Game theory

I've done a lot of algebra again on the game theory where 2 players ante $1 and get a real number between 0 and 1.

Seeing as some people are posting a solution, eg where player A bets with .86 or more and with .07 or less; I modelled 2 variables, ah and al, being the high point and low point respectively where A bets.


There is one other variable there, called b, which is where player B calls.


If b >= ah then...


EVA = ah squared - al squared - 2.al + 4.al.b + 3.(b squared) - 2.b - 4.ah.b + 2.ah + 1


EVB = 2 - EVA (because they both anted $1, and I didn't take it into account as -ve in the betting)


To work out b, you differentiate, and see when that = 0. It comes out to... b = 1/3 + 2/3(ah-al)


I have more formulae if b evaluates to less than ah, so be careful, this is a restriction on the above formulae.


So if A bets at .86 or above, and .07 and below; then I calculate b to be .86 (same as ah, but that is ok).


I haven't worked out whether A's EV is negative, yet, but I will be very surprised if it isn't. I think A's EV will always be negative if B calls according to his calculated maximum value.


I'll be back later and I will look at this further.

ah= 7/9 . al= 1/9 . b= 5/9 ... assuming A is betting the pot ($2).

Well I plugged the formula and values into my spreadsheet; and I have to admit that all those people who said betting is correct were correct.

A's EV comes to 1.0959; which is more than the 1 if they both always checked.


All I can say is that I did all this stuff for the draw game I usually play, which is 5, 10 blinds and 20 to play, and raising turned out to be bad.


Well I've learned something.

If anyone is interested

the formula for al 1; then b = al (the lower bound)

if (2.al + ah)

* A's EV comes to 1.0959 ... *


EVA= 1,111111111

**********

EVA= 1,111111111


Try again.

@#$#@ All my text has gone.


I'll put it up again later.

For those that are interested what the formulae are if b 1; then b = al (the lower bound)

If (2.al + ah)

For those that are interested, I have the formulae for when b lte ah. lte means "less than or equal to", and for those that haven't worked it out yet, this noticeboard won't let me write that in the way that would be synonymous with writing >=. So not only do I have to be smart and work all this stuff out, I also have to be smart and get it through this parser.


EVA = -2.(ah squared) - (al squared) - 2.al + 4.al.b - 2.b + 2.ah.b + 2.ah + 1

EVB = 2-EVA (as before)


Differentiating EVB with respect to b to find a maximum for b reveals...

If (2.al + ah) > 1; then b = al (the lower bound)

If (2.al + ah) lt 1; then b =ah (the upper bound)

If (2.al + ah) = 1; then b can be anything between al and ah inclusive.


So in our example with al = 0.07, and ah = 0.86, the values fit option 3, and b can be anything in between, so 0.5 is acceptable.

EVA = 1.0959 regardless of b


Next is to differentiate EVA with respect to ah and al and to see if we can come up with the optimal values for those, now that we know how b will play.


What fun I'm having.

And would you 2+2 people be so kind as to change the parser to accept signs like

... about ?


Do you say that: * al = 0.07, and ah = 0.86, the values fit option 3, and b can be anything in between, so 0.5 is acceptable. * is optimal strategy ?


Optimal strategy is: ah= 7/9 . al= 1/9 . b= 5/9 .


Prove me wrong !?

I just gave you the #$#@ formulae and how it was calculated. I can do no more. My formulae are correct. They check with other things that I've done.


What about give your working, and we'll pull that to pieces and find the mistake.

The values that I'm using to test with come from a post by David Sklansky titled "Re: Two Quick Questions for David" on 18th Feb 2002. The first level post on the thread is titled "Master's Thesis Game Theory Problem".


I don't just want to know values, I am about finding out how to do these problems mainly. I will get to finding proper values for ah and al all in good time.

... answer my question.


Do you say that: al = 0.07, and ah = 0.86, b = 0.5 is optimal strategy ?


* What about give your working, and we'll pull that to pieces and find the mistake. *


I can't believe it !? Are you really implying that I made a mistake ;-)

Both ev's have to be between 0 and 2, since there is $2 dead money in the pot (the antes).

This is incomplete. What does B do if A checks. How does A respond to a bet? You don't even say how B responds to A's bet.


If that's A's betting strategy, then B should call with 1/9 and above, fold otherwise.


Why must A always bet with hands 7/9 and above. Couldn't he bet them say 90/% of the time and check otherwise. Now B's betting strategy will be more complicated because he may be betting into a sandbagger (it may not be optimal for A to do that, but you need to look at it). So B would have to bluff less in response to A's check.


I think this problem is more complicated than you gave it credit for.

>>I think this problem is more complicated than >>you gave it credit for.


I worry about this myself. The restrictions placed on solving this problem preclude strategies that, intuitionally, may be optimal. The widely-held belief is that you shouldn't ever do the same thing every time in the same situation, so as to obscure the strength of your hand. The initial restrictions demand that the same thing always be done in the same situation with the same 'cards.' As George points out, sandbagging is not permitted in this model. I put a link up regarding the analysis of a similar game- the author states certain assumptions about bluffing and then at the end says "I was somewhat surprised to see bluffing is actually part of the analytical optimal strategy." It could have come out there are no proper numbers to bluff with, but he discovered that blufing is part of OPT *because* he thought to include it in the first place.


So howzaboutit? Does Sandbagging enter into OPT if you allow it? I doubt I'd be able to put my money where my mouth is and figure it out... I got the hell away from a math major, comp sci treats me much better.


2ndGoat


P.S. Another idea that interests me- how much value is gained by either player having a "tell", defined as once every x hands, one player can determine the other's hand within a certain precision with a certain degree of confidence.

Since check-raising isn't an option in this, and Sklansky's, problem, then sandbagging may not increase expectation. You may be giving up more than you get in return.


I gave your answers a quick run through and it looks like they may be right. A's counter-stragey to B's betting (5/9 and above, I assume) doesn't matter, as it will work out the same whehter A calls or folds. I will give it some more thought in the future.


I tried working on Sklansky's problem a little and came to the conclusion that the distribution of the final hand ranks can be represented by a trapezoid, as they won't be linear as in the problem you solved. The exact dimensions of the trapezoid will be dependant on the range of cards that are played on the first card. Interesting.

B betting with 5/9 or better is wrong. Perhaps 5/9-7/9 and 0-1/9 is best, with A calling with 5/9-7/9 and doing whatever with 2/9-5/9. I haven't worked it all out yet, but the result is something like that, I think. ;-)

This problem is very complicated, but with time we'll get through it all.


I'm very interested in counter strategies also, etc etc.


I've put all this stuff on a spreadsheet, and it looks like "The poker player formerly known as Jack" is correct with the optimal strategy values. This math that we've done so far only assumes that B knows what A is doing. The are probably all sorts of things to look at.

When I get a bit further I will start another thread, but it all takes time for me. So much time. But if we take the time and do it properly, then we should learn a lot from it.