Game Theory Problem Of The Week
 

For this week's game theory problem we will take a look at another situation .

There are two players who pick numbers from 1-100 without replacement . Each player posts a $1 ante but player one must always check even though he's first to act . Player two has the option of betting the pot or checking behind . Given this knowledge , what strategy must player two employ to maximize his EV ?

We may make the assumption that player one and two are playing optimally aside from the stipulation placed on player one .

Re: Game Theory Problem Of The Week
 

Player one can only check and call or check and fold . There is no raising in this game .

Re: Game Theory Problem Of The Week
 

lemee guess, ur taking a math class and trying to get us to do ur homework for u? joking obv.

good stuff, keep it comin.

Re: Game Theory Problem Of The Week
 

lol, thx

Re: Game Theory Problem Of The Week
 

Gave up on the previous question so quick?

Re: Game Theory Problem Of The Week
 

I'm using this problem as a stepping stone into tackling the more daunting task .

We will get there in due time .

Be patient [img]/images/graemlins/smile.gif[/img]

Re: Game Theory Problem Of The Week
 

Player B should be betting [0-.5]*(1/1+p)(bluffs) so if the pot is 1 he should be betting for value the top 50% of his hands times 1\2 of of his value hands so that he is betting a total of 75% of his hands. This is a guess btw.

Re: Game Theory Problem Of The Week
 

Ok I got it .

I think .

Re: Game Theory Problem Of The Week
 

P2
[1,11] Bet
[12,78] Check
[79,100] Bet

P1
[1,11] Fold
[12,100] Call

Gives P1 an EV of -1/9

Re: Game Theory Problem Of The Week
 

Ok so far I have something like P2 bets with 67-100 and checks behind everything else . Player 1 calls with 79-100 and folds everything else .

Re: Game Theory Problem Of The Week
 

Oh sorry i thought 0-50 was better than 51-100. And I'm way off with my guess anyway. But should he not bet more hands than 33%?

Re: Game Theory Problem Of The Week
 

Whatever hand player two bets with means that the lowest number in player 1's calling range must be able to beat one third of player 2's range .

ie 2 bets with [0.5,1] then the best strategy for player 1 is to check call with [0.666666,1]; 0.66666 beats one-third of the numbers from [0.5,0.66666]and loses to (0.66666,1]

Re: Game Theory Problem Of The Week
 

[ QUOTE ]
I finally spotted another error that we were all making .

In the last example , if player 1 bets with any number , then player 2 must call with any number !

ie , Hero bets everything from 0-1 . We mistakenly suggested that villain should call with 1/3-1 . This is blatantly wrong !!

If villain hadn't posted the ante , then this assumption is certainly valid . However , since he posted an ante , calling with any number is better than foregoing the loss of that ante .

To make this question and the previous question more interesting , we should add the stipulation that only Hero posts the ante and that he acts first .

I can't believe , nobody picked up on this .

[/ QUOTE ]

I think the reason you have such problems is you don't look at a fold as -EV. It's EV=0 for that decision, but that choice makes the overall EV negative.

In a hypothetical game where we both ante $1, you are dealt in the range of [1,50] and I'm dealt in the range of [51,100] When I bet $2, you should always fold.

You consider folding EV = $0.

Your strategy has an EV of -$1

Re: Game Theory Problem Of The Week
 

[ QUOTE ]
For this week's game theory problem we will take a look at another situation .

There are two players who pick numbers from 1-100 without replacement . Each player posts a $1 ante but player one must always check even though he's first to act . Player two has the option of betting the pot or checking behind . Given this knowledge , what strategy must player two employ to maximize his EV ?

We may make the assumption that player one and two are playing optimally aside from the stipulation placed on player one .

[/ QUOTE ]


Optimal is:

P1
[1,56] Fold
[57,100] Call

P2
[1,11] Bet
[12,78] Fold
[79,100] Bet

P1 EV = -1/9

P1 can't improve by making any changes against P2
P2 can't improve by making any changes against P1

Re: Game Theory Problem Of The Week
 

Here is the EV set up using the cash game definition of EV .

Player two bets with [x,1] and player one check calls with [(2x+1)/3,1] . Notice that the answer to this will be the same as in the discrete case .

