Hi, anyone knows where can I find the proof of the Von Neumann poker model?
(not the solution)
The game is [0,1] deck with 2 players. Players 1 c/c or c/f, player 2 bets or checks

Also, is there any paper out there about draw games. Like a [0,1] where the 1st event is assign hand values x, y in [0,1] randomly (uniform distribution) for players X and Y, then a phase where they decide to pat or change the card, and finally a ronund of betting

Thanks in advance

For your first question, Mathematics of Poker. For your 2nd, I dunno. I feel like it should be solvable with pencil and paper but it's not my specialty.

interesting toy game. I'll think about it soon.

Yes I think is the game in page 115. Ferguson says in his book that Von Neumann found all solutions (all nash equilibriums) of the game but using the fixed bluff region as a certain bottom subset of hands is the most reasonable (easy to use or best of all equilibrium points?). Im curious about all other solutions. I tried to take a look at the Von Neumann's book, but the math notation is very hard. Seems like there's some functional analysis. It'll take me a while to read it and find where's the game discussed
In the mathematics of poker, it looks like the authors are looking for a continuous way to reach the equilibrium by expanding the valuebetting and bluffing subintervals and the calling range untill it reaches equilibrium

Im going to train my mind seeing if for a certain union of arbitrary disjoint intervals (a reasonble measured set in [0,1]) for the bettor, exist a unique calling range for the caller.

Thanks for the replies

here's the Von Neumann model with betting B(x) in the positive real numbers, where x is a random hand in the [0,1]
http://www2.denizyuret.com/ref/newman/newman59.pdf
what I dont understand is why he decides in proof I that for a certain bet B there are exactly 2 hands x1 and x2 that should be B(x1) = B(x2) = B

A way to understand why every hands have an optimal betting value is studing the [AKQJT] vs [AK432] game, with 1st player checking and the other player that can check back or bet; if bet, 1st player can only call or fold; every player have is own deck, so AvA have the same probability of Av3.

There are only few case to analyse: 2nd player betting the same ammount with [A,K and some bluffs]. Or 2nd player betting a certain ammount with [A and some bluff], and another ammount with [K and some bluff].
For every sub-case, you have to find the bluffing frequencies*, the calling range for first player, and the bet that maximize the EV.

If we set the EV of folding to 0 and EV checking to 0.32, the result should be 0.374 with only a betting range while 0.391 with two betting range. Now increasing the number of bucket (eventually approaching the [0-1] distribuition) there are situation in which betting a different ammount with every hands is optimal**.


*= it is possible to do the calculation without expliciting the bluffing frequencies, it depends on what is the reference system
**= side question, exist situation in which you should bet the same ammount on the last street/without re-raising? (apart from being limited by rules, or not having enough stack so all-in is the only solution)

carlop did you find these results to the [AKQJT] vs [AK432] game yourself or are they published somewhere?

Yeah, where do I find this [AKQJT] game analysis?

I solved the game by hand + few query of wolframalpha for the maximization, but I have choose the game ad hos so there were less complication possible.

For example, this is a subgame: IP bet an ammount b with A + some bluff, OOP call with 100% A and y combo in T-K (y can be greater than 1)
indifference of bluffing for IP
(-b) (1+y) + (+1) (4-y) = 0
-b +4 = y(b+1)
y = (4-b)/(b+1)
equity for IP (bluffing have 0 equity, so only A equity is needed)
EV = (1 * 0.5 + y * (1+b) + (4-y) * 1)/5
EV = (1 * 0.5 + (4-b)/(b+1) * (1+b) + (4-(4-b)/(b+1)) * 1)/5
Then using: http://www.wolframalpha.com/input/?i...%29+*+1%29%2F5, the maximum appear to be with a betsize of 1.23 (where the pot is worth 1 unit), and with every A IP player win 1.205 (since the bluffing increase the value of the top of the range)

Then the other two cases, summing up, comparing the two possible solution ([AK bluff] or [A bluff] + [K bluff]) and you have the result.


This method can be used only for small instance, a better approach is programming a generic solver, something that does all the math above automatically. I have it almost done, but seems hard to find any use from it (a part from solving simple half-street games or full-street games with predefined betsize).

For OP problem, for solving the draw version, a step needed is solving [0-x] vs [0-y] half-street games.

The tree of your half-street drawing game is like this:
1st player, if hand<threshold1 stay, otherwise draw
---if stay: 2nd player if hand<threshold2 stay, otherwise draw
------if stay: the game is [0-threshold1] vs [0-treeshold2]
------if draw: the game is [0-threshold1] vs [0-1]
---if draw: 2nd player, if hand<threshold3 stay, otherwise draw
------if stay: the game is [0-1] vs [0-threshold3]
------if draw: the game is [0-1] vs [0-1]

The last game is known, the second one have the same betting structure of the [0-1]vs[0-1] (every hands in 2nd player range worst than a bluff have 0 ev). The other two games need to be solved in a general way.

Solving this drawing game can be doable, with few (?) hours of work, but a prerequisite is being able to solve the [0-1] vs [0-1] game (and I feel this can't be done without variational calculus, I have to check Mathematics of Poker).

Pat bluffing doesnt exist? Or you bluff with bottom of pat range maybe?

u using the free version of wolframalpha? I have big formulas to maximize/minimize :P

IP bluff with the bottom of his pat range.
In real poker things can be more complicated, but here exist an order of the hands (that doesn't depend on other player range/different way to hitting flot/etc). So in this game exist a threshold, you pat everything better and draw with anything worse.

And yes, the free online version works good for most of the situation but it have some limit on the size of the input.