Win rate with optimal strategy against limit raise bot

[Abbaddabba]

[ QUOTE ]
utting in an extra $3 gave you and extra $0.02 worth of hot and cold equity from the flop action, but it gave up chances to fold with as low as 10.1% equity getting $12:$2 on the river. That would save you $0.46-$0.59 on the 96 out of 1081 turn and river combinations where the board stays unpaired and both the turn and river are below a 7. That is worth over 4 cents, more than the value you obtained on the flop.

[/ QUOTE ]

what about the remainder of the combinations where the larger pot makes something into an easy call, where in a smaller pot it would be a close (or difficult) call? in those cases you benefit from a bloated pot. it works both ways. and i think they tend to be evenly distributed on both sides of the coin.

[ QUOTE ]
But if it's a HU match, for the bot's stack then I can't see the point of investing too heavily in 'marginal' hands like 1 pair pre-flop; when you can have a much better spot by being more patient.

You know it's going to raise all the time, so take advantage of it, when you have a winner.

[/ QUOTE ]

Because you can do both, and win more by not having the mentality of a loose/passive.

[brian64]

On the U of Alberta site there is a paper (http://www.cs.ualberta.ca/~darse/Papers/kan.msc.pdf) which says:

"An always raise bot can be exploited to a maximum degree of around 3 sb/g."

I think this refers to a game with 4 bet caps. They have probably calculated the optimal strategy or something very close to it.

[pzhon]

[ QUOTE ]
[ QUOTE ]
Putting in an extra $3 gave you an extra $0.02 worth of hot and cold equity from the flop action, but it gave up chances to fold with as low as 10.1% equity getting $12:$2 on the river. That would save you $0.46-$0.59 on the 96 out of 1081 turn and river combinations where the board stays unpaired and both the turn and river are below a 7. That is worth over 4 cents, more than the value you obtained on the flop.

[/ QUOTE ]

what about the remainder of the combinations where the larger pot makes something into an easy call, where in a smaller pot it would be a close (or difficult) call? in those cases you benefit from a bloated pot. it works both ways. and i think they tend to be evenly distributed on both sides of the coin.


[/ QUOTE ]
Then you should think about it some more. You regret bloating the pot both when you turn a close call into an easy call and when you turn an easy fold into a close call. It's not balanced at all.

When your intuition disagrees with logical arguments and calculations, it's time to learn something. Don't state your intuition again and again in the hopes that the math will change.

[mykey1961]

[ QUOTE ]
On the U of Alberta site there is a paper (http://www.cs.ualberta.ca/~darse/Papers/kan.msc.pdf) which says:

"An always raise bot can be exploited to a maximum degree of around 3 sb/g."

I think this refers to a game with 4 bet caps. They have probably calculated the optimal strategy or something very close to it.

[/ QUOTE ]

They make that statement, but there isn't anything in the paper that backs that up.

The "always raise" bot was something called Teddy, it wasn't strictly an always raise, because in the setting, it would try to raise even when the pot was capped, and since that was "illegal" the dealing software would cause it to fold instead of call.

It's opponent "Monash" had to learn "Teddy's" propensity to raise, it didn't know in advance. I have to believe "Monash" loved the idea that if it capped, "Teddy" would fold, assuming "Monash" knew not to cap until the river.

So the win rate "Monash" acheived (1.1678sb/g) is probably not too close to maximal against a true always raise, and call if capped bot.

And the paper doesn't give a example of a maximal explotive bot vs a true raise as much as possible, then call bot.

[DarkMagus]

[ QUOTE ]
If you want to earn 2% on your money, put it in a bank, if you want to make some real ROI, you need to take some risks.

You don't have a 100% certain strategy, what you have is some situations that can come up where you have a 100% chance of winning.

There is no guarentee that those situations will come up while you are at the table.

I'd put 70% of my bankroll on an 85% shot paying even money, every day of week, and twice on sundays, unless I think the game isn't conducted fairly.

[/ QUOTE ]

If there's no cap, why not just put your whole roll into the game, wait until you get the nuts on the river, and then cap it? I mean you double your roll every time, and it's not like having the nuts on the river is so rare that you drastically kill your win rate, especially since you can still win a ton even on the hands where you don't have the nuts.

[mykey1961]

You can't have it both ways, either you wait until you get the nuts on the river, or you win a ton on the hands where you don't have the nuts.