EV(P2)= 1*(1-x)*(2x+1)/3 + 3*(1-x)*(2-2x)/3*1/3 -3*(1-x)*(2-2x)/3*2/3 - (1-x)/3

There are 4 different product terms :

The first is your EV|player 1 folds .
The second is your EV|player 1 calls and you win
The third is your EV|player 1 calls and you lose
The fourth is your EV|player 2 checks

After simplifying of the EV formula you should get

EV(P2)=(-4x^2+6x-2)/3

EV' = -8x/3 +2
So x=3/4 .

This means that player two should bet with 75-100 and player one should call with 84-100 . This is better than my previous attempt .

Re: Game Theory Problem Of The Week
 

Ok here is what I just don't seem to understand.

You apply a formula, get an answer. But do you do anything else to verify it actually answers the question?

Your P1
[1,83] fold
[84,100] call

Your P2
[1,74] Check
[75,100] Bet

My P1
[1,56] Fold
[57,100] Call

My P2
[1,11] Bet
[12,78] Check
[79,100] Bet


My P1 vs Your P2
EV = -0.094545

Your P1 vs My P2
EV = -0.118182


My Strategy wins 0.011815 Ante's per hand from yours while rotating positions.



Maximal against your P1 "P1MO"
[1,65] Bet
[66,92] Check
[93,100] Bet

Maximal against your P2 "P2MO"
[1,83] Fold
[84,100] Call

Your P1 vs Your P2
EV = +0.023636

Your P1 vs P1MO
EV = -0.447879

P2MO vs Your P2
EV = +0.023636

Maximal would win 0.2357575 Ante's per hand from your strategies while rotating positions

Re: Game Theory Problem Of The Week
 

[ QUOTE ]
Here is the EV set up using the cash game definition of EV .

Player two bets with [x,1] and player one check calls with [(2x+1)/3,1] . Notice that the answer to this will be the same as in the discrete case .


[/ QUOTE ]

Why do you assume P2 bets with [x,1]?

why not [0,x] and [1-2x,1]?

For this problem you if you have P2 bet only with 1, 99, and 100:
Player 1 would optimally fold 1, call with 99, 100 and be indifferent to calling or folding with 2 thru 98.

Since calling and folding would both produce the same EV for P1, folding means losing the ante, then P2's EV when betting with a 1 is (-3*2+1*97)/99 = 0.919191

That is clearly better than if P2 checked with a 1.


[ QUOTE ]

EV(P2)= 1*(1-x)*(2x+1)/3 +
3*(1-x)*(2-2x)/3*1/3
-3*(1-x)*(2-2x)/3*2/3
-(1-x)/3

There are 4 different product terms :

The first is your EV|player 1 folds .
The second is your EV|player 1 calls and you win
The third is your EV|player 1 calls and you lose
The fourth is your EV|player 2 checks

After simplifying of the EV formula you should get

EV(P2)=(-4x^2+6x-2)/3

EV' = -8x/3 +2
So x=3/4 .

This means that player two should bet with 75-100 and player one should call with 84-100 . This is better than my previous attempt .

[/ QUOTE ]

Again, you get an answer, but not the correct answer. It's correct for your formula, but your formula isn't correct for the question.

Re: Game Theory Problem Of The Week
 

mykey1961 could you explain where you get the numbers (-3 and 2+1 in your equation below. thank you

(-3*2+1*97)/99 = 0.919191

Re: Game Theory Problem Of The Week
 

[ QUOTE ]
mykey1961 could you explain where you get the numbers (-3 and 2+1 in your equation below. thank you

(-3*2+1*97)/99 = 0.919191

[/ QUOTE ]

Assume P2 has a value of 1:

If P1 has a value of 99, or 100, P1 calls and beats P2 for a 3 ante profit. That's a 3 ante loss for P2. 99 and 100 are 2 numbers

If P1 has a value of 2 thru 98, P1 has the same EV if it calls or folds. If it folds, it's loss is 1 ante, and that's a 1 ante win for P2. 2 thru 98 represents 97 numbers.

