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Wizard
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Wizard
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August 12th, 2012 at 9:54:08 AM permalink
Yes, if X has 0.6 that is better than average. However, maybe he is trying to sucker Y into calling more, enabling his bluffs to be more powerful. You have to consider that how X's strategy influences Y's behavior.


I would add that all points where the strategy changes should be on exactly borderline cases. That is the key to how I solved the problem. I plan to explain my solution tomorrow.

I also suspect that my answer of Y calling at 0.4 may better take advantage of non-optimal X strategies than other points between 0.1 and 0.7.
It's not whether you win or lose; it's whether or not you had a good bet.
MangoJ
MangoJ
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August 12th, 2012 at 10:21:20 AM permalink

I somehow agree that X's bluff serves him an advantage.
What I don't understand is: why does X bluff on 0-0.1, but not - say - on 0.5-0.6. If he would bluff with the higher "hand" wouldn't that give him an extra advantage when the bluff is called ?
buzzpaff
buzzpaff
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August 12th, 2012 at 12:24:26 PM permalink
I got x + 1/2 y Am I close ?
charliepatrick
charliepatrick
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August 12th, 2012 at 1:27:34 PM permalink
Quote: MangoJ


I somehow agree that X's bluff serves him an advantage.
What I don't understand is: why does X bluff on 0-0.1, but not - say - on 0.5-0.6. If he would bluff with the higher "hand" wouldn't that give him an extra advantage when the bluff is called ?

Case 1: Y is only going to call with .7 or more - in which case it doesn't matter what score you bluff with as if called it will lose. If you don't bluff your chances of winning are the score you have, therefore why not bluff with 0-0.1 where you had little chance of winning, and now have a .7 chance, rather than 0.5-0.6 where you were favourite to win.

Case 2: Y finds out you only bet with a hand greater than x (in your case x=.5). It is illogical for him to call with any hand less than x and technically should only call with hands greater than x+0.1 or .7. Thus this reverts to case 1 as any bluff loses if called.
dwheatley
dwheatley
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August 13th, 2012 at 11:42:16 AM permalink
Quote: weaselman

I don't understand. :( Need some help please!



Quote: weaselman


How is it you guys are saying the first player should check between .1 and .7?
Suppose, I have .6. If I check, my EV is 1.2. If I raise, and the other guy folds, it is 2. if he calls, it is 1.4. Either way, I get more than 1.2. Why would I not want to bet?


Your EV if Y calls isn't 1.4, it depends on his strategy. In this case, we can fix Y's strategy to be call greater than .4 to do the math. Then your EV is only 1/3. Even if Y changes his strategy to other optimal Y responses, the EV of raising with a mid hand is still too low.
Wisdom is the quality that keeps you out of situations where you would otherwise need it
Wizard
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Wizard
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August 13th, 2012 at 12:22:22 PM permalink

My method of solving this problem comes from The Mathematics of Poker by Bill Chen and Jerrod Ankenman.

The crux of their solution is to find indifference points where each player perceives the same expected value between two options. Letís look at this problem as an example.

Letís define some variables:

x1 = Point at which X is indifferent to bluffing and checking.
x2 = Point at which X is indifferent to checking and raising.
y1 = point at which Y is indifferent to folding and checking after a raise by X.

Take it on faith that this would be the strategy structure for both players. If you can't, buy the book mentioned above, or take up a college-level game theory class.

Next, we must decide which number will be greater, x1 or y1. It seems obvious to me that y1 will be greater. If it isn't to you, I suppose you could work out the problem both way, and see which results in a greater expected value for player X.

Now we're ready to get our hands dirty with some math. Let's start by finding an equation for x1. We do this by examing the possible outcomes for all possible values of Y, and how much X will win whether he bluffs or checks with exactly x1 points.


Y Bluff Check Difference Exp. Val.
1 to y1 -2 -1 -1 y1-1
y1 to x1 1 -1 2 2y1-2x1
x1 to 0 1 1 0 0



The expected value column is the product of the probability that Y will have a value in the stated range and how much X stands to gain to bluff if Y is in that range.

The sum of the expected value column is 3y1 - 2x1 -1. We set this equal to zero because at exactly x1 player X should be indifferent to bluffing and checking. So, 3y1 - 2x1 -1 = 0.

Next, do the same thing for x2.

Y Bluff Check Difference Exp Val
1 to x2 -2 -1 -1 x2-1
x2 to y1 2 1 1 x2-y1
y1 to 0 1 1 0 0


For x2 the expected value equation is 2x2 -y1 -1 = 0.

Next, do the same thing for y1, but consider every possible number of points for player X.

Y Bluff Check Difference Exp Val
1 to x2 -2 -1 -1 x2-1
x2 to y1 -1 -1 0 0
y1 to x1 1 1 0 0
x1 to 0 2 -1 3 3x1


For y1, the expected value equation is 3x1 + x2 -1 = 0.

