Hi,
I have been studying the game theory problems in both Kill Everyone and Mathematics of poker.
It seems to me that there is a fairly large error in one of the examples in Kill Everyone, (or maybe I'm missing something :-))

The problem given in the section Bet Sizing on Bluffs and Monsters on page 34-35 seems wrong to me.
The example considers how much you can call on the turn based on your implied odds on the river, calculated using game theory.

The example in short is:
Pot is $100, both players stacks are $1200.
You have a draw that will come in 24% of the time. You also hold enough hands to bluff the correct amount on the river.
Your opponent is ahead on the turn.

The book suggests you can call a $200 bet on the turn because that will be your implied odds from shoving the river.
The reasoning is as follows:
After the $200 bet and call on the turn the pot is $500.
So if you make your hand on the river and shove you are betting $1000 to win $500 so your opponent must call you 1/3 of the
time to make you break even on your bluffs. So equally you win $500 2/3 of the time, and of course you break even on your bluffs.

Your Ev is then 24% x 1/3 x 1500 + 24% x 2/3 x 500 = $200
So you make $200 on the river and can therefore call a bet that size on the turn.

However, if we look at example 20.2 on page 249 of the Mathematics of Poker then we can see that your opponent does not have to
call you 1/3 of the time on the river. There are bets on two streets so we need to take both streets of action into account.
Your opponent needs to make your entire play of calling the turn with nothing and bluffing the river break even. Not just the river
bluff. 76% of the time you will have called a bet and will not be able to bluff on the river.

So copying the example given in MOP my calculations are:

if x is the percentage of time your opponent must call you on the river, then

76% you lose your $200 call = -$152
24% you bluff and get called x% and lose $1200 = -$288x
24% you bluff and your opponent folds (1-x)% and you win $300 = $72(1-x)

So, even if your opponent never calls your bet on the river (x=0) your EV is -$80.
This is a losing play to call $200 on the turn.

I am only just studying this stuff, but it seems like the mistake made in the whole section of the book, is to assume the
single street solution is the same as the multi street solution, which it is not, if I understand MOP.

Some feedback from the authors would ve greatly appreciated.

Thanks,
Alan

The optimal play on the river is to bet 40% of your hands - the 24% of the time you've hit & another 16% of the time as bluffs. This makes your opponents actions "irrelevant". If he never calls, then 40% of the time you take $500, making the $200 call okay. If he always calls, you have $2500 24% of the time and $1000 60% of the time, adding up averaging $1200, the same as folding. Similarly if he calls/folds 1/3 & 2/3 you also break even. So the book is correct.

In your second example, if the opponent is calling less than 1/3 of the time then you are making money on your bluffs. This makes up for the lost money. Try your math again assuming the player on the draw bluffs 100% of the time and solve for x. You'll see that it's profitable for all x < 1/3.

Tysen

Hi,
Thanks for the quick answers guys. That's what I love about this forum.
I am learning from a book, ask a question and get an answer from the author within minutes. Please understand that I am just learning this stuff and am trying to work it out properly.

I still have some confusion about how to reconcile the solution given by Chen and Ankenman in the Mathematics of Poker and your example. First of all, thanks Pokerfarian for pointing this out. I had indeed misinterpreted the problem. The example involves calling with one hand, a draw, and betting or
bluffing based on the cards that come on the river. However, I was assuming that the player called the turn with a range of hands and when a draw card hit and the player shoved, he either had it or didn't (bluff). So when the draw card hit 24% of the time, 2/3 of the time the player had the draw, the other 1/3 he was bluffing.

The example however is, correct me if I am wrong,

The player has a flush draw for example (that will come in 24%)
If the flush comes in, the player always shoves.
If another credible draw, say a straight card, comes in the player shall also shove 16% of the time.
So with one hand you either value bet or bluff depending on the river card.

This is subtly different to calling on the turn with a range of flush and straight draws and some other weak draws and shoving when the flush comes with your flush draws and some of your straight draws. The first is one hand played differently for every river card, the second is many hands played the same for one river card!