When P2 has a 1, there are 99 possible numbers for P1

P2's expected profit is (-3 antes * 2 numbers + 1 ante * 97 numbers) / 99 numbers = 0.919191 ante per number

This calculation is just for the case when P2 has a 1.

Re: Game Theory Problem Of The Week
 

Why not avoid betting 1-11 and bet with 67+ instead ?

I'm not sure the significance in why you're betting with 1-11 here .

Re: Game Theory Problem Of The Week
 

[ QUOTE ]
Why do you assume P2 bets with [x,1]?

why not [0,x] and [1-2x,1]?

[/ QUOTE ]

Because in this game, where lower hands *never* beat higher ones, there is no reason to bluff with the absolute lowest cards, as opposed to the cards just below the range where you're betting for value.

For example, lets say P2 is betting 50% of the cards, and that P1 is calling with 75%. (Yes, I realize this is not anywhere near optimal for either player, it is an example to illustrate a point)

It simply *does not matter* if P2's actual betting range is 1-25,75-100 or 50-100, or even 1-10,21-30,41-45,75-100 for that matter. P2's EV is exactly the same in all three of these situations, because P1 *never* calls with a card lower than 75. So P2 can only win when P1 folds, in which case it doesn't matter if P2's card was a 25 or a 74.

This seems to be a common mistake through these problems, based on the fact that in holdem it is often better to bluff with the bottom of the range and fold the middle, because you're far less likely to be dominated when you do call, giving you an extra 10 or 15% chance to win.

With that in mind, let's consider your "optimal" for a moment.

[ QUOTE ]

Optimal is:

P1
[1,56] Fold
[57,100] Call

P2
[1,11] Bet
[12,78] Fold
[79,100] Bet


[/ QUOTE ]

This is *clearly* not optimal for P2, because P2 could gain value without changing the total number of cards he's betting simply by betting 67-100 instead of 1-11,79-100. You have P1 calling with many numbers lower than 67, so 1-11 has absolutely no chance of winning when P1 calls, but 67-78 definitely do have a decent chance of winning.

And if P2 knows P1's strategy, he can increase his EV even further by betting a much wider range, somewhere in the neighborhood 30-something to 100.

Your strategy is also clearly sub-optimal for P1. P2 is betting 33 numbers, so it is absolutely silly for P1 to call with anything lower than 67, because it has absolutely no chance of winning.

An equilibrium (or "optimal", take your pick) strategy is one where neither player can adjust their strategy to gain EV, *even if they know the other player's strategy*.

In the "optimal" strategy you've given, if P2 told P1 what his strategy was, P1 could clearly adjust to increase his EV, and vice-versa.

That alone should be enough to show that your strategy is not optimal.

*EDIT*

You also seem to be solving the wrong problem (and making a mistake that I also made in reading the problem).

[ QUOTE ]
P2
[1,11] Bet
[12,78] Fold
[79,100] Bet

[/ QUOTE ]
"Fold" is not an option for P2.

Re: Game Theory Problem Of The Week
 

[ QUOTE ]
[ QUOTE ]
Ok here is what I just don't seem to understand.

You apply a formula, get an answer. But do you do anything else to verify it actually answers the question?

Your P1
[1,83] fold
[84,100] call

Your P2
[1,74] Check
[75,100] Bet

My P1
[1,56] Fold
[57,100] Call

My P2
[1,11] Bet
[12,78] Check
[79,100] Bet


My P1 vs Your P2
EV = -0.094545

Your P1 vs My P2
EV = -0.118182


My Strategy wins 0.011815 Ante's per hand from yours while rotating positions.



Maximal against your P1 "P1MO"
[1,65] Bet
[66,92] Check
[93,100] Bet

Maximal against your P2 "P2MO"
[1,83] Fold
[84,100] Call

Your P1 vs Your P2
EV = +0.023636

Your P1 vs P1MO
EV = -0.447879

P2MO vs Your P2
EV = +0.023636

Maximal would win 0.2357575 Ante's per hand from your strategies while rotating positions

[/ QUOTE ]

Mykey , you got to be kidding me . Why on earth would you restrict player one to check call with only 57-100 ?



