Question on Strategy Calculation (from Bill Chen's Book Math of Poker)

Question

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?

Answer 1

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!

Answer 2

I have been struggling over the same example for the same reason.

Let P(A) = Prob A has nuts = 0.2,

P(A') = Prob A has dead hand = 0.8,

P(B) = Prob A bets,

P(A'|B) = Prob A has dead hand given A bets 

P(A|B) = Prob A has nuts given A bets

Then P(A|B) = [P(A)P(B|A)]/[P(A)P(B|A)+P(A')P(B|A')] = (0.2*1)/(0.2*1 + 0.8x)

and P(A'|B) = [P(A')P(B|A')]/[P(A)P(B|A)+P(A')P(B|A')] = (0.8x)/(0.8x + 0.2*1)

so = P(A|B)(-1) + P(A'|B)(5) = (4x)/(8x + 0.2) - 0.2/(0.8x + 0.2)

        = (4x - 0.2)/(0.8x + 0.2)

Therefore 4x - 0.2 > 0

      x > 0.05

This answer is very close to but not quite the answer of 0.04 given in the book. Can anyone explain why this is, or perhaps this is confirmation that what's written in the book isn't as mathematically rigorous as it could be?

Answer 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.

Answer 4

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.

Answer 5

I think I've got it:

The formula used in the book MUST assume the player A has NOT already bet, which I don't think is made very clear in the book at all. I don't think this is what was intended, since the book clearly says that B's calling expected value is conditional on A already betting, in which case the book is wrong.

This gives us < B,call > = P(A AND B)(-1) + P(A' AND B)(5)

= 0.2*(-1) + 5x

So I think the book is wrong on the basis that this formula doesn't use a conditional probability.

Answer 6

"The Mathematics of Poker" is better suited for some sort of dynamic programming. It's like you're holding the Ace of Spades directly up to your eye to try and see the suit...you do NOT need to get that close to the card. If you want to bog down your brain while you're playing poker though, great book : )