[0,1] game exercise

[well]

[ QUOTE ]
I get the following solution to this game:

B: value-raises [2/3,1]
bluff-raises [0,1/9]
check-calls [8/27,2/3]

A: calls a raise [1/3,1]
value-raises [13/27,1]
bluff-raises [0,14/81]

And, as PTB predicted, the game has a value to A of 34/729.


[/ QUOTE ]

Hmmm, I got some different results.

I found that B should check-call in [1/3,2/3] and for A I found:
bluff in [0,1/16] and raise in [1/2,1]!

With this, A's gamevalue would be 1/18.

Next Time.

[Aisthesis]

Well, if my results are now correct on #4.1, this is actually starting to take shape in a way that one might (with the usual "grain of salt" necessary with this kind of thing) consider applying to real-life heads-up situations.

But before getting into that discussion (which I'm actually itching to do!), one more "simulation." The background for this is that in a heads-up game, sometimes I get into a standard PF raise of 2 BB, sometimes 3, sometimes variable, sometimes 4 (against some players who like quick decisions and/or with enormous blind pressure at the end of a tourney, sometimes also a lot of all-ins pre-flop)--all depending a lot on just feel.

So, what about the following game: Same as before with $1 SB and $2 BB, but any raise must be to $6 (minimum raise is not allowed). How does that influence hand selection here? Of course, if anyone wants to also try and solve it with raises to $8 (4xBB), that would be most interesting.

If no one else has already done it, I'll give it a try on the 3xBB scenario here in a little while.

[Aisthesis]

Without going through the tricky part of it again (the check-call threshold for B), your results for A's bluff-raising can't be right: Since B is getting 3:1 on the call, A has to bluff once for every 3 times he value-raises. Hence, if it were correct that A raises [1/2,1], he should have to bluff [0,1/6].

Again, without going through the math on the tricky part, you come up with a better EV for A. So, if the 1/16 was a typo and we've both calculated EV's correctly, then B is better off if he sets his check-call threshold a little lower, namely at 8/27 rather than 9/27 (=1/3).

[well]

[ QUOTE ]
Without going through the tricky part of it again (the check-call threshold for B), your results for A's bluff-raising can't be right: Since B is getting 3:1 on the call, A has to bluff once for every 3 times he value-raises. Hence, if it were correct that A raises [1/2,1], he should have to bluff [0,1/6].

[/ QUOTE ]

I do not agree, since I am getting the same results as I did before after checking.

My approach to the problem is quite different from yours, so I am not sure why we disagree.

Here's why I think I am right:

The strategies -

Player A:

[0,a] - bluff or fold
[a,b] - check or fold
[b,c] - check or call
[c,1] - raise or call

Player B:

[0,w] - bluff
[w,x] - limp, then fold
[x,y] - limp, then call
[y,1] - raise

Then assume 0 <= w <= a <= b <= x <= c <= y <= 1

Writing out all weigthed possible outcomes results in the following EV for player A

EV(a,b,c,w,x,y) = -b+x-w-3xa+ya+3wb+a^2+yb-xc-yc+c^2

Partial deratives:

d/da EV = -3x+y+2a
d/db EV = -1+3w+y
d/dc EV = -x-y+2c
d/dw EV = -1+3b
d/dx EV = 1-3a-c
d/dy EV = a+b-c

We have an equilibrium when all these are zero.
That's exactly when:
{a,b,c} = {1/6,1/3,1/2} and
{w,x,y} = {1/9,1/3,2/3}.

You see?

Next Time.

[well]

It's B's EV writtin out, not B's...

[Aisthesis]

Well, the problem with the differentiating the EV equation, as I tried at first when David proposed the first 2 of these, is really just that it's there are so many individual parts of it that it's very easy to make a mistake. That's why I had a whole bunch of wrong answers even on David's #3 before coming up with the right one: I was always messing up a plus or a minus sign somewhere or forgetting something. So, after David's cautionary "don't use fancy math" post, I just tried to look very carefully at the way he and Jerrod solved the problems.

Anyhow, my EV is better for B, so that would seem to be an improvement, right? I'll double-check the proportions and see if I can find a further explanation for the difference.

[Aisthesis]

Ok, I'm not even going to get into the derivatives simply because I consistently make a clerical error when adding up the EV equation that way. (also, I'm assuming now that the 1/16 was a typo--you're saying A should bluff-raise [0,1/6], right?)

But, basically, I'm claiming that you don't have A raising enough. So, that actually gets us into another question I was interested in exploring but haven't gotten around to: Exploiting (purported) sub-optimal strategies (particularly interesting in the poker context!!!).