[/ QUOTE ]
If P2 is playing optimal
[1,11] bet
[12,78] check
[79,100] bet

then P1 is indifferent to calling with [1,78] and calls with [79,100]

To find optimal for P1, you need to find which [x,78] x to maximize P1's EV against any strategy for P2.

x happens to be 57.

[ QUOTE ]

If player two is betting [0,11] and [89,100], then player one "could" check-call with any number from [12,88] since he will always be able to beat 11 numbers and lose to 11 numbers .


[/ QUOTE ]

You can't fine tune a strategy to just one opponent (even if that opponent is optimal) and call it optimal against all.

[ QUOTE ]

Your strategy for player one is NOT optimal !!

Here is a better strategy for player one than the one you proposed .

Player one check calls with [12-88] given that player two bets with [0,11] and [89,100] .

[/ QUOTE ]


Your P1 [12,88] Check else fold vs my P2 [1,11] and [79,100] bet else fold gives:
P1 EV = -0.226263

My P1 [57,100] call else fold vs your P2 [1,11] and [89,100] bet else check gives:
P2 EV = +0.090909

My strategy for P1 and P2 vs your's for P1 and P2
Your total EV = -0.067677 per hand when rotating positions

Can you find a strategy against my P1 which gives P2 an EV > 1/9?
Can you find a strategy against my P2 which gives P1 an EV > -1/9?

Re: Game Theory Problem Of The Week
 

[ QUOTE ]
You can't fine tune a strategy to just one opponent (even if that opponent is optimal) and call it optimal against all.

[/ QUOTE ]
If you can fine-tune your strategy in a way that is +EV, then one of the two of you isn't playing optimally at all.

That's the whole point of trying to find these equilibrium strategies.

***NINJA EDIT ADD***

[ QUOTE ]
Can you find a strategy against my P1 which gives P2 an EV > 1/9?
Can you find a strategy against my P2 which gives P1 an EV > -1/9?

[/ QUOTE ]
I already gave examples of adjustments vs your P1 *and* your P2 that are quite clearly +EV.

Re: Game Theory Problem Of The Week
 

[ QUOTE ]
[ QUOTE ]
Why do you assume P2 bets with [x,1]?

why not [0,x] and [1-2x,1]?

[/ QUOTE ]

Because in this game, where lower hands *never* beat higher ones, there is no reason to bluff with the absolute lowest cards, as opposed to the cards just below the range where you're betting for value.

For example, lets say P2 is betting 50% of the cards, and that P1 is calling with 75%. (Yes, I realize this is not anywhere near optimal for either player, it is an example to illustrate a point)

It simply *does not matter* if P2's actual betting range is 1-25,75-100 or 50-100, or even 1-10,21-30,41-45,75-100 for that matter. P2's EV is exactly the same in all three of these situations, because P1 *never* calls with a card lower than 75. So P2 can only win when P1 folds, in which case it doesn't matter if P2's card was a 25 or a 74.

This seems to be a common mistake through these problems, based on the fact that in holdem it is often better to bluff with the bottom of the range and fold the middle, because you're far less likely to be dominated when you do call, giving you an extra 10 or 15% chance to win.

With that in mind, let's consider your "optimal" for a moment.

[ QUOTE ]

Optimal is:

P1
[1,56] Fold
[57,100] Call

P2
[1,11] Bet
[12,78] Fold
[79,100] Bet


[/ QUOTE ]

This is *clearly* not optimal for P2, because P2 could gain value without changing the total number of cards he's betting simply by betting 67-100 instead of 1-11,79-100. You have P1 calling with many numbers lower than 67, so 1-11 has absolutely no chance of winning when P1 calls, but 67-78 definitely do have a decent chance of winning.


[/ QUOTE ]

If P2 changes his strategy to betting [67,100] then his EV goes from +0.111111 to +0.068687 against my P1

[ QUOTE ]


And if P2 knows P1's strategy, he can increase his EV even further by betting a much wider range, somewhere in the neighborhood 30-something to 100.



[/ QUOTE ]

if P2 changes his strategy to:

betting [30,100] P2's EV = -0.169091
betting [31,100] P2's EV = -0.165455
betting [32,100] P2's EV = -0.161616
betting [33,100] P2's EV = -0.157576
betting [34,100] P2's EV = -0.153333
betting [35,100] P2's EV = -0.148889
betting [36,100] P2's EV = -0.144242
betting [37,100] P2's EV = -0.139394
betting [38,100] P2's EV = -0.134343
betting [39,100] P2's EV = -0.129091

That takes P2 from winning +0.111111 (my way) to losing instead.


