Classic Type Game Theory Problem

[Jerrod Ankenman]

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If y < .5 and A is better than .5, A should be pat and B should draw.

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This doesn't seem right. If y = .0001 & A has .0002, surely A would draw.

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Er, .0002 is not better than .5.

Jerrod

[BBB]

As was pointed out by someone else, I think you made a mistake in B's EV when pat. I fixed the your post, and found a strategy boundary (for a different reason) at y=0.586, if I figured correctly.

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Ok, now the interesting case: y > .5.

Now A should obviously call with hands that are better than y, since by doing so he locks up equity equal to his hand's value. He may or may not "pat-bluff."

If he does do so, however, since there is always a showdown, it's dominated for him to do so with any hands that are weaker than hands he would throw away and draw.

Hence A's strategy is:

(draw region) -> (patbluff region ?) -> y -> (valuepat region)

The only relevant threshold here is between the draw region and the patbluff region. Call this value x.

B doesn't have regions, only a mixed strategy of drawing and being pat to make A indifferent to patbluffing. Call B's % of hands that draw z. B will select this value such that A is indifferent to drawing or patbluffing at x.

At x, here are A's equities (in pots):

<draw> = (1-y)
<pat> = (z)(x)

1-y = zx, or
z = (1-y)/x (1)

Now let's consider A's play. A will choose x such that B is indifferent to being pat or drawing when A is pat.

B's EVs:
<draw>: (1-x)/2 (draw equity against the range [x,1])
&lt;pat&gt;: <font color="red">(</font>y-x<font color="red">)/(1-x)</font> (wins when A patbluffs, otherwise 0<font color="red">. Note that A's hand must be on the interval [x,1].</font>)

(1-x)/2 = <font color="red">(</font>y-x<font color="red">)/(1-x)</font>
<font color="red">(1-x)^2 = 2(y-x)
1-2x+x^2 = 2y-2x
1+x^2 = 2y
x^2 = 2y-1
x = sqrt(2y-1)</font>

Now there's a lower bound to the validity of this equation. For example, consider y = .55. Then x = <font color="red">.31</font>! <font color="red">This is clearly wrong; note that if A stands pat at .31, then if B stands pat, A loses, and if B draws, A has .31 equity, whereas if A draws, B will stand, and A has .45 equity. B is still indifferent, since he does not know A's card, but A has made a -EV decision in order to give B a neutral EV decision.</font>

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In order to avoid such situations, A must draw whenever A's holecard is &lt; 1-y. Setting y-1 equal to sqrt(2y-1) yields y = 2-sqrt(2), or about 0.586.

Thus, the cutoff x can only be sqrt(2y-1) when y &gt; 0.586. For lower values of y, a higher cutoff must be established, since A is obligated to draw when his holecard is &lt; 1-y.

So, for 0.5 &lt; y &lt; 0.586, B must draw if A stands pat, since A's patbluffing cutoff will be higher than what it would need to be to make B's decision EV neutral. Thus, A should stand pat with any card larger than 0.5 if 0.5 &lt; y &lt;0.586.

EDIT: Also, if 0.5 &lt; y &lt;0.586, A should stand pat with a holecard &gt; 1-y, since B will then draw, and A's equity will be his holecard value (A's equity will be 1-y if he draws, since B will then stand).

So, to summarize:

PLAYER A:
If y &lt; 0.5, stand pat if holecard is better than 0.5, draw if worse than 0.5
If 0.5 &lt; y &lt; 0.586, stand pat if holecard is better than y-1, draw if worse than y-1.
If y &gt; 0.586, stand pat if holecard is better than sqrt(2y-1), draw if worse than (2y-1)

PLAYER B:
If A draws, stand pat if y &gt; 0.5, draw if y &lt; 0.5
If A stands pat, draw if y &lt; 0.586, randomly draw z = (1-y)/sqrt(2y-1) of the time if y &gt; 0.586.

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If y &lt; .5 and A is better than .5, A should be pat and B should draw.

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This doesn't seem right. If y = .0001 &amp; A has .0002, surely A would draw.

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Er, .0002 is not better than .5.

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Doh. I misread that. I thought it said if A was better than y. :| My bad.

