Maximally exploitive strategy question

[SeanC]

Hello,

I'm reading Mathematics of Poker and I've hit a snag with one of the examples of determining the maximally exploitive strategy. To paraphrase, the situation is as follows:

Pot has 3 units. Play A has a flush draw that he gets one last (closed) card to draw to. He will make the flush 20% of the time. B has two-pair. A will bet his flush always and will bluff a certain percentage of the time based on the pot. We need to determine B's MES.

The book says B's MES is to fold all the time if A bluffs less than 5%, obtaining an ex-showdown equity of 3x, where x is A's bluffing frequency.

My first question is why is the pot size * the bluffing frequency the equity of the strategy in the case of bluffing less than 5% of the time?

Next it says that if A is bluffing more than 5% of the time, B's MES switches to calling all the time. Why? The expectation of this call is given as x-.2. Why does this expectation function change and how is it determined that it changes at 5%?

MOP has been very good at fully explaining everything else so far, but I'm having trouble with this one.

Any help is appreciated.

[DrVanNostrin]

5% is the "indifference threshold"--it's the point at which the EV(calling) = EV(folding). Player B is indifferent to folding/calling if player A bluffs 5% of the time.

Player B will be getting 4:1. Player B is has him beat 20% of the time, so if player A bluffs 5% of the time player B will win once for every 4 times he losses.

20:5 = 4:1

If player A bluffs less than 5% of the time player B will not win often enough to call; if he bluffs more than 5% of the time player B will win often enough to call.

If player A bluffs less than 5% of the time player B will fold everytime. A's EV is equal to the amount he makes with his bluffs*the frequency he makes them at. That is 3*x.

If player A bluffs more than 5% of the time player B will call everytime. A's EV is the amount he makes on his flushes - the amount he loses with his bluffs. He'll make a flush 20% of the time and make a bet with it each time, that's worth 0.2. Then he loses a bet everytime he bluffs with is with frequency x. That's 0.2 - x.

If you graph these two functions you'll notice they intersect at x = 0.05.

[RustyBrooks]

Well said. This is covered nicely in the Theory of Poker, also.

Note that for game theory reasons, you should proportion your bluffs accordingly, so that if you are the one with the draw, you bluff the optimal (the indifferent) amount. Sklansky recommends doing this by choosing a set of river cards that bluff on. So if you need to bluff 5% of the time, and there are, say, 36 unknown cards to come, and any spade makes your hand, then you want to choose 2 cards where, if you see that card on the river, you bluff. In this case, perhaps the Kh and Qh if neither are in view.

I think this is
bluffcards : unknowncards = 1 : pot/drawodds
so for this case
2 : 36 ~= 1 : 4/.2
(there's some approximatin goin on)

[SeanC]

[ QUOTE ]
5% is the "indifference threshold"--it's the point at which the EV(calling) = EV(folding). Player B is indifferent to folding/calling if player A bluffs 5% of the time.

Player B will be getting 4:1. Player B is has him beat 20% of the time, so if player A bluffs 5% of the time player B will win once for every 4 times he losses.

20:5 = 4:1

If player A bluffs less than 5% of the time player B will not win often enough to call; if he bluffs more than 5% of the time player B will win often enough to call.

If player A bluffs less than 5% of the time player B will fold everytime. A's EV is equal to the amount he makes with his bluffs*the frequency he makes them at. That is 3*x.

If player A bluffs more than 5% of the time player B will call everytime. A's EV is the amount he makes on his flushes - the amount he loses with his bluffs. He'll make a flush 20% of the time and make a bet with it each time, that's worth 0.2. Then he loses a bet everytime he bluffs with is with frequency x. That's 0.2 - x.

If you graph these two functions you'll notice they intersect at x = 0.05.

[/ QUOTE ]

Great reply, thanks. Very well explained and makes perfect sense. I'm surprised to see how little someone has to be bluffing for calling to be correct.

I have a couple more questions, if you don't mind.

With the 3x function, why isn't it 4x? The pot is 3, he adds 1 with his bluff...thus 4? I'm sure there's a good reason, I just don't know it.

That aside, the way to find the indifference threshold is to graph the fold and call function and find the intersection, right?

In the book, it's assuming the viewpoint of player B (the caller), as opposed to A (the bettor). It's also dealing with the defensive value of a strategy (as opposed to EV).

Now, a table is given to show the functions of the value of B's folds as 3x and B's calls as x-.2. X, by the way, represents A's bluffing frequency. The table is as follows:

A's bluffing % |------| <B, MES>
0 |-------------------| 0
2 |-------------------| -.06
4.8 |-----------------| -.114
5 |-------------------| -.15
5.2 |-----------------| -.148
10 |------------------| -.1
20 |------------------| 0

Now, why are these values negative? Also, I'm not getting the right result when graphing the functions of the book(3x; x-.2). When I graph yours (3x; .2-x), I do get an intersection at x=.05 and y=.15. I'm just curious if it's a typo in the book or if I'm just missing something.

