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davet
09-10-2005, 06:39 PM
The situation, typical no- fold'em hold'em. Many family pots, many more 6+ to the flop.

flop: 66T

This is completely arbrirary, the flop could be anything, so long as it is paired.

What is the chance that someone is sitting on trips?

And also, where do I learn to do all these fun little math equations?

AaronBrown
09-17-2005, 07:12 PM
Let me answer a simpler question first. Suppose you take 66T out of a deck, then shuffle the remaining 49 cards and deal two cards. What is the chance those two cards will form three of a kind with the 66T board?

The answer is there are Combin(49,2) = 1,176 possible combinations of two cards you can deal from the deck. 1 of them is 66, which gives four of a kind. 3 of them are TT and 6 of them are 6T, both of which give a full houses. 88 of them are 6 paired with a card other than 6 or T, which gives three sixes. Depending on which of these hands you want to count as a set, you can add them up and divide by 1,176 to get the probability.

In a real Hold'em game, it's not so simple. The basic probability is the same, but there is more than one person who can be holding the set. On the other hand, most hands with 6 in them will fold preflop. In a real game you are much more likely to see a set with a AAT flop than a 66T flop.

Cobra
09-18-2005, 10:09 AM
The probability that someone at your table was dealt one or more sixes after the flop is relatively easy to determine. The usefulness of this information is suspect because you don't know wether that person stayed to see the flop. That put aside the probability that one or more people at your table was dealt at least one six assuming you were not is:

9 opponents 1-(combin(45,18)/combin(47,18)) = 62.4%

4 opponents 1-(combin(45,8)/combin(47,8) = 31.5%

Now in trying to answer the question: is that player still there after the flop lets come up with some reasonable hand selections they may have played and determine the probability someone was dealt that hand. This is called inclusion - exclusion. Lets say your opponenet would stay with 66, A6 suited, 76, 65. And lets assume you do not have a A, 7, 6, 5.

Total combo's 19.

Hand possibility 1081

Opponents 9

Probability on one or more hands:

= 9*18/1081 - (9c2)*146/(47c2)/(45c2) = 15.8%.

Obviously the probability that someone would have three of a kind depends on how many playable combinations of hand that card makes.
A 6 makes few combo's
A 10 makes more
A ace makes a lot.

Cobra

09-19-2005, 03:43 PM
[ QUOTE ]
Let me answer a simpler question first. Suppose you take 66T out of a deck, then shuffle the remaining 49 cards and deal two cards. What is the chance those two cards will form three of a kind with the 66T board?

The answer is there are Combin(49,2) = 1,176 possible combinations of two cards you can deal from the deck. 1 of them is 66, which gives four of a kind. 3 of them are TT and 6 of them are 6T, both of which give a full houses. 88 of them are 6 paired with a card other than 6 or T, which gives three sixes. Depending on which of these hands you want to count as a set, you can add them up and divide by 1,176 to get the probability.

[/ QUOTE ]

You totally lost me but I am trying my best to understand. Why take the 66T out of the deck? How is this making the problem simpler?

"The answer is there are Combin(49,2) = 1,176 possible"

How in the world did you come up with 1,176? What did you do? 2 x 49 = 98 so that's not it. 49 squared is 2,401 so that's not it. What does "combin (49,2) mean? Does it mean multiply the two numbers? Squard the first number and divide it by the second number? How are you calculating numbers? Please show the math? Please, please, please.

Thank you

09-19-2005, 03:53 PM
here's a simpler answer. could be totally bogus. I found it on a learn-texas-holdem.com. if a paired board - if 5 guys in the hand there is a 43% chance one has trips, if 4 guys in a hand 34% one has trips, 3 - 26%, and 2 17%.

seems close to the last guys calculations.

LetYouDown
09-19-2005, 03:55 PM
Combin(49,2) = 49!/(47! * 2!) = 49 * 48/2 = 1176

...as a formula:

C(N,R) = N!/[(N-R!) * R!]

09-19-2005, 04:13 PM
[ QUOTE ]
Combin(49,2) = 49!/(47! * 2!) = 49 * 48/2 = 1176

...as a formula:

C(N,R) = N!/[(N-R!) * R!]

[/ QUOTE ]

Make it simpler please. Spell it out. What does the "C" stand for? What does "N" stand for? What does "R" stand for?
Is "Combin" short for "possible combinations" or does it mean to "combine two numbers or probabilities" You need to define your variables.

LetYouDown
09-19-2005, 04:37 PM
[ QUOTE ]
Make it simpler please. Spell it out. What does the "C" stand for? What does "N" stand for? What does "R" stand for?
Is "Combin" short for "possible combinations" or does it mean to "combine two numbers or probabilities" You need to define your variables.

[/ QUOTE ]
You're awfully demanding! LOL...I don't *need* to define anything, C(N,R) is standard notation. C stands for Combinations. C(N,R) is the Number of ways to choose R items from N items. So C(49,2) is the number of combinations you can have if you select two items from a group of 49. Order doesn't matter...so item #12 then item #13 is the same as item #13 then item #12.

09-19-2005, 10:58 PM
[ QUOTE ]
[ QUOTE ]
Make it simpler please. Spell it out. What does the "C" stand for? What does "N" stand for? What does "R" stand for?
Is "Combin" short for "possible combinations" or does it mean to "combine two numbers or probabilities" You need to define your variables.

[/ QUOTE ]
You're awfully demanding! LOL...I don't *need* to define anything, C(N,R) is standard notation. C stands for Combinations. C(N,R) is the Number of ways to choose R items from N items. So C(49,2) is the number of combinations you can have if you select two items from a group of 49. Order doesn't matter...so item #12 then item #13 is the same as item #13 then item #12.

[/ QUOTE ]

Thanks very much. I just want to learn badly. Sorry if I sound demanding. could you please show the math now.

Thanks again.