Important No Limit Math Problem

[David Sklansky]

One and two dollar blinds. You are in the small blind. Everyone folds. (Infer nothing about card distributions from that). Your hand is A8 offsuit. Your opponent in the two dollar big blind has more money than you. He is an expert.

You have only two options. Fold, or turn your cards face up and move in. He sees your A8 before he acts. Whether you should fold or move in depends on how much more money, besides the dollar blind, you have in front of you. Hopefully it is obvious that you should not move in if your bankroll is above x dollars. What's x

[Mike Haven]

doesn't this question get to the very essence of gambling?

if he sees your cards before he bets he knows what his odds are to beat your hand

if he has AA then he will call whatever you bet, so you would want to bet zero here, but still be in for the 7% chance of winning - so you would "want" your bankroll to be zero and to have already called all in for the small blind

if he has some other hand it is simple to work out what you should have to enable him to win slightly less than he will win by calling the bet, in the long run (assuming that is a possibility with the cards he holds)

but as you don't know his hand, if you want to gamble that he has less than an average hand against you, again it is simple to calculate the figure to bet to ensure he doesn't win more than he loses by calling an average hand, in the long run

"do you feel lucky, punk?"

[1800GAMBLER]

one dolla; holla.

[TAFKAn]

I'm going on pure intuition here.

It's around 11 to 12 bucks. I certainly wouldn't consider it if I had more than that, so that must be right. /images/graemlins/smile.gif

[cbloom]

David, this question seems ill-posed. For example, it depends on exactly how big his stack is compared to yours, and how you play compared to each other, and the size of your stack compared to the blinds. Your behavior depends on his behavior, and his behavior depends on how he expects to do against you in the future. For example, if he is much better than you, he won't want to "mix it up" as much and will play this safer, which will increase your EV but decrease his variance. Clearly in no-limit you don't just play fo EV, you must also be aware of variance and future chances, as you yourself have stressed many times. eg. he shouldn't take even a 75% shot today if he can take a 90% shot tomorrow. Similarly, if x is small compared to his stack, he's going to call anything to try to knock you out (in a tournament). Your hand is about a 60% shot against random hands, so assuming he doesn't even look at his hand, he'll call as long as x <= 9. So, certainly if x <= 9 you should move in.

In another case, he might have a much bigger stack than you and he may be a much better player than you. In that case, he will only call if he clearly has the best of it. "Having the best" of A8 occurs for these holes -
88 - AA
A9 - AK
There are 96 of these holes (accounting for the 8 and A that are taken). There are 1275 possible holes he can have, so these are about P = 0.07529 of holes. When he has one these, he wins about 75% of the time. So, assuming he calls with any of these holes and wins 75% when he does have them -

your EV is better for moving in with any x <= 147.

(there's a small error in this - I didn't account for him having A8, in which case he should call and you'll likely split it; this helps him a tiny bit, so x should maybe be 146).

Also, if he's really conservative and only calls with AA, you should move in with x <= 803 (just based on EV).

[Robk]

He'll call with AA (3 combos), AK-A9 (60 combos), KK-88 (36 combos), 77-22 (36 combos), and A8 (9 combos). There are 50c2 = 1225 possible hands, he'll fold the other 1081 hands.

Let x be your bankroll. Your EV against each group is
AA -- .07x-.93x = -.86x
AK-A9 -- .28x - .72x = -.44x
KK-88 -- .28x - .72x = -.44x
77-22 -- .43x - .57x = -.14x
Everything else: you win $3 because he folds.

So your EV = 3*(1081/1225) - .86x(3/1225) - .44x(60/1225) - .44x(36/1225) - .14x(36/1225) = 2.65 - .041x > 0

=> x < 64.6

[Ikke]

[ QUOTE ]
He'll call with AA (3 combos), AK-A9 (60 combos), KK-88 (36 combos), 77-22 (36 combos), and A8 (9 combos).

