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caretaker1
12-29-2004, 07:28 PM
Is there a website, or for the more mathematically inclined, formulas, which will show the percentage chance that at least one other player will have a third card on a paired flop (for example: flop TT7, what's the % that someone has at least a third T). Also, formulas including additional players would be helpful (in other words, % if there are 2 opponents, 3 opponents, etc.). Finally, I realize this is unrealistic, but assume that each third card is equally likely (2's as likely as A's; as if opponents were playing random hands).

gaming_mouse
12-29-2004, 07:51 PM
[ QUOTE ]
Finally, I realize this is unrealistic, but assume that each third card is equally likely (2's as likely as A's; as if opponents were playing random hands).

[/ QUOTE ]

Given this assumption, which you correctly note is unrealistic, the calculation is fairly easy. Say you are against three players. There are two T's left in the deck, out of a total of 47 unseen cards (we substract the 3 cards on the board, as well as the 2 cards in your hand). We'll now calculate that chance that NONE of your opponents has a T.

The chance that your first oppo does not hold a T is:

(45 choose 2)/(47 choose 2)

Given that your first oppo does not hold a T, the chance that your second oppo does not hold T is:

(43 choose 2)/(45 choose 2)

And the third:

(41 choose 2)/ (43 choose 2)

Cancelling terms, we get (for the chance that no oppo holds a T):

(41 choose 2)/(47 choose 2)= .7585

So there is about a 24% chance that at least 1 of 3 opponents holds a T. The calculation is similar for more players.

gm

caretaker1
12-29-2004, 10:44 PM
Thank you for the formula.

Cobra
01-01-2005, 12:05 PM
I am new to the probability area so hopefully someone will check my numbers to insure I am not misleading anyone. I have come up with some formula's that will give you some information on what your opponents may hold given a certain flop. As said this may not be useful because its saying your opponents after the flop have random hands. That said here goes.

N = Number of oppenents

No one has a ten = C(45,2n)/C(47,2n)

Exactly one person has a ten = 2*C(45,2n-1)/C(47,2n)

One person has two tens = N/C(47,2)

Two players have a ten = (C(45,2n-2)-N)/C(47,2)

One or more players have a ten = 1-C(45,2n)/C(47,2n)

These numbers could be used if the Flop has AAx , you don't have an A, and are wondering if one of your opponents has one. Or if the Flop has an Axx, you have an Ax, and are wondering wether one of your opponents was dealt the other Ace's.

Cobra