nomdeplume
07-24-2005, 04:05 PM
I'm reading Matthew Hilger's book, 'Internet Texas Hold'em'. In it, he suggests that beginners don't call with KJs in EP unless there is 1 limper already in. So if you're UTG, for example, you should fold according to his charts.
This started me wondering what this would cost in the long run. I've done some calculations and I'd be grateful if someone could check their correctness for me.
I've used PokerRoom's EV stats for this. I know that these are for the average player, but it's a start.
These stats say that KJs UTG has an average profit of 0.29 big bets. Therefore, in a $5/$10 game (for example), KJs is worth on average 0.29*$10 = $2.90.
However, there are only 4 possible KJs hands out of a possible 1326, and you'll only be UTG 1/10th of the time (assuming a 10 handed game).
So the fraction of the time you can expect to be dealt KJs UTG is (4/1326)*(1/10) = 1/3315, or 1 in every 3315 hands. Am I correct so far?
Therefore, the cost of folding KJs UTG is $2.90*(1/3315) = $0.0009. In other words, it costs almost nothing. This surprised me as I thought it would be more than this. So I think I may have got my calculations wrong.
Anyone help me out?
This started me wondering what this would cost in the long run. I've done some calculations and I'd be grateful if someone could check their correctness for me.
I've used PokerRoom's EV stats for this. I know that these are for the average player, but it's a start.
These stats say that KJs UTG has an average profit of 0.29 big bets. Therefore, in a $5/$10 game (for example), KJs is worth on average 0.29*$10 = $2.90.
However, there are only 4 possible KJs hands out of a possible 1326, and you'll only be UTG 1/10th of the time (assuming a 10 handed game).
So the fraction of the time you can expect to be dealt KJs UTG is (4/1326)*(1/10) = 1/3315, or 1 in every 3315 hands. Am I correct so far?
Therefore, the cost of folding KJs UTG is $2.90*(1/3315) = $0.0009. In other words, it costs almost nothing. This surprised me as I thought it would be more than this. So I think I may have got my calculations wrong.
Anyone help me out?