Bill Chen Type Pure Math/Poker Problem
 

This will not help your poker game. And the question has probably been asked and answered somewhere out in math land. But just in case it hasn't, it seems that it is fundamental enough that someone should take a crack at it.

N players ante x dollars each. Each player is dealt a card with a real number between zero and one. Without seeing what anyone else does, each player must fold or bet y dollars. Best hand wins the pot. If everyone folds they all get their money back. In terms of n, x and y, how high a number do you need to be dealt to you, to bet the y bucks?

Re: Bill Chen Type Pure Math/Poker Problem
 

How about narrowing it down to x=1 and n<=10?

Re: Bill Chen Type Pure Math/Poker Problem
 

Everyone bets the same amount?

Re: Bill Chen Type Pure Math/Poker Problem
 

First: VOCABULARY
NP: Number of player
NL: Number of limpers
P: My position (1<P<=NP)
Z: Number of hand possibilities
V: The power of my hands (1 is the worse, Z is the best)
X: Ants
Y: A bet

F is the probability to be confronted to a better value for the people AFTER ME:
F = (1-((V/Z)^(NP-P))
This is the 1-the proba of having not a better hands than me
(the proba is my ranks divide by the number of total hands multiply N times, N = number of people after me)
G is the probability to be confronted to a better value for the people who has limped BEFORE ME
G = ((Z/(2*NP))*NL)/(Z) = (0.5*NL)/NP
It's 0.5 (the ration of hands better than the average) multiplied by the number of limpers and divided by the number of players.

The probability to loose: H=F+G
If H >= 1, you don't BET !!!!!!!!
The probability to loose money is H*(X+Y)
The probability to win is I=(1-H)
The probability to win money is I*(NP*X+NL*Y)
if WonMoney-LooseMoney>0, you need to bet [img]/images/graemlins/wink.gif[/img]

Some numeric application.
10 players, Y=5*X, 100 possible hands.

Case1.
I'm on the button, my hands ranks is 70, everybody folds.
F = (1-((70/100)^(10-8))=1-(0.7)^2=0.51
G = 0
H = F+G = 0.51
Loose money (0.51)*(X+Y) = 0.51*(6X)=3.06X
Won Money = (1-0.6) = 0.4*10X=4.0X
So your profit is +0.94X, you need to BET.

Case2.
I'm in position 5 with 75, 2 limpers.
F = (1-((75/100)^(10-5))=1-(0.75)^5=0.864
G = (0.5*2)/10 = 0.1
H = F+G = 0.964
Loose money H*(X+Y)=0.964*6X=5.784X
I = 0.044
Won money = 0.044*(10X+2Y)=0.044*20X=0.88X
You don't bet !!!!

Cas3: Like Cas2 but you have 92 ranked and 3 limpers
F = (1-((92/100)^(10-5))=1-(0.92)^5=0.34
H = (0.5*3)/10 = 0.15
H = F+G=0.49
Loose money: 0.49*6X=3.0X
Won money: 0.51*(10X+3Y)=0.51*25X=12.75X
YOU NEED TO BET [img]/images/graemlins/laugh.gif[/img]

Case4: You are first with 85
F = (1-((85/100)^(10-1))=1-(0.85)^9=0.80
G = 0
H=0.80
Loose money: 0.80*6X=4.8X
Won Money: 0.20*10X=2.0X
You don't BET

Case5: You are in position 9 with 52, 1 limpers
F=1-0.52=0.48
G=0.05
Loose money=0.53*6X=3.18X
Won money=0.47*(10X+Y)=0.47*15X=7.4X
So you BET

Re: Bill Chen Type Pure Math/Poker Problem
 

First of all, the game is symmetrical with respect to all
players, i.e., the optimal strategy that each player uses is
identical with each other and it looks like this: bet $y
when h>z(x,y,N) where z is the minimal hand to show a profit
which is obviously a function of x, y and N.

Let's assume that the higher number wins and that the
probability that anyone gets a hand less than r is just r
(or that each real number anyone is dealt is an independent
identically distributed random variable with uniform
distribution over (0,1] ).

For the real number z which is minimal for which you bet $y,
the chances that nobody else gets a hand better than z (and
hence won't play) is z^(N-1) since there are (N-1) other
hands and the chances that one is worse is z. Here, you
"pick up the antes" which nets you +$(N-1)x.