[/ QUOTE ]

Havent got time to do any calculations myself, but this seems wrong. The hands he'll call your call all-in raise with are a function of x. For instance, with SB $1 and BB $2 and x very small (like 1.5 after posting blinds) he'll call with any hand. So you should make his calling standards also a function of x, with that function given by his known EV equation.

Hope I made myself clear...typed fast/

Regards

[well]

Here are some of my thoughts.

Let w denote the probability of winning for me if I knew what II had and d the probability of a draw.

On a draw, II wins 2 with probability d.
If I win, II wins 1-x with probability w.
If II wins, II wins 3+x with probability 1-d-w.
So, on a call II's EV would be -(2w+d)(x+1)+x+3.
On a fold, II's EV would be zero.
II will call if 2w+d <= (x+3)/(x+1).

Back to me.
For readability, let c denote prob{2w+d <= (x+3)/(x+1)}.
Let W denote the probability of winning for me, D the probability of a draw, 1-W-D the probability of losing for me, all three given 2w+d <= (x+3)/(x+1).

Let's see what happens if I move all-in:

If II folds - and he will with probability 1-c, I win 3 with probability 1.

If II calls - and he will with probability c, then
on a draw, I win 1 with probability D,
if I win, I win x+2 with probability W and
if II wins, I win -x with probability 1-D-W.
So, when I move all-in, my EV would be (1-2W-D)cx + 3(1-c) + (2W+D)c.
When I fold, my EV would be zero.
Hence, I would move in if x <= (3c-3-(2W+D)c) / (2W+D-1)c.

Obviously, w,d,W,D and c all depend on the hand I have.
The values of these parameters could - I presume - be calculated or estimated by simulation.

Next Time.

[irchans]

I think RobK got it right, but there are a couple of typos in his post:

1) KK-88 (36 combos) should be KK-88 (33 combos).
2) he'll fold 1225-3-60-33-36-9 = 1084 hands.

Let x be your bankroll before you post the small blind. The EV against each group is
AA -- .07(x+1) - .93(x-1) = -.86x +1
AK-A9 -- .28(x+1) - .72(x-1) = -.44x +1
KK-88 -- .28(x+1) - .72(x-1) = -.44x +1
77-22 -- .46(x+1) - .54(x-1) = -.08x +1
A8 -- .5(x+1) - .5 (x-1) = 1
Everything else: you win $3 because he folds.

So your EV > 0 when

3*(1084/1225)+ (-.86x+1)(3/1225)+ (-.44x+1)(60/1225)+ (-.44x+1)(33/1225)+ (-.08x+1)(36/1225)+ (1)(9/1225) >0.

2.77 - 0.0379 x >0

Or approximately when x <= 73. Great Post RobK! /images/graemlins/laugh.gif

[irchans]

Ikke,

Your objection, "The hands he'll call your call all-in raise with are a function of x" is correct. All of the hands that are not on RobK's list win less than 48%. All the hands on RobK's list win 50% or more of the time. When x is high enough, I think the opponent will only call with hands on RobK's list.

[Copernicus]

It is 1 if "bankroll" is defined as the starting bank, 0 if "bankroll" means after the SB.

It doesnt matter what two cards DS named. BB has perfect knowledge and will never bet into a situation where he has negative EV. If he never has negative EV you can never have positive EV, so your best position is to be all in after the SB. If you cant be forced to put more money in, then you realize maximum value of the blinds based on the probability that your cards win.

One of the basic tenets of poker is that you gain EV when you can induce the opponent to do something different than he would if he knew your cards...he does know your cards, so you cannot gain EV.

