The Triton poker series has popularized short deck hold-em. I am sure that everyone knows by now that Short Deck is played with a 36-card deck since all the 2's, 3's, 4's, and 5's have been removed (of course, this is why Short Deck is also called Six-Plus poker).
It is a wildly fast-moving game with frequent large pots and tons of action. Pre-flop equities run much tighter than in Long Deck (regular) hold-em. Similar to PLO, the in-hand swings are wild it being common for each card (flop, turn, river) to greatly change who is ahead in the hand.
In a previous thread, I looked at how often does a flush occur in Short Deck vs Long Deck. Here I want to do the same thing for quads. Since there are fewer ranks in each suit, quads are more common in Short Deck than in Long Deck.
I want to look into the differential frequency of quads in Short Deck vs Long Deck. In particular, I will look at how often quads are made by at least one player at a six-max table if each hand goes to showdown on each deal. Also, to keep the comparison "apples-to-apples", I will ignore the possibility of straight flushes in both variants (i.e., both quads and a straight flush being made on the same deal).
Long Deck (regular 52-card hold-em):
I will take the total number of deals of 6-Max Long Deck to be:
= C(52,5)*C(47,12)*(1*3*5*7*9*11)
= 1,411,633,731,355,657,009,200
Now let's go through all the ways that one or more players can make quads.
Case 1: Quads on board
= C(13,1)*C(4,4)*C(48,1)*C(47,12)*(1*3*5*7*9*11)
= 338,927,666,591,994,480
Case 2: Trips plus two singletons on board
= C(13,3)*C(3,1)*C(4,3)*C(4,1)*C(4,1)*[C(47,12)-C(46,12]*(1*3*5*7*9*11)
= 7,615,055,657,896,727,040
Case 3: Full house on board
Case 3A: Make quads with board trips, do not make quads with board pair
Case 3A1: 0 of board pair in hole cards
= C(13,2)*C(2,1)*C(4,3)*C(4,2)*C(1,1)*C(44,11)*(1*3* 5*7*9*11)
= 298,482,089,357,612,160
Case 3A2: 1 of board pair in hole cards
= C(13,2)*C(2,1)*C(4,3)*C(4,2)*C(1,1)*C(2,1)*C(44,10 )*(1*3*5*7*9*11)
= 193,135,469,584,337,280
Case 3A3: 2 of board pair in hole cards
= C(13,2)*C(2,1)*C(4,3)*C(4,2)*C(1,1)*C(2,2)*C(44,9) *(1*3*5*7*9*10)
= 25,082,528,517,446,400
Case 3B: Make quads with board trips and make quads with board pair
= C(13,2)*C(2,1)*C(4,3)*C(4,2)*C(1,1)*C(2,2)*C(44,9) *(1*3*5*7*9)
= 2,508,252,851,744,640
Case 3C: Make quads with board pair, do not make quads with board trips
= C(13,2)*C(2,1)*C(4,3)*C(4,2)*C(2,2)*C(44,10)*(1*3* 5*7*9)
= 8,778,884,981,106,240
Case 4: Two pair and one singleton on board
Case 4A: Make quads with exactly one board pair
Case 4A1: 0 of other board pair in hole cards
= C(13,3)*C(3,2)*C(4,2)*C(4,2)*C(4,1)*C(2,1)*C(2,2)* C(43,10)*(1*3*5*7*9)
= 447,723,134,036,418,240
Case 4A2: 1 of other board pair in hole cards
= C(13,3)*C(3,2)*C(4,2)*C(4,2)*C(4,1)*C(2,1)*C(2,2)* C(2,1)*C(43,9)*(1*3*5*7*9)
= 263,366,549,433,187,200
