Here's another thread about poker math and probabilities that I am working through with my daughter.

Everybody knows that Razz is a lowball variant of 7-card stud. Best hand in Razz is 5432A. Straights and flushes do not count against a player (and pairs do not "counterfeit" a low card). A person's hand is the lowest five distinct ranks (ace is low) ordered from high-to-low. If they don't have five distinct ranks, this is even worse than having five to a King though herein we will not delve into this case.

Our challenge was to tally how many different Razz hands are in each "category" where category means the highest of the lowest five ranks. So categories are: K's, Q's, J's, T's, 9's, 8's, 7's, 6's, and 5's (clearly you can never make a 4, 3, 2, or A). As stated above, there is also an additional category of "do not have five distinct ranks".

We think we have cracked this using logic and combinatorics. For a hand to be in category X, the hand must have a card of rank X and exactly four distinct lower ranks. There are four different possible counts for these five ranks: {[1,1,1,1,1], [2,1,1,1,1], [2,2,1,1,1], [3,1,1,1,1]}. Call this set Q which will be used below.

Let the detailed elements of Q be designated as Eiq where i goes from 1 to 5 and q goes from 1 to 4. Let Uq be the number of 1's in the qth element of Q. Let Sq be the sum of the elements in the qth element of Q.

Of course, to make the math work we need to designate a King as rank 13, a Queen as rank 12, and a Jack as rank 11.

For combinations, we use the notation C(A,B) to be the number of ways you can choose B items out of A objects.

Then, for X = 5, 6, 7, 8, 9, 10, 11, 12, or 13, we have:

Tally(X) = Sum [over Q from 1 to 4] of [C(1,1) * C(X-1,4)] * C(5,Uq) * C(52-4*X,7-Sq) * {Prod [over I from 1 to 5] of C(4,Eiq)}.

Letting Z denote the "Do not have 5 distinct ranks" category:

Tally(Z) = C(52,7) - Sum [over X from 5 to 13] of Tally(X).

I know the formula for Tally(X) looks like a mess, but in reality it is pretty straightforward. We need to sum over the four distinct count "cases" of the five distinct lowest ranks. C(1,1) means that the hand must have rank X. C(X-1,4) means that the hand must have exactly 4 ranks less than X.

Of these five ranks, there are C(5,Uq) ways to select the ranks that have exactly 1 card (the other counts are forced in each Q).

There are C(52-4*X,7-Sq) ways to select the cards of higher ranks in the hand.

Finally, there are C(4,Eiq) ways to select the cards in the rank represented by Eiq (case Q and rank I).

Multiplying all these terms together, and summing over the Count Cases, yields the tally of all the 7-card hands that will fall into Category X.

Using this formula, we derived the following tallies:

Category	Tally
Ace	0
Deuce	0
Three	0
Four	0
Five	781,824
Six	3,151,360
Seven	7,426,560
Eight	13,171,200
Nine	19,174,400
Ten	23,675,904
Jack	24,837,120
Queen	21,457,920
King	13,939,200
Fewer than 5 Ranks	6,169,072
GRAND TOTAL	133,784,560

If these tallies are correct, we can see that the most common category (mode) is a Jack, whereas the median category is a Ten.

Again, all comments/refutations/confirmations are welcome.

P.S. Feel free to comment on the application of the Principle of Inclusion-Exclusion (PIE) to any of the above.

We might gain some simplicity by using PIE to derive a tally from the previous tally, which would eliminate redundant calculations. For instance, N(6) = 5*N(5) - (something).

I don't know when I'll have time to attempt that.