Hard to win a ton, if you're waiting.

If I put my whole bankroll in, I would reraise preflop until I got 70% of it in with AA.

With KK I would get 64% of my bankroll in preflop.
With QQ I would get 59% of my bankroll in preflop.
etc


Bottom line is I would be getting/giving a lot more action, and therefore building my bankroll faster than the occasional double up.

[Abbaddabba]

[ QUOTE ]
Then you should think about it some more. You regret bloating the pot both when you turn a close call into an easy call and when you turn an easy fold into a close call. It's not balanced at all.


When your intuition disagrees with logical arguments and calculations, it's time to learn something. Don't state your intuition again and again in the hopes that the math will change.

[/ QUOTE ]



getting 5:1 when you are 1 : 4 to win
vs
getting 10:1 when you are 1 : 4 to win

explain how a smaller pot is preferable.

[mykey1961]

Looking at the 5:1

You would have put in 2 of that 5, and need to put in 1 more to make it 6.

and with 1:4 to win, you've got 20% equity.

Looking at it strictly from a decision perspective.

EV = Pot * Equity - Cost. 6 * 0.20 - 1 = +0.20

Looking at it from a strategy perspective.

EV = Pot * Equity - Total Cost. 6 * 0.20 - 3.0 = -1.80

What that says is while the last decision is correct, somewhere along the line you got unlucky, or made some bad decisions.

Now the 10:1

You would have put in 4.5 of that 10, and need to put in 1 more to make it 11.


EV = Pot * Equity - Cost. 11 * 0.20 - 1 = +1.20


EV = Pot * Equity - Total Cost. 11 * 0.20 - 5.50 = -3.30

You're still right to call on this street, but you've made bigger mistakes getting to this point.

[mykey1961]

[ QUOTE ]
NUMBERS! [img]/images/graemlins/smile.gif[/img]

mykey1961, I would be curious what the calculation looks like. I guess you have written some small application. If you don't mind I would like to see the code. With this app it seems one can determine the EXACT maximum BB/100 against this bot.

[/ QUOTE ]

Code?

This isn't exactly what I used to create the previous set of numbers, but I think this is faster and am running it on all preflop combos to get a complete answer.

I'm guessing you with either have a ton of questions, or simply say "what the hell are you doing".