[BBB]

Oops, I meant:

So, to summarize:

PLAYER A:
If y &lt; 0.5, stand pat if holecard is better than 0.5, draw if worse than 0.5
If 0.5 &lt; y &lt; 0.586, stand pat if holecard is better than <font color="red">1-y</font>, draw if worse than <font color="red">1-y</font>.
If y &gt; 0.586, stand pat if holecard is better than sqrt(2y-1), draw if worse than (2y-1)

PLAYER B:
If A draws, stand pat if y &gt; 0.5, draw if y &lt; 0.5
If A stands pat, draw if y &lt; 0.586, randomly draw z = (1-y)/sqrt(2y-1) of the time if y &gt; 0.586.

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[Trantor]

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I just thought of this problem recently when a player mistakingly exposed his pat lowball hand too soon in a triple draw game against an all in opponent.

I realized that the situation rephrased in rigorous terms is a simply stated classic type game theory problem, perhaps never addressed before. Just in case that's so, we'll call it the Sklansky Exposed Pat Hand Problem. It goes like this:

Player A and Player B are both dealt a real number from zero to one. Higher number wins. No betting except for antes. Player A looks at his downcard and decides whether to keep it or replace it. If he replaces it he gets that second card face down. After Player A acts, Player B has the same option. And of course his decison will be based partially on what A did. But the thing about this game is that Player B's first card is face up. So B knows that A's decision to replace was based on what A saw.

If there is $100 in the pot and both players play perfectly what is the EV for both players? What is the optimum strategy?

I'm going to put this question in the Poker Theory, Probability, and Science Math and Philosophy Forums at the same time.

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A variation on old theme?

* Just for fun a gambler approaches you and asks you if you want to make an even bet with him. Sure you say, being a gambler yourself. Now he says he's going to pick two different random numbers from a distribution not known to you. He will then write the two numbers of two small pieces of paper and put each one in each of his enclosed hands. You then pick a hand and he will reveal the number in that hand. You then place your money on which hand you think contains the larger number. He will match your money, betting on the other hand, the numbers will then be revealed and whoever was right keeps the sum of the money. You only get one shot at this, and the bet is $10,000. The question is, is there a strategy to beat 50:50 odds and if so how?

http://www.azillionmonkeys.com/qed/hints.html

[BillChen]

I believe this is right. Also notice that the threshold is x = sqrt(2) - 1, where if y = 1-x, x stands pat at 0.414. The value of r shows up a lot.

Bill Chen

[Darryl_P]

If there is some nonzero probability that the distribution has finite bounds, then you can achieve positive EV by choosing an arbitrary number and deciding to go with the first number if it's higher, otherwise switching.

[BBB]

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PLAYER A:
If y &lt; 0.5, stand pat if holecard is better than 0.5, draw if worse than 0.5
If 0.5 &lt; y &lt; 0.586, stand pat if holecard is better than 1-y, draw if worse than 1-y.
If y &gt; 0.586, stand pat if holecard is better than sqrt(2y-1), draw if worse than (2y-1)

PLAYER B:
If A draws, stand pat if y &gt; 0.5, draw if y &lt; 0.5
If A stands pat, draw if y &lt; 0.586, randomly draw z = (1-y)/sqrt(2y-1) of the time if y &gt; 0.586.

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Darryl P posted in the SM&amp;P forum that Player A's EV is .5062 for this strategy. (I later calculated the same answer, so I believe he is correct.).

[punter11235]

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Player A and Player B are both dealt a real number from zero to one.

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This game doesnt exist. Because its impossible to deal real numbers out of all real numbers. The set of cards must be finite.

Best wishes

[David Sklansky]

"This game doesnt exist. Because its impossible to deal real numbers out of all real numbers. The set of cards must be finite."

For those who don't get it, he only means the above in an irrelvant technical way due to the way infinity is handled. If the question stipulated that the cards ranged from zero to one. in jumps of one billionth, and I asked for the answer to four decimal places, the answer would be just what was calculated.

Meanwhile one wonders why the point is brought up at all. Could it perhaps be a gut reaction to an inability to do the problem even if it was rephrased in a way to meet the objection?