Game Theory Problem

[Patrick Sileo]

[ QUOTE ]
Intuitively, I think the strategies for part A and part B converge with time and that the fact of whether player 2 knows about the xray vision becomes unimportant.

[/ QUOTE ]

Since time does not explicitly appear in the game, any notion of convergence over time does not apply.

[jogsxyz]

Forgot. You must bluff occasionally,
else opp will be able to exploit you.

You bet x>=1/2 and fold x<1.
Opp knows you are only betting x>=1/2.
Then opp only needs to call y>2/3.
In the x>=1/2 and y>=1/2 quadrant, the
$EV for opp becomes
1>x>2/3 and 1>y>2/3 = 0
1/2=<x<2/3 and 1>y>2/3 = $3*1/18 = $1/6
1/2=<x<2/3 and 1/2=<y<2/3 = $-1*1/12 = -$1/12
Opp improves his $EV by $1/12 in this quadrant.
No effect on his $EV in the other sectors.
Opp game $EV becomes 1/4 + 1/12 or $1/3.

You can either bluff whenever x<1/4 or
when x<1/2 use some randomizing mechanism
to bluff half the time. This forces opp to
keep calling at y>1/2.

Correction. You pay $2 to see opp2's hand.
Opp2 sees a $6 pot. He can call for $2.
That's the same as a bet of half the size
of the pot. From the TWC Ankenman showed
that opp2 should call with y>.4.
The top left quadrant should be
$1*4/5 + $3*1/5 = $1.4
The game $EV is now
$7/5*1/4 + 0 - $1(1/2) = -$0.15
Your Game $EV is -$0.15 against opp2.

Your Game $EV was -$0.25 against opp1.

[Utah]

[ QUOTE ]
[ QUOTE ]
Intuitively, I think the strategies for part A and part B converge with time and that the fact of whether player 2 knows about the xray vision becomes unimportant.

[/ QUOTE ]

Since time does not explicitly appear in the game, any notion of convergence over time does not apply.

[/ QUOTE ]

Sure it does. I need to find the strategy for which I am 100% indifferent to whatever actions my opponent takes. I should be able to tell my opponent my exact strategy and he shouldnt be able to exploit that knowledge to win even 1 penny more assuming optimal play.

The strategy must be time independent. When both players know of the xray vision the game starts out as time independent - the first hand strategy is played the same as the millionth hand strategy with optimal play. There is no additional knowledge that can be exploited, unless one player starts playing sub-optimally.

This does not hold for the second case where the xray vision is not known. The play of player 1 can be exploited by player 2 once he learns player 1 strategy - which he will be able to figure out at some point. It does not matter whether player 2 ever learns of the xray vision because with enough hands he will be able to tell the exact strategy player 1 is using and thus player 2 will change his play and this will change the value of the game for player 1 as he is not indifferent to the change in strategy by player 2. So, player 1 will adjust as well. Over time, I am guessing, it should equal the same strategy as the xray vision being known.

What I think is interesting is that without the knowledge of the xray vision that the game is a different game every time it is played because player 2 has more knowledge. Thus, each game has its own strategy equation until such time as enough hands are reached so that player 2 knows player 1s strategy and the back and forth strategy adjustments reach equalibrium.

Now, if you assume that I must devise my strategy before play starts and I cant change it or that I have no knowledge of previously played hands then they will not converge obviously.

Of course, I may be wrong and I havent spent a lot of time thinking about it, but the above appears to be correct.

[Patrick Sileo]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
Intuitively, I think the strategies for part A and part B converge with time and that the fact of whether player 2 knows about the xray vision becomes unimportant.

[/ QUOTE ]

Since time does not explicitly appear in the game, any notion of convergence over time does not apply.

[/ QUOTE ]

Sure it does. <snip>

Of course, I may be wrong and I havent spent a lot of time thinking about it, but the above appears to be correct.

[/ QUOTE ]

Please don't take my <snip> of your response as dismissive -- your thinking about the real world seems to be on point. It is simply that so far the problem has been formulated as a two-player, one-shot, simultaneous move game. The variable "t" is glaringly absent from all of the formal analysis.

