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I have a question regarding an example from the book Mathematics of Poker by Chen and Ankenman. There on p. 48 he gives

Example 4.1 Two players play headsup limit poker on the river. Player A has either the nuts (20% of the time) or a valueless (or dead) hand (80% of the time), and Player B has some hand of mediocre value - enough to beat dead hands, but which loses to the nuts. The pot is four big bets, and A is first. Lets us first consider what will happen if A checks. B could bet, but A knows exactly when he has B beaten or not; hence he will raise B with nut hands and fold at least most of his bluffing hands. B cannot gain value by betting; so he will check. As a result, A will bet all of his nut hands. A might also bet some of his dead hands as a bluff; if B folds, A can gain the whole pot.

We'll call the % of total hands that A bluff with x. A's selection of x is his strategy selection. B loses one bet for calling when A has a nut hand, and wins five bets (the four in the pot plus the one A bluffed) when A has a bluff. B's calling strategy only applies when A bets, so the probability values below are conditional on A betting. The expection of B'S hand if he calls is:

[B, call] = P(A has nut)(-1) + P(A has bluff)(+5)

[B, call] = (0.2)(-1) + 5x

[B, call] = 5x - 0.2

If B folds, his expection is simply zero.

[B, fold] = 0

Remark: [B,call] denotes the expection if B chooses to call, likewise [B,fold] if he chooses to fold.

I guess this analysis is not correct, according to the sentence "so the probability values below are conditional on A betting" we should use conditional probabilities instead. Assume A has 20% of the time the nuts, x*100 % of the time he bluffs and the rest he simply checks. Then the probabilty that he has the nuts if he bets is

P(A has nuts | A bets) = P(A has nuts and A bets) / P(A bets) = ( 0.2 * 1 ) / (0.2 + x)

and accordingly the probability that he bluffs if he bets is

P(A bluffs | A bets) = P(A has dead hand and A bets) / P(A bets) = x / (0.2+x)

where

  • P(A has nuts and A bets) = 0.2
  • P(A has dead hand and A bets) = x
  • P(A has dead hand and A checks) = 1 - (0.2+x)
  • P(A bets) = 0.2 + x

so that the correct expression for B calling should be

[B, call] = P(A has nuts | A bets) * (-1) + P(A bluffs | A bets) * 5

[B, call] = - 0.2/(0.2+x) + 5x/(0.2+x)

[B, call] = (5x-0.2)/(0.2+x)

Am I right or is the book right?

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Do you mean the example from the book, or my explanation thereafter? –  Stefan Apr 1 at 15:17
    
I thought about it and I believe the book is right. –  Valentin Grégoire Apr 1 at 16:12
    
@Stefan both :D –  Gaz Winter Apr 1 at 17:22
    
@Valentin: can you explain? why not using conditional probabilites here... –  Stefan Apr 1 at 17:24
1  
@GazWinter: yes, probabilities can sometimes be mind-boggling... –  Stefan Apr 1 at 17:26

3 Answers 3

The book is right. You have not given us the expected value given that A bets. To do that, you would need the following formula:

[B, call] = (P(A has nuts | A bets) * (-1) + P(A bluffs | A bets) * 5) * P(A bets)

Distribute the P(A bets) and think about each term in isolation. That gives you the true expectation:

[B, call] = P(A has nuts | A bets) * P(A bets) * (-1) + P(A bluffs | A bets) * P(A bets) * 5

Looking at it like this should hopefully be easier to understand.

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I do not understand your point. Given that A bets I calculate the conditional expection for B if he call, and in general EV(X|Y=y) = sum x_i P(X = x_i |Y = y). So there is no multiplication with P(Y) involved? –  Stefan Apr 2 at 10:39

I am pretty sure that your calculations are right and the formula you got is more mathematically accurate than the books.

This doesnt necessarily mean that the book is wrong though. I believe that they just wanted to explain the situation without introducing the concept of conditional probabilities and they made up their explanation with:

B's calling strategy only applies when A bets, so the probability values below are conditional on A betting.

so they wouldnt need to do more advanced math. As I pointed out in my comment both formulas lead to the same logical conclusion - if x>4% B should call.

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In the chapter before conditional probability, even Bayes rule and some statistics like confidence interval is introduced. And the authors even mention that they will took an rigouros approach, so I do not believe that they use wrong formulae on purpose. –  Stefan Apr 2 at 10:42
    
The thing is that to me the sentence "so the probability values below are conditional on A betting." sounds more like an explanation that is added to the calculations below rather than a reminder that they are conditional since they are clearly calculating them as independent variables. Its kinda like "They are conditional but lets forget that for these calculations(because it doesnt really matter)". That said it does sound a bit like an excuse to simplify the math and if everything else is extremely rigorous I might be wrong. –  Daniel Apr 2 at 12:14

I just read this section of the book and was surprised by the error addressed here. Agree with Stefan's analysis.

Whilst it is true that to solve 5x - 0.2 = 0 is the calculation which matters - whether to call or fold, it certainly doesn't determine the expectation.

The graph on page 49 is also wrong. If A bluffs 10% of the time, for example, then B has a positive expectation of 1-bet, not 0.3:

A's strategy: 70% checks; 20% bets with the nuts; 10% bluffs. So 2/3 of the time (20% of the 30%) when betting, A has the nuts and B loses one bet (if calling) ; the other 1/3rd B gains 5 bets. So EV is (2/3 * -1) + (1/3 * 5) = 1.

And of course, if he never bluffs, every time B calls he loses one bet, not 0.2!

A rather mystifying mistake, especially since it was rubber-stamped by the graph!

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