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

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 '14 at 15:17
I thought about it and I believe the book is right. – Valentin Grégoire Apr 1 '14 at 16:12
@Stefan both :D – Gaz Winter Apr 1 '14 at 17:22
@Valentin: can you explain? why not using conditional probabilites here... – Stefan Apr 1 '14 at 17:24
@GazWinter: yes, probabilities can sometimes be mind-boggling... – Stefan Apr 1 '14 at 17:26

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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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 '14 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 '14 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 '14 at 12:14

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?

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This all looks correct to me. The 0.05 value of x represents the breakeven point of how often the bettor is betting when they have a dead hand. I don't have the book in front of me for reference, but I'm guessing the 0.04 represents the percent of ALL hands that the bettor is betting and has a dead hand (.05 * .8). So .20 of all hands he bets and has the nuts (you lose 1 for calling), and .04 of all hands he bets with nothing (you win 5 for calling). That works out to an EV of 0 for the caller, as desired. – Dr.DrfbagIII Feb 11 at 14:56
I've only just read this. I agree with what you've said. So although the book says that x is A's bluffing frequency (the frequency with which he should bet GIVEN he has dead hand), it's actually the frequency he bluffs and AND has a dead hand, which isn't such a useful statistic. Thanks for your input. – Charles 2 days ago

In summary

• Player A has the nuts 20% of the time
• Player A will bet for sure if player A has the nuts
• Player A will also bet x% of the time with air
• pot is 4
• player A bet is 1

From the perspective of Player A if he/she only bets with the nuts and Player B knows that and will fold to any bet then Player A gets nothing. Player A has to bluff a certain x% of the losing hands to get proper action from player A.

So bl = bluff loosing
How often should Player A bluff a loosing hand to give Player B 0 EV to call or (or fold)

I think Chen has it set up wrong
If you only charge the bet on nuts then winning is only 4 (not 5)
And even with that I don't think it would be correct

Player B EV = 0 = -1 + 0*1/5*5 + 4/5*bl*5

-1 call the bet
0 if Player A is betting with the nuts (1/5)
4/5 ratio of air
bl - bluff loosing (only betting a portion of the loosing hand)
5 pot to pick up

1 = 4 bl
bl = 1/4 = 0.25
So should bluff a loser 25% of the time

Absolute bluff rate = 4/5*1/4 = 1/5 = 0.20

So Player A should bet the nuts 20% and bluff 20% for a bet rate of 40%
Player A should bluff as much as bet for value

From the perspective of Player B EV to call is = -1 + 1/5*5 = 0
And EV to fold is 0
You have put then in a lose lose (or 0 0) situation
Given you only have the nuts 20% if you can give 0 0 that is the best you can get

Let look at 10 rounds of Player B EV

Case A Player B only bets the nuts and Player A folds every time

``````nuts    4
nuts    4
air     0
air     0
air     0
air     0
air     0
air     0
air     0
air     0
SUM     8
``````

Case B Player A will bluff 20%
Player B will call exactly 1/2 the time as got them totally confused

``````nuts call        5
nuts fold        4
air bluff call  -1
air bluff fold   4
air check        0
air check        0
air check        0
air check        0
air check        0
air check        0
SUM 12
``````

Case C Player A will bluff 20%
Player B will fold to every bet

``````nuts fold       4
nuts fold       4
air bluff call  4
air bluff fold  4
air check       0
air check       0
air check       0
air check       0
air check       0
air check       0
SUM            16
``````

Case D Player A will bluff 20%
Player B will call to every bet

``````nuts call        5
nuts call        5
air bluff call  -1
air bluff call  -1
air check        0
air check        0
air check        0
air check        0
air check        0
air check        0
SUM 8
``````

Case D is proof the 20% bluff rate it correct
If Player A turns Player B into a calling station (the desired outcome) then if Player A bluffs more than 20% Player A loses money

Once you turn them into a calling station then bluff less
You need make them know you are capable of bluffing to manipulate them

If they are folding then bluff more until you turn them into a calling station

How often should Player B call to make Player B EV 0 is more interesting question

Let's look at a bigger bet - bet the size of the pot
So now that 5 is a 2

0 = -1 * 4/5*bl*2
1 = 8/5 bl
bl = 5/8
Absolute bluff rate = 4/5*5/8 = 4/8 = 1/2 = 0.5
Yes I know it seems wrong but mathematically you should actually bluff a bigger bet more

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`Player B EV = 0 = -1 * 4/5*bl*5` How could this ever be equal to zero? When you say bl, I assume that's the same as P(B|A') (probability of betting given player A holds a dead hand). Thanks for the detailed response, but I'm having trouble understanding how 4 non zero real numbers could multiply to make zero?!? – Charles Feb 11 at 11:23
Also I've just realised that the long handed way of what I originally wrote can be restated <B,call> = 0.2*(-1) + 0.8*x*5. What's written in the book is <B,call> = 0.2*(-1) + 5*x. Why wouldn't you include 0.8? Surely we want the probability that player A has a dead hand and is bluffing, instead of the probability that player A is bluffing given he has a dead hand? – Charles Feb 11 at 11:45
This seems more like a comment on @Frisbee's answer instead of an answer to the question itself. – Chris Farmer Feb 11 at 14:09
@Charles The X can be what you define it. If you don't put in 0.8 then it is absolute bluff rate. If you put in 0.8 then it is bluff rate of a loser. It is still the same thing. – Frisbee 8 hours ago

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.

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