ITH Poker Forum

[seanof30306]

So, this is being discussed at a table I'm dealing at last night. The consensus of the math wizards at the table is the solution is 47 unknown cards on the flop * 46 unknown cards on the turn, or 2162. Another chimes in that no, it would be 2161:1. Another chimes in and says no, you have to divide 2162 by 2, then subtract 1 to get the odds

(2162 / 2) - 1:1
1081 - 1:1
1080:1

I said, in this instance, the shotgun method would be accurate, since there is only one combination that delivers quads, but the correct way to solve problems over two cards is to use combinations. I did this proof:


Hold a pocket pair (or flop a pair); determine the odds of making runner-runner quads (turn and river)


1: Determine known/unknown cards:

5 known cards (2 you hold + 3 on flop)

47 remaining unknown cards (52 cards in the deck - 5 known cards)


2: Determine the number of possible two card combinations from the remaining 47 unknown cards:

47 * 46 / 2 * 1
2162 / 2
1081 = number of possible two card combinations from the remaining 47 unknown cards


3: Determine the number of two card combinations that will make quads:

1 = No math needed, there is only one combination that makes quads.


4: Divide the number of combinations that will make quads by the total number of combinations:

1 combination makes quads

1081 possibe two card combinations

1/1081
0.00093 = % chance of making runner-runner quads from the flop to the river


5. Convert percentage to odds:

(1081 - 1):1
1080:1 = odds of making runner-runner quads from the flop to the river


Second, since making quads require both of the remaing outs, and since there is a 0% chance of making quads if any of the active opponents in the hand hold one, or both of those outs, would you not have to add cards held by active opponents to the known cards, and subtract them from unknown cards? I understand you normally would't do that, but in this case, you can only make quads if they don't hold one or both of your outs.

[the_hawk]

You have 2 outs on the turn and 1 on the river if you hit the turn, so your numbers look right (2/47 * 1/46).

Unknown cards are unknown cards, whether they are in the deck or in opponents' hands.

[Bugsbunny]

Easy way (I think), assuming you have a calculator or google:
1/(2/47*1/46)=1081
Odds are 1080 to 1

And as the_hawk states, unknown cards are unknown cards. If you start to calculate the chance that an active opponent has a given card you'll eventually end up with the same odds (there's examples somewhere using flush draws). If you start taking the actual cards in opponents hands into account then you've changed the parameters of the question, since those cards are no longer unknown, but then the answer is different on a case by case basis. If it's a 10 handed game and you know that none of the other players have one of the cards then you no longer have 47 and 46 unknown but 29 and 28 unknown, making the odds: 1/(2/29*1/28)=406, or 405 to 1. So you could say that in a 10 handed game the exact odds are either 0 or 405 to 1, with the average odds being 1080 to 1