<font class="small">Code:</font><hr /><pre>
__for g1 := _32s to _AA do
__begin
____p1 := PreflopIndex[g1,0];
____BnC2 := BitsAndCards2[p1];
____PF[0] := -0.5;
____PF[1] := 0;
____PF[2] := 0;
____with BnC2 do
____begin
______u12 := Bits;
______ps1 := State0^[c1];
______ps2 := ps1^[c2];
______write(CardName[c1],CardName[c2],' ');
____end;
____CountF := 0;
____fillchar(F,sizeof(F),0);
____F[1,0] := -2.0;
____F[2,0] := -4.0;
____for b123 := 0 to 22099 do
____begin
______BnC3 := BitsAndCards3[b123];
______if (u12 and BnC3.Bits) = 0 then
______begin
________inc(CountF);
________with BnC3 do
________begin
__________u345 := u12 or BnC3.Bits;
__________ps3 := ps2^[c1];
__________ps4 := ps3^[c2];
__________ps5 := ps4^[c3];
__________os1 := State0^[c1];
__________os2 := os1^[c2];
__________os3 := os2^[c3];
__________write(CardName[c1],CardName[c2],CardName[c3],' ');
________end;
________CountT := 0;
________fillchar(T,sizeof(T),0);
________T[1,1,0] := -3.0;
________T[1,2,0] := -6.0;
________T[2,1,0] := -5.0;
________T[2,2,0] := -8.0;
________for b4 := 0 to 51 do
________begin
__________BnC1 := BitsAndCards1[b4];
__________if (u345 and BnC1.Bits) = 0 then
__________begin
____________inc(CountT);
____________with BnC1 do
____________begin
______________u6 := u345 or BnC1.Bits;
______________ps6 := ps5^[c1];
______________os4 := os3^[c1];
// write(CardName[c1],' ');
____________end;
____________fillchar(R,sizeof(R),0);
____________CountR := 0;
____________R[1,1,1,0] := -5.0;
____________R[1,1,2,0] := -11.0;
____________R[1,2,1,0] := -8.0;
____________R[1,2,2,0] := -14.0;
____________R[2,1,1,0] := -7.0;
____________R[2,1,2,0] := -13.0;
____________R[2,2,1,0] := -10.0;
____________R[2,2,2,0] := -16.0;
____________for b5 := 0 to 51 do
____________begin
______________BnC1 := BitsAndCards1[b5];
______________if (u6 and BnC1.Bits) = 0 then
______________begin
________________inc(CountR);
________________with BnC1 do
________________begin
__________________u7 := u6 or BnC1.Bits;
__________________ps7 := ps6^[c1];
__________________os5 := os4^[c1];
// write(CardName[c1]);
________________end;
________________CountO := 0;
________________W := 0; L := 0; D := 0;
________________for o1 := 0 to 1325 do
________________begin
__________________BnC2 := BitsAndCards2[o1];
__________________if (u7 and BnC2.Bits) = 0 then
__________________begin
____________________with BnC2 do
____________________begin
______________________os6 := os5^[c1];
______________________os7 := os6^[c2];
____________________end;
____________________if ps7 &gt; os7 then inc(W)
____________________else
______________________if ps7 = os7 then inc(D)
______________________else
________________________inc(L);
__________________end;
________________end;
________________E := (W-L)/(W+D+L);
________________O[1,1,1,1] := E*7.0;
________________O[1,1,1,2] := E*13.0;
________________O[1,1,2,1] := E*13.0;
________________O[1,1,2,2] := E*19.0;
________________O[1,2,1,1] := E*10.0;
________________O[1,2,1,2] := E*16.0;
________________O[1,2,2,1] := E*16.0;
________________O[1,2,2,2] := E*22.0;
________________O[2,1,1,1] := E*9.0;
________________O[2,1,1,2] := E*15.0;
________________O[2,1,2,1] := E*15.0;
________________O[2,1,2,2] := E*21.0;
________________O[2,2,1,1] := E*12.0;
________________O[2,2,1,2] := E*18.0;
________________O[2,2,2,1] := E*18.0;
________________O[2,2,2,2] := E*24.0;
________________for SPF := 1 to 2 do
________________for SF := 1 to 2 do
________________for ST := 1 to 2 do
________________for SR := 1 to 2 do
__________________R[SPF,SF,ST,SR] := R[SPF,SF,ST,SR] + O[SPF,SF,ST,SR];
// write(#8,#8);
______________end;
____________end;
____________for SPF := 1 to 2 do
____________for SF := 1 to 2 do
____________for ST := 1 to 2 do
______________T[SPF,SF,ST] := T[SPF,SF,ST] + Max(Max(R[SPF,SF,ST,0],R[SPF,SF,ST,1]),R[SPF,SF,ST,2])/ CountR;
// write(#8,#8,#8);
__________end;
________end;
________for SPF := 1 to 2 do
________for SF := 1 to 2 do
__________F[SPF,SF] := F[SPF,SF] + Max(Max(T[SPF,SF,0],T[SPF,SF,1]),T[SPF,SF,2])/CountT;
________write(#8,#8,#8,#8,#8,#8,#8);
______end;
____end;
____for SPF := 1 to 2 do
______PF[SPF] := PF[SPF] + Max(Max(F[SPF,0],F[SPF,1]),F[SPF,2])/CountF;
____write(#8,#8,#8,#8,#8);
____assignfile(tf,'vsAllRaiseBot.txt');
____{$I-} append(tf); {$I+}
____if IOResult &lt;&gt; 0 then rewrite(tf);
____writeln(tf,PF[0]:11:8,' Fold ',BitsToNames(u12));
____writeln(tf,PF[1]:11:8,' Call ',BitsToNames(u12));
____writeln(tf,PF[2]:11:8,' Cap4 ',BitsToNames(u12));
____closefile(tf);
__end;
</pre><hr />

[Abbaddabba]

you didnt make a mistake.
you put the money in when you had an equity edge. of course when you isolate the times you miss your draw and dont pair your equity will not be favorable in hindsight.
the fact that the pot is bigger is a consequent of you putting in money with a positive edge.

the "easy" call is more profitable than the close call.


But obviously it's ridiculous to use that as a compelling reason to make the pot bigger. im just using it as an example to challenge phzons rationale.