[Mike Haven]

yes - i can see that

if you put in X more then if he only calls when he has a 50% chance of winning, as he has to put in (X-1) he will win X/2 and you will win (X-1)/2 in the long run

[rivaridge]

If you assume that he will fold if he has a worse hand than yours and stay inif he has an equal or better hand and if you assume that when he stays in his hand will win roughly twice as many times as yours (Remember he is equal or bettor if he stays in) then the answer is close to $3.00. I wont get too deep in to my math but if you go all in with $3 then 50% of the time you (when he folds) you will win 2$ or an expected result of plus 1. 33% of the time you will win $5 or an expected result of +$1.65 and 66% of the time you will lose $4 or negative $2.64. If you add these all up you are at an expected result of 0 in other words you are breakeven at this point. If you plug in 4$ your expected result is negative.
Consequently if my above assumptions are correct (he calls 50% of the timeand when he calls the opponent is a 2-1 favorite I also assuume A8 o is a 50-50 hand) than my answer is 3$.
Sorry my math gets sloppy please point out errors.

[Copernicus]

If you add up your 50% of the time one thing happens, 33% another thing happens and 66% another thing happens, then youve got 150%...nice result in bizarro world!

[rivaridge]

Good point the math is wrong and I am too tired to fix it( I think if you correct it the answer is closer to 6$) but If I am onthe right track please finish it. If not forget what I said.

[George Rice]

Of course you can have positive ev, there's dead money in the pot!

If you had $1 left you can call the BB and you are getting 3:1 on that call. A8o will win more than 25% against a random hand so this is positive ev.

If you had $2 left you can raise the BB because you are getting 2:1 (assuming he calls the $1) and you hand is better than 33% against hands he will call with (he's getting 5:1 on the call and that is enough to call with anything). This is also positive ev.

I don't know what the winning % is for A8o against a random hand, but I think it's a favorite. The BB will be able to call with anything that gives him 3:1 or better to call (I think). So if the SB moves in with $3 more, BB can call with anything, and SB is favorite getting 5:3.

At higher amounts BB will start to fold some hands and SB will be getting less and less odds on a call. BB will call with many hands because of his pot odds and SB will win a percentage of these calls as well as $3 when BB folds.

At an amount like $10, BB will be getting 13:9 on a call and probably 75% or more of his hands will be worth a call. Let's say that SB wins 45% of the times he is called. So he wins (.45 x .75 x 12) + (.25 x 3) = $4.80. He loses .55 x .75 x 10 = $4.13. So if these estimates are correct, then $10 would be okay to go all in on.

I don't have the inclination to figure out exactly what percentage of hands BB should call with at any given bet. Perhaps I'm missing something that overrides all of this and renders this an easier problem. But there is clearly a number for x which is more than a few dollars. I'm guessing it's somewhere between $9 and $12.

[Copernicus]

The equations are not quite correct, I believe, because when he calls and you win, you win x-1+3=x+2, not x+1. That doesnt change the answer that much, and you probably wind up around 74.5.

Note, however, that your equations are maximized when x=0...your greatest EV comes when you are all-in, which, I guess, is how you reconcile the math with my "thought experiment" approach.

There is a fallacy in my original "zero" answer, though, which was based on the premise that he wont bet if it gives him negative EV implying that you will be negative EV in all situations where you bet. The problem with that is the $3 pot gives you positive equity whenever he folds, so he cannot always put you into a negative EV situation.

[Bozeman]

(RobK and) Irchans solved it with x being your BR including your SB, not your BR after the SB is in. I really thought that you'd need to include some dogs that BB would call with, but the fact that A8 is a favorite over so many hands makes x large enough that the probability of winning BB needs to call, Pw=(x-2)/2x=1/2-1/x, is only slightly less than 50%, and there are no dogs better than JTs (different suit than A and 8) which is about 48%, so you only need to consider the hands that are favorites. BTW, BB should call with any two cards as long as x<~3.6.

Craig

[rivaridge]

Quote
"If he never has negative EV you can never have positive EV, "

This is not true. There are many scenarios in poker where it is correct for each player to play relative to pot odds (even if you knew each others cards).

Also, isnt this a double negative? /images/graemlins/smile.gif

[Ikke]

Let a be the amount of the SB, and 2a be the BB.