NOTE: I just changed their 3x function to -3x and it graphed correctly. Is that a typo then?

[SeanC]

[ QUOTE ]
Well said. This is covered nicely in the Theory of Poker, also.

Note that for game theory reasons, you should proportion your bluffs accordingly, so that if you are the one with the draw, you bluff the optimal (the indifferent) amount. Sklansky recommends doing this by choosing a set of river cards that bluff on. So if you need to bluff 5% of the time, and there are, say, 36 unknown cards to come, and any spade makes your hand, then you want to choose 2 cards where, if you see that card on the river, you bluff. In this case, perhaps the Kh and Qh if neither are in view.

I think this is
bluffcards : unknowncards = 1 : pot/drawodds
so for this case
2 : 36 ~= 1 : 4/.2
(there's some approximatin goin on)

[/ QUOTE ]

Cool, thanks for the reply. I actually remember that in ToP, but I didn't understand the context. Now I do and that's a great way to standardize it.

Can you explain your math in the end a little more? I'm not exactly following what you're solving and from whose viewpoint (though I assume the bettor's).

[RustyBrooks]

When you are the one with the draw, you will have to bluff some times on the end. As we see from your original question, if you bet too often, or not enough, your opponent can play optimally by always folding or always calling. It's best for you to bluff right at the margin between the two of them, although realistically as long as your close I think it's OK. The problem most people have, when they can do A sometimes and B sometimes is they almost always do A or almost always do B.

I think I probably made it more complicated than it needs to be. You already know how to find the percentage of times to bluff optimally, that's just 1 in pot/drawodds. So in your initial example with a 4 unit pot, 20% drawing odds, that's 1 in 20 or 5%

The other part of the equation is just figuring out how many cards to use as your "bluff outs" - these are cards that you will use to randomize your bluffs. This is really most relevant in games like draw or stud where your final card is private and will not affect your opponent's hand.

If you're drawing to, say, a spade flush, and the math says you should bluff 5% of the time, you want to pick a number of bluff cards, so that bluffcards/unseencards = bluff percentage. For seven card stud, on 6th st you usually know:
your 6 cards, 4 of your opponent's cards, 6 upcards of other players and then whatever other cards were dealt to other people on 4th, 5th etc. So at a full table at most there are 36 unknown cards, there may be as few as 25 but that's not common. Anyway, you count those at the time. Assume for a moment that it was heads up from 3rd on, so there are 36 unknown cards.

So 2/36 is about as close to 5% as we can get, that's 5.5%. Just pick 2 cards that have not been face up in this hand yet and make those your bluffing cards. Make them easy to remember like, the black fours, or Ac 2c or something like that.


This is, by the way, nice and all, but you really should be resorting to game theory like this only when your opponents are so tough that you can't get the best of them through hand reading, betting patterns, etc. If that's the case I'd recommend moving tables, really [img]/images/graemlins/wink.gif[/img]

In other words, there are some players you will almost never bluff, because they will always call even though they should fold. Against other players a bluff will work so often that you should try it much more often that is expected. There are lots of meta-game implications... if you've been caught recently you should not bluff. If you've shown down a lot of strong hands you might consider bluffing. If you have trouble getting paid off, try bluffing. If opponents have made very light calldowns, consider not bluffing, and so forth and so on.

[DrVanNostrin]

[ QUOTE ]
With the 3x function, why isn't it 4x? The pot is 3, he adds 1 with his bluff...thus 4?

[/ QUOTE ]
A successful bluff is only 3 bets better than a check. He doesn't really win the bet he put in since that would have been his bet had he checked.

[ QUOTE ]
That aside, the way to find the indifference threshold is to graph the fold and call function and find the intersection, right?

[/ QUOTE ]
That works, but it's easier to find just find the answer using algebra. Find the EV(calling) and the EV(folding), set them equal to each other and solve. In the book's example you could just solve for x in the following equation:

3x = 0.2 - x


[ QUOTE ]
Now, why are these values negative? Also, I'm not getting the right result when graphing the functions of the book(3x; x-.2). When I graph yours (3x; .2-x), I do get an intersection at x=.05 and y=.15. I'm just curious if it's a typo in the book or if I'm just missing something.

[/ QUOTE ]
The values in the book are negative because it's refering to player B's EV, the answer I gave was for player A's EV. Based on the way they defined an EV of 0 player A's EV = -player B's EV (what player A gains is equal to what player B loses). From the sounds of it there's a typo. Player B's EV is:

-3x for 0 < x < 0.05
x - 0.2 for 0.05 < x < 1

[SeanC]

[ QUOTE ]
When you are the one with the draw, you will have to bluff some times on the end. As we see from your original question, if you bet too often, or not enough, your opponent can play optimally by always folding or always calling. It's best for you to bluff right at the margin between the two of them, although realistically as long as your close I think it's OK. The problem most people have, when they can do A sometimes and B sometimes is they almost always do A or almost always do B.

