Heads Up Game Theory exercise

[jay_shark]

If you bet with 44+ then your opponent should call with 63+.

You will fold with 43 numbers out of 100 which gives you EV=0 . 57 numbers out of 100 you will shove .

EV: 0*43/100 + 57/100*[1.5*62/99 + 3.5*1/3 -2.5*2/3]= 0.2504545

I'm going to create a new thread and revise some of my work .I fixed a typo above .

My Ev(x) function should read as :

EV(x)= (3-3x)/2*[1.5x + 3.5*1/3 -2.5*2/3]
EV(x) = (3-3x)/2*[1.5x-0.5]
EV(x)=1/2*[4.5x -1.5 -4.5x^2+1.5x]
EV(x) = 1/2*[-4.5x^2 +6x -1.5]
EV'(x) = 1/2*[-9x+6]
EV'(x)=-4.5x + 3
-4.5x + 3 =0
x=2/3
so a=0.5

[TNixon]

[ QUOTE ]
EV: 0*43/100 + 57/100*[1.5*62/99 + 3.5*1/3 -2.5*2/3]= 0.2504545

[/ QUOTE ]

Your probabilities are off here. You have a probability total inside the part that's all multiplied by 57/100, that adds up to more than 100% of the time when you bet. (62/99 + 1/3 + 2/3 = 1.626, which is significantly greater than 1)

It should be:

When you bet (57/100):
you win 1.5 when opponent folds (62/99)

If your opponent calls: (38/99)
you lose 2.5 when you have under 63 (19/57 or 1/3)
you lose 2.5 half the time when you have 63+ (1/2 of 2/3)
you win 3.5 half the time when you have 63+ (1/2 of 2/3)

Your EV formula is wrong.

It should be
57/100( 1.5*62/99 + 38/99( 3.5 * 1/6 - 2.5 * 1/6 - 2.5 * 1/3 ) )
...
.57( .939 + .383( .583 - .416 - .833 ) )
...
= .3898

Which is slightly different than my previous figure because I used 44/64 instead of 44/63.

But regardless, even .25 is > .167, so betting 44+ is more +EV than betting every hand, correct?

And if your other calculation:
[ QUOTE ]
x=2/3
so a=0.5

[/ QUOTE ]
Is correct:

When you bet: (50/100)

you win 1.5 when opponent folds (66/99) (is it even proper to take unseen cards out of the calculation here? since we don't know if player's card is greater or less than 66? Shouldn't this be 66/100)

If your opponent calls: (33/99)
you lose 2.5 when you have under 67 (13/50)
you lose 2.5 half the time when you have 67+ (1/2 of 26/50)
you win 3.5 half the time when you have 63+ (1/2 of 26/50)

leading to

.5( 1.5*66/99 + 33/99( 3.5 * .26 - 2.5 * .26 - 2.5 * .26 ) )
...
.5( 1 + 1/3( .91 - 1.3 ) )
...
= .435, which is higher than our previous best-calculated EV, and according to your maximization problem, should be the maximum EV possible, right?

So, according to your formula, the optimal play would be pushing 50+, correct?

Whoohoo!

[TNixon]

[ QUOTE ]
you win 3.5 half the time when you have 63+

[/ QUOTE ]
That was a cut/n/paste typo that I can no longer fix, but the actual calculations are correct.

*AGAIN*
Wow, I brutalized the crap out of that in a couple places. The probabilities for the times you win and lose when you have a hand greater than or equal to the big blind's are all wrong. Not horrendously wrong, but wrong, nonetheless. They should be 37/57 (not 2/3) in the first problem and 33/50 (not 26/50) in the second.

And the lesson from this is?

Checking your probability total, making sure it always equals 1, is good. Had I done that, like I did in my first few calcs, I would have caught this error before hitting submit the first time.

I suck at math. Still, they're not *massively* far off. You'd just have to calculate the exact right numbers before being able to compare the two. They're both pretty clearly higher than .1667 though. [img]/images/graemlins/smile.gif[/img]

By the way. I suspect that if you took the small blind into account (and I would still argue that you should, since not taking it into account leads to very misguided conclusions), you'd actually end up with an optimal betting range of somewhere very close to 44+.

There are many situations (specifically when there are multiple betting rounds) where it's simply too difficult, or not worthwhile, to calculate the overall EV of a hand. This is not one of those situations, and only considering the EV of the bet itself leads to incorrect conclusions about how to play, so overall EV should be considered, just as you would in a jam-or-fold situation in NLHE.

[jay_shark]

EV should be :

EV= 0*43/100 + 57/100*[1.5*62/99 + 38/99*3.5*1/3 -38/99*2.5*2/3]= 0.4260606060

You have a few errors . The probability your opponent calls you when you shove is 38/99 . Fair enough .
Given that he calls you , you will win 2/3*1/2 = 1/3 of the time . So we have 3.5*38/99*1/3

The probability you lose given your opponent calls is 2/3 .
Check 1/3+2/3=1

[Klyka]

Given that there is no bluffing (which seems to have been the assumption so far) BB calls with 2/3 of the hands that SB bets with, since he is getting 2 to 1 and therefore needs 1/3 probability to win for a call to be correct.

I'll turn the game around and assume that the lowest card wins the pot at showdown. It will still be the same game, but it makes the calculations easier.

