I think what I am about to ask is within the ability of today's computers.

One Card Poker. Dealt from a 100 card deck with a number from one to a hundred on each card. 100 is the best hand.

A five handed game with a one and two-dollar blind. No rake.

To come in you must put in precisely fifteen dollars (fourteen more if you are the small blind. Thirteen more if you are the big blind and didn't' get a walk.)

No raises. If someone already bet, you can only call.

If you understand the rules, you know that the first player has a simple decision, but the second player's decision depends on whether the first player played. The third player faces four possible scenarios and the small blind faces eight. The big blind faces 16

With what hands do you enter the pot under each scenario? For example, you could say "if the first and third player enter the pot the big blind needs above x to call.

If you can get these answers, it should be easy to answer the question as to what percentage of hands are walks and what percentage does a bet take it down without a showdown. For extra credit tell me the EV of each position.

This is a version of the [0, 1] toy game.

Edit - I guess that doesn't really apply here because that math is for an IP river bet with a polar range, but David is asking about a linear push-fold strategy

While waiting for someone to find the general solution, how bout some guesses as to what percentage of hands would have a showdown between two or more players. And a second guess if there was a three dollar rake. And a third guess if there was no rake but a 60 cent ante per player.

I can try and pseudo solve this using some game theory principles.

Firstly, showdown values are pure preflop which would make ranges tighter than in holdem since there isn't variability with post flop play.

A $15 RFI from HJ would be risking 15 to win 3, needing to succeed 15 / (15 + 3) = 83.3% of the time. Which means everyone must defend at least 16.6% of the time. The burden of defense is divided between the remaining 4 players for about 4.2% of the time. Perhaps the BB defends slightly more than the other players, but we're going to ignore that for simplicity. That means the players later to act should defend with a 96 or better. This automatically makes 96 and better a profitable hand for HJ to open with. Also note that HJ having those hands will block the later players from having them. BB's odds to call are 13/(13 + 15 + 3) = 41.9% so this is close to our bluff to value ratio. The best bluff hands don't matter, but in practice the higher the better so the next 4 cards down 95 to 92 are a solid choice. We could do better with a mixed strategy with folding some 92 combos, but I don't want to go that in depth. We can repeat this process for all the other player's RFI.

RFI>>>
HJ: 92+
CO: 91+
BU: 87+
SB: 69+

Overcalling is going to be the same as the value portion of the opener's range...

Calls Vs>>>
HJ: 96+
CO: 95+
BU: 92+
SB: 83+

Further overcalling is going to be tighter yet.

A bet should take the pot down 83.3% of the time. The number of walks is the inverse of the RFI's % multiplied together. 1 - ( .91 * .90 *.86 *.68) = 52%. EV of each position should be close to 0 if playing perfectly, with a slight disadvantage to the earlier positions and a slight profit to the SB and BB player.

How does the early positions have a disadvantage if always folding is an EV of 0?

I just assumed that the players in the blinds already have a chip out in front of them so they have a higher EV situation since they are more priced in. I could very much be wrong though. It's just a guess. If we are talking EV before blinds are posted it would actually be the other way around, where everyone else has a little more EV and the blinds are at a deficit.

I think I can (almost) perfectly solve the [0,1]-version of the game where each player receives uniformly at random and independently a value ("card") from the real interval [0,1]. Below please find a (near) Nash equilibrium of this game in which each player enters the game (by raising or calling) if their card is above a certain threshold that depends on the actions of the previous players. The thresholds are rounded so they are, of course, only near an equilibrium. Notice that the thresholds are relatively close (but not equal) to those given by TheGodson in the previous post.

PS: I solved this game by setting up a system of multivariate polynomial equations with a variable for every threshold and expressing that the players are indifferent between entering the game and folding at the thresholds. I ran this system through a numerical solver that quickly computes a solution when starting from a "reasonable" initial solution.

PPS: I can also do 6max, rake, or other raise / blind sizes.

PPPS: Of course, I cannot guarantee the absence of errors. If you wish to reference the strategy, please cite this post. Setting up the equation system and coding it required some effort.

