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Old 03-02-2019, 11:22 PM   #1
MadADX
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Stud odds of boat+ with no pair on board

Posted this to the stud forum too but doesnít seem to be too much on action there so posting here too if thatís alright..

What are the statistical odds of a player on the river to have a full house or better with no pair on board showing? (Meaning rolled up trips plus boating/quadding up on 7th street) TIA!
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Old 03-03-2019, 02:07 AM   #2
whosnext
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Re: Stud odds of boat+ with no pair on board

I am including making any full house or quads. The reason I mention this is that you can have 2-pair before 7th street and then make a full house on the river. If you don't want to include that possibility, let us know.

I also am treating this as a "pure" probability question, purely random cards and no other information such as other players' holdings or upcards, any player actions, etc.

Quads (you can make quads with any of your four upcards, and doing so, of course, requires the remaining three of that rank to be your three downcards):
= C(4,1)*C(3,3)
= 4

Full House (you can make a full house with any two of the ranks of your four upcards, one of which you'll have 3 of and the other of which you'll have 2 of -- so the "3" requires 2 of the 3 remaining of that rank and the "2" requires 1 of the remaining of that rank):
= C(4,2)*C(2,1)*C(3,2)*C(3,1)
= 108

TOTAL = 4 + 108 = 112

There are C(48,3) total 3-card holdings that you can hold, which is 17,296. So the probabiility:

= 112 / 17,296
= 0.6475% or around once every 154 times.

I hope this is correct and helpful.
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Old 03-06-2019, 02:24 AM   #3
MadADX
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Re: Stud odds of boat+ with no pair on board

Thank you very much! That is awesome. So this math is for when none of the 4 upcards are paired, correct?
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Old 03-06-2019, 02:46 AM   #4
whosnext
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Re: Stud odds of boat+ with no pair on board

Correct.
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Old 03-06-2019, 04:16 PM   #5
MadADX
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Re: Stud odds of boat+ with no pair on board

And do we know how the odds change if two of the four upcards are paired? (Meaning one single pair, ex: K994)
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Old 03-06-2019, 06:13 PM   #6
whosnext
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Re: Stud odds of boat+ with no pair on board

This can also be done using logic and combinatorics (again, doing this as a "pure" probability question). Using your example of K994 upcards.

Quad 9's (two Nines down):
= C(2,2)*C(46,1)
= 46

Quad 4's (three Fours down):
= C(3,3)
= 1

Quad K's (three Kings down):
= C(3,3)
= 1

Full House with exactly one Nine and one King and no Fours down:
= C(2,1)*C(3,1)*C(40,1)
= 240

Full House with exactly one Nine and one Four and no Kings down:
= C(2,1)*C(3,1)*C(40,1)
= 240

Full House with exactly one Nine and two Kings down:
= C(2,1)*C(3,2)
= 6

Full House with exactly one Nine and two Fours down:
= C(2,1)*C(3,2)
= 6

Full House with exactly one Nine and one King and one Four down:
= C(2,1)*C(3,1)*C(3,1)
= 18

Full House with exactly one Nine and a pair of some other rank (not Kings or Fours) down:
= C(2,1)*C(10,1)*C(4,2)
= 120

Full House with No Nines, 2 Kings down:
= C(3,2)*C(43,1)
= 129

Full House with No Nines, 2 Fours down:
= C(3,2)*C(43,1)
= 129

Full House with No Nines, trips of some other rank (not Kings or Fours) down:
= C(10,1)*C(4,3)
= 40

TOTAL = 976.

So the probability:

= 976 / 17,296

= 5.642923219% or around once every 18 times.

Last edited by whosnext; 03-06-2019 at 06:19 PM.
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