Tough, Important, General Case, Game Theory Problem

[trillig]

Why don't the players just either have a chip in their hand or not to indicate in or out at the showdown?

(1/n) * 1.1 = # of times I'd want to be in pot.
[10% more than average of players divided # of hands to come out reasonable ahead.]

At least that is what I would shoot for...

I played Home Poker quite a bit 10+ years ago and we played GUTS a lot, which was match the pot if you lose and hold chip if in, don't if out.

We played Nickel/Dime/Quarter and once got the pot to $40, I was the poor sap that paid it and I took it down the next hand with J/7.

The good old days, we loved that game!

-t

[M.B.E.]

Let m be the number of your opponents (=n-1), and let c be the mth root of b/(ma+b). It's been established that "optimal" play is to bet any card > c, and otherwise fold.

Suppose one of your opponents is playing too tight, but the rest are playing "optimally". That is, one opponent will bet iff dealt a card greater than r, with c < r < 1. Then your most profitable strategy is to always bet, and your EV per hand will be

[(r^m - c^m)ar + (r-c)(m-1)(br^m)]/m.

[M.B.E.]

Suppose you have m opponents (=n-1) and one of them is playing too loose. The other opponents play "optimally", i.e. they bet iff dealt a card greater than c = [b/(am+b)]^(1/m).

The loose opponent will bet iff dealt a card greater than r, with 0 < r < c.

As others have said, you can take advantage of the loose opponent by loosening up yourself, but not as much. You will maximize your EV by betting iff dealt a card greater than s, for some s with r < s < c.

The question is, how do you compute s?

The value I've come up with is

s = [c(ma+b) + r(a+b)] / (ma + a + 2b)

I don't know if this is correct, and I also don't know whether, if it is correct, it can be expressed in simpler form, using the identity c^m=b/(am+b).

[huzitup2]

If there's $3 in the pot (in the form of "antes") and the 2nd bet is $2 . . .

*

*

*

Shouldn't I bet with ANY hand that has at least two chances in seven of winning ?

- I used 2/7 since I am getting 5-2 odds if someone else calls; obviously I'm getting better odds if both of my opponents call.

Given the nature of the game there is no correct bluffing frequency - since there is no bluffing !

I hope to find the time to disect this problem - I did take the required math courses to have a go at it but it's 5:30 am and I've been playing all night - but off the top of my head this seems like a reasonable "guess"; I cannot imagine the solution being this simple, but could it possibly be off by much ?

- H

P.S. Your posts - and the responses to them - receive a great deal of scrutiny. If I am way off I am CERTAIN I'll hear about it (and I have a pretty good idea of whom the loudest respondents will be :-).

[huzitup2]

I also played this game ALOT; my results were very good due mostly to the fact that the those I played against were "knuckledraggers", a/k/a lower forms of primates :-)

This was a GRRRRRRRRREAT game for spotting tells; I laid down some very good hands when one or more of my opponents was squirming and was right far more often than wrong.

BTW, we played 3-card hands - with the CORRECT rankings used - which made things a bit more complex since ocasionally one of the others would get excited about a hand that was not as good as HE thought it was, but it still allowed me to get [pretty] good at "reading" players.

I mucked trip Aces in a nice sized hand against a player whose hand was sweating AND trembling only to have a squable break out after he rolled over his Ac Kc Qc.

- The "rule" was you could look at any hand after the result was in.

Sadly, once the group agreed that I had made a fantastic laydown (and wasn't cheating) "GUTS" was seldom played - at least not on the nights that I showed up.

C'est la vie.

- H