Posted Mon Aug 14, 2006 12:17 pm GMT by Cmoneyt8ker
Heres a senario in our home game last night and I want to know if any of yall know the math on this...
I have J10 my Opp. has K10
Flop is 3 8 9
He bets I call
Turn is Q 
He bets and I go all in...we are both chip leaders and have good stacks.
he calls and of course hits his one out.....J on river
What percentage was I to win that hand on the turn...I thought it was like over 90%
P.S. does anyone know of a poker site where you can play private SNG??
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Posted Mon Aug 14, 2006 12:24 pm GMT by JackKingOff
ur a bit over 90% to win since hes down to 3 or less outs... but anyhow yes there are private sngs on PS.. but u must have earned 2000 FPPs then you email support and then you can set up private tournies w/ PW's
Posted Mon Aug 14, 2006 12:44 pm GMT by shorn7
44 cards left, 3 help him, so 41/3, 1/x, 92.7% winner for you.
Posted Mon Aug 14, 2006 12:49 pm GMT by JackKingOff
shorn how do u get that can u explain it
Posted Mon Aug 14, 2006 12:52 pm GMT by shorn7
Sure. On the turn when all the $$ goes in and you turn over your cards, you have now seen 8 exposed cards (your two, his two, and the 4 on the baord). So, that leaves 44 cards left in the deck. 3 of those cards help him and 41 do not. So, he is a 41-3 dog to make his hand. Divide 41/3 and you get 13.66666. Convert that to a percentage by hitting the 1/x button on your calculator and you get 7.3% or his chance of winning. You could actually just divide 3/41 too and get the same result.
Hope that helps.
Posted Mon Aug 14, 2006 1:07 pm GMT by groton
yes but there is ten dead cards
since the Burn for the flop and Turn should be added in right?
Posted Mon Aug 14, 2006 1:25 pm GMT by MJJ
| groton wrote: | yes but there is ten dead cards
since the Burn for the flop and Turn should be added in right? |
Nope- those were not exposed, they don't count (just like peoples folded hands)
Now if another player showed a (folded) J that would enter the equation
Posted Mon Aug 14, 2006 1:48 pm GMT by mooseontheloose
Yeh it isn't an exact science. Technically, every single on of your outs could've already been folded/burned on any given hand. However, since we have no idea which cards were folded/burned, we have to do outs counting all cards in the deck with the assumption that all our cards are live. It's also a reason why more outs is better. Not only because you have a better chance of hitting an out than your opponent, but also a better chance that you have outs left in the deck.
Posted Mon Aug 14, 2006 2:45 pm GMT by efram
3 outs,
1 card left (river)
use the 2x rule
3 outs mulitped by 2, = 6% roughly
he's 6% your 94% to win
when compared to the math 41/3 that method is close enough to determine the rough percentages.
for the turn & river combined, its 4 x # of outs. The more outs there are the more skewed the percentage is but its always in the ball park.
Posted Mon Aug 14, 2006 3:05 pm GMT by Gogie
| mooseontheloose wrote: | | Yeh it isn't an exact science. Technically, every single on of your outs could've already been folded/burned on any given hand. However, since we have no idea which cards were folded/burned, we have to do outs counting all cards in the deck with the assumption that all our cards are live. It's also a reason why more outs is better. Not only because you have a better chance of hitting an out than your opponent, but also a better chance that you have outs left in the deck. |
Ah, but it is an exact science. You are calculating the exact odds that one of the cards he needs will hit the river. The only "knowns" are the cards that have been exposed, so you have to assume there are still 44 cards left to come (yes, there really are only 44 cards minus the burned cards minus the other players mucked hole cards, but those cards don't "count" since you don't know what they are).
| efram wrote: | 3 outs,
1 card left (river)
use the 2x rule
3 outs mulitped by 2, = 6% roughly
he's 6% your 94% to win
when compared to the math 41/3 that method is close enough to determine the rough percentages.
for the turn & river combined, its 4 x # of outs. The more outs there are the more skewed the percentage is but its always in the ball park. |
Sorry efram, but the correct answer is that he has a 6.81% chance of hitting his J, which is calculated as 3 (jacks remaining) divided by 44 (total cards remaining). In other words, out of the 44 unseen cards, 3, or 6.81%, are jacks and 41, or 93.19%, are not jacks. You should have been using 44/3, not 41/3 in your calculation. Also, instead of using 44/3 and then taking the inverse, why not just use 3/44? One less step to get the same answer.
Posted Mon Aug 14, 2006 3:25 pm GMT by Jauron
See post below.
Posted Mon Aug 14, 2006 4:01 pm GMT by Adam Marshall
Aw, Jauron! Screw that thing, this one is better...
http://www.texasholdem-poker.com/odds-calculator/texas-holdem.php
Here's the pre-generated result in that caculator, right here.
Moose, the whole folded/burned cards thing isn't how it's done. There's an article at...
http://www.texasholdem-poker.com/odds_outs.php
...that touches on it. If you looked through other cards, it kinda defeats the purpose. Might as well just look at the river card and say that Cmoney was a 0% to win going to the river.
Posted Mon Aug 14, 2006 4:23 pm GMT by Jauron
You know I had it in my head that was the one linked for some reason. I'll stop pimping someone elses calculator. 
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