Player B knows player A's hand, and the only decision B has to make is whether to call an all-in bet by A who he has covered.

EVB = P1*(3a+x) - (1-P1)*x = P1(2x+3a) - x

where P1 is the probability that player B wins.

So, B will call an all-in bet when:

P1>= x/(2x+3a)

Or: Given that player B calls, he will win at least x/(2x+3a) times.

The following step is more difficult. You should now find out, given A's hand, which percentage of the hands will win at least x/(2x+3a) times against A's given hand. Problem is that we don't want to define x yet, because we need it as a variable in player A's EV equation. So there is some kind of function F=F(P1) which gives the probability of player B calling an all-in raise. This function might be obtainable through simuation but I imagine it would be quite hard to do so.

But lets go on:

Then:

EVA=(1-F)*3a + F(P1*(x+3a) - (1-P1)*x)=0

Now this equation is easy to solve because a, P1 (function of x) and F (function of x) are all given. So there follows x. But how to find F is the real difficulty of this problem.

I hope I didnt make any typo's in the equations.

Regards

[Bozeman]

You made one mistake, if BB calls and loses, he only loses x-a since SB has put in x+a, but BB has already put in 2a. (This is using the convention that x is SB's bankroll after posting, unlike that used by irchans).

EVB = P1*(3a+x) - (1-P1)*(x-a) = P1(2x+2a) - x + a
so EV is positive when P1>(x-a)/2(x+a).

It was not clear to me either that RobK and irchans solution was correct (and they should put in exact winning percentage for all the calling hands instead of approx. averages), but since BB is usually a dog, x is very large, so P1->1/2. The cutoff for BB to play any underdog hands is x<49.


Craig

[George Rice]

What about smaller values of x. There will come a point where it is correct for the BB to play more hands because of the odds he's getting. I'm thinking that there may be a range below your value for x where it may be incorrect for the SB to go all-in. In other words someting like x<15, 35<x<73. When the value gets lower the BB can play hands like KQ or maybe even K7, and some suited hands may be playable where some off-suit hands may not be for certain values of x. When this happens SB loses many of his blind steals. Will his winnings on the additional hands cover this? Perhaps. I'll let someone else figure that. /images/graemlins/wink.gif

[Ikke]

Thanks for the correction.

[ QUOTE ]
The cutoff for BB to play any underdog hands is x<49.


[/ QUOTE ]

If this is true, then this situation seems very relevant. Because often, for instance tournaments, x is way less than 49 times the SB. Yes, if x is very large compared to a then RobK is right. But this is a huge simplifcation and you rule out a lot of situations by assuming this. I think Sklansky wasnt looking for that answer, but for a more general solution.

Regards

[The Gift Of Gab]

If you are using x as your stack before you post the SB, then your EV when he folds is 2, not 3, as your stack increases by 2 dollars. Also, the break-even point is when your EV is > -1, not 0, since by RobK's convention you lose a dollar when you fold your SB.

Also, RobK's original EV numbers are correct by this convention and irchans' are not.

If x is your stack size, then the total pot size after he calls is 2x. Your EV is your equity EQ times the pot size (how much you end up with,) minus your starting stack size x (what you started with.)

EV = 2x*EQ-x = x(2*EQ-1)

Your EV vs. AA, for example, is
x(2*0.07 - 1) = -0.86x

as RobK had originally.

By this convention your EV when he has A8 should be 0 (except for A8s, which no one's taken into account yet.)


Redoing the calculations in light of this:

-1 < EV
-1 < 2*(1084/1225) + (-.86x)(3/1225) + (-.44x)(60/1225) + (-.44x)(33/1225) + (-.08x)(36/1225) + (0)(9/1225)

2.768 > 0.0396x

x < 69.9 (or 68.9 the way Sklansky meant it)


These are just the hands with which he will call regardless of x. Are there others?

