L
LetoAtreides82
Rising Star
Bronze Level
I'm reading the book Poker Math That Matters by Owen Gaines and in the section where he's explaining how to calculate expected value in Poker he gives the following example:
You: 5 of spades and 6 of spades
community cards: K of spades, T of spades, 2 of hearts
Pot= $25
Your opponent goes all in with his $24, making the pot $49. You have enough chips to cover this bet. Should you call or fold?
We have a flush draw so we have 9 outs. If we call we'll see two cards so using the 4/2 rule (multiply by 4 if we get to see the turn and the river, 2 if just one card) we multiply 9 by 4 to get 36%. We have a 36% chance of winning this showdown.
Our pot odds are x / (x+y) where x is the amount to call and y is the pot amount before we call, so in this case it would be $24 / ($24+$49) which comes out to about 33%.
Since we want to make money we should only bet if our chances of winning the showdown are higher than our pot odds. In this case we have a 36% of winning which is higher than the 33% pot odds so the correct decision in the long run would be to call.
To calculate our EV we multiply the pot amount after we call by our chances of winning, then we subtract the cost to call. So in this case it would be .36 ($73) - $24 = $2.28. The EV in this case would be $2.28.
But why are our chances of winning only 36%, why is it just 9 outs? If our opponent just has for example J of diamonds and 2 of clubs, he only has a pair of 2's which could be beaten if the turn or river is a 5 or a 6. Using a showdown calculator it tells me I have a 47% of winning a showdown against a player with a random range of cards.
So wouldn't the EV actually be .47 ($73) - $24 = $10.31? That's a big difference and I don't understand why I would need to assume I would need the best possible hand (in this case a flush) when I could possibly win with a higher pair or two pairs, or three of a kind or the very lucky 6 high straight.
You: 5 of spades and 6 of spades
community cards: K of spades, T of spades, 2 of hearts
Pot= $25
Your opponent goes all in with his $24, making the pot $49. You have enough chips to cover this bet. Should you call or fold?
We have a flush draw so we have 9 outs. If we call we'll see two cards so using the 4/2 rule (multiply by 4 if we get to see the turn and the river, 2 if just one card) we multiply 9 by 4 to get 36%. We have a 36% chance of winning this showdown.
Our pot odds are x / (x+y) where x is the amount to call and y is the pot amount before we call, so in this case it would be $24 / ($24+$49) which comes out to about 33%.
Since we want to make money we should only bet if our chances of winning the showdown are higher than our pot odds. In this case we have a 36% of winning which is higher than the 33% pot odds so the correct decision in the long run would be to call.
To calculate our EV we multiply the pot amount after we call by our chances of winning, then we subtract the cost to call. So in this case it would be .36 ($73) - $24 = $2.28. The EV in this case would be $2.28.
But why are our chances of winning only 36%, why is it just 9 outs? If our opponent just has for example J of diamonds and 2 of clubs, he only has a pair of 2's which could be beaten if the turn or river is a 5 or a 6. Using a showdown calculator it tells me I have a 47% of winning a showdown against a player with a random range of cards.
So wouldn't the EV actually be .47 ($73) - $24 = $10.31? That's a big difference and I don't understand why I would need to assume I would need the best possible hand (in this case a flush) when I could possibly win with a higher pair or two pairs, or three of a kind or the very lucky 6 high straight.
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