Happy Extinction
10-27-2003, 10:01 PM
I tried to come up with a couple of quick rules to help me figure when to stay in in NL hold'em. I pretty much made this up.
This is for after the flop.
First of all, I tried to figure out the chances of getting one of my outs after the flop. I came up with
6 outs = around 25%
8 outs = around 33%
14 outs = around 50%
So for example, if I have A K and the flop comes out 2 6 9 (let's say suits aren't a factor in this hand) I might say i have a decent chance of winning if I pair up at least one of my hole cards. That means I have six outs and around a 25% of winning the hand.
I calculated this as (1-(41/47)*(40/46) = 24.1%
Is this close to being correct?
Okay, if this is correct. Then how do I bet. Lets say there is 200 in the pot right now and somebody leads with a bet of 50.
My first instinct is to say a bet of 50 is close to an even bet. but that can't be right since if I don't get one of my outs on the 4th card, i will have to bet more money.
So I might assume that i will have to bet twice which means, and I would be guessing, that my total outlay to see two cards would be 100. But I would also assume that the pot would be bigger. Say three people are in, besides me, i might assume and ending pot of 200 + 200 + 100 (assuming 2 people fold at the end) so my math after the flop becomes:
(500 ending pot) * (25% chance of winning) / 2 bets = around 60
60 is my indifference bet, so I will bet the bet of 50.
I notice that the more people staying in, my indifference bet goes up. I realize that the more people in the more chance for good hands that beat me.
The whole decision tree changes after the 4th card.
This is for after the flop.
First of all, I tried to figure out the chances of getting one of my outs after the flop. I came up with
6 outs = around 25%
8 outs = around 33%
14 outs = around 50%
So for example, if I have A K and the flop comes out 2 6 9 (let's say suits aren't a factor in this hand) I might say i have a decent chance of winning if I pair up at least one of my hole cards. That means I have six outs and around a 25% of winning the hand.
I calculated this as (1-(41/47)*(40/46) = 24.1%
Is this close to being correct?
Okay, if this is correct. Then how do I bet. Lets say there is 200 in the pot right now and somebody leads with a bet of 50.
My first instinct is to say a bet of 50 is close to an even bet. but that can't be right since if I don't get one of my outs on the 4th card, i will have to bet more money.
So I might assume that i will have to bet twice which means, and I would be guessing, that my total outlay to see two cards would be 100. But I would also assume that the pot would be bigger. Say three people are in, besides me, i might assume and ending pot of 200 + 200 + 100 (assuming 2 people fold at the end) so my math after the flop becomes:
(500 ending pot) * (25% chance of winning) / 2 bets = around 60
60 is my indifference bet, so I will bet the bet of 50.
I notice that the more people staying in, my indifference bet goes up. I realize that the more people in the more chance for good hands that beat me.
The whole decision tree changes after the 4th card.