67.9 is around 48.6% of 2*69.9=139.8, which means that he shouldn't call with JTs (48.0%EV.) This is close enough that to kill this thing for good someone should redo it using more accurate EV numbers and include A8s. You'd have to calculate multiple EVs per hand to take suits into account and weight them.

I would much rather go play cards then do this myself, so, uh, I'll let others elaborate.

[The Gift Of Gab]

Actually, as Bozeman points out, x would have to be 49 (or 50, depending on who you're talking to) for him to play JTs. Tweaking some EV numbers obviously won't change x enough to come close to that, so it's still going to be around seventy bucks.

[Bozeman]

Yes, when x is smaller, BB is correct to play KQ, JT, etc.; however, this doesn't change SB's play, because SB makes money (even on the call alone) when BB calls as an underdog. For example, BB should call with KQo (4 suits), if x<13.5, but then SB makes 3. (I used irchans' convention here)
Craig

[Bozeman]

I think you are incorrect: you still make 3 wrt folding when you raise and are not called.

[Bozeman]

I know I don't see all the relevance of this, but I have to say this is one of the best questions you have posted, David,
Craig

[The Gift Of Gab]

I still think it's right using the RobK/irchans convention. The inequality represents folding and losing your dollar SB or shoving in and either winning his two dollar BB or playing a big pot. My guess is that if you used x as the amount left in your stack after posting (as Prof. Sklansky intended) that the result would be the same but the equations would be uglier. Or have I just confused myself?

[David Sklansky]

The opponent should not call big bets with worse than A9 or a pair. Disregard A8 for simplicity. That's about one tenth of all hands. When he calls he will win about about 70%. The bettor, with x more dollars to bet, thus has an expectation of .9 times 2, plus .03 times (x+1), minus .07 times (x+1). Folding gives him an expectation of minus one.

Seting them equal and multiplying by 100 gives us:

180 + 3x + 3 -7x -7 = -100

4x = 276

x = 69

[Bozeman]

There are many conventions one could use. I nominate x to be the answer to Sklansky's question, and y=x+1 to denote the answer found by RobK, etc.. Also, one can say that folding is 0 EV or include the SB and say it is -1 EV (as you do). Both will give the same answer, provided one doesn't make any mistakes. I didn't follow your post closely enough in my earlier reply, and you did do it correctly (almost) for your conventions, and, not surprisingly, since conventions are only conventions, your inequality:

-1 < 2*(1084/1225) + (-.86x)(3/1225) + (-.44x)(60/1225) + (-.44x)(33/1225) + (-.08x)(36/1225) + (0)(9/1225)

is the same as Irchans':

3*(1084/1225)+ (-.86x+1)(3/1225)+ (-.44x+1)(60/1225)+ (-.44x+1)(33/1225)+ (-.08x+1)(36/1225)+ (1)(9/1225) >0

But then you must have made a math error (the .0396 is wrong), because I get (multiplying out the 1225):

46.38y<3393, so y=73.16 (the same as irchans) to an excessive 4 significant places.

But the real answer, using enumerated results for all possible matchups, will be 70+-10.

Craig

PS Which #'s in the approximate average winning percentages are definitely relatively incorrect? Why? Which one should be higher than the other?

PPS What did you suspect the answer would be before doing the calculation? I thought it would be about 10, oops.

PPPS How would things change if your opp. couldn't see you cards (but you still could only fold or go allin)?

[cbloom]

I think this is wrong. If he has a pocket pair less than 8's, he is not a 70% favorite, in fact, he is just about a 50/50 shot. In fact, calling with pocket pairs less than 8's is questionable and perhaps wrong in most practical situations. I think the right answer is closer to x = 100.

[aces961]

First of all, The 70% figure is a average figure for all the hands that he could call you with. Some hands he may only be a 52 to 48 favorite and some he may be a 93 to 7 percent favorite with. The 70 percent figure is just stating that if you take a weighted average of his winning percentages on all the hands he is correct to call you on the result will be about 70 percent.

