August 10th, 2012 at 4:08:30 PM
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Recently I purchased the book The Mathematics of Poker by Bill Chen and Jerrod Ankenman. It has a lot of math exercises of simplified versions of poker to try to get at the essence of the game. Here is a problem along those lines that I made up.

What is the optimal strategy for both players? What is the expected value for X under mutual optimal strategy?

If sufficient interest is show in this exercise, then problem 2 will be the same, but allow Y to bet after a check, after which X can call or fold.

- Player X and Y each ante $1.
- Both are given a random number uniformly distributed from 0 to 1. The higher number wins.
- Player X may bet $1 or check.
- If player X checks then Y is forced to check.
- If player X bets then Y may call or fold.

What is the optimal strategy for both players? What is the expected value for X under mutual optimal strategy?

If sufficient interest is show in this exercise, then problem 2 will be the same, but allow Y to bet after a check, after which X can call or fold.

It's not whether you win or lose; it's whether or not you had a good bet.

August 10th, 2012 at 5:03:30 PM
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I would imagine the best strategy for player X is to raise with a score higher than 0.5 and check anything else.

Then I would imagine the best thing for player Y is to call with anything .75 or higher, and fold the rest. Well assuming he knows that player X is playing with optimal strategy above. But then after throwing in the Ante's it probably changes the strategy right?

I'm too lazy to try to work out the math, since I'm sure I wouldn't get it right anyway.

Then I would imagine the best thing for player Y is to call with anything .75 or higher, and fold the rest. Well assuming he knows that player X is playing with optimal strategy above. But then after throwing in the Ante's it probably changes the strategy right?

I'm too lazy to try to work out the math, since I'm sure I wouldn't get it right anyway.

August 10th, 2012 at 5:10:46 PM
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I would guess off the top of my head that X should raise a percentage of the time equal to his number. If he got .73, he should raise 73% of the time. This should throw off Y enough to not be able to determine X's hand.

But actually, that sounds like X could raise with 0.01 and really no chance to win or not raise with 0.99 and really no chance to lose. So maybe X should raise a percentage of time equal to (Value+1)/2 min 0% max 100%.

But actually, that sounds like X could raise with 0.01 and really no chance to win or not raise with 0.99 and really no chance to lose. So maybe X should raise a percentage of time equal to (Value+1)/2 min 0% max 100%.

August 10th, 2012 at 6:24:22 PM
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I really like the idea of this game.. actually I have no experience with poker math at all, but if I get this right on this try, I might be interested in a purchase of the book.

My first ansatz for this game would be as followed (tried to make it as general as possible):

Player X will bet with probability b(x), where x is X's random number of distribution p(x) (check otherwise).

Player Y will call with probability c(y), where y is Y's random number of distribution p(y) (fold otherwise).

Both players know this strategy, especially each one knows both b(x) and c(y).

The expectation EVX for player X would be EVX = int dxdy p(x) p(y) [ $2 * b(x) * c(y) * (x>y) - $2 * b(x) * c(y) * (x<y) + $1 * b(x) * (1 - c(y)) + $1 * (1 - b(x)) * (x>y) - $1 * (1 - b(x)) * (x<y)].

The expectation EVY for player Y would be EVY = int dxdy p(x) p(y) [ $2 * b(x) * c(y) * (y>x) - $2 * b(x) * c(y) * (y<x) + $1 * (1 - b(x)) * (y>x) - $1 * (1 - b(x)) * (y<x) - $1 * b(x) * (1 - c(y)) ]

Here the symbol (a>b) is 1 if a > b, 0 otherwise. Hence (a>b) + (a<b) = 1.

The uniform probabilies p(x) and p(y) are constant, let's see how far I will get with this ansatz.

Player Y choses a strategy c(y) such that - for a given X's strategy b(x) - it will maximize EVY.

