explanation for exercise solution ?
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November 1st, 2013 at 5:06:59 PM
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I'm currently dipping into some poker math, and reading Chens book "The mathematics of poker" the Wizard recommended some time ago around here.
This is one of the first exercises mentioned. While I agree with the solution, I find the printed way more elegant and easy, yet I don't understand why it is valid from a mathematical point of view.
So here is the exercise reduced to the significant parts (4.1, for those who have the book).
"The game is some limit poker, heads up. B knows A has only two possible hands, 20% A has nuts, and 80% A has a dead hand. The pot size is 4 units. A bets (1 unit).
What percentage of his hands should A bluff ?"
First of all, the solution I found (P(X|Y) is the probability of X conditional Y):
EV(B calls) = P(nuts|bets) * (-1) + P(dead|bets) * 5
with P(dead|bets) = P(bets|dead) * P(dead) / (P(bets|nuts) * P(nuts) + P(bets|dead) * P(dead)) = 0.8*y / (.8*y + .2)
and P(nuts|bets) = P(bets|nuts) * P(nuts) / (P(bets|nuts) * P(nuts) + P(bets|dead) * P(dead)) = .2 / (.8*y + .2)
where I assumed A will bet all nuts hands, and will bluff at probability y his dead hands.
Resulting in
EV(B calls) = (4*y - .2) / (.8*y + .2).
The correct bluffing rate should be EV(B calls) = EV(B folds) (which is 0), or y = 5%.
I.e. A should bluff 5% of his dead hands, or 5%*80% = 4% his total hands.
Now here is the solution of the book:
EV(B calls) = P("A has nuts") * (-1) + P("A has a bluff") * 5 (looks good to me)
= .2 * (-1) + x*5 (this line I don't understand)
with solution x = 4%.
I know this solution is most elegant and easy computation, but I don't understand it fully. Okay, x is the bluff rate with respect to the total hands.
But if this is a true EV, shouldn't all P's add up to one ?
Either there must be some term missing with P("A checks with dead hand") - but this should not contribute to the call-EV, as there is no call,
or the author deliberately does not use the correct Bayesian probabilitites, given that A bets ?
Whats wrong here? Although both results are eventuall identical, where is the cheat ?
This is one of the first exercises mentioned. While I agree with the solution, I find the printed way more elegant and easy, yet I don't understand why it is valid from a mathematical point of view.
So here is the exercise reduced to the significant parts (4.1, for those who have the book).
"The game is some limit poker, heads up. B knows A has only two possible hands, 20% A has nuts, and 80% A has a dead hand. The pot size is 4 units. A bets (1 unit).
What percentage of his hands should A bluff ?"
First of all, the solution I found (P(X|Y) is the probability of X conditional Y):
EV(B calls) = P(nuts|bets) * (-1) + P(dead|bets) * 5
with P(dead|bets) = P(bets|dead) * P(dead) / (P(bets|nuts) * P(nuts) + P(bets|dead) * P(dead)) = 0.8*y / (.8*y + .2)
and P(nuts|bets) = P(bets|nuts) * P(nuts) / (P(bets|nuts) * P(nuts) + P(bets|dead) * P(dead)) = .2 / (.8*y + .2)
where I assumed A will bet all nuts hands, and will bluff at probability y his dead hands.
Resulting in
EV(B calls) = (4*y - .2) / (.8*y + .2).
The correct bluffing rate should be EV(B calls) = EV(B folds) (which is 0), or y = 5%.
I.e. A should bluff 5% of his dead hands, or 5%*80% = 4% his total hands.
Now here is the solution of the book:
EV(B calls) = P("A has nuts") * (-1) + P("A has a bluff") * 5 (looks good to me)
= .2 * (-1) + x*5 (this line I don't understand)
with solution x = 4%.
I know this solution is most elegant and easy computation, but I don't understand it fully. Okay, x is the bluff rate with respect to the total hands.
But if this is a true EV, shouldn't all P's add up to one ?
Either there must be some term missing with P("A checks with dead hand") - but this should not contribute to the call-EV, as there is no call,
or the author deliberately does not use the correct Bayesian probabilitites, given that A bets ?
Whats wrong here? Although both results are eventuall identical, where is the cheat ?
