Classic Type Game Theory Problem

[alThor]

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I came to the conclusion that the ideal strategy for A is to stand pat on all hands where 1.0 - B < A.

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That is part of the optimal strategy in some cases, when b > [sqrt(2)-1], which is around .4142. There, B always draws when A does.

But the interesting cases are when b < [sqrt(2)-1], where it turns out A stands pat when his hand exceeds [1 - sqrt(1-2b)]. This value is above b, so it does contain an element of bluffing. Furthermore, B randomizes between standing and drawing. Use the usual indifference conditions for these cases.

alThor

[JoshuaD]

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You have to find the optimal solution.

It doesn't matter wheter it is a repetitive game or a one time/occasional only.

BB King’s ... formerly known as the pokerplayer formerly known as Jack

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It does matter as to whether you should always just use the mathematically optimal solution if there is one or a mixed strategy with opponents who have previously seen how you play. This is similar to knowing in triple draw lowball the frequency with which your opponent will attempt to freeze you on a worse hand or stand pat on a bluff on the 2nd draw and bet out on the end.

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I'm pretty sure D.S is looking for the game-theory sound way of playing. You can always adjust from that basic strategy according to your opponents tendencies.

[Darryl_P]

Are you sure? As b approaches 0.5 from the right, Y approaches infinity. Isn't Y supposed to be <=1?

What if you simply use Y=(1-b)/b ?

With this condition, player A will just barely have to decide to never bluff -- the typical condition in these types of problems.

When A has the ideal bluffing hand (ie. just a tiny fraction less than B's hand), his EV from bluffing (ie. standing) is Y*b

From drawing his EV is 1-b so the idea is to make these equivalent, ie Y*b = (1-b), or Y=(1-b)/b

Or am I missing something?

Here's my take on it, show me where I'm wrong.

Terminology:

Avalue is Player A's card. Bvalue is player B's card.

Axioms:

First off- your odds of winnign a hand is 1-value. A 1 has no chance, a 0 has 100%, a .5 is even money.

Secondly- a randomly drawn card, assuming even distribution, has an average value over the long term of .5.
This means if a player draws, his hand basicly becomes a .5 (averaged over many many hands).



Strategy no bluffing:

Starting hand 1: A>B, B<.5. A should redraw. B should stand pat, as he is expected to win over A's .5. B wins 1-Bvalue percent of the time. Bvalue can be assumed to be the average value for the bottom half- .25. So B wins 75% of the time. This combo occurs .5*.75=.375 of the hands. So B gets .375*.75=.28125= $28.125 of expectation.

Starting hand 2: A<B, B<.5. A should stand pat. B sees the stand pat and redraws. A wins 1-Avalue percent of the time, B wins Avalue percent of the times. THis occurs .5*.25=.125 of the hands. B value's average hand is .25. A is less than him, so his average hand is .125. So B here gets .125*.125=.015625, or 1.5625 in expectation

Starting hand 3: B>.5. This is the tricky one. If A stands pat, B will redraw as he is beat. If A redraws, B has to redraw as A is now a .5. So B has to redraw either way. This puts him at .5 A will decide what to do based on his card. If he is <.5, he stands pat, if >.5, he redraws. When he redraws (50% of the time, or 25% of the total of the game), he has a .5 and thus a 0 expectation. When he is under .5, B will win 1-Avalue of the time. A here over the long run is .25 (middle of the range). This occurs .25 of the time. So B's expected value here is .25*.25=.0625=$6.25 in expected value.


In short, without bluffing B will win 35.9375% of the time. Huge advantage for A.


Adding in bluffing is more complicated, I'll try and post later. It may need to wait untila fter work.

Bleh, work is boring today. Besides, the more I thought about it, bluffing is irrelevant.

Ok, adding in bluffing.

If B>.5, he will draw. So his expected card value is .5. If A is <.5, drawing on a bluff will hinder him. If A>.5, standing pat on a bluff will hinder him. So A should not bluff if B>.5. Thats ok, we're already huge favorites here. Lets not get too greedy [img]/images/graemlins/smile.gif[/img]


If A<B and B<.5, A shouldn't bluff. A redraw will take A up to .5, and make B a heavy favorite. A already has the high advantage here, can't waste it. B won't redraw on the bluff anyway, as chances are A redrew and lost.


If A>B and B<.5, A shouldn't bluff. He won't get B to redraw, as chances are he'd redraw to a worse hand.


In short, unless you add a round of betting at the end, bluffing is useless in this game. So I refer to my previous analysis.

Umm, I just realised I played my hand for lowball, as the original post mentioned it. Reverse all the conditionals, but the math and logic work with the conditions reversed.

[Darryl_P]

I just calculated B's EV under this strategy and it turns out to be 0.4924 -- determined as follows:

If B's hand is less than .5, his EV will be .375 (avg. of .25 and .5, straightforward to show why)

If B's hand is some number b >=.5 his EV will be given by:

b^2 + (1-b)*((1-b)/b)*((1-b)/2)

If A's first card is less than b, he draws and will miss with a probability of b -- that's the first way B can win and that's the first term in the above.