As to the fact that you consider the calling with pocket pairs less than 8's questionable and perhaps wrong in most practical situations. Consider the fact that he knows exactly what your hand is. Consider the worst of these pocket pairs the pair of twos of the same suits as your a8. This has a .521 ev against your a8o. Clearly a call is correct since he is risking x-2 dollars to win 2x dollars. (x-2)/(2x) is less than 1/2 for all positive x, since his ev on any pocket pair under 8's is above 1/2 it is always correct for him to call.

On a side note that .521 ev is about the house ev on some blackjack variations, and I don't think you would argue that its not profitable for casinos to deal blackjack.

[Ikke]

Was way too quick with my response and didnt think it through. A has to move in with A8 anyways when x<49 (so when B is right to call with inferior hands as well) so keeping his opponents calling standards a function of x is just making things way more complex then is necesarry.

Nice problem!

Regards

[Duke]

He could correctly call with worse hands depending on how many chips he has compared to your 69 (or however many you have left), and also depending on the stack sizes of the other players involved (if it were a tournament).

If you raise to 5 all in, well, he's getting 7-3 on his call with 95s. You like the call, but it's correct for him as well. If he has $1M in chips, your 69 means nothing and he will call with anything just to eliminate you.

Extreme examples, but the point is that he's correct to call with a lot of hands depending on how many chips you both have, or how many you have relative to the blind. This does not change whether or not it's correct for you to push with the A8, either. But wait, it does.

If this is a tournament and he has a ton of chips, well, he's calling with anything, and you're risking elimination as what is at best a slight favorite. With 1-2 blinds, this makes no sense if you have $69 left.

It would be different in a cash game.

~D

[The Gift Of Gab]

That's strange. Now I can't figure out how I got that number...my slide rule must be broken.

I think I was a little confused by irchans mixing Xes midstream, but it looks like he was right all along.

Some of the EV numbers are off by a few percent, but that won't make much difference to the result or what the opponent should call with.

I thought the answer would be around 20 before I saw any numbers.

In a vacuum, I think most opponents would call much less than they would if they could see your A8o, because with many hands it's easy for them to believe that they're in a classic small favorite/big dog situation. But I play mostly NLHE cash games, not tournaments.

[muck_nutz]

[ QUOTE ]

This is close enough that to kill this thing for good someone should redo it using more accurate EV numbers and include A8s. You'd have to calculate multiple EVs per hand to take suits into account and weight them.

I would much rather go play cards then do this myself, so, uh, I'll let others elaborate.


[/ QUOTE ]

Boredom got me. Internet provider was having problems. No game close by. Reading material staring back at me wasn't ineresting. And I had to be in the house. So I wrote the program to do the detailed sum (used the source forge poker-eval lib also used by twodimes.net). I ended up with 73.92. Thats higher then your estimate which suprises me. Probably wrong. Oh well, internet is back up and life intrudes on silly little amusements.

[johnnyhearts]

I have two questions with the solution. Firstly, it seems to me that the "expert" probably won't gamble with pocket pairs smaller than 88 but would stick to pocket pairs higher than this and, of course, dominating A-high selections. This would have the effect of lowering his call percentage from 10% to about 8% and also boosting him to approximately a 3/1 favorite when he does.

Secondly, why aren't the blinds considered to be the pot that already exists? Perhaps I misunderstand the equation.
My equation would set EV when the "expert" folds equal to expected loss when he calls. (ie.)

.92 x 3 = .08[.75(x+3) - .25(x+3)]

2.682 = .04x

x = $67

Perhaps you can help me to better understand these equations and, by the way, a lot of posters would love to see me proven stupid!

Johnny Hearts/ Team Leisure

[johnnyhearts]

It seems I made an error. On the expected loss side of the equation the preexisting pot isn't a part of the loss, however, it is a part of the compensation when he sucks out! Therefore I will amend my equation and answer to:

.92 x 3 = .08[.75x - .25(x+3)]

x = $67.50

[muck_nutz]

[ QUOTE ]

Firstly, it seems to me that the "expert" probably won't gamble with pocket pairs smaller than 88 but would stick to pocket pairs higher than this and, of course, dominating A-high selections.