I think we need to directly address what is or might be meant by one player having x-ray vision, but the other being ignorant of this fact. Are we to assume that the latter believes they both have information sets consisting of only their individual "hands"? Or might the player know the other has different/superior information, but of unknown content/reliability. If it is the first case, introducing time shouldn't matter. If it is the second, we will first need to make some more concrete assumptions before a temporal learning model can be formalized -- if that's what you want to do...

[Utah]

[ QUOTE ]
Please don't take my <snip> of your response as dismissive

[/ QUOTE ] I did not. My response was meant to be taken jovially [img]/images/graemlins/smile.gif[/img]

[Jerrod Ankenman]

Btw, the solution to the game if the guy can't do the clairvoyance thing is 2/9,4/9,8/9, not all these wrong answers involving 1/2 that people have said in this thread. I've worked on this problem a bit; it's not very easy - or at least I messed up on the parameterization the first time through.

[well]

[ QUOTE ]
Btw, the solution to the game if the guy can't do the clairvoyance thing is 2/9,4/9,8/9, not all these wrong answers involving 1/2 that people have said in this thread. I've worked on this problem a bit; it's not very easy - or at least I messed up on the parameterization the first time through.

[/ QUOTE ]

To my opinion, you have misunderstood the original question. You give three numbers, which I interpret as incorporating "bluffs" for player I. This is indeed essential to most [0,1]-games we have discussed so far, but here player I has no option to "check" for a cheap show-down:

[ QUOTE ]
You both ante one dollar. You are first and can bet two dollars or fold. If you bet he can call or fold only.

[/ QUOTE ]

If you did not misread that part, then let me explain how I got those numbers and I would like you to point out where I went wrong...

X,Y ~ Unif[0,1] i.i.d. these are the "cards".

If bet into, we may assume without any strategic loss that if player II will call a certain fraction of his range, this fraction will be [y,1]. So y will be the single strategic variable for player II.

Now if player I bets and does not get called, it doesn't really matter what card he has as there won't be a showdown. The same holds for a fold and there is no check possibility. If he bets and is called, however, the strength of the card does matter as long as it is within player II's calling range, which is [y,1]. The values of the cards player I does bet in [0,y] are again meaningless. This means that if player I is, say, to bet half of his cards below y, he might as well use [y/2,y] for this.

Hence, in the optimum, player I could bet any card in the range [x,1] and fold otherwise.

Now it's a matter of calculating the EV. I will do this from the moment the antes are out and the cards are dealt. So this is a constant sum (2), two-player game.

Since we are looking for an equilibrium, these strategies cannot be exploited even if a player knows the strategy of his or her opponent. From this we can derive the following inequality:

x <= y (smaller or equal),

as it is pointless for player II to call with a hand which is positive to be lower than player I's card.

A fold by player I contributes 0 to his EV.
If player I bets, and player II folds, player I will win the $2, so this contributes:

2(1-x)y

to his EV.

If player I bets a card in [x,y] and is called by a card in [y,1], he is bound to lose his bet:

-2(y-x)(1-y)

If player I bets a card in [y,1] and is called by a card in [y,1], half of the time he will lose the bet (-2), and half of the time he will win the pot plus the extra bet (4), this nets 1. This brings the contribution of the last scenario to:

(1-y)(1-y)

The EV for player I will therefore be:

1+2x-2y-4xy+3y^2

By letting both partial deratives be 0, we might find a solution:

d/dx EV = 2(1-2y)

d/dy EV = 2(3y-2x-1)

This yields:

y=1/2 and x=1/4

This solution satisfies the inequality (x<=y). Let's test for optimality now...

Now if player I knows y=1/2 he cannot do anything about the situation as EV/.{y->1/2} equals 3/4.

If player II knows x=1/4, the EV becomes 3/2 (1-2y+2y^2), which is minimized by y=1/2.

So, we've found the equilibrium:

(x*,y*) = (1/4,1/2)

e*=3/4 (which means that player I loses a quarter each game, since he hase to come op with $1 ante as well).

Do you agree? And then, do you agree too with my solution to part B? Any comment/question regarding the calculations are welcome.

Regards,

Well.

