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Old 03-10-2012, 12:54 PM   #1
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Has The Basic Simplest NL Toy Head Up Game Been Solved

You and your opponent have a stack of n dollars. You are both dealt a real number between zero and one. Highest number wins. A stranger starts the pot off with a dollar of his money. No other blinds or antes. The first player bets or checks. If he bets, the second player calls or folds. If he checks there is a showdown. (First player obviously has the edge because he breaks even by checking everything.)

Say the player has been dealt the number x. In terms of x and n what is his strategy? I assume there will be cases where he sometimes checks and sometimes bets and if he bets he sometimes bets one amount and sometimes bets at least one other alternative
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Old 03-10-2012, 02:28 PM   #2
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Re: Has The Basic Simplest NL Toy Head Up Game Been Solved

You bet a different amount for every value bet worthy hand you hold. Yes, that means an infinite number of bet sizes is optimal. This would seem to telegraph your hand strength, however, each of these value hands is paired with the appropriate frequency of bluffs for each bet size, so that your opponent does not know whether they should really be calling or folding.

If you want details on this NL half street game, it's found beginning on p154 in MOP.
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Old 03-10-2012, 05:46 PM   #3
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Re: Has The Basic Simplest NL Toy Head Up Game Been Solved

When you say breaks even, I'm guessing you mean to say he realizes 50% of the equity in the pot because his opponent will never have the opportunity to bet given that he can only call or fold. He wins on average at least 50 cents every time he plays the game because a stranger put the initial dollar into the pot.

The first player's pot equity will always be the real number that he is dealt - i.e. if he is dealt .854327, if he checks he will win 85.4327% of the time.
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Old 03-10-2012, 11:13 PM   #4
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Re: Has The Basic Simplest NL Toy Head Up Game Been Solved

I'm not sure you can actually "solve" this toy game at all. Not without defining N and alot of other conditions first. I think the solution to this game with a stack of $1 each would look very different than the solution to the game with $10,000.00 each.
But, because you'll have +/-0 EV just checking I think it would make sense only to bet if you have a number that's nowhere near .5. I think the game could be broken down into bet sizes based on your number when your number is .75-1.0 only, and .25 or less. So, your equity is going to be your number times $1.00. But you don't want to bet 40 cents when you have a .4 because you'll always be better off checking in that situation. Now it turns into a game theory problem. Does your opponent understand the game and does he understand that you understand it, etc? If you're on the same page he's not going to pay you off with .6 because of the GAP concept unless you make a very small wager and I think we're dealing with infinities here and its not a "Basic or Simple Toy game" at all.
Without thinking too much into this it feels like a broken game to me.
I think you're playing rock paper scissors to some degree here and the game needs to be better defined for meaningful discussion. I mean it seems like you can EITHER play this GTO OR try to make the most money but not both.
I think you could play an unexploitable and profitable game simply by betting when you have a hand in the top .75 and checking if not. I think the GTO version of this game would have you starting out with only one bet size say $1 into the $1 pot and always betting when you have a hand in the top 75%, else checking and then you'd have to start to adjust from their to make the most money against a particular opponent ballancing and levelling as you went. If he seems to be folding to much start bluffing with hands in the bottom 1/3 of your range, if he's calling too much start betting the top 60% of your range.
I think you could find different amounts for different numbers that would yeild the same results, like deciding what to value bet. A 50 unit bet with a 50% chance of being called is the same as a 25 unit bet with a 100% chance of being called.
I've been trying to wrap my head around this and come up with a concrete set of rules and I don't think it's just a limitation of my math knowledge. I don't think one exists.

Essentially it's this; the most sure way to profit is to only bet when you have a 1. You'll never lose money long term and will surely profit. The more hands you start to value bet and even bluff with the more you stand to profit but the more exploitable you become. Identifying the proper strategy would require first identifying your goal, the lowest risk or highest reward, and it's still going to be very dependant on your opponent and what he thinks you'll do plus the actual amount of your stack sizes. Is it really better to wager one amount than another when you have a certain hand? The more you wager the less likely you are to be called and visa versa. With only the info given in the toy game I don't think it makes any more or less sense to bet $1 than it does $100 with a top 5% hand.
But maybe I just found the limits of my logic skills and I'm full of it.
I'll think about this all night while trying to sleep and likely have a totally different take on it tomorrow.

BTW, OP, you are my aboslute HERO.,
I love you, in a purely plutonic, non-homosexual, male bonding sort of way.
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Old 03-11-2012, 07:37 AM   #5
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Re: Has The Basic Simplest NL Toy Head Up Game Been Solved

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Originally Posted by Donovan View Post
Essentially it's this; the most sure way to profit is to only bet when you have a 1
With a range of "real numbers between zero and 1" you will be making this bet so infrequently (If you played this game for a year straight its likely neither of you get dealt a 1 I'd guess) that the opponent would never call your bet. Your strategy is too face up (he can fold .99999)...so only betting 1 is the same as checking 100% essentially.

You will need to be overbetting the pot a lot to maximize here I'd guess....mostly as bluffs, because he will have to have a very high number to call your overbet bluffs, and your big winrate would come from the few times you do get dealt a higher number and get a call.

