Important, Simple, Hard, Game Theory Problem

[David Sklansky]

I'm pretty sure I could do this if my life depended on it but there are too many smart people on this forum who could save me the exertion. It may have already been done for all I know.

Two players are both dealt a real number from zero to one. There is a one dollar small blind and a live two dollar big blind. Only one raise, total, is allowed. Thus the small blind can fold, call one dollar, or raise to a total of four dollars. If he raises, the big blind can only call. If he calls, the big blind can raise two dollars more and the small blind can call or fold.

Firstly, who has the best of this?

Secondly, what's the optimum strategy for both players along with their EV.

[IcarusJam]

So just to specify, each player gets one card, with a decimal number on it? The player with the higher card wins?

[AaronBrown]

I may take a crack at this, but right now I'll post some pointers if someone else wants to do the work.

There are nine possible outcomes. I've listed them below. The payout afterwards, like (0,3) means small blind gets $0 and big blind gets $3. I measure starting after the blinds are posted, that is I don't subtract $1 from SB and $2 from BB to account for the blinds they've paid. Therefore, in each case the two outcomes add up to $3, the amount in the pot at the start.

SB folds (0, 3)

SB Calls, BB Calls, SB wins (3,0)

SB Calls BB Calls, BB wins (0,3)

SB Calls, BB Raises, SB Folds (-1, 4)

SB Calls, BB Raises, SB Calls, SB wins (5, -2)

SB Calls, BB Raises, SB Calls, BB wins (-3, 6)

SB Raises, BB Folds (3, 0)

SB Raises, BB Calls, SB wins (5, -2)

SB Raises, BB Calls, BB wins (-3, 6)

In each scenario, you have to work backwards. For example, if SB raises, BB needs at least a 25% chance of winning to call. So BB will call if her number is larger than the 25%-point of SB's raising range.

SB knows this so his raising range will be noncontiguous. It will look something like 0 to 0.1 and 0.7 to 1. BB will call a raise with any number from 0.7 to 1, because that's 75% of SB's raising range. Her average call will be with 0.85. If SB was bluffing (0 to 0.1) he loses, if his raise was honest (0.7 to 1) it's a 50/50 coin flip who wins. The reason SB bluffs with 0 to 0.1 instead of, say, 0.6 to 0.7, is he loses equally with both ranges, so he'd rather save the more valuable 0.6 to 0.7 for calling instead of raising.

If you lay out all the outcomes and work backwards, you can come up with a strategy that will look like:

For SB, you need the range of cards for which he will:

(a) Fold
(b) Call and fold to a raise
(c) Call and call a raise
(d) Raise

These ranges need not be contiguous, but they must cover the entire interval from 0 to 1 with no overlaps.

For BB you need the ranges where she will:

(a) Check after a call
(b) Raise after a call
(c) Fold to a raise
(d) Call a raise

The first two must cover the interval from 0 to 1 with no overlaps; and the second two must do the same. Here we know the shape of the ranges. The range (b) will look like 0 to x and 1 - 8*x/3 to 1, while (a) will be x to 1-8*x/3 for some x. (c) will be all numbers below some critical value y, (d) will be all values above y.

If you solve for x and y, assuming you know the four ranges in SB strategy, you can then adjust SB's strategy, knowing what x and y are.

This requires some careful accounting and algebra, but no advanced mathematics, nor conceptual breakthroughs.

[bxb]

I think that the answer will be more complicated. In some situations you will raise X% and fold (1-X)%.

[wax42]

[ QUOTE ]
I think that the answer will be more complicated. In some situations you will raise X% and fold (1-X)%.

[/ QUOTE ]This kind of strategy would be dominated by never folding the better hands and always folding the worse hands. So there must be optimal strategies that do not do this.

[AaronBrown]

[ QUOTE ]
I think that the answer will be more complicated. In some situations you will raise X% and fold (1-X)%.

[/ QUOTE ]This is true in general, but not in this case (as wax42 already said). Randomized or "mixed" strategies are more common when you have discrete probabilities. In this problem, you can set the ranges to get exactly the probabilities you want, and you can often improve the strategies by using that freedom. For example, if the strategy dictates SB folds 10% of the time, he should do it on numbers 0 to 0.1. There's no advantage to folding with higher numbers, and in some cases those higher numbers will be useful for calling and raising strategies.

If instead of getting a number from 0 to 1 each player got, say, a card from Two to Ace, you might need to randomize. For example, if you still wanted to fold 10% of the time, you'd fold all your Twos and 30% of your Threes.

General game theory problems can be very hard, even simplified poker games exceed the capacities of modern computers. But this particular problem is reasonably simple, it can be solved by hand (although not so simple that I've actually done it).

[PokrLikeItsProse]

[ QUOTE ]
So just to specify, each player gets one card, with a decimal number on it? The player with the higher card wins?

[/ QUOTE ]

I assume the rules are the same as the [0,1] game in The Mathematics of Poker, where the lowest number wins.

[JaredL]

[ QUOTE ]
[ QUOTE ]
So just to specify, each player gets one card, with a decimal number on it? The player with the higher card wins?

[/ QUOTE ]

I assume the rules are the same as the [0,1] game in The Mathematics of Poker, where the lowest number wins.

[/ QUOTE ]

It's the same either way. Do you see why?

[emerson]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
So just to specify, each player gets one card, with a decimal number on it? The player with the higher card wins?

[/ QUOTE ]

I assume the rules are the same as the [0,1] game in The Mathematics of Poker, where the lowest number wins.

[/ QUOTE ]

It's the same either way. Do you see why?

[/ QUOTE ]

The crack at it I took above assumes high number wins.

[jogsxyz]

This very simple game is non-trivial. To solve it the game must be broken into it's parts. I don't have the energy to solve at this time. I'll get you guys started.
Low values win. Makes the math easier.
Y is SB and X is BB.
Treat the 3 chips in the blind as free money. After solving the game readjust the values for SB and BB.

Part 1. Y may raise to 4 or fold. X may call or fold.
Part 2. Y may raise, complete the blind, or fold. If Y completes the blind, game over X has no options.
Part 3. Y completes the blind. X may raise or check. Y may call or fold. This is a variation of the Weideman Challenge. Only X's space is [0,1] and Y's space is [y_r,y_c], where Y completes the blind.

Part 1. Before you attempt this part solve for Y's bluffing frequency. Too hard to type in the steps to the solution on this site.

Y bluffs
(2/11)(f/(1-f))

f is the frequency of Y raises.

Someone else with more energy can do the rest. Use Excel.