What am I missing here?

As David Sklansky demonstrated earlier, determining the odds of flopping two pair is fairly simple. My question is: how do you determine the odds of flopping two pair or better?


Example: 83o. There is no possible straight or flush on the flop, for simplicity's sake. You need two cards to improve to two pair or better, and you really don't care whether the third card makes you a boat or quads.


I can see three possible ways to calculate this, but I am not sure which is correct.


1) 6/50 * 5/49 * 3 ~= 3.67%

2) 6/49 * 5/48 * 3 ~= 3.83%

3) (6/50 * 5/49) + (6/50 * 5/48) + (6/49 * 5/48) = 3.75%


Granted, .16% isn't the end of the world, especially when you're dealing with probabilities this small (25.1/1 vs. 26.25/1), but I'm a perfectionist, I guess.


Thanx in advance,

Big John

Number 1 covers everything except quads which only contributes another .02%. That is, it covers all permutations of 83x where x can be anything including 8 or 3.


For quads, just add 2(1/50)(1/49)(1/48) to cover 333 and 888.

Sorry, they are all wrong. It's like this:


(3/50)(3/49)[6(44/48) + 3(4/48)] = 2.11%


+ .02% for the quads = 2.13%.


The first term in the brackets just gives Sklansky's 2 pair result. The other term is for 838 or 833 which only contributes a small amount. We only multiply by 3 in the second term since there are only 3 permutations. We multiply by 6 in the other case since we start we have permutations of 83x.

Bruce -


I think you missed one vital step...


I asked for two pair _or better_, which includes trips. There are 6 cards that can make a pair with any two non-paired cards. After that, there are 5 more cards (two of the pair, and three of the other card) that make two pair or better (3 for two pair, 2 for trips).


I also thought I had accounted for the possibility of a boat or quads. If the first two cards of a flop improve the hand, it will fill the criteria regardless of the last card, which is why I left that term out of the equation.


Then again, my probability skills aren't what they should be, so I may be missing something really simple.


Big John

When I have a question like this, I look up the answer in Mike Petriv's Hold'em's Odds Book. I may not improve my math skills by looking up the answer- but I know that answer. If it's important, here are the figures.


You are holding two offsuit cards (83o). There are 19,600 possible flops. The chances of you flopping:


two pairs (8's and 3's): 396/19,600

trips: 264/19,600

full house: 18/19,600

quads: 2/19,600


total: 680/19,600 or 3.47% or 27.8:1

I didn't consider trips. I misinterpreted your question as asking for 2-pair not caring if you got a boat or quads. You also have to multiply my quad calculation by 6.


Your method counts some combinations twice. The problem comes in when you multiply by 3. For example, look at 8s 3s 3c and 3s 3c 8s. When you multiply by 3, the second combination generates the first one.


Here is the correct solution breaking it into 4 cases:


hole cards: 83o


2-PAIR (83x or 38x *3, x not 8 or 3):


(6/50)(3/49)(44/48)*3 = 2.02%


TRIPS (88x or 33x *3, x not 8 or 3):


(6/50)(2/49)(44/48)*3 = 1.35%


BOAT (838 or 833 *3):


(3/50)(3/49)(4/48)*3 = .091%


QUADS (888 or 333 *3):


(2/50)(1/49)(1/48)*6 = .01%


TOTAL: 3.47%

There are 50*49*48 = 117,600 possible flops

I would do this by breaking it down into cases card by card.


There are 6 ways to get your first 8 or 3 and 5 ways to get an 8 or 3 once you already have one.


getting an 8/3 on first two cards: 6/50 * 5/49

getting an 8/3 on first and last: 6/50 * 44/49 * 5/48

getting an 8/3 on last two cards: 44/50 * 6/49 * 5/48


Summing these three terms yeilds 3.47%

There are 50*49*48 = 117,600 possible flops


You are making the mistake of sextuple(?)-counting (6x) many flops. Using your approach, all of these flops are different.


Ac,5d,2s

Ac,2s,5d

5d,Ac,2s

5d,2s,Ac

2s,Ac,5d

2s,5d,Ac


Of course, all "six" of these flops are the same.


The proper formula is (50*49*48)/3*2 = 19,600

You're right