PPPPS: Here's the solution:

HJ raises first in for cards >= 0.9177

CO raises first in for cards >= 0.8923
CO calls after [ raise ] for cards >= 0.9579

BTN raises first in for cards >= 0.8445
BTN calls after [ raise fold ] for cards >= 0.957
BTN calls after [ fold raise ] for cards >= 0.9446
BTN calls after [ raise call ] for cards >= 0.9768

SB raises first in for cards >= 0.6886
SB calls after [ raise fold fold ] for cards >= 0.9546
SB calls after [ fold raise fold ] for cards >= 0.9411
SB calls after [ raise call fold ] for cards >= 0.9759
SB calls after [ fold fold raise ] for cards >= 0.916
SB calls after [ raise fold call ] for cards >= 0.9755
SB calls after [ fold raise call ] for cards >= 0.9684
SB calls after [ raise call call ] for cards >= 0.9862

BB calls after [ raise fold fold fold ] for cards >= 0.9522
BB calls after [ fold raise fold fold ] for cards >= 0.9375
BB calls after [ raise call fold fold ] for cards >= 0.975
BB calls after [ fold fold raise fold ] for cards >= 0.9097
BB calls after [ raise fold call fold ] for cards >= 0.9746
BB calls after [ fold raise call fold ] for cards >= 0.9671
BB calls after [ raise call call fold ] for cards >= 0.9858
BB calls after [ fold fold fold raise ] for cards >= 0.8235
BB calls after [ raise fold fold call ] for cards >= 0.9738
BB calls after [ fold raise fold call ] for cards >= 0.966
BB calls after [ raise call fold call ] for cards >= 0.9855
BB calls after [ fold fold raise call ] for cards >= 0.9513
BB calls after [ raise fold call call ] for cards >= 0.9853
BB calls after [ fold raise call call ] for cards >= 0.981
BB calls after [ raise call call call ] for cards >= 0.9915

That seems right. Thank you. Could you now perhaps do the same calculation with one additional rule? There is still only one round of betting after the first card is dealt but when there is a showdown the active players get a second card which they add to their first one. Best total wins.

Also, I would like to see your solution when called pots are raked three dollars.

I did come up with similar values using fictitious self play. @joker I am curious about what kind of multivariate polynomial equations you solved - any reference on this approach?

I believe this game could, in principle, be solved in a similar fashion as the original game. However, the equation system will be significantly more complex. I'll give it a try but it might take me a while. Using heuristic methods might give an (approximate) answer more quickly.

Let me explain it for three players for simplicity. I use variable names r, rc, rcc, and rfc for the thresholds above which BTN raises, SB calls, BB calls after SB called, BB calls after SB folds, respectively. At equilibrium we'll have r<rc<rcc and r<rfc. Also players are indifferent between folding and entering the game at the thresholds. Consider, for example, player SB at threshold rc conditioned on BTN raising before. SB wins the pot if BTNs card is below rc and if BB folds. The success probability of calling is therefore (rc-r)/(1-r)*rcc, which must equal the pot odds 14/(14+15+1+2) to make SB indifferent. Similar considerations for the other thresholds give the below four equations (which can be turned into polynomial equations by multiplying with the denominators).

rc*rfc=15/(15+1+2)
(rc-r)/(1-r)*rcc=14/(14+15+1+2)
(rcc-r)/(1-r)*(rcc-rc)/(1-rc)=13/(13+15+15+2)
(rfc-r)/(1-r)=13/(13+15+1+2)

One can solve this numerically using free online solvers. The system has multiple solutions but only one with all thresholds in [0,1] and r<rc<rcc and r<rfc. This must be an equilibrium.

For five player I wrote a small script to generate the equation system automatically and solved it with an built-in solver.

I am not aware of a reference for this approach but it seems elementary to me.

Concerning your approach based on fictitious play: Is there an existing software or library for this that can be flexibly configured or did you program it yourself? This approach might be easier and quicker to get approximate solutions than the approach above.