I think I probably made it more complicated than it needs to be. You already know how to find the percentage of times to bluff optimally, that's just 1 in pot/drawodds. So in your initial example with a 4 unit pot, 20% drawing odds, that's 1 in 20 or 5%

The other part of the equation is just figuring out how many cards to use as your "bluff outs" - these are cards that you will use to randomize your bluffs. This is really most relevant in games like draw or stud where your final card is private and will not affect your opponent's hand.

If you're drawing to, say, a spade flush, and the math says you should bluff 5% of the time, you want to pick a number of bluff cards, so that bluffcards/unseencards = bluff percentage. For seven card stud, on 6th st you usually know:
your 6 cards, 4 of your opponent's cards, 6 upcards of other players and then whatever other cards were dealt to other people on 4th, 5th etc. So at a full table at most there are 36 unknown cards, there may be as few as 25 but that's not common. Anyway, you count those at the time. Assume for a moment that it was heads up from 3rd on, so there are 36 unknown cards.

So 2/36 is about as close to 5% as we can get, that's 5.5%. Just pick 2 cards that have not been face up in this hand yet and make those your bluffing cards. Make them easy to remember like, the black fours, or Ac 2c or something like that.


This is, by the way, nice and all, but you really should be resorting to game theory like this only when your opponents are so tough that you can't get the best of them through hand reading, betting patterns, etc. If that's the case I'd recommend moving tables, really [img]/images/graemlins/wink.gif[/img]

In other words, there are some players you will almost never bluff, because they will always call even though they should fold. Against other players a bluff will work so often that you should try it much more often that is expected. There are lots of meta-game implications... if you've been caught recently you should not bluff. If you've shown down a lot of strong hands you might consider bluffing. If you have trouble getting paid off, try bluffing. If opponents have made very light calldowns, consider not bluffing, and so forth and so on.

[/ QUOTE ]

Great post, thanks. When calculating as you said, what units are used in a normal pot? Let's say it's $1-2NL and there's $50 in the pot. Is that a 50 unit pot or a 25 (big blinds), etc.?

How does that math work on the flip side--the person with the decent hand determining the optimal calling percentage?

[DrVanNostrin]

[ QUOTE ]
How does that math work on the flip side--the person with the decent hand determining the optimal calling percentage?

[/ QUOTE ]
It works the same way. Take the example from the OP. Let y equal the frequency with which player B calls. Again there will be an indifference threshold. If player B calls less than this player A should bluff everytime; if player B calls more than this player A should never bluff.

EV(bluffing) = 3(1-y) + (-1)*y
EV(not bluffing) = 0

Set them equal and solve. y = 0.75.

If B will call less than 75% of the time A should bluff everytime; if he calls more than 75% of the time A should never bluff; if he calls exactly 75% of the time it doesn't matter if a bluffs or not.

This result is consistant with common sense pot odds way of thinking about it, which may make a little more sense. Player A needs to risk 1 bet in order to steal 3 bets. If he can steal the 3 bets 1 of 4 times he'll break even. Player B must catch him 3 out of 4 times (3/4 = 0.75) .

[SeanC]

[ QUOTE ]
[ QUOTE ]
With the 3x function, why isn't it 4x? The pot is 3, he adds 1 with his bluff...thus 4?

[/ QUOTE ]
A successful bluff is only 3 bets better than a check. He doesn't really win the bet he put in since that would have been his bet had he checked.

[ QUOTE ]
That aside, the way to find the indifference threshold is to graph the fold and call function and find the intersection, right?

[/ QUOTE ]
That works, but it's easier to find just find the answer using algebra. Find the EV(calling) and the EV(folding), set them equal to each other and solve. In the book's example you could just solve for x in the following equation:

3x = 0.2 - x


[ QUOTE ]
Now, why are these values negative? Also, I'm not getting the right result when graphing the functions of the book(3x; x-.2). When I graph yours (3x; .2-x), I do get an intersection at x=.05 and y=.15. I'm just curious if it's a typo in the book or if I'm just missing something.

[/ QUOTE ]
The values in the book are negative because it's refering to player B's EV, the answer I gave was for player A's EV. Based on the way they defined an EV of 0 player A's EV = -player B's EV (what player A gains is equal to what player B loses). From the sounds of it there's a typo. Player B's EV is:

-3x for 0 < x < 0.05
x - 0.2 for 0.05 < x < 1

[/ QUOTE ]

Thanks for the reply. Regarding the 3 or 4 unit pot, I was thinking from the viewpoint of the made hand deciding on whether to call or not. If there are 3 units in the pot and the opp bets 1, wouldn't the function be 4x as you are missing 4*bluff frequency of value, not 3?

Good point on the algebraic solution. :P

Yeah, must be a typo.