[ QUOTE ]

Your strategy is also clearly sub-optimal for P1. P2 is betting 33 numbers, so it is absolutely silly for P1 to call with anything lower than 67, because it has absolutely no chance of winning.


[/ QUOTE ]

against my P2, P1 is indifferent to calling or folding with 12 thru 78.

The thing is, P1 also has to defend against non-optimal P2's as well, so it can't always fold when it's indifferent to the optimal.

[ QUOTE ]

An equilibrium (or "optimal", take your pick) strategy is one where neither player can adjust their strategy to gain EV, *even if they know the other player's strategy*.

In the "optimal" strategy you've given, if P2 told P1 what his strategy was, P1 could clearly adjust to increase his EV, and vice-versa.

That alone should be enough to show that your strategy is not optimal.


[/ QUOTE ]

That's exactly what I've done here

I'm telling you my P1, and P2.

Can you do better than +1/9 against my P1?
Can you do better than -1/9 against my P2?

The answer in both cases is No.

[ QUOTE ]


*EDIT*

You also seem to be solving the wrong problem (and making a mistake that I also made in reading the problem).

[ QUOTE ]
P2
[1,11] Bet
[12,78] Fold
[79,100] Bet

[/ QUOTE ]
"Fold" is not an option for P2.

[/ QUOTE ]

[12,78] Fold is a typo, should be check
Bouncing back and forth between P1 and P2 can get confusing.

P1's options are call and fold
P2's options are bet and check

Re: Game Theory Problem Of The Week
 

[ QUOTE ]
Why not avoid betting 1-11 and bet with 67+ instead ?

I'm not sure the significance in why you're betting with 1-11 here .

[/ QUOTE ]

I believe it's because when P2 checks:
[12,78] vs [1,100] gives P2 an EV of -0.074444
[1,66] vs [1,100] gives P2 an EV of -0.226667

When P2 Bets with
[1,11] and [79,100] gives P2 an EV of +0.185556
[67,100] gives P2 an EV of +0.295354

+0.185555-0.074444 = +0.111111
+0.295354-0.226667 = +0.068687

Re: Game Theory Problem Of The Week
 

[ QUOTE ]
If P2 changes his strategy to betting [67,100] then his EV goes from +0.111111 to +0.068687 against my P1

[/ QUOTE ]

Then you are calculating EV incorrectly, because against a P1 who plays this strategy:

P1
[1,56] Fold
[57,100] Call

It is clearly obvious that betting 67+ is better than betting 1-11,79+

Just think about what happens when you bet 67-78 instead of 1-11. If P1 calls, 67-78 has a chance to win, where 1-11 does not. If P1 folds, it doesn't matter what P2's number was, so there's no difference.

If there's extra value when P1 calls (because you win more often), and the same value when P1 folds, betting 67-78 instead of 1-11 is clearly +EV.

This:
[ QUOTE ]

betting [30,100] P2's EV = -0.169091
betting [31,100] P2's EV = -0.165455
betting [32,100] P2's EV = -0.161616
betting [33,100] P2's EV = -0.157576
betting [34,100] P2's EV = -0.153333
betting [35,100] P2's EV = -0.148889
betting [36,100] P2's EV = -0.144242
betting [37,100] P2's EV = -0.139394
betting [38,100] P2's EV = -0.134343
betting [39,100] P2's EV = -0.129091


[/ QUOTE ]

leads me to believe that you're running this through a simulator program. Is that correct?

If so..

I'm going to have to say there's clearly a bug in your code, because you're getting incorrect answers to *very* obvious problems.

[ QUOTE ]
The answer in both cases is No.

[/ QUOTE ]

Actually, the answer in both cases is Yes, and the fact that you disagree means that you're doing something wrong, probably in your EV calculation, because it is so blatantly obvious that betting 67+ is better than betting 1-11 and 79+ that I don't even need to run the math, because simple deduction shows that it's true.

