Quote:
Originally Posted by abanger
Just checking in with your progress. Looking forward to putting these solutions to work
Well, it just so happened I was typing this up just now!
Finally, the complete set of twelve Ferguson indifference equations, non-simplified and usable for any size bets, raises, and check-raises.
These are verified and double-checked by the solutions given for limit poker at the end of the paper, as well as via Gauss-Jordan elimination method for solving the corresponding 12 equations in matrix form.
First, a few notes on potential pit-falls when working with equations of this type, then the equations themselves, and finally the values for half-pot bets and raises.
1) Absolutely use the variable P instead of the numerical value for the size of the pot when deriving the equations. If you don't you will mix up some pots with a number ratio that just happens to equal pot.
2) The wording used in section 2 for player 2 uses the complete expression for probability, in the form (P+B) with probability (x-y)/(1-y) etc. This is done only to make the sentence grammatically correct according to game theory, since every result player 2 achieves is in reaction to the dual purpose bets/raises of player 1. So, technically, every outcome of player 2 is a ratio of the differing purposes of player 1. None of this is required to solve the indifference equations, provided the relationships of the variables stay in the order given (which is required by GTO).
3) Absolutely remember that every raise also has a corresponding "call" simultaneous to the raise. If P1 bets 2, and P2 Raises 6, Player 2 is actually putting B+R = 2+6 into the pot.
example:
Player 1 bets, player 2 raises, player 1 calls the raise...
If Player 1 wins, he wins a net increase of P+B+R which is the pot plus the money Player 2 added to the pot (B+R).
However, if Player 1 loses, then he has a net decrease of -B-R which is the money he put into the pot and lost (The pot itself belongs to neither player and is never a loss)
4)When error checking with the limit-poker values at the end of the paper, the pot size is always 2 for all values computed in the table. The Bet/Raise sizes are what changes. Leftmost column is P=2,B=2,R=2. Next left-most column is P=2, B=.6666, R=.6666, next column is P=2, B=.3333, R=.3333.
5)As with (4), be careful when choosing a number for R, to take into account the correct method of computing a raise. A raise is a call, plus a raise. For half-pot bets and raises, both the bets and the raises should lay 3:1 odds for a player to call.
Example:
If Pot =2, then a half-pot bet of B=1 will give 3:1 odds on a call.
If Pot=2 and B=1, then a raise of R=2 gives the correct 3:1 odds to a call.
For the expression Pot+Bet+Call+Raise to give 3:1 odds, then it is required that P=2, B=1 (the call=B), and R=2
To call the raise at the end of this action is to risk 2 to win 6, 3:1 odds.
The equations below are the complete indifference equations from the paper:
UNIFORM(0,1) TWO-PERSON POKER MODELS
Chris Ferguson, TiltWare, Los Angeles
Tom Ferguson, Mathematics, UCLA
C´ephas Gawargy, Mathematics, Universit´e Paris 1
located at:
https://www.math.ucla.edu/~tom/papers/poker2.pdf
Section 4. Two Rounds of Betting With Sandbagging.
One complete street of (0,1) poker with bets, raises, and check-raises, the indifference equations, for any pot size, any bet size, and any raise size (raise must be equal or greater than bet):
Player 1 indifferent at a:
k = B/(P+B)
Player 1 indifferent at b:
(P+B)g - (P+B)h + (P+2B+R)j = B+R
Player 1 indifferent at c:
-(P+2B)h + (P+2B+R)j = R
Player 1 indifferent at d:
Bg – Bh - (P+B)k + (P+2B)m = 0
Player 1 indifferent at e:
(P+2B+R)m - (P+2B+R)k + Rn = R
Player 1 indifferent at f:
Bg – Bh – Rj + (B+R)k – Rm +Rn = 0
Player 2 indifferent at g:
(P+B)b – Bd + Bf – Pg = B
Player 2 indifferent at h:
(P+B)b - (P+2B)c – Bd + Bf +2Bh = B
Player 2 indifferent at j:
-(P+2B+R)b + (P+2B+R)c +Rf = R
Player 2 indifferent at k:
(P+B)a - (P+B)d + (P+2B+R)e - (B+R)f = 0
Player 2 indifferent at m:
(P+2B)d - (P+2B+R)e + Rf = 0
Player 2 indifferent at n:
e + f – 2n = 0
Solving the above 12 equations in 12 unknowns for half-pot limit, P=2, B=1, and R=2 yields these results:
a = 3/48 = .0625
b = 17/48 = .354
c = 18/48 = .375
d = 36/48 = .750
e = 39/48 = .8125
f = 45/48 = .9375
g = 6/48 = .125
h = 30/48 = .625
j = 36/48 = .750
k = 16/48 = .3333
m = 18/48 = .375
n = 42/48 = .875