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.

[emerson]

I believe the SB has the edge here because he has the choice of folding, calling, or raising. But the BB can't re-raise.

I believe the SB needs to raise any time he has the advantage, .5 or better. He has the advantage on this bet and takes any fold equity that the BB might gain by playing back.

When the SB raises, he offers the BB 3 to 1 to call. So he should include some bluffs. Bluffs with 1/4 the frequency of the raises would seem optimal, and he could do this with his worst hands. So we'll say that the SB raises with anything >=.5, or <.125.

The SB is getting 3 to 1, so should at least call with any hand >=.25.

So this has the SB raising with hands >.5 or <.125, folding hands >.125 <.25, and just calling with hands >.25 <.5.

BB now has few options. He should call all raises with hands >=.5, which would be zero ev. He can't gain by calling with worse hands, in anticipation of a bluff, because SB is bluffing optimally so the pot odds would not be correct.

What is left for the BB is betting $2 back at the SB when he just calls. If BB anticipates the strategy of the SB, he knows that the SB calling range is .25 to .5. The BB should raise with any hand greater than the median between these two points. So the BB raises with >=.375.

Now, some bluffing for the BB. The SB will be getting 3:1 to call, so the BB should also bluff 1/4 as often as he makes legitimate bets. He is betting for value with hands >=.375 up to 1. One fourth of this range lets him bet an additional .15625. So, with bluffs included, when the SB just calls, the BB will bet hands >=.34375
up 1.

BB just checks with hands <.34375.

Back to the SB after the BB has raised. The SB should now call with any hand >=.375, having anticipated the BB's strategy, and fold lesser hands.

I think this is the strategy, or something very close to it. I'll be suprised if the SB does not have the edge. It seems he should have the edge for the same reasons that he has the edge in heads up limit holdem.

Anyway, that's my guess.

When I have more time I'll put these frequencies on a spreadsheet and try to determine the edge for each with this strategy.

[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.