The [0-1] game is a toy game where each player draws a random hand strength between 0-1, (e.g. 0.125, 0.634, 0.879, etc). The highest hand wins. The range is uniform.

The SPR is 1.

The game tree is simple, IP can bet or check, OOP can call or fold:

XX
XBC
XBF

Here's the hard part - IP can bet any size between 0%-100% pot with any part of their range.

Is this game solvable?

Given OOP can't XR, IP doens't need to use the same betsize for deifferent value hands. the higer the equity(closer to 0) the bigger IP will bet. I think there will be infinite sizings used. maybe it's possible to find a function that gives optimal betsize for every equity.

You can find oop xc frequency given the bet size. Then you write down EV of betting size b with hand that has equity q. By finding maximum of ev function you'll find optimal sizing.

You can first try under the assumption that oop xc mdf.

My guess that solution will look something like bet pot with hands 0.85-1 then use different sizing for every hand between 0.5-0.85 and balance that with bluffs.

im not great at math so i'm not sure if you can 100% accurately solve for an infintite amount of betsizes and hand values.

You could definitely pragmatically solve it though if you abstract betsizings and hand values discretely.

No, the more equity you have the bigger you want to bet, so the solution would actually be to use infinite betsizes. Also OOP doesn't XC mdf because IP bluffs have some equity.

SPR is 1 so pot is biggest possible bet.

I didn't say it would defend mdf.

It should be possible to solve this with discrete values for ranges right? 0.1, 0.2, 0.3 etc.

If so, then hypothetically you could just solve smaller and smaller increments to approximate the continuous solution. Or am I missing something here?

right if u do it discretely then yea, i thought u meant with a math equation of some sort

Tombos21,

The answer is that the optimal betsize is full pot. You already know this.

Are you really wanting to discuss (0,1) poker theory?

I can’t tell whether or not you are trolling.

Yes you can definitely solve it. I don't know how but I would strongly suspect that one of the infinite amount of optimal solutions doesn't use any mixed frequencies and only uses full pot. Of course, it would be betting the weakest/strongest hands and checking hands near the middle (much like real poker!)

Something along the lines of always pot with .6 or higher and .2 or lower (so value:bluff ratio is 2:1), while always checking anything in between. I'm sure my numbers are off but I think one optimal strategy basically looks like that

My intuition says we wouldn't see mixed strategies with the continuous distribution in this game. But who knows. In your example, with respect to the defenders range, anything between .2-.6 becomes equally strong, so you'd think there would be a mixed strategy in there somewhere.

Honestly I'd love to play with a 0-1 solver. I think it could expand poker theory in a lot of ways. It's often hard to interpret solver outputs because there's so much noise and card removal and unstable equities etc etc. But a 0-1 solve would presumably give really clean outputs that could be used to understand the underlying game theory a bit better.

OOP can't raise so I think we would just see every digit above a certain threshold betting any amount it wants to bet for value, from pot to 1% and smaller. It wouldn't mix, it would just choose whatever sizing that digit wants.

Then bottom of range would mix bluff sizing frequencies. Not that it would matter since there are no blocker effects, but solver would probably mix each bluff digit as a default, even if some equilibriums could have pure betting strategies with the lower digits

This isn't true. Maybe you should be more careful to be accurate when calling someone out.

This game should be solvable by hand. IMO it's actually easier to solve when SPR is infinite, though.

Before doing math, IP will bet anything >.5 for value, and use all sizings from (0, 1], depending on hand strength.

I'm unsure but I don't think IP would bet anything >.5, I mean lets say we bet with .5000001 for example, then we lose money on the bet even if opponent calls 99.9% of their range with [.001-1.0]

Apologies, I did not intend to offend you.

Yes, it is solvable by hand. The solution you are thinking of doesn't work with finite stacks plus card removal using a deck of cards. So if you want the solution to be something you can implement in 'real' poker, that will resemble a solver, use the other solution provided in MoP. I think they called it a half street game.

Then you just bet small enough to get called by 99.999999999999999999999999999999999999999% of range

Tombos21,

Here is the solution of the [0,1] game you described.

It works for all bet sizes. It also discretizes nicely to a deck of cards, with removal and ties etc.


The range of the IP player is this:


0|---bluff--|a|--check--|b|--vbet--|1



The range of the OOP player is this:

0|-----fold----|c|----call----|1


There are three points of indifference, a, b, & c.

Bet size is variable, call it B.

The pot can be whatever, but making it 2 (one ante from each player) seems useful.

Using the principle of indifference, there is only one equation for each point (a, b, c).


At point |a|: 2c − B(1 − c) = 2a

At point |b|: 2b − c = 1

At point |c|: (B + 2)a − B(1 − b) = 0


The value of the game for IP is positive: V(IP) = B/(B+1)(B+4)

Technically speaking, OOP is always calling with a strategy that minimizes this. However, since the game favors IP his creates an inflection point. Taking the derivative and setting equal to zero yields an optimal bet size: B = 2

Here are some results for different bet sizes:

B = 0.5 raw EV=1.074074 net EV=0.074074 (~7.4bb per 100 hands)

B = 1 raw EV=1.1 net EV=0.1 (10bb per 100 hands)

B = 1.5 raw EV=1.1090909 net EV=0.10909 (just under 11bb per 100 hands)

B = 2 raw EV=1.1111 net EV=0.1111 (just over 11bb per 100 hands)

B = 3.236067977 raw EV=1.105572809 net EV=0.105572809 (10.5bb per 100 hands)

I found this paper which expresses the game analytically. Are you getting similar results?

I wonder how this could be extended to multistreet games.

Why do you assume it is optimal to use 1 betsizing instead of multiple (or infinite)

That is the best paper on (0,1) I have ever read. It is the recipe of how to construct all the possible extensions of the model as well.

As for multistreet games, this is modelling the very end of a hand of holdem. To get to multistreet we have to work backwards, toward earlier streets by adding in the real poker decisions that holdem offers.

Then, once working backwards far enough it is possible to marry the model with a tabular solution to preflop and flop. That would not be perfect GTO, since it is a subgame perfect tabular solution, plus a working model for the rest. Very close though.

All in my opinion, of course.

Having worked with these models before, it just becomes apparent that a model which chooses bet size based on dealt hand does not work with card removal.

For example, if the game is draw two cards and best hand wins, then you bet 100000000bb on both AA and your 0.5000001 combo. If I hold any Ax hand, I can call and print.

What about the model of real optimal poker, in which multiple betsizings is the norm on most spots, rather than the exception. Also no one talked about betting an infinite amount, but about having a strategy composed of infinite betsizings (which would be the case in optimal 0-1 game)

Real poker features asymmetric ranges (including polar ranges, capped ranges, etc.) which bring in all sorts of complications, including multiple bets sizes.

This game, in the most basic form presented here features only symmetric uniform ranges.

As for infinite bet sizes in this (0,1) game, you can bet in pennies into a pot of 100 bucks if you want. The betsize is determined by calculus which takes all the betsizes from 0 to infinity into account.

What im simply saying is that "pot is the best sizing" seems nonsensical to me, since theres no best sizing, all of them would be used from 0 to all in

All I can say, is that I showed several bet sizes and the corresponding EV's. The paper Tombos21 linked is pretty advanced stuff, but solving for betsize is the final step in every version of this game. Always with calculus, and always has a single answer (a saddle point).