Peter-23
02-16-2005, 04:59 PM
Hi,
Im not a wiz at this so please dont call me /images/graemlins/confused.gif if you think Im out of my mind.
Lets say we play Texas holdem and I at the river had a hand that could only be beat by 10 different combinations of the remaining 991 possible ones.
Situation 1.
At a ten handed table where everyone play anything til the end then the probability I would lose is.
(10 * 9)/991 = 9% giving me about 0.1 to 1 against.
Situation 2
Now lets say Im still at a ten handed table but now people fold before the flop. But somehow I know that if anyone of my opponents have one of the ten hands that can beat me they will always play them to the river.
This time 5 opponents fold preflop and the rest play to the river giving me 4 opponents at the river.
First, are my calculation in the first situation correct?
Second, if it is, what are my winning odds in the second situation?
I think it still is 0.1 to 1, (This is an ideal situation) am I correct, mathematically I mean?
Peter
Im not a wiz at this so please dont call me /images/graemlins/confused.gif if you think Im out of my mind.
Lets say we play Texas holdem and I at the river had a hand that could only be beat by 10 different combinations of the remaining 991 possible ones.
Situation 1.
At a ten handed table where everyone play anything til the end then the probability I would lose is.
(10 * 9)/991 = 9% giving me about 0.1 to 1 against.
Situation 2
Now lets say Im still at a ten handed table but now people fold before the flop. But somehow I know that if anyone of my opponents have one of the ten hands that can beat me they will always play them to the river.
This time 5 opponents fold preflop and the rest play to the river giving me 4 opponents at the river.
First, are my calculation in the first situation correct?
Second, if it is, what are my winning odds in the second situation?
I think it still is 0.1 to 1, (This is an ideal situation) am I correct, mathematically I mean?
Peter