So the functional derivative must vanish (for every point y0)

0 = ðEVY / ðc(y0) = int dx p(x) p(y0) [ $2 * b(x) * (y0>x) - $2 * b(x) * (y0<x) + $1 * b(x)]

or 0 = $2 * int_0^y0 dx p(x) * b(x) - $2 * int_y0^1 dx p(x) * b(x) + $1 int dx p(x) * b(x)

Inserting this back into EVY(all c-linear terms vanish in the integral):

EVY = int dxdy p(x) p(y) ($1 * (1-b(x)) * ((y>x) - (y<x)) - $1 b(x) )

For any (consistent) strategy EVX + EVY = 0, so player X choses his strategy b(x) to maximize EVX:

EVX = - EVY = int dxdy p(x) p(y) ($1 * (1-b(x)) * ((y<x) - (y>x)) + $1 b(x) )

Maximum for

0 = ðEVX/ðb(x0) = int dy p(y) p(x0) $1 * ( (y>x0) - (y<x0) ) + $1 * p(x0)

Again inserting that back into EVX:

EVX = int dxdy p(x) p(y) $1 * ( (y<x) - (y>x) ) = 0.

Hm.... for mutual optimal strategy I get EVX = EVY = 0. Such a stragey would be to always check. Maybe I'm missing something (i.e. I didn't take into account that b(x) and c(y) are between 0 and 1.). I will try again tomorrow (or maybe someone can give a hint ...)

Edit:

Some thoughts about the exercise setting itself. I could guess that this exercise should serve an educational example on poker play. Information asymnetry would suggest that player Y holds a positional advantage (since he is to act last). However if Y systematically refuses to bet or raise on his position - he cannot use that advantage.

My first ansatz for this game would be as followed (tried to make it as general as possible):

Player X will bet with probability b(x), where x is X's random number of distribution p(x) (check otherwise).

Player Y will call with probability c(y), where y is Y's random number of distribution p(y) (fold otherwise).

Both players know this strategy, especially each one knows both b(x) and c(y).

The expectation EVX for player X would be EVX = int dxdy p(x) p(y) [ $2 * b(x) * c(y) * (x>y) - $2 * b(x) * c(y) * (x<y) + $1 * b(x) * (1 - c(y)) + $1 * (1 - b(x)) * (x>y) - $1 * (1 - b(x)) * (x<y)].

The expectation EVY for player Y would be EVY = int dxdy p(x) p(y) [ $2 * b(x) * c(y) * (y>x) - $2 * b(x) * c(y) * (y<x) + $1 * (1 - b(x)) * (y>x) - $1 * (1 - b(x)) * (y<x) - $1 * b(x) * (1 - c(y)) ]

Here the symbol (a>b) is 1 if a > b, 0 otherwise. Hence (a>b) + (a<b) = 1.

The uniform probabilies p(x) and p(y) are constant, let's see how far I will get with this ansatz.

Player Y choses a strategy c(y) such that - for a given X's strategy b(x) - it will maximize EVY.

So the functional derivative must vanish (for every point y0)

0 = ðEVY / ðc(y0) = int dx p(x) p(y0) [ $2 * b(x) * (y0>x) - $2 * b(x) * (y0<x) + $1 * b(x)]

or 0 = $2 * int_0^y0 dx p(x) * b(x) - $2 * int_y0^1 dx p(x) * b(x) + $1 int dx p(x) * b(x)

Inserting this back into EVY(all c-linear terms vanish in the integral):

EVY = int dxdy p(x) p(y) ($1 * (1-b(x)) * ((y>x) - (y<x)) - $1 b(x) )

For any (consistent) strategy EVX + EVY = 0, so player X choses his strategy b(x) to maximize EVX:

EVX = - EVY = int dxdy p(x) p(y) ($1 * (1-b(x)) * ((y<x) - (y>x)) + $1 b(x) )

Maximum for

0 = ðEVX/ðb(x0) = int dy p(y) p(x0) $1 * ( (y>x0) - (y<x0) ) + $1 * p(x0)

Again inserting that back into EVX:

EVX = int dxdy p(x) p(y) $1 * ( (y<x) - (y>x) ) = 0.

Hm.... for mutual optimal strategy I get EVX = EVY = 0. Such a stragey would be to always check. Maybe I'm missing something (i.e. I didn't take into account that b(x) and c(y) are between 0 and 1.). I will try again tomorrow (or maybe someone can give a hint ...)