But since other threads have shown that people who are confused simply don't accept the most simple logical chains, I'll go ahead and run the math.

In the meantime, why don't you show us how you're calculating EV, because you're obviously doing something wrong.

Re: Game Theory Problem Of The Week
 

[ QUOTE ]
[ QUOTE ]
You can't fine tune a strategy to just one opponent (even if that opponent is optimal) and call it optimal against all.

[/ QUOTE ]
If you can fine-tune your strategy in a way that is +EV, then one of the two of you isn't playing optimally at all.

That's the whole point of trying to find these equilibrium strategies.


[/ QUOTE ]

That's only true if the there is no positional advantage in the game.

You can find strategies for P2 that are +EV because this game has a 1/9 ante advantage for player P2.


[ QUOTE ]

***NINJA EDIT ADD***

[ QUOTE ]
Can you find a strategy against my P1 which gives P2 an EV > 1/9?
Can you find a strategy against my P2 which gives P1 an EV > -1/9?

[/ QUOTE ]
I already gave examples of adjustments vs your P1 *and* your P2 that are quite clearly +EV.

[/ QUOTE ]

That's not true.

Optimal EV for P1 against my P2 is -1/9
Optimal EV for P2 against my P1 is +1/9

None of your strategies reached or exceeded those EV's

Re: Game Theory Problem Of The Week
 

[ QUOTE ]
[ QUOTE ]
If P2 changes his strategy to betting [67,100] then his EV goes from +0.111111 to +0.068687 against my P1

[/ QUOTE ]

Then you are calculating EV incorrectly, because against a P1 who plays this strategy:

P1
[1,56] Fold
[57,100] Call

It is clearly obvious that betting 67+ is better than betting 1-11,79+

Just think about what happens when you bet 67-78 instead of 1-11. If P1 calls, 67-78 has a chance to win, where 1-11 does not. If P1 folds, it doesn't matter what P2's number was, so there's no difference.

If there's extra value when P1 calls (because you win more often), and the same value when P1 folds, betting 67-78 instead of 1-11 is clearly +EV.

This:
[ QUOTE ]

betting [30,100] P2's EV = -0.169091
betting [31,100] P2's EV = -0.165455
betting [32,100] P2's EV = -0.161616
betting [33,100] P2's EV = -0.157576
betting [34,100] P2's EV = -0.153333
betting [35,100] P2's EV = -0.148889
betting [36,100] P2's EV = -0.144242
betting [37,100] P2's EV = -0.139394
betting [38,100] P2's EV = -0.134343
betting [39,100] P2's EV = -0.129091


[/ QUOTE ]

Lead me to believe that you're running this through a simulator program. Is that correct?

If so..

I'm going to have to say there's clearly a bug in your code, because you're getting incorrect answers to *very* obvious problems.

[ QUOTE ]
The answer in both cases is No.

[/ QUOTE ]

Actually, the answer in both cases is Yes, and the fact that you disagree means that you're doing something wrong, probably in your EV calculation, because it is so blatantly obvious that betting 67+ is better than betting 1-11 and 79+ that I don't even need to run the math, because simple deduction shows that it's true.

But since other threads have shown that people who are confused simply don't accept the most simple logical chains, I'll go ahead and run the math.

In the meantime, why don't you show us how you're calculating EV, because you're obviously doing something wrong.

[/ QUOTE ]

P1Fold := [1..56];
P2Bet := [67..100];
Sum := 0;
for v1 := 1 to 100 do
for p2 := 1 to 100 do
if (v1 <> p2) then
begin
..p1 := v1;
..Net := 1;
..if p2 in P2Bet then
..begin
....Net := 3;
....if p1 in P1Fold then
....begin
......p1 := 0; Net := 1;
......// this makes P1 always lose when it folds and makes P2 only net 1 unit
....end;
..end;
..if p1 > p2 then inc(Sum,Net) else dec(Sum,Net);
end;
writeln(Sum/9900.0:9:6);

Re: Game Theory Problem Of The Week
 

[ QUOTE ]
[ QUOTE ]
If P2 changes his strategy to betting [67,100] then his EV goes from +0.111111 to +0.068687 against my P1

[/ QUOTE ]

Then you are calculating EV incorrectly, because against a P1 who plays this strategy:

P1
[1,56] Fold
[57,100] Call

It is clearly obvious that betting 67+ is better than betting 1-11,79+



[/ QUOTE ]
That's your opinion


[/ QUOTE ]

Just think about what happens when you bet 67-78 instead of 1-11. If P1 calls, 67-78 has a chance to win, where 1-11 does not. If P1 folds, it doesn't matter what P2's number was, so there's no difference.