So, now we have three equations and three unknowns. We all know from algebra class that is all we need to solve the problem! I'm going to assume the reader knows how to basic matrix algebra. So I'll skip all that and jump to the solution which is:

x1=0.1
x2=0.7
y1=0.4
It's not whether you win or lose; it's whether or not you had a good bet.
weaselman
weaselman
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August 13th, 2012 at 3:48:59 PM permalink
It still does not make too much intuitive sense to me, I am afraid, so I wrote a simulation to convince myself ...
It did not work. :(

When X and Y play according to Wizard's strategy, I am getting EV of -0.005 and +0.005 for them respectively. First of all, X can definitely do better than this by never betting at all. Now, changing Y's raising threshold to 0.5, makes the EV -/+ 0.035, which seems to suggest that raising between 0.4 and 0.5 is not optimal play for Y. Moreover, raising threshold to 0.7 does even better for Y at 0.095.

What am I missing here? For what it's worth, here is the simulation (you can play around with $xbet and $ybet settings to adjust the strategies. everything else is, I think, pretty straightforward, but I can explain if needed).

#!/usr/bin/perl
while(1)
{
my ($x, $y) = (rand(), rand());

my $xbet = ($x < 0.1 || $x > 0.7) ? 1 : 0;
my $ybet = $xbet && ($y >= 0.4) ? 1 : 0;

my $val = $x > $y ? 1 : -1;
if($xbet)
{
$val *= 2 if($ybet || $val < 0);
}

$xwin += $val;
$ywin -= $val;
$N++;

printf ("%8d: X: %6.3f, Y: %6.3f\n", $N, $xwin/$N, $ywin/$N) unless $N % 1000000;
}
"When two people always agree one of them is unnecessary"
Wizard
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Wizard
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August 13th, 2012 at 4:13:05 PM permalink
Quote: weaselman

When X and Y play according to Wizard's strategy, I am getting EV of -0.005 and +0.005 for them respectively.



I think there is something wrong with your program. At least three people have found strategies where the EV of X is 0.1, including my strategy.
It's not whether you win or lose; it's whether or not you had a good bet.
charliepatrick
charliepatrick
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August 13th, 2012 at 4:45:56 PM permalink
Quote: Wizard

...strategies where the EV of X is 0.1. ....

I agree with the .1. What is interesting about your solution is that while X has an optimum (initial) strategy, which means Y actually has a range of replies, the answer you give for Y prevents X from adopting a better strategy knowing Y's answer.
One way or working it out (cheating) is to create a spreadsheet where you put X's value (use .0005 to represent 0-0.001 etc.) and then decide whether he bets or not. No decide how often Y calls/folds factored in. Then you can play around and vary the numbers.
What turns out interesting is Y's optimum strategy prevents X gaining too much. For instance if Y calls too often then you're better off not bluffing and raising with lower hands, and if Y rarely calls then you bluff more, both which give X a better EV.
Bluff Bet Bet ? Y Folds Y Calls X Bets / Wins X Bets / Loses X Checks / Wins X Checks / Loses
Values: 0.1 0.7 0.4 Totals: 430 -210 240 -360 100
.000 500 1 0.4 1.2 0.00 0.00
.001 500 1 0.4 1.2 0.00 0.00

.099 500 1 0.4 1.2 0.00 0.00
.100 500 0 0 0 0.10 0.90
.101 500 0 0 0 0.10 0.90

.698 500 0 0 0 0.70 0.30
.699 500 0 0 0 0.70 0.30
.700 500 1 1.001 0.599 0.00 0.00
.701 500 1 1.003 0.597 0.00 0.00

.997 500 1 1.595 0.005 0.00 0.00
.998 500 1 1.597 0.003 0.00 0.00
.999 500 1 1.599 0.001 0.00 0.00

I guess it's the same as the dealer hits 16 and stands on 17; otherwise players might adopt a better, and more profitable, strategy.
weaselman
weaselman
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August 13th, 2012 at 5:38:41 PM permalink
Quote: Wizard

I think there is something wrong with your program.


Indeed, I found the error :)
The fixed version, shows X's expectation 0.1 with your strategy.

#!/usr/bin/perl
while(1)
{
my ($x, $y) = (rand(), rand());

my $xbet = ($x < 0.1 || $x > 0.7) ? 1 : 0;
#my $xbet = $x > 0.8;
my $ybet = $xbet && ($y >= 0.4) ? 1 : 0;

my $val = $x > $y ? 1 : -1;
if($xbet)
{
$val *= 2 if($ybet || $val < 0);
$val = 1 unless $ybet;
}

$xwin += $val;
$ywin -= $val;
$N++;

printf ("%8d: X: %6.3f, Y: %6.3f\n", $N, $xwin/$N, $ywin/$N) unless $N % 1000000;
}
"When two people always agree one of them is unnecessary"

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