Anyone want to take a crack at it? (Assuming it hasn't already been done.)

https://www.twoplustwo.com/magazine/...r-toy-game.php

Aww bless, you really like these little puzzles don't you Sklansky

Looks like a variant of Nash-Shapley poker. I shall endeavor to provide the solution, plus I was planning to do so anyway. Simply a fantastic article btw!




Should be solvable for any n-person game.

Why only 9 players?

Wait, you want THE SOLUTION to n-person pre-flop Hold’em poker?

I will only provide the answers up to 9 players, with 4 significant digits. More precision may give away the solution, which I have not found published, and do not wish to publish if I am successful.

Yeah, I ninja edited that, jg.

Oh, no raises allowed. <whew!>

First guess, on a napkin....

Let r be the rank of the highest previous limper from the ranks [1,...n]

Any player should limp with hand x: (0,1) where x >= r/(r+1)

Oops, the blind...

So now it should read where x >= r+1/(r+2)

Test for player 9 after 7 limpers (player 1 was forced in)

7+1/(7+2)=8/9=pot odds offered to player 9 player if all had limped.

Test for player 8 after 3 previous limpers

3+1/(3+2)=4/5=pot odds offered to player 8 versus 4 players.

So much for napkins....

Lets try it this way.

Let n be the number of previous limpers already in the pot...

Player 1 is the blind, forced in and not counted as a limper.

Hands are x: (0,1) where 1 is the nuts

Any player requires a hand such that x >= 1/(2+n) to limp

+EV vs folding (no raises allowed) for any limper should be (n+1)x

No guarantee the above is correct. Can we get a “yay or nay” from anyone? I may have precisely missed the point of the gto solution, since there is no value in “limp bluffing”.

-Rob

Also, the nature of the infinite game played by infinite participants is quite informative, imho.

I left out the antes. Probably belongs next to n in the above solution. n(amount of ante)

If n is only dependent on previous limpers, then that cannot be correct. Take the 3 player example from the article.

B's calling situation is not solely based on whether no callers have entered the pot - he has to assume that C will enter the pot some of the time and thus cannot call as often as if A had been the only other player.

Some how I want to say Bernoulli trials and by proxy the binomial dostribution would be helpful with this. You could model each player as calling success and folding failure. Not sure though would have to put more thought into it.

Ahh, yes. Are we to assume how many players are seated at the table? This will be part of the solution, and if all have antes it gets quite interesting!

You can use the notation you were using for an in determinant amount of callers.

Each caller is dependent on how many callers in front of them and how many potential callers behind them because they all contribute to how often they have to win.

Basically player n can have any number of calling scenarios where the number of callers varies from 0-(n-1).

Luckily if player n doesn't have n-1 callers in front of her, her calling decision reduces to the same decision as a player in the spot (n-(n-(k+1))) where k is the number of callers in front of player n.

So for example say 9 players total with 6 callers in front. Player 9's calling decision is the same as player (9-(9-(6+1))) = 9-2 = 7 when all players have called before. Yeah I think that should be right.

Almost seems like there is a series construction in there, but not sure how you account for callers behind.

Potential callers and their antes must be predetermined, so the final solution will be a function of the number of players seated.

True. Sorry all over the place today.

The only real dislike I have is that antes are a reason to raise, but the toy game forbids this, but will serve to illustrate the value of raising I suppose.

As a matter of fact, we should only solve it with the addendum that illustrates how badly the EV of a limper is affected by a potential raise from a player ahead.

Why not start with the simplest case? One blind. No ante. Three players. If the hands goes from zero to one then player B's minimum hand is x and player C's minimum hand (when B plays) is y, then:

xy =1/2 and y[(y-x)(1-x)] = 1/3

I have used f_i to denote the minimum calling hand of the ith player, A is sum of the antes. Skipping some steps because typing this out is tedious. Basically I just setup the system of equations and solved by substitution.

f₂f₁ = 1/(A+2)
f₁ = 1/(f₂(A+2))
0 = f₂(f₂-f₁)/(1-f₁)(A+3)-1
0 = f₂³(A+2)(A+3)-f₂(2A+5)+1
There are not always nice solutions when A != 0. To get exact solutions when the equation doesn't factor nicely we'd need to use the cubic formula.
With no antes there are nice solutions.
f₂ = (3+sqrt(3))/6
f₁ = (3-sqrt(3))/2

For more players it quickly gets messy, as the calling frequency of each player will depend on which players have already called. For example a fourth player has different calling frequencies depending on whether two alone has called, three alone has called or both two and three have called. The equations get too complicated to be worth attempting to solve by hand. I'd just use a computer if approximate solutions are wanted for more players.

EDIT: BTW, I have not checked my partial solution for anything other than A = 0. Feel free to check and point out if it's in error.

I think it will suffice if we arrive at an expression of the calling hand, which accounts for a known number of players behind yet to act, but also takes into account the limps that have already happened, most importantly the most recent limp and how many players were behind that player when that player limped.

So, the pot size is total limpers +1

The hand to beat, is the last limp hand, taking into account the number of players behind THAT player at that moment.

Finally, we need to increase on that previous best hand by just enough to beat it, and still break even versus every player left behind us. The closer we are to the end of the line, then the closer we can be to the previous last best hand that called.

Presuming browni3141's equations are correct (they look good to me) for the 3-person case, here are the numerical solutions from solving the cubic equation with various antes.

Ante	Min Call for B	Min Call for C if B Calls
0.0	0.634	0.789
0.1	0.614	0.775
0.2	0.597	0.762
0.3	0.580	0.750
0.4	0.565	0.738
0.5	0.550	0.727
0.6	0.537	0.716
0.7	0.524	0.706
0.8	0.513	0.697
0.9	0.502	0.687
1.0	0.491	0.679

Not shown is the min call for C if player B folds. Clearly this is the standard simple pot odds calculation (which I think means the min call is given by 1/(3A+2)).

I hope these are correct but mistakes are always possible.

Let's denote the solution for 4 player game as GT4

We know the complete solution for the 3 player game..let it be GT3

Blind player is always P0

P1 acts first..p3 acts last in a 4 player game.

If P1 folds, game is reduced to 3 player game ( assuming no antes)

So if P1 folds, follow the solution of GT3

If P1 calls, we have a game.

Because P1 calls first to act..his calling strategy is pure. It can't be based on actions of other players (the math behind it is..but the final result is one number)

Let it be c1.

Calling strategy for p2 once P1 has called = c2

Calling strategy for p2 once P1 has folded = follow the solution GT3

Calling strategy for p3 is not pure.

In the branch of the decision tree that P1 calls, p2'so action affects p3's decision

P3 calls with c3a when p2 folds after P1 calls
P3 calls with c3b when p2 calls after P1 calls
P3 calls with the strategy given by GT3 when P1 folds

So

For P1

C1 * c2 * c3a = 1/2

For p2

C2 * c3b * (c2-c1) / (1-c1) = 1/3

For p3 when p2 folds

C3a * ( c3a -c1) /(1-c1) = 1/3

For p3 when p2 calls

C3b * (c3b-c1)/(1-c1) * (c3b-c2) /(1-c2) = 1/4

Some intuitive guesses

C3a < c2 ( p3 can play looser when p2 has folded even though he is getting the same price because no one is left to act behind him)

I fiddled around with these equations for the four-player game both analytically and numerically. I think I have found the solution as follows:

C1 = 0.71645219

C2 = 0.84053218

C3B = 0.90625195

C3A = 0.83028740

A very neat solution confirming the above intuitive guess.