Or if A's first card is more than b (probability of 1-b), then B draws with probability (1-b)/b and wins with probability (1-b)/2 -- that's the second term in the above.

Now that we know B's winning probability as a function of his hand b, we just need to integrate the function from 0.5 to 1 (which gives the area under the graph) and multiply by 2 to scale it up since our interval has length 0.5 and not 1.

Evaluating that integral produces 0.609814 if I didn't make any arithmetical mistakes. Seems reasonable, though, since it basically says B's expectation will be 0.609814 if he's dealt a card higher than 0.5. If bluffing were not allowed it would be 0.625 and so being slightly less seems reasonable.

The final answer of 0.4924 is just the average of 0.6098 and 0.375

[BB King's]

<font color="red">Are you sure? As b approaches 0.5 from the right, Y approaches infinity. Isn't Y supposed to be &lt;=1 </font>

You got a point there. Back to HQ !

<font color="red"> When A has the ideal bluffing hand (ie. just a tiny fraction less than B's hand), his EV from bluffing (ie. standing) is Y*b </font>

No, it is Y*x ... We have to calculate the situation when A is in marginal bluffing-position.

<font color="red"> From drawing his EV is 1-b so the idea is to make these equivalent, ie Y*b = (1-b), or Y=(1-b)/b </font>

I somewhat aggree ... if we substitute Y*b with Y*x !?!

<font color="red"> Or am I missing something? </font>

Maybe not - maybe a little - but not much.

Thanks !!!

[Xhad]

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In short, unless you add a round of betting at the end, bluffing is useless in this game. So I refer to my previous analysis.

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You are making a huge error in that you fail to realize that if A never "bluffs" by checking a loser, B should always draw no matter what his hand is whenever A stands pat. If B adopts this strategy, A can exploit it by intentionally checking if B has a pat hand if the probability of B making a worse hand is greater than the probability of A improving to beat B. I posted a more detailed analysis in the Poker Theory forum.

[gumpzilla]

Very interesting question. I haven't read the analysis of anybody else yet. I'll call the numbers of players A and B lowercase a and b, respectively. There are 2 antes in the pot, each player has paid 1.

B should never switch if A has switched and b &gt; .5, since then b will be likely to outperform A's second number. Likewise, if A has switched and b &lt; .5, B should always switch. So when A switches, B's strategy is easy.

It is when A stands pat that things get complicated for B, because this is the only area where A can "bluff." The simple strategy of switch whenever A stands pat fails to situations like a = .65, b = .7 where A benefits by tricking B into switching a winning number.

My guess is that most likely some kind of mixed strategy is going to be necessary here, and I don't have the experience to work out what's optimal in that regard. Let's assume that B adopts a slightly more complex strategy than above: B switches only when b &lt; .5 or when A stands pat and .5 &lt; b &lt; beta, where beta is some threshold value above which B always stands pat. If A knows B will employ this form of strategy, and happens to know beta, how does A proceed? We can break down into three regimes:

b &lt; .5: EV of 0 when A switches, since b will switch too, and EV of 2a - 1 when A stays, since b will switch. So, when b &lt; .5, A should stand pat if and only if a &gt; .5. Averaging over all a, A has an EV of .25 whenever b &lt; .5.

.5 &lt; b &lt; beta: B will stand pat if A switches, so switching nets A an EV of (1 - 2b)for switching and an EV of (2a - 1) for staying. 2a - 1 exceeds 1 - 2b for a &gt; 1 - b, so a should stand pat for all a &gt; 1 - b and switch for a less than that. Averaging this one is more complicated, so it's quite possible that I've made a mistake here, but I get that averaged over all a for a given b, a picks up an EV of (1 - b)^2 here. This analysis is only valid for b &gt; .5, so the best A can do over this range is .25, and the larger b gets the worse A does, which makes sense, so this looks plausible. We'll average over b at the end.

When b &gt; beta, A's choice is simple. B will never switch, so A's strategy becomes stand pat when winning, change when losing. The EV when switching is (1 - 2b), and the EV when standing pat is 1 (because A only stands pat when winning.) Averaged over all a, the EV in this range is (1 - 2b)b + (1-b) = 1 - 2b^2. As a sanity check, this gives EV of -1 for A when b = 1, as it should.

Okay, so now we know how A does on average for any given b. We need to average over all b for a given beta to figure out the value of the game for a particular strategy. I get (on the off chance that anybody has actually been following all of this thus far, this is another integral you probably want to check me on) that the average over all b for a given strategy is 1/6 + (beta^2)*(beta - 1).

At beta = 2/3, this EV gets pushed down to 1/54, which is the minimum. So for a fixed strategy, beta = 2/3 is the best B can do, and he's a slight loser there, just under 2 cents on the dollar by my calculation. I'm actually surprised he's not more of a loser. And mixed strategies are beyond me at the moment.