[/ QUOTE ]

Why would you say that?

This reminds me of a conversation that took place in a NLHE game I was playing in one night. One fish says to the other about a WPT final hand (layne vs howard): "I can't believe he would go bust with 33. I mean, I know its a favorite to A9 preflop, but I still wouldn't go bust with it." I (silly me) ask what the blinds, antes, and action was. He ask, "what does it matter". I nod my agreement and go back to playing.

[johnnyhearts]

I will graciously assist you with your query. The blinds are $1 and $2 and the stacks are: small blind = $65-70 depending on whose solution you agree with, big blind = covers the small blind. The action is folded to the blinds. Couldn't you have read Mr. Sklansky's original post for this info though?

Beep-Beep Hearts

[Copernicus]

He wasnt asking about the blinds, dummy, he was commenting on your statement that he would fold hands below 88. They are favorites over A8 and if you had bothered to read DS's posts you wouldnt have bothered to eliminate them.

And not only did you set up the problems wrong but your arithmetic was wrong in both. Good thing you are (supposedly) a GoL, because you certainly couldnt cut it in job that uses math.

[johnnyhearts]

During a discussion with a fellow gentleman of leisure it was discovered that Mr. Sklansky wasn't referring to a tournament, therefore, the expert will take the edge on smaller pairs. I stand corrected on that issue.

I will also admit that the last term in my equation should be (x + 2) since the BB has an extra chip in his blind and is compensated $1 on his call of x. I also suspect I might have made a rather small "bookkeeping" error in my calculations!

Nevertheless, I challenge somebody to explain what is wrong with this equation!

.89(I round 10.7% up) x 3 = .11[.7x - .3(x+2)]

x = $64 approx.

For you doubters, how about this? The EV on the dollar amount cutoff needs to be set at 0. In other words, it wouldn't matter whether you pushed in x or folded the hand.
Therefore the equation would be:

EV when BB folds + -EV when he has AA + -EV when he has 88-KK + -EV when he has 77-22 + -EV when he has A9-AK = 0 (I agree that A8 can be tossed since the only situation that matters is when the big happens to have A8s.)

.89 x 3 + .002[.07(x+2) - .93x] + .027[.25(x+2) - .75x] + .029[.45(x+2) - .55x] + .049[.25(x+2) - .75x] = 0

x = $64 approx.

Now, I would like to know what is wrong here. Perhaps Mr. Sklansky would be so kind to explain or perhaps one of the Johnny Shout-down Club would like to take a crack at it. Prove Johnny's a dummy or hold your peace, gentlemen!

Johnny Hearts/ Team sofas and brewskies

[Wake up CALL]

" Prove Johnny's a dummy or hold your peace, gentlemen!"

Now remember I am a member of your fan club Johnny but why would anyone bother to prove you are a dummy when you have done such a fine job of providing proof yourself in this thread? /images/graemlins/smile.gif

[johnnyhearts]

I won't be mean to Wake Up Call since he's a fan of mine but I will comment that nobody has told me why folding gives an EV of -1 automatically irrespective of what is happening situationally in this problem! (ie.) Folding with $20 is a -EV while folding with $500 is a +EV. That's why I set EV at 0. If the -1 is referencing the $1 small blind it would seem this needs to be considered part of the pot. You don't lose it by folding. In addition, why is it .9 x $2 in Mr. Sklansky's problem since when the BB folds you win $3 not $2? Once again, maybe I don't understand but unless someone can explain or, in the event you're not that smart, run a computer simulation then I will now start assuming I'm right!

Send it!

Johnny "Dummy" Hearts

[Bozeman]

See my reply to giftofgab above for a long ago answer to this