Head Up Theory Question

[xpsyuvz]

Marv, good job with that 8-card optimal stategy link.


But I am wondering, what happens when the game allows for an infinite amount of bets/re-reraises (as opposed to just four)?

I assume there is still an optimal strategy right? What would it be?
And then, how many re-raises should a person with the 7-card make?


I ask because I find David Sklansky's original question way too intractable. If anyone can answer the above questions (or even using a deck with less cards) then I would find it interesting to read.

But if nobody can answer these simpler questions, then I doubt anyone will be able to know for sure what the answer to DS's original question is.

[Stephen H]

[ QUOTE ]
The $1 ante also deviates from the [0,1] game.

[/ QUOTE ]

I disagree. One justification for no folding was "infinite pot"; and the size of the pot never comes into play in any of the equilibrium equations. The ante matters not at all. The only difference I see is that the [0,1] game is played on the real line, and the 1,000,000 card game is discrete.

[ QUOTE ]

"What this means is that when infinite bets are allowed, you should generally put in another raise with about 41% of the hands your opponent can have to have raised you this far."

From the indifference equation:
x_n = r^(n-1)/(1+2r)
r = 0.414 "Golden Mean of Poker"
x_n = point where opponent is indifferent to checking or betting
n = number of bets to be put in before stopping


[/ QUOTE ]

Of course! I had it mostly right, except for the starting bet - it's special, since you don't have to call a bet, but could check and get checked behind or bet to (where you have to just call). This changes the range for the initial bet.
Incidentally, I wouldn't go through the trouble of inverting this equation (while rounding to 3 sig figs in the process) when what you really want is a chart of what x_n is for varying values of n. Also of note is that the equation for y_n where n is even (player y goes 2nd) is identical to x_n, so you can create the chart with both even and odd n, and use it for going first and second.

I created a chart based on those equations, and now I get the same answers you have (although mine aren't rounded to 3 sig figs). Which puts the answer at "1 bet and 7 raises" for the game with no folding.

I think the logic on my extrapolation into the game with folding is quite faulty, and I no longer have any confidence in it. However, I'm fairly sure it's within an order of magnitude or so.

Thanks for the reply!

[marv]

[ QUOTE ]
If you play optimum game theory betting strategy, your opponent does equally well by folding or calling. Optimum calling strategy is to fold no more or less often than needed to keep it from being right for your opponent to always bluff.Thus to make calculations in multi round situations easier, assume one always at least calls, even though bluffs happen.

[/ QUOTE ]

So what problem do you actually want us to solve?

What would be the solution to that problem in the 8-card game where every optimal strategy folds the 2nd nuts some of the time?

Marv

[Slade]

To solve the pot limit problem first you need to find out how good of a hand you need to raise. If you take a [0,1] game you can find the point where your raise will break even against an opponent with the same raising standards in a pot limit game with the following equation. (Note that this equation assumes all bets are pot sized)

x - (1-x)*x - (1-x)*(1-x)*4 = 0

You will get x=(2/3) which means you should raise with the top (1/3) hands. The same equation modified can be used to find the sqrt(2)-1 magic number for the limit problem.

Using this the standards needed for each raise should be,

1) 666667
2) 888890
3) 962964
4) 987655
5) 995886
6) 998629
7) 999544
8) 999849
9) 999950
10) 999984
11) 999995
12) 999999
13) 1000000

Since you are putting in the first bet, the last bet you will put in is bet number 11. If you were to act second then you would put in bet number 12 last.

[SA125]

[ QUOTE ]
I haven't made any attempt to figure out the exact answer. Just watching you guys.

[/ QUOTE ]

Coupled with his quote that he wrote HPFAP basically to make money for himself (something so obvious yet, because of the devout nature of members here seeking the inner capacities to the reasons for successful understanding and applications to such a complicated game), along with seeing how David played in the invitational (not so great yet he caught well at the end and won), leads one to believe he sometimes does all of this for sh*ts and giggles.

[JackWilson]

Not directed at SA125.

I can't believe I just read this whole thread expecting a solution somewhere only to get to the end and realize there is none (yet)

[reubenf]

[ QUOTE ]
leads one to believe he sometimes does all of this for sh*ts and giggles.

[/ QUOTE ]

Is there any other reason to do anything?

[xpsyuvz]

Even if DS was just posing the question “for sh*ts & giggles”, I did find this thread helpful -- as I can now see how the “optimal” strategy is probably to call much earlier than I would have initially imagined.

As after looking again briefly at Marv’s 8 card strategy, it seems to support the idea of fewer raises. So I’m more inclined to believe that ideas like Underlord’s could be pretty accurate -- like only raising about 8 times (or about 15 times total for both players) and then calling. (Of course this is if you are playing against someone that would try to exploit you if you weren’t playing “optimally”. But if you were playing against normal people, you could probably exploit their wrong intuitions and keep raising for a while.)

[thejameser]

[ QUOTE ]
I never studied much game theory, but the answer seems clear enough to me: none. Same for pot limit.

Why?

Because here you've got unlimited money, and yet you're playing an ultra-low-stakes version of the most boring card game I've ever heard of -- one where you only get one card in the first place and it's going to take the dealer like a week to shuffle the deck between hands. Ever seen a 1,000,000-card deck? It's as big as a house!

Come on! Why do you care? Just fold and go play Parcheesi or something.

I didn't need me no fancy equations to figure this one.

[/ QUOTE ]

we've got another einstein on our hands!

[cheapsuit]

well, i dont think infinite betting would be right because of the same reason that inifinte betting in skalnsky's question would be wrong even if youre getting 99.9999% equity. at some point, you have to consider that youre going to be beat and you dont want to lose an infinite amount of money even if the odds are in your favor

as for the difference between 110 bets and 111, i dont really know. it seems like such a silly thing to think about but its definiteyl a question that has weight when it comes to poker theory. i am going to lay low and not make any more speculations and in the meantime a math-buddy of mine is working up some calculus for the KK vs random question by me and the one sklansky posed. so ill let you all know once he gets down figuring anything out