However, the problem I see with the Kill Everyone example is that it doesn't take into account the full range of hands you have to call the turn bet of 200 with. If you are going to shove when the straight card comes in, you have to call the turn with straight draws as well. If you are going to bluff when any card comes in i.e. 100% then you have to call with a wide range of hands on the turn to make your bluff credible. Clearly it will be highly exploitable if you only call the turn with flush draws and bluff when a straight card or any card comes in. Your bluff must be credible, and so to bluff a higher percentage, you must call the turn with more hands and that will cost you money when you don't bluff on the end.

So Tysen, your question of recalculating the math when you bluff 100% doesn't make sense to me because a bluff won't be credible for most river cards. To make it credible you have to call with a wider range on the turn which costs more money. I think you need to take this extra cost into account.

Chen and Ankenman's example was expressed like this and they showed that the one-street solution (where your opponent will call 1/3 of the time to make your bluffs break even) doesn't apply to the two street game. Even though the river betting and pot look identical, the previous action on the turn changes the correct solution. They have said that your opponent must call enough to make your entire bluff, of calling on the turn and shoving the river, break even. This is surely different. In their example your opponent could call less on the river because he made you pay for your draw, and other calls, on the turn. So what your bluffs gain on the river, they are only making up for what you lost by calling the turn with more hands. I think your example doesn't take into account this extra cost on the turn.

Like I say, I am only trying to learn this stuff, and reconcile Chen and Ankenman's examples with yours. I really appreciate that you take the time to answer my (probably stupid) questions. I think the Mathematics of Poker and Kill Everyone are two of the best books on poker.

I will keep on with the math as well.

Thanks,
Alan

Hi everyone,

this is my first post but I thought I just jump into the deep end.

The problem with multiple street models is the state of where you take your calculations. Its a classic conditional probability argument. The factors which are missing in this problem is out of the pot $100 what portion was contributed by each player since this affects our calculations. Who was SB, BB or neither of them was???

Alan to explain the answer in Kill Everyone:

Since the pot is $100 and both players stack sizes are $1200 if we take the condition that neither player invested any money in the $100 initial pot (Dead money) then we can solve the equation like this.

P = Pot
B = Bet size
PP = Pot size on river (P+2xB)

k1 = 33.33%*1200 [Profit when Draw hits and he calls]
k2 = 66.66%*(PP-B) [Profit when Draw hits and he folds]
k3 = 33.33%-1200 [Profit when Draw misses and he calls]
k4 = 66.66%*(PP-B) [Profit when Draw misses and he folds]

E = 0.24*(k1+k2)+0.76*(k3+k4)

Then solve for E by changing B in excel. Using goal seek you can find out that B = $200.

Thats how the $200 bet size is worked out using a decision tree method.

Also just for completeness:

33% 103.9896
24% 67% 47.9952
76% 33% -303.9696
67% 151.9848
E 0

25% 78
24% 75% 54
76% 25% -228
75% 171
E 75

30% 93.6
24% 70% 50.4
76% 30% -273.6
60% 136.8
E 7.2

40% 124.8
24% 60% 43.2
76% 40% -364.8
60% 136.8
E -60

using 4 different scenarios if opponent calls 25% E is 75, if opponent calls 30% E is 7.2 and if opponent calls 40% E = -60.

Hi All - This is a tangential question. Also on p. 35, there's a comment about Casino Match Play Chips, that using them for "big risky" bets gets the most EV. However, when I calculate the EV for using a $1 match play for betting on "Red" vs. "#12", the EV's are the same:


Assume $1 bet + $1 match play in Roulette

EV for Betting on "Red" = (18/38) * (2) * ($1+$1) = $1.89

EV for Betting on "#12" = (1/38) * (36) * ($1+$1) = $1.89



An alternative interpretation is that the authors are suggesting that one should use the Match Play Chips for as much as they will match - i.e. if the Match Play Chip will Match a bet between $1-$10, one should use it to Match a $10 bet.

Can anyone explain this in more detail? I'm just making sure there's not some detail about Match Play Chips that I'm missing (Since the casino I go to most often has many Match Play promotions). Thanks!