[0,1] game exercise
 

This question is more for novice [0,1] gamers such as myself, just as an exercise in applying the principles recently learned (I hope!) in mulling over David's problems. It will definitely be totally easy for those who had no trouble with #4 but (for me, at least) seems slightly more difficult than #3.

The game is another simplification of David's #4: A bets $1 blind. B can then limp or fold (B cannot raise). If B limps, A can then check or raise. Finally, B can call A's raise or fold his hand. So, basically, the game is the same as #4 except that B loses his raising option.

What are the optimal strategies for A and B? And what is the value of the game?

I'll post my own answer separately to see if others agree or if I'm still missing something on this.

Re: [0,1] game exercise
 

My answer:

B: limps on [11/20,1]
calls a raise on [7/10,1] (otherwise folds)

A: bluff-raises [0,1/20]
value-raises [17/20,1] (otherwise checks)

The value of the game is on my calculation 81/400 for B.

My reasoning (I was unable to completely avoid "accounting" here but hope that I was able to simplify enough to keep that part to a minimum):

Let's say x is the limp threshold for B. To make A indifferent to bluffing, B will have to call 2/3 of the time that A raises because the bluff costs A only $1 to potentially gain $2.

So, B will have to call a raise with hands above (2/3)*x + 1/3

Thus, A will value-raise on the top half of those hands--i.e., over (1/3)*x + 2/3.

Since B is getting 3:1 on the call, A will have to bluff once for every 3 value raises. So, A will bluff raise up to 1/9 - (1/9)*x

Now, we want to solve for x so that B is indifferent to limping or folding at x; hence when the value of limping at x is 0.

A's hand result for B
[0,1/9-x/9] -1
[1/9-x/9,x] 1
[x,1] -1

So, we get the indifference equation: 0 = -1/9 + x/9 + x - 1/9 + x/9 - 1 + x = 20x/9 - 11/9

So, x = 11/20

Just plugging this value into the formulae for the other values should give the results indicated above as to the indifference thresholds.

So, now we just have to calculate the value of the game. A's bluffs all just break even. B wins a bet when he has [11/20,1] and A has [1/20,11/20]. That has a value of 9/40.

If A couldn't value raise, that would be it, because then they would also break-even on the remaining hands. But B also loses 1 additional bet (he actually loses 2 bets, but 1 of these he would have lost anyway even if A couldn't raise) when he has [7/10,17/20] and A has [17/20,1]. That happens with a probability of 9/400.

So the value of the game for B is 9/40 - 9/400 = 81/400.

Re: [0,1] game exercise
 

I guess when compared too Sklansky's #4, this would show the value to B of the raising option. Or put another way, how much does B lose out on in #4 if he adopts the strategy of either folding or just limping?

PairTheBoard

Re: [0,1] game exercise
 

True, and it looks to me like it's not much--although I really put this problem out there without any heuristic intentions with regard to poker. I'm just trying to learn how this blasted game works! [img]/images/graemlins/wink.gif[/img]

Question for Jerrod
 

An interesting game working toward the NL scenario would be the following: A and B both have stack-sizes of, say, $4 (also assuming immediate rebuy after a loss and cash-out on anything over $4 just to keep it simple). And to keep it simple, let's initially make it like David's #3, except that B can limp or make any raise in increements of $1 (3 different raising options). A has no raising option but can only call or fold to a raise.

Is this one pretty doable? I went through just enough of your "primer material" to get me started and noticed that you do have some NL-type problems (I did read through the infinite raise question), and it may also be covered there.

My inclination would be to work backwards from the $4 raise (very top hands plus some bluff hands), then to the $3 raise (slightly less premium hands plus presumably a few more bluffer hands), then the $2 raise. The limp threshold (since A has no raise option) would again have to be 1/2, as in David's #3, I would think.

I guess it really doesn't matter what order you do them in, though, since the only difference on the $2 raise hands is going to be that part of them are going to be raised more. So, essentially, wouldn't one just be solving the same problem with different pot odds, depending on the size of the raise?

Re: [0,1] game exercise
 

P.S.: Just doing some approximation, it looks like the difference in value between this game and the real #4 is just a bit over 1/36--not surprizing, since 1/36 was the gain for B when given the raise option in #3.

Re: Question for Jerrod
 

Actually, this simplified NL case doesn't look quite as difficult as I had thought (or am I fooling myself?). In fact, one could make the following change with a presumed interesting result: If B raises, he must put in at least $2 but can put in any real number value up to $4.

In that case, I would think, if he raises $2, he will have either exactly 5/6 or the comparable bluff hand. And up to the $4 threshold, there should be a one-to-one mapping on value-raises between his hand and the size of his bet. It's just that, given a raise-size, A will always have to figure that B either holds THE value-raise hand (on values strictly between 2 and 4) or a bluffing hand.

Re: [0,1] game exercise
 

If this is correct then what we have is:

No Raise - .25
B Raises - .278
A&B Rais - .236
A Raises - .203

A acts last in these games, essentially having both the blind and the Button. It looks like what these results show is that the Power of the Raise is something like 67% more valuable to the player in last position.

PairTheBoard

Testing your hypothesis
 

That definitely seems to be the case for all of the games proposed.

Consider the following game, where the playing field is completely level except that A is last to act: BOTH A and B put in $1 blind. B can check or raise $1. If B raises, A can only call or fold. But if B checks, A can either check or raise (on which B can choose only to call or fold).

A should win this game if your hypothesis is correct.

Re: Testing your hypothesis
 

I would hope so. It should prove the value of position for games like this.

PairTheBoard