On the other hand, with a probability of 1-z^(N-1), someone
does pick up a hand better than z and since that person is
playing optimally and bets, you end up losing $(x+y).

Thus, you need the critical value of z to satisfy that the
expectation

z^(N-1)*[+(N-1)x] + (1-z^(N-1))*[-x-y] is zero.

(for a hand higher than z, then the expectation >0 since the
LHS is an increasing function in z with N>=2, x>0, y>0).

This reduces to

z^(N-1) = (x+y)/(Nx+y)

or z = [(x+y)/(Nx+y)]^(1/(N-1))


The more interesting question is this: what if position is
important, i.e., the first player to the left of the button
acts first, etc. It's clear that if everyone is playing
optimally, because of the "gap principle", the first player
doesn't quite need as strong a hand since the other players
"know" that it is very unlikely that they can win if they
hold a hand slightly better than the utg player's minimal
opening hand (as long as x,y and N are reasonable).

Re: Bill Chen Type Pure Math/Poker Problem
 

[ QUOTE ]
The more interesting question is this: what if position is important, i.e., the first player to the left of the button acts first, etc. It's clear that if everyone is playing optimally, because of the "gap principle", the first player doesn't quite need as strong a hand since the other players "know" that it is very unlikely that they can win if they hold a hand slightly better than the utg player's minimal opening hand (as long as x,y and N are reasonable).

[/ QUOTE ]

I haven't had a chance to check your math for the original problem, but I am pretty sure this intuition is wrong. While the fact that you cannot be reraised after coming in in early position allows you to play more hands, I still think you would want to be tighter than average because of information leakage (we're assuming everyone is playing optimally, so everyone knows everyone else's strategy).

If y is much larger than Nx, it is true UTG can often "steal the antes", but he will pay a significant price those times someone else has a stronger hand.

If y is small compared to Nx, then others players will simply be able to call with most numbers that are above UTGs minimum because of the good pot odds they will get.

Re: Bill Chen Type Pure Math/Poker Problem
 

Firstly, we can normalize the size of the ante to 1 since the only thing that matters is the ratio of bet size to ante size.

The first thing that we need to do is find a cutoff. This cutoff is the highest point where the EV(bet) = EV(fold). At this cutoff, we will get no value from others also betting with a worse hand (think of it like value betting), because it's the smallest hand that anyone will bet.

At this point, which I'll call z, the EV of betting assuming that others will bet if and only if their hand is better than z is:
n*z^(n-1) - y(1-z^(n-1))

The EV of folding is:
1*z^(n-1)

Setting them equal gives:
z = ((y+n-1)/y)^(1/(1-n))
or z = (y/(y+n-1))^(1/(n-1))

Again, my y here is different from your y, since my y is the ratio of bet size to ante size.

Re: Bill Chen Type Pure Math/Poker Problem
 

Right. Need the expectation of playing as greater than the
expectation of folding and the expectation of folding is
going to be (-x)(1 - z^(N-1)) and not zero.

Then, instead,

z = [ y/((N-1)x+y)]^[1/(N-1)].

Re: Bill Chen Type Pure Math/Poker Problem
 

Actually, bigpooch, I think you may be right about UTG betting in a positional game without reraising. He should bet a little bit more to either steal more blinds or get more value out of his premium hands (when others use optimal strategies, he will get enough of both effects to mitigate for the times he is called by a better hand). He will still, of course, lose EV as other players will no longer call UTG bets with their weakest hands, which others players bet in the original game primarily because of the chance to steal the antes (combined with the small chance they beat a weaker hand).

If there were reraising allowed, obviously UTG would have to tighten up considerably and would lose a lot more EV in the game.

Re: Bill Chen Type Pure Math/Poker Problem
 

Pooch,

When discussing position, how would it work? A key thing here is that if everyone folds then everyone gets their ante back. In the sequential game, would the last player get the antes, or would they have the option to bet, and if they do then the earlier players could call (in other words early players may check, they don't have to fold)?

Solving such a game for 3 or more players is a fairly tedious exercise because the strategy space is already pretty large - player 1 would have 4 possible decision points, 2 would hae 3 possible, 3 would have 4 possible.

However, in the 2 player case if player 1 checks, then 2 bets a larger number of hands than 1 would bet - both more bluffing hands and more value betting hands.