If there's extra value when P1 calls (because you win more often), and the same value when P1 folds, betting 67-78 instead of 1-11 is clearly +EV.


[/ QUOTE ]

As I've shown elsewhere, the bigger difference is when P2 Checks

[ QUOTE ]


This:
[ QUOTE ]

betting [30,100] P2's EV = -0.169091
betting [31,100] P2's EV = -0.165455
betting [32,100] P2's EV = -0.161616
betting [33,100] P2's EV = -0.157576
betting [34,100] P2's EV = -0.153333
betting [35,100] P2's EV = -0.148889
betting [36,100] P2's EV = -0.144242
betting [37,100] P2's EV = -0.139394
betting [38,100] P2's EV = -0.134343
betting [39,100] P2's EV = -0.129091


[/ QUOTE ]

leads me to believe that you're running this through a simulator program. Is that correct?


[/ QUOTE ]

Calculate instead of simulate

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If so..

I'm going to have to say there's clearly a bug in your code, because you're getting incorrect answers to *very* obvious problems.



[/ QUOTE ]

Again, that's your opinion

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The answer in both cases is No.

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Actually, the answer in both cases is Yes, and the fact that you disagree means that you're doing something wrong, probably in your EV calculation, because it is so blatantly obvious that betting 67+ is better than betting 1-11 and 79+ that I don't even need to run the math, because simple deduction shows that it's true.



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I think it's time you run the math, and give your deductionator a checkup.

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But since other threads have shown that people who are confused simply don't accept the most simple logical chains, I'll go ahead and run the math.

In the meantime, why don't you show us how you're calculating EV, because you're obviously doing something wrong.

[/ QUOTE ]

Again for the 3rd time, that's your opinion.

Re: Game Theory Problem Of The Week
 

Mykey here is where you're mistaken .

If player two bets with [0,11] and [79,100] but checks everything else , then player two is indifferent to calling or folding with [12,78] .

ie, 12 beats 11 numbers but loses to 22 numbers . Since he is getting 2:1 odds , any call he makes is neutral EV, even with the number 12 . Similarly , if player one checks calls with 78 , then he can only beat precisely 11 numbers and STILL lose to 22 numbers ;namely , [79,100] . In both situations , any call he makes from [12,78]is neutral EV .
So saying that player one should check call with [57,100]is wrong !!

Re: Game Theory Problem Of The Week
 

One bug in your code

[ QUOTE ]
..if p1 > p2 then inc(Sum,Net) else dec(Sum,Net);

[/ QUOTE ]

This line increases the sum if *PLAYER 1* wins, which means you're tracking P1's equity here.

If you haven't taken that into account, then you're actually seeing P1's EV decrease, and while P1's EV decreases, P2's EV *increases*.

Still working out the calculations on my own.

Re: Game Theory Problem Of The Week
 

Durr.

If you read this before I tried to erase the record of my stupidity, revel in the knowledge that sometimes I'm a retard.

[img]/images/graemlins/smile.gif[/img]

Re: Game Theory Problem Of The Week
 

[ QUOTE ]
Mykey here is where you're mistaken .

If player two bets with [**1**,11] and [79,100] but checks everything else , then player two is indifferent to calling or folding with [12,78] .

ie, 12 beats 11 numbers but loses to 22 numbers . Since he is getting 2:1 odds , any call he makes is neutral EV, even with the number 12 . Similarly , if player one checks calls with 78 , then he can only beat precisely 11 numbers and STILL lose to 22 numbers ;namely , [79,100] . In both situations , any call he makes from [12,78]is neutral EV .
So saying that player one should check call with [57,100]is wrong !!