Case 4A3: 2 of other board pair in hole cards
= C(13,3)*C(3,2)*C(4,2)*C(4,2)*C(4,1)*C(2,1)*C(2,2)* C(2,2)*C(43,8)*(1*3*5*7*8)
= 30,099,034,220,935,680
Case 4B: Make quads with both board pairs
= C(13,3)*C(3,2)*C(4,2)*C(4,2)*C(4,1)*C(2,2)*C(2,2)* C(43,8)*(1*3*5*7)
= 1,881,189,638,808,480
Case 5: One pair and three singletons on board
= C(13,4)*C(4,1)*C(4,2)*C(4,1)*C(4,1)*C(4,1)*C(2,2)* C(45,10)*(1*3*5*7*9)
= 3,310,893,764,302,924,800
GRAND TOTAL = 12,535,934,221,413,242,640
PERCENT = 0.888%
Short Deck (36-card Six Plus hold-em):
I will take the total number of deals of 6-Max Short Deck to be:
= C(36,5)*C(31,12)*(1*3*5*7*9*11)
= 553,027,606,647,516,000
Case 1: Quads on board
= C(9,1)*C(4,4)*C(32,1)*C(31,12)*(1*3*5*7*9*11)
= 422,480,982,924,000
Case 2: Trips plus two singletons on board
= C(9,3)*C(3,1)*C(4,3)*C(4,1)*C(4,1)*[C(31,12)-C(30,12]*(1*3*5*7*9*11)
= 9,158,297,436,288,000
Case 3: Full house on board
Case 3A: Make quads with board trips, do not make quads with board pair
Case 3A1: 0 of board pair in hole cards
= C(9,2)*C(2,1)*C(4,3)*C(4,2)*C(1,1)*C(28,11)*(1*3*5 *7*9*11)
= 385,731,246,700,800
Case 3A2: 1 of board pair in hole cards
= C(9,2)*C(2,1)*C(4,3)*C(4,2)*C(1,1)*C(2,1)*C(28,10) *(1*3*5*7*9*11)
= 471,449,301,523,200
Case 3A3: 2 of board pair in hole cards
= C(9,2)*C(2,1)*C(4,3)*C(4,2)*C(1,1)*C(2,2)*C(28,9)* (1*3*5*7*9*10)
= 112,786,914,240,000
Case 3B: Make quads with board trips and make quads with board pair
= C(9,2)*C(2,1)*C(4,3)*C(4,2)*C(1,1)*C(2,2)*C(28,9)* (1*3*5*7*9)
= 11,278,691,424,000
Case 3C: Make quads with board pair, do not make quads with board trips
= C(9,2)*C(2,1)*C(4,3)*C(4,2)*C(2,2)*C(28,10)*(1*3*5 *7*9)
= 21,429,513,705,600
Case 4: Two pair and one singleton on board
Case 4A: Make quads with exactly one board pair
Case 4A1: 0 of other board pair in hole cards
= C(9,3)*C(3,2)*C(4,2)*C(4,2)*C(4,1)*C(2,1)*C(2,2)*C (27,10)*(1*3*5*7*9)
= 578,596,870,051,200
Case 4A2: 1 of other board pair in hole cards
= C(9,3)*C(3,2)*C(4,2)*C(4,2)*C(4,1)*C(2,1)*C(2,2)*C (2,1)*C(27,9)*(1*3*5*7*9)
= 642,885,411,168,000
Case 4A3: 2 of other board pair in hole cards
= C(9,3)*C(3,2)*C(4,2)*C(4,2)*C(4,1)*C(2,1)*C(2,2)*C (2,2)*C(27,8)*(1*3*5*7*8)
= 135,344,297,088,000
Case 4B: Make quads with both board pairs
= C(9,3)*C(3,2)*C(4,2)*C(4,2)*C(4,1)*C(2,2)*C(2,2)*C (27,8)*(1*3*5*7)
= 8,459,018,568,000
Case 5: One pair and three singletons on board
= C(9,4)*C(4,1)*C(4,2)*C(4,1)*C(4,1)*C(4,1)*C(2,2)*C (29,10)*(1*3*5*7*9)
= 3,663,318,974,515,200
GRAND TOTAL = 15,612,058,658,196,000
PERCENT = 2.823%
Summary
We see that quads are more common in 6-max Short Deck than in 6-max Long Deck (2.823% vs 0.888%) when all hands go to showdown on all deals.
I hope this is somewhat illuminating and prompts other analyses of Short Deck, which may be the new poker frontier?
Comments welcome.
Last edited by whosnext; 05-26-2019 at 02:00 PM.
