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Old 12-04-2017, 06:25 PM   #1
David Sklansky
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Elaborating On My Multiplayer Toy Game Article

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

https://www.twoplustwo.com/magazine/...r-toy-game.php
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Old 12-05-2017, 08:41 AM   #2
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Re: Elaborating On My Multiplayer Toy Game Article

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

Last edited by Yadoula8; 12-05-2017 at 08:48 AM.
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Old 12-05-2017, 12:47 PM   #3
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Elaborating On My Multiplayer Toy Game Article

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!

Elaborating On My Multiplayer Toy Game ArticleElaborating On My Multiplayer Toy Game ArticleElaborating On My Multiplayer Toy Game ArticleElaborating On My Multiplayer Toy Game ArticleElaborating On My Multiplayer Toy Game ArticleElaborating On My Multiplayer Toy Game ArticleElaborating On My Multiplayer Toy Game Article


Should be solvable for any n-person game.

Last edited by robert_utk; 12-05-2017 at 12:56 PM.
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Old 12-05-2017, 01:03 PM   #4
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Re: Elaborating On My Multiplayer Toy Game Article

Quote:
Originally Posted by robert_utk View Post
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!

Elaborating On My Multiplayer Toy Game ArticleElaborating On My Multiplayer Toy Game ArticleElaborating On My Multiplayer Toy Game ArticleElaborating On My Multiplayer Toy Game ArticleElaborating On My Multiplayer Toy Game ArticleElaborating On My Multiplayer Toy Game ArticleElaborating On My Multiplayer Toy Game Article


Should be solvable for any n-person game.
Why only 9 players?
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Old 12-05-2017, 01:04 PM   #5
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Re: Elaborating On My Multiplayer Toy Game Article

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.
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Old 12-05-2017, 01:07 PM   #6
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Re: Elaborating On My Multiplayer Toy Game Article

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Originally Posted by just_grindin View Post
Why only 9 players?


Yeah, I ninja edited that, jg.
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Old 12-05-2017, 01:17 PM   #7
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Re: Elaborating On My Multiplayer Toy Game Article

Oh, no raises allowed. <whew!>
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Old 12-05-2017, 01:40 PM   #8
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Re: Elaborating On My Multiplayer Toy Game Article

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.

Last edited by robert_utk; 12-05-2017 at 02:06 PM. Reason: Using n* was misleading/confusing so “r” instead
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Old 12-05-2017, 05:47 PM   #9
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Re: Elaborating On My Multiplayer Toy Game Article

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
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Old 12-06-2017, 11:49 AM   #10
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Re: Elaborating On My Multiplayer Toy Game Article

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
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Old 12-06-2017, 11:52 AM   #11
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Re: Elaborating On My Multiplayer Toy Game Article

Also, the nature of the infinite game played by infinite participants is quite informative, imho.
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Old 12-06-2017, 01:35 PM   #12
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Re: Elaborating On My Multiplayer Toy Game Article

I left out the antes. Probably belongs next to n in the above solution. n(amount of ante)
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Old 12-06-2017, 02:10 PM   #13
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Re: Elaborating On My Multiplayer Toy Game Article

Quote:
Originally Posted by robert_utk View Post
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
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.
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Old 12-06-2017, 02:12 PM   #14
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Re: Elaborating On My Multiplayer Toy Game Article

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!
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Old 12-06-2017, 03:02 PM   #15
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Re: Elaborating On My Multiplayer Toy Game Article

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Originally Posted by robert_utk View Post
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.
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Old 12-06-2017, 03:07 PM   #16
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Re: Elaborating On My Multiplayer Toy Game Article

Potential callers and their antes must be predetermined, so the final solution will be a function of the number of players seated.
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Old 12-06-2017, 03:10 PM   #17
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Re: Elaborating On My Multiplayer Toy Game Article

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Originally Posted by robert_utk View Post
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.
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Old 12-06-2017, 04:23 PM   #18
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Re: Elaborating On My Multiplayer Toy Game Article

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.
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Old 12-06-2017, 04:44 PM   #19
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Re: Elaborating On My Multiplayer Toy Game Article

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.
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Old 12-06-2017, 10:16 PM   #20
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Re: Elaborating On My Multiplayer Toy Game Article

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
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Old 12-07-2017, 04:31 AM   #21
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Re: Elaborating On My Multiplayer Toy Game Article

I have used fi 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.

Last edited by browni3141; 12-07-2017 at 04:38 AM.
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Old 12-07-2017, 02:57 PM   #22
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Re: Elaborating On My Multiplayer Toy Game Article

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.
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Old 12-07-2017, 10:10 PM   #23
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Re: Elaborating On My Multiplayer Toy Game Article

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.

AnteMin Call for BMin Call for C if B Calls
0.00.6340.789
0.10.6140.775
0.20.5970.762
0.30.5800.750
0.40.5650.738
0.50.5500.727
0.60.5370.716
0.70.5240.706
0.80.5130.697
0.90.5020.687
1.00.4910.679

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.
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