Let x be the highest (worst) card that we (SB) bet (again, no bluffing).

First of all, we fold 100-x of our cards, or (100-x)/100 of the time. Our ex-showdown(*) EV from that is 0, so I'll just omit that from the calculation.

When we bet (which we do x/100 of the time) the following will happen:

BB will call (2/3*x)/99 of the time. 2/3 of those times we will have a hand within his calling region, and we will average winning half of those hands, making for a profit of 1/2 BB (1/2*3.5 - 1/2*2.5). 1/3 of the time we will have a hand that he always has beaten, costing us 2.5 BB. So over all, when he calls we will average a loss of 1/2 BB.

BB will fold (99-2/3*x)/99 of the time, giving us a profit of 1.5 BB.

So our EV is:

EV = x/100 * (1.5*(99-2/3*x)/99 - 1/2*(2/3*x)/99)

This equation peaks at x=56. Turning it around back to being a game where the highest card wins, we have 101-56=45 being the worst (lowest) card that we should bet. So TNixon is pretty close to the truth.

However, this is not very relevant, since there will be bluffing, which changes things quite a bit.

---

(*) I think you guys have really been messing things up by not making a difference between showdown EV and EX-showdown EV.

---

[insert a bunch of disclaimers]

[TNixon]

[ QUOTE ]
You have a few errors .

[/ QUOTE ]
Yes. Errors which I already explained and took credit for.

:P

[ QUOTE ]
EV= 0*43/100 + 57/100*[1.5*62/99 + 38/99*3.5*1/3 -38/99*2.5*2/3]= 0.4260606060

[/ QUOTE ]
Which is *way* closer to what I calculated (with my slight probability errors) than what you calculated, with your incorrect formula.

[mykey1961]

I think you have a problem with these lines.

[ QUOTE ]
<font class="small">Code:</font><hr /><pre>
c1 = c2 = (int)(100.0f * (double)rand() / (double)RAND_MAX);
while( c2 == c1 )
c2 = (int)(100.0f * (double)rand() / (double)RAND_MAX);

</pre><hr />

[/ QUOTE ]


c1 = c2 = rand() assigns the same rand to both c1 and c2

since they are always the same, you then assign a new rand to c2, which 1% of the time will be the same as c1 again.

A possible solution:

<font class="small">Code:</font><hr /><pre>
do {
c1 = (int)(100.0f * (double)rand() / double)RAND_MAX);
c2 = (int)(100.0f * (double)rand() / double)RAND_MAX);
until (c1 &lt;&gt; c2);
</pre><hr />

[mykey1961]

[ QUOTE ]
[ QUOTE ]
Raise 3x BB would be 3bb, no?

[/ QUOTE ]
No. Raising 3x BB would add 3BB to what's already in the pot (the small blind), up to a total of 3.5xBB.

Raising *to* 3x is a raise of 2.5bb, for a total of 3bb.

[/ QUOTE ]

Actually Raise 3xBB is a combination of a call of 0.5BB, and a raise of 3BB. so the pot after the raise would be 5BB with 3BB for the opponent to Call.

Raise to 3xBB is a combination of a call of 0.5BB and a raise of 2xBB, so the pot after the raise would be 4BB with 2BB for the opponent to Call.

[mykey1961]

[ QUOTE ]
Ok here is the answer to the first problem . I'll let others think about the second .

1) The sb can either raise an additional 2.5 bb's which means the BB is getting 2:1 on his call . If the sb folds at any point for any specific hand then his EV =0 . I'll show that pushing with any number is positive EV .

If you have card #1 , then it's clearly the lowest number form the deck . However , the bb is not aware of this . He must call you if he believes his hand can beat at least a third of yours . Since he's getting 2:1 on his call , he should call with numbers 34,35,36,...100 . Notice that 34 beats precisely 33 numbers and loses to 66 numbers . So , if the sb pushes with any card , then he actually increases his EV . Since this is the case , there is no bluffing frequency for the sb .

The probability that the BB wins given that he calls you will converge to 2/3 as the numbers approach infinity . In this case , the numbers stop at 100 but it still converges to 2/3 fairly quickly .

Just show that 1/3 + 2/3*1/2 = 2/3 . Simply reason that the BB will beat one third of the hands when he calls and the sb shows 1-33 . However two thirds of the time , he will win half of the hands (2/3*1/2) .

Ev(sb) = 1/3*1.5 + 2/3*(3.5*1/3 - 2.5*2/3)
Ev(sb) = 0.166666666

This shows that raising with any number is better than folding , even if your first card is a 1 .

The second problem is a bit harder and algebra intensive but it is pretty neat . The solution hinges primarily on ideas expressed in the first problem but it's still interesting to work out .

[/ QUOTE ]

I gotta disagree with this result.

BB should always fold 1 thru 62, raise 1/8 of the time with 63, and always raise 64 thru 100

By doing this, the SB is indifferent to raising or folding with 1 thru 62, and should raise with 63 thru 100.

[mykey1961]

In the case where the SB can raise to 10xBB, the BB should fold with 1 thru 82, and call with 83 thru 100.

This leaves the SB indifferent to raising with 1 thru 82, and should raise with 83 thru 100.

On average the SB would lose about .17BB per hand

In the 3x version the SB would lose about .07BB per hand