[well]

[ QUOTE ]
[...] You and your opponent are dealt the proverbial real number between zero and one. You both ante one dollar. You are first and can bet two dollars or fold. If you bet he can call or fold only.

But before you bet you can throw two more dollars into the pot to see his hand.

With what hands would you do this and what is your overall strategy assuming:

A. He knows the rules and turns his hand face up when you toss in the two bucks (before possibly betting two more)?



[/ QUOTE ]

Here is my solution:

Player I does not use x-ray vision if his "card" is in [1/10,7/10].

He will bet any hand in this range and player II will call only his upper half: [1/2,1].

Player I will use x-ray vision outside this range.

If the revealed card is in [0,1/10], player I will bet and player II will fold.

If the revealed card is in [1/10,7/10], player I will bet and player II wil call with probability 2/3.

If the revealed card is in [7/10,1], player I will bet if he can beat this card, and will bet with probability 5(1-y)/(15y-9) if he cannot. Here y is the value of the revealed card. Player II will call with probability 2/3. This optimal bluffing strategy could also be realized if player I will bet whatever card in the range [0,(1-y)/3].

The whole x-ray vision part can be realized by the following protocol:

Player I will bet if he can beat the revealed number y.
If he cannot beat this number, he will bet any card in the range [0,(1-y)/3].
If bet into, player II will fold any number in [0,1/10] and call any number in [1/10,1] with probability 2/3.

After the antes are posted, the expected return for player I is 5/4. Since he had to ante $1, his net win per game is $0.25.

For now, I will not bother any of you with the calculations. Yet, I dare any of you to show me a strategy for player II that exploits this strategy of player I, or a strategy for player I that exploits player II's strategy, provided that player I will not use his x-ray vision only in a certain interval.

Regards,

Well.

[Jerrod Ankenman]

[ QUOTE ]
[ QUOTE ]
Btw, the solution to the game if the guy can't do the clairvoyance thing is 2/9,4/9,8/9, not all these wrong answers involving 1/2 that people have said in this thread. I've worked on this problem a bit; it's not very easy - or at least I messed up on the parameterization the first time through.

[/ QUOTE ]

To my opinion, you have misunderstood the original question. You give three numbers, which I interpret as incorporating "bluffs" for player I. This is indeed essential to most [0,1]-games we have discussed so far, but here player I has no option to "check" for a cheap show-down:


[/ QUOTE ]

Oh, yes, I did misread the question and thought it was bet-check. So yes, 3/4, 1/2 is right for that game. I'll look at the other question later.

jerrod

[jogsxyz]

Found a different method and a same solution.
Found a minimax equation in a calculus book.
The first derivative of a curve set to zero is
the maximum value of that curve.

Only solved the first part of the game.
No clairvoyance for player Y.
[0,1] game, low values win.
Both X and Y are dealt number from [0,1].
Pot has one chip. Y may bet one chip or fold.
X may call or fold.



Y bets =<y for value and bluffs >=z.
X calls =<x and folds >x.
0<y=<x=<z<1.

Y should bluff bet/(pot + bet) part of the time
1-z=y/2
X should call 2/3 of Y's betting range, since
X win 2 chips when right and loses one when wrong.
x=y.

The expected value for Y is the sum of the areas
in the unit square. The payoffs for Y are:
grey area equal zero,
blue areas equal one,
pink area equal minus one,
red area equal minus three.

****EV=y(1-x)-(z-y)-3(1-z)x+(1-z)(1-x)
substitute for x and z
y(1-x)= y(1-y)=y-y²
-(z-y)=-(1-y/2-y)=3y/2-1
-3(1-z)x=-3(y/2)y=-3y²/2
(1-z)(1-x)=y/2(1-y)=y/2-y²/2
Now
****EV=y-y²+3y/2-1-3y²/2+y/2-y²/2
........=-3y²+3y-1

The 1st derivative is
........=-6y+3
set to zero
y=1/2
x=1/2
z=3/4

Replug these values back into the EV equation.
EV=(1/2)(1/2)-1/4-3(1/4)(1/2)+(1/4)(1/2)
EV=1/4-1/4-3/8+1/8
EV=-1/4

The EV for Y is -0.25.