Edit:

Some thoughts about the exercise setting itself. I could guess that this exercise should serve an educational example on poker play. Information asymnetry would suggest that player Y holds a positional advantage (since he is to act last). However if Y systematically refuses to bet or raise on his position - he cannot use that advantage.

August 10th, 2012 at 8:24:24 PM
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If X checks, he wins with probability equal to his number, x. The EV of this decision is 2x (the money in the pot is a sunk cost).

X should bet whenever the EV of betting is greater than 2x. This depends on Y's strategy. So we consider what Y does with his number y in response to a bet. Folding is 0 EV, calling is -1 + 3*p(winning). The p(winning) depends on y, and the conditional distribution of x given that X bets.

But it's bed time where I live, so maybe tomorrow.

X should bet whenever the EV of betting is greater than 2x. This depends on Y's strategy. So we consider what Y does with his number y in response to a bet. Folding is 0 EV, calling is -1 + 3*p(winning). The p(winning) depends on y, and the conditional distribution of x given that X bets.

But it's bed time where I live, so maybe tomorrow.

Wisdom is the quality that keeps you out of situations where you would otherwise need it

August 11th, 2012 at 8:41:26 AM
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MagnoJ at least gets points for the best effort thus far, but nobody has produced what I (perhaps incorrectly) think is the right answer.

As a hint, I don't think randomizing is necessary. That is more appropriate for games with discrete possibilities. For example I asked a similar question to this, but with a three-card deck earlier, which entailed a random strategy for the lowest card.

As a hint, I don't think randomizing is necessary. That is more appropriate for games with discrete possibilities. For example I asked a similar question to this, but with a three-card deck earlier, which entailed a random strategy for the lowest card.

It's not whether you win or lose; it's whether or not you had a good bet.

August 11th, 2012 at 9:12:48 AM
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I don't have time to even attempt to solve this, but here's a general hunch.

Because Player X will control the action (whether there's betting, or not) you'd obviously always want to bet with higher hands, but you'd also want to throw in enough very low hands that will almost always lose if you check.

EG: If you have .00, you're going to lose by checking. But if you bet, there's a chance you'll win if they fold, so you have more equity by betting, as long as Player X folds at least 33% of the time. (There is 2 dollars in the pot already from the antes, your dollar makes it 3%, and if they fold 1/3 of the time, then your EV is 1 dollar for betting, rather than 0 dollars for checking.) I don't have time to find what the proper ranges would be.

It also seems to me that "optimal strategy" would be impossible, since each players optimal strategy would be based off the other players current strategy. In other words, because player Y's optimal strategy is going to based off of what Player X's strategy is, it could change. (although I guess that's more accurate in actual Poker)

Because Player X will control the action (whether there's betting, or not) you'd obviously always want to bet with higher hands, but you'd also want to throw in enough very low hands that will almost always lose if you check.

EG: If you have .00, you're going to lose by checking. But if you bet, there's a chance you'll win if they fold, so you have more equity by betting, as long as Player X folds at least 33% of the time. (There is 2 dollars in the pot already from the antes, your dollar makes it 3%, and if they fold 1/3 of the time, then your EV is 1 dollar for betting, rather than 0 dollars for checking.) I don't have time to find what the proper ranges would be.

It also seems to me that "optimal strategy" would be impossible, since each players optimal strategy would be based off the other players current strategy. In other words, because player Y's optimal strategy is going to based off of what Player X's strategy is, it could change. (although I guess that's more accurate in actual Poker)

August 11th, 2012 at 9:20:45 AM
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My gut says that the first player should always raise.

Vote for Nobody 2016!

August 11th, 2012 at 11:29:19 AM
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I have a feeling it's wrong, but here is my guess:

Player X should bet with 0.0625 or higher, otherwise check.

Player Y should call with 0.25 or higher, otherwise fold.

Player Y should call with 0.25 or higher, otherwise fold.

August 11th, 2012 at 12:12:39 PM
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Quote:JBI have a feeling it's wrong, but here is my guess:

I disagree. Click the spoiler button for details.

If X checks on 0.0625 I show that the optimal strategy for Y would be to fold with 19/64 or less, assuming X raised.

It's not whether you win or lose; it's whether or not you had a good bet.