Ok, if A is being too selective about his raising hands, as I'm claiming, then B can exploit that by calling less frequently.

In this case, I come up with check-calling for B on [1/2,2/3] as best counter to A's strategy. If B does that, he (B) actually WINS 1/36 of a bet by my calculation, so A is losing money.

What's kind of weird here is that there seems to be a big jump in B's check-call threshold depending on the particular sub-optimal strategy of A (I think a lot of the time, B will want to start calling right where the value-raises begin). If A didn't bluff at all, then B would want to go above 1/2 for the limp-call. It would actually be pretty interesting to explore how this kind of thing works (David's #3 might be a good one to use for this, just to see how A needs to react if B bluffs too much, bluffs too little, or doesn't make enough value-raises).

But for now, I'm pretty convinced that your EV equation misses some cases. The derivatives always turn out really easy once the EV equation is correct.

So, if I'm wrong, I'd at least like to know why... [img]/images/graemlins/smile.gif[/img]

[Aisthesis]

I'm having some doubts about having B call the raise only at 1/2, but here's my main objection:

First, an observation: The situation is completely symmetrical, so if B checks, A will adopt the same raising strategy against B's range of hands that B adopted against A's random hand ([0,1]).

But B has checked on only 5/9 of his possible holdings--not on 2/3 of them. So, if B check-calls on [1/3,2/3], he is calling A's raise after checking only 3/5 of the time rather than 2/3 of the time. That should mean that A's bluffs will actually make a (net*) profit for A on your scenario, although A's value raises will make slightly less profit.

Is it possible that you missed taking out the bottom 1/9 of B's hands somewhere in your EV equation when B just checked?

That's clearly where the difference lies.

*the bluffs will actually still lose money for A, but they should lose less than the $1 he has already put into the pot blind.

[well]

Ok, let's assume you're right about B's strategy.

[ QUOTE ]
I get the following solution to this game:

B: value-raises [2/3,1]
bluff-raises [0,1/9]
check-calls [8/27,2/3]

A: calls a raise [1/3,1]
value-raises [13/27,1]
bluff-raises [0,14/81]

[/ QUOTE ]

Now focus on the part where B limps.

Suppose A will
bluff in [0,a] and
valueraise in [b,1].

Suppose you're right about 1/9 <= a <= 8/27 <= b <= 2/3

Now let's see what EV we get for A given that B checks.
This means B is always in [1/9,2/3]
Notice that the solution we get here for a and b when maximizing this A's EV,
will be the same when maximizing A's whole EV given B's strategy.

[0,a] - The bluff

Since a <= 8/27, A will lose 2 when called, otherwise win 1.
The probability of B folding is (8/27-1/9)/(2/3-1/9) = 1/3
This yields a*(1*1/3 - 2*2/3) = -a

[a,b] - The check

A will win 1 if B is in [1/9,a] and lose 1 if B is in [b,2/3], otherwise they will break even.
The probability of B being in [1/9,a] is (a-1/9)/(2/3-1/9) = 9/5a-1/5
The probability of B being in [b,2/3] is (2/3-b)/(2/3-1/9) = 6/5 - 9/5b
So this results in (b-a)*(9/5a-1/5-(6/5 - 9/5b) = -9/5a^2-7/5b+7/5a+9/5b^2

[b,2/3] - The value-raise in B's range
A will win 1 if B folds, or 1/3 of the time.
A will win 2 if B is in [8/27,b] - or (b-8/27)/(2/3-1/9) = 9/5b-8/15 of the time - and otherwise break even.
Here we have (2/3-b)*(1/3+2*(9/5b-8/15)) = -22/45+47/15b-18/5b^2

[2/3,1] - The unbeatable value-raise
A will win 1 if B folds, and win 2 if B calls.
This means (1-2/3)*(1/3+2*2/3) = 5/9

Summing these four gives
2/5a-9/5a^2+26/15b-9/5b^2+1/15

d/da : 2/5 - 18/5a
d/db : 26/15 - 18/5b

Hence, player A will choose a=1/9 and b=13/27

So A will reply with a 1/9 bluffing solution, which is in conflict with your 14/81.

I think the problem might lie in the the fact that the middles of the intervals [0,1] and [1/9,2/3] are not the same.

Of course I might have made some mistakes, so you have a task to do again!

Next Time.

[Aisthesis]

According to my calculations, these values give an EV for A of 239/900. So, A does win, as expected.

But the strange numbers in the EV scheme (which don't cancel each other out very well) make me wonder if there's something wrong here. Anyone care to come up with something better?