[/ QUOTE ]


There are a lot of strategies for P1 that are maximal against P2's [1..11,79..100] Bets

All of them have P1 fold from [1..11,79..100] in common

Looking at just the range of [12,78]:

P1 must defend against a strategy that is maximal against P1.
To do that we need to choose the right range for P1 in [12..78] to call with.

It works out that P1 needs to fold with [12,56] and call with [57,78] to be optimal against any opponent.

That makes the full strategy for P1
[1,56] fold
[57,100] call



I haven't confirmed that no mixed strategies are required, but it doesn't appear they are.

Re: Game Theory Problem Of The Week
 

[ QUOTE ]
One bug in your code

[ QUOTE ]
..if p1 > p2 then inc(Sum,Net) else dec(Sum,Net);

[/ QUOTE ]

This line increases the sum if *PLAYER 1* wins, which means you're tracking P1's equity here.

If you haven't taken that into account, then you're actually seeing P1's EV decrease, and while P1's EV decreases, P2's EV *increases*.

Still working out the calculations on my own.

[/ QUOTE ]

if there is a showdown where the cards actually matter, both players will have invested the same amount 1 or 3 antes.

I track P1's EV. Whatever P1 loses, clearly P2 wins.

Re: Game Theory Problem Of The Week
 

[ QUOTE ]
Durr.

If you read this before I tried to erase the record of my stupidity, revel in the knowledge that sometimes I'm a retard.

[img]/images/graemlins/smile.gif[/img]

[/ QUOTE ]

I saw it, but I didn't get back to it in time to quote it, therefore I forgot exactly what it said.

I'm thinking you'll agree that's a good thing. [img]/images/graemlins/smile.gif[/img]


Oh yeah, something about one side only losing 2 compared to 3 for the otherside.

Re: Game Theory Problem Of The Week
 

[ QUOTE ]
I track P1's EV. Whatever P1 loses, clearly P2 wins.

[/ QUOTE ]

Yes, but if you're looking at value sum directly, seeing it decrease when you plug in a different strategy for P2, and saying "P2's EV is lower in this other situation, then you're misinterpreting the number.

If the number decreases, P1's EV has decreased, and P2's EV has *increased*.

[ QUOTE ]
if there is a showdown where the cards actually matter, both players will have invested 3 antes.

[/ QUOTE ]
Yes, after a moment of stupidity, I realized the net gain and loss really was 3. Guess you read that before I tried to ninja out my public display. [img]/images/graemlins/smile.gif[/img]

I still think you're reading your numbers backwards though. I wrote my own code that does exactly what yours does, tracking P2's EV instead, and mine shows P2's EV increasing when I switch from 1-11,79+ to 67+.

After a second look, *my* code was buggy. More in a minute. lol

Re: Game Theory Problem Of The Week
 

When I refer to P2's EV, I state the negative amount my program produces, since the program produces P1's EV.

Re: Game Theory Problem Of The Week
 

Ok, it appears I have a lot of crow to eat. After fixing the bugs in my program, it spits out .111 EV for player 1 when betting 1-11,79+, and .0686 when betting 67+.

I don't understand how this can possibly be the case, since betting 67-78 has the same chance of getting the other player to fold as betting 1-11, and a much greater chance of winning when they do call.

Can you explain this to me?

Re: Game Theory Problem Of The Week
 

You might have missed this piece I posted earlier which explains why "simple deduction" is wrong.

I believe it's because when P2 checks:
(M) [12,78] vs [1,100] gives P2 an EV of -0.074444
(J) [1,66] vs [1,100] gives P2 an EV of -0.226667

When P2 Bets with
(M) [1,11] and [79,100] gives P2 an EV of +0.185556
(J) [67,100] gives P2 an EV of +0.295354

P2(M) EV = +0.185555-0.074444 = +0.111111
P2(J) EV = +0.295354-0.226667 = +0.068687

This means that betting with [67,100] is better when only considering the betting part of the game.

Re: Game Theory Problem Of The Week
 

I get 0.1176 but I reduced the problem to [0,1] . You should get 1/9 in the discrete case .

1*0.32*0.79 + 3*0.32*0.21*1/3 - 3*0.32*0.21*2/3

+1*0.68*0.11 - 0.68*0.21 = 0.1176