I'm trying to solve a particular half-street Toy Game.

Two Players
Pot Size = 1
Hands uniformly distributed (0..1)
High hand wins
Stacks are infinite

Player A can raise any positive bet size or check
Player B can check or fold

I thought this would not be too tough game to solve and while I have made progress the algebra and calculus has gotten real ugly.

Anyway, I figured that the solution to this problem would have one two possibilities:

PROPOSITION A: Player A will always bet the same amount when he bets. In that case I'm pretty sure the optimal bet size is 1 (pot) and the solution is easy. He raises top 2/9 hands, bluffs bottom 1/9 and the caller calls top 4/9.

PROPOSITION B: Player A will raise an infinite number of different betsizes depending on his hand. As hands ---> 1 his bets ---> infinity. As hands ---> 0 his bets will also ---> infinity. For each betsize b there is a value-betting hand and a bluffing-hand.

My approach was to assume PROP B is true (which I think it is) and work from there. If it leads to contradction then obv. it's not true and can be dropped.

That leads to two strategy functions:
PLAYER A: Bet(h) takes in a hand (0..1) and returns a bet size (0 = check)
PLAYER B: MinCall(b) takes in a bet size and returns the worst calling hand.

Here are some of the things I've derived assuming PROP B is true. I'll leave off any proofs for now since they are kind of gory.

MinCall(b) has the form (b + k) / (b + 1) where k is a (as yet unkown) constant (experiments in excel show it is about 1/7).

Player A will bluff all hands in the range (0..k)
Player A will check all hands in the range (k..r) where r = (1 + k) / 2
Player A will value bet all hands in the range (r..1)

So I have Bet(h) partially definied:

Bet(h) for h in (0..k) is as yet unknown
Bet(h) for h in (k..r) = 0 (check)
Bet(h) for h in (r..1) = sqrt( (1-k) / [2(1-h)] ) - 1

My experiments in Excel also show that the EV for Player A is about .571 while the EV assuming constant bet sizing for Player A is 0.556 so that test looks reasonable. I don't need to know the bluffing part of Bet(h) to compute EV because it turns out the EVBluff(h,b) is just k and can not depend on h (the bluffing hand) or b (the size of the bluff). So what I did was let Excel minimize the EV function by goal seeking the unkown k to minimize the EV of Player A (since Player B gets to choose k). That lead to k being about 1/7 and EV being about .571.

I may will have made some grand mistake(s) early on in my reasoning and all of this may be built on quicksand.

I'd be interested if this toy game has been solved (or similar) or if anyone has any any insights or comments.

Thanks.

I believe this game was solved by Chen and Ankenman in MoP. See ex 14.3, p. 154. Cliff notes: the bettor bets a different amount for every hand value.

Ok... thanks, I'll take a look when I get a chance.

hmmm.... but then caller would know our hand and only call with better hands. I would think that each bet size would have to correspond to at least one good hand and one bad hand.

That seems fair to me.

It does. I was imprecisely trying to say that your Prop B was correct. And you are very much on the right track ... bluffing region = 1/7.

Oh cool.... maybe I won't look it up for a while.

Another thing I know (I think I know) but didn't state, because it's kind of hard to express.... there is a relationship between the slope of Bet(h) in the bluffing region and the slope of Bet(h) in the value betting region. This is to make the caller indifferent to calling a given bet when his hand in in between.

Bet'(ValueHand) = -Bet'(BluffHand) x B / (B + 1) where B = Bet(ValueHand) = Bet(BluffHand)

In other words when I bet B caller does not know if I have ValueHand or BluffHand and the probability that I hold either hand needs to be "just right" to make him indifferent to calling or folding middling hands.

CalllerEV = (B + 1) x Prob(BluffHand) - B x Prob(ValueHand) = 0

The relative slopes of Bet() function at ValueHand and BluffHand determine the odds of me holding ValueHand or BluffHand. That's kind of wierd in an of itself since for any bet B there are exactly two hands I can hold, yet the probabilites that I have either hand are not 1/2 each! That's very strange to me.

So basically if my Bet(h) is correct on the value betting side

BetValue(h) = sqrt((1 - k) / [2 (1- h)] ) - 1

then

BetBluff'(h) = BetValue'(h) x (BetValue(h) + 1) / BetValue(h)

but I haven't been able to solve that differential equation into BetBluff(h).

