/club.gif[/img] 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.

[img]/images/graemlins/spade.gif[/img] Eight permutations rotate 3 suits, e.g., [img]/images/graemlins/spade.gif[/img] -> [img]/images/graemlins/heart.gif[/img] -> [img]/images/graemlins/diamond.gif[/img] -> [img]/images/graemlins/spade.gif[/img]. 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.

[img]/images/graemlins/spade.gif[/img] Six permutations rotate 4 suits, e.g., [img]/images/graemlins/spade.gif[/img] -> [img]/images/graemlins/heart.gif[/img] -> [img]/images/graemlins/club.gif[/img] -> [img]/images/graemlins/diamond.gif[/img] -> [img]/images/graemlins/spade.gif[/img]. 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 [img]/images/graemlins/spade.gif[/img] X [img]/images/graemlins/club.gif[/img], X [img]/images/graemlins/heart.gif[/img] X [img]/images/graemlins/diamond.gif[/img]. 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 [img]/images/graemlins/spade.gif[/img] K [img]/images/graemlins/heart.gif[/img]. 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.