Probably there's a much easier way to look at this.

You are totally on the right track.

Thanks... that helps me alot to want to keep plugging away at this even though I'm at a bit of an impasse. I think I'll get it eventually.

Ok, I was finally able to solve for k by minimizing bettors ev.

My problem before was, I did not simplify the evraise expression enough and thought I could not integrate it (wolfram alpha couldn't which dissuaded me from trying very hard). But after further manipulation I was able to simplify enough to be able to integrate it.

0 --------- k --------------------- r ----------------------1
|---bluff--- | ------- check -------|-------- raise ---------|

c = (b+k)/(b+1) (minimum calling hand given bet size, found by ensuring all bluffs have the save ev = k)
b = sqrt((1-k)/(2(1-h))) – 1 (bet size in raising region, found by maximimizing the ev of a given hand in r..1)
r = (1 + k) / 2 (start of raising region)

evraise(h) = prob(fold) + prob(call) [-b + prob(win) (1 + 2b) ]
= c + (1-c) [-b + (h-c)/(1-c) (1 + 2b)]
= h (1 + 2b) – b(1 + c)
after substituting in c and b further manipulation leads to
= 3 – h – k – sqrt(8 (1-k)(1-h))

evbluff(h) = k
evcheck(h) = h
evraise(h) = 3 – h – k – sqrt(8 (1-k)(1-h))

ev bluff region = k^2
ev check region = (r-k)(r+k)/2 = 1/8 + 1/4k – 3/8k^2
ev raise region = integral [3 - h - k - sqrt(8 (1-k)(1-h) ) dh] from r to 1

ev raise region =
= 3h – (1/2)h^2 – kh + (2/3) sqrt(8) (1-k)^(1/2) (1-h)^(3/2) from r to 1
= 5/2 – k – (3/2)k – 3/2 + (1/8)(k^2+2k+1) + (1/2)k^2 + (1/2)k - (2/3) sqrt(8) (1-k)^(1/2) [(1-k)/2]^3/2
= 1 – 2k + (1/8)k^2 + (1/4)k + (1/8) + (1/2)k^2 - (2/3)(1-k)^2
= 1 – 2k + (1/8)k^2 + (1/4)k + (1/8) + (1/2)k^2 - (2/3)k^2 + (4/3)k - (2/3)
= 1 + (1/8) - (2/3) – 2k + (1/4)k + (4/3)k + (1/8)k^2 + (1/2)k^2 - (2/3)k^2
= 11/24 – (5/12)k - (1/24) k^2

ev total = ev bluff region + ev check region + ev raise region
= k^2 + [1/8 + 1/4k - 3/8k^2] + [11/24 - 5/12k – 1/24k^2]
= 7/12k^2 – 1/6k + 7/12

ev total’ = 7/6k – 1/6
7/6k – 1/6 = 0
k = 1/7

So I still have to try to find a Bet(h) in the bluffing region.

I was pondring why it comes out to 1/7. We have three regions with relative sizes 1|3|3. It’s easy to show that the check and raise regions are equal. My guess is there must be a way to show that the raising region is 3x the size of the bluffing region based on the relative slopes of bet(h) in those two regions. I have a feeling that may have been a simpler approach to finding k.

b = sqrt((1-k)/[2(1-h)]) – 1

since k = 1/7

b = sqrt(3/7)(1-h)^(-1/2) -1

so given any bet b, the corresponding value hand is

v = 1 – (3/7)(b + 1)^(-2)

dv/db = 6/(7 (b+1)^3)

and since (I think, not totally sure here) df/db = -dv/db (b)/(b+1) then

df/db = -(6 b)/(7 (b+1)^4)

f = (3b+1)/(7(b+1)^3)

I don’t know how to get the inverse of the above equation which would be needed to compute a bet from a hand in the bluffing region.

so....

better bets hand h is follows:
if h <= 1/7 he bets b such that (3b+1)/(7(b+1)^3) = h
if h is between 1/7 and 4/7 he checks
if h is >= 4/7 he bets sqrt(3/7)(1-h)^(-1/2) -1

caller calls a bet b when his hand is >= (b + 1/7) / (b + 1)

Here is a sideways graph with hands on the y-axis and bets on the x-axis. Bottom is bluffing region, top is value betting, middle is calling function.

graph here