(say you overbet to 100, he has to be good .990099% of the time to call , so he will/can only call you with .99009901 or higher. I'd guess you should definatley be betting a LOT of numbers under .500 along with the top 50% of numbers between .990099011 and 1 to this amount.......so he almost always folds but when he calls you he's coolered over half the time)the bets in between should have some more conventional sizing probably.
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Old 03-11-2012, 01:19 PM   #6
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Re: Has The Basic Simplest NL Toy Head Up Game Been Solved

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With a range of "real numbers between zero and 1" you will be making this bet so infrequently (If you played this game for a year straight its likely neither of you get dealt a 1 I'd guess) that the opponent would never call your bet. Your strategy is too face up (he can fold .99999)...so only betting 1 is the same as checking 100% essentially.

You will need to be overbetting the pot a lot to maximize here I'd guess....mostly as bluffs, because he will have to have a very high number to call your overbet bluffs, and your big winrate would come from the few times you do get dealt a higher number and get a call.

(say you overbet to 100, he has to be good .990099% of the time to call , so he will/can only call you with .99009901 or higher. I'd guess you should definatley be betting a LOT of numbers under .500 along with the top 50% of numbers between .990099011 and 1 to this amount.......so he almost always folds but when he calls you he's coolered over half the time)the bets in between should have some more conventional sizing probably.
The problem here is semantics I think. Usually we're looking for GTO or unexploitable play, but you can literally x all hands unexploitably so we're looking to max profit I guess. Like I said, i think it needs to be clarified somewhat.
I came up with a new approach to this game but I am still not sure that any bet sizing is actually best becaue the amount of your bet has an inverse relationship with the hands you'll be called with. If you bluff, the higher you bet the more it succeeds but the more costly it is when it fails and if you value bet the higher it is the more you win when called but less likely you are to be called. I don't know if there's any reason to vary your bet size until you have history with opp and I think this toy game is really ALL about capatilizing on opp mistakes, that is; adjusting.

Here's my example strategy; assume $10 stacks and an unknown opponent.

To begin with I will always bet $1 into the $1 pot. My opponent should call with anything >.33 because he'll be getting 2 to 1 and his hand will have more than 33% equity vs random hand and has no reads on me. He should quickly realize I'm checking all else and only ever betting for value. Once he starts folding enough hands that it's likely he has adjusted and is now only calling with the top 2/3 of my raising range (2/3 is fine becaue he's getting 2 to 1) I will adjust. He won't often be calling with middle hands or folding stronger than average hands. I can start betting the bottom of my range, say 0-.25 because he will fold better hands, checking the middle of my range because he generally only calls with better/folds worse, and I will want to start value betting tighter, maybe .75+. So after he's conditioned to call with only top 1/3 of his hands, I polarize my range to very strong hands and bluffs. After a bit, my opponent can and should adjust by adding in the middle of his range as bluff catchers (I will be betting 2/3 of my hands but 1/2 bluffs and 1/2 for value so he can call with my checking range getting 2 to 1 and at least break even). Now that he sees what I'm doing I can stop betting the weak hands and start betting anything >.5 for value again.

I think that's the right approach. Start betting any hand that is a favorite but giving him odds to call with worse until he stops paying off with worse. At that point I drop the bottom of my value bet range and only value bet very strong hands (top 33%? depending on how he adjusts) and I can add in the bottom of my range 0-.33 as bluffs, and just check the middle 3rd of my hands. He'll eventually see my range is polarized (and SUPER exploitable) and should adjust by calling with the top 2/3 of his hands knowing I'm only betting hands that beat even the bottom of that range 1/3 of the time and he's getting odds to boot. At that point he will start calling VERY whide and I can value bet him to death with anything greater than average. The tricky part is to identify when he's adjusting ASAP, especially once I start polarizing my range and make it profitable to call whide, but the benifits of him calling whide again are hugely exploitable. So, basically, If he adjusts by calling whide then I just start over.
Chances are though that in reality he will fail to adjust properly to one or more of my gear changes so, not only do we profit by being in +EV strategy until he adjusts but we profit on every adjustment phase for as long as it takes him to adjust but, better still; we're likely to find a weakness and be able to exploit his failure to properly adjust.
We should get the best of him right out of the gait, and as long as we can reckognize his reactions better than he catches on to ours we should own him hard. After all, we have a plan and we have the initiative; we control the tempo and he has to react to us. If we can't win this game we should quit poker for sure.

Thoughts????
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Old 03-11-2012, 01:50 PM   #7
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Re: Has The Basic Simplest NL Toy Head Up Game Been Solved

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Thoughts????
I think that you don't have full understand the problem, when we talk about game theory optimal (GTO) we are looking for the Nash equilibrium for the game, that are a couple of strategy such that no one can exploit the other player.

So there is no need to "adjust", you keep playing the same strategy in every hand, something like:
player 1: bet to 1 with any hand in 0-0.1111 or 0.77778-1, check everything else
player 2: call with any hand in 0.555556-1.
(this is optimal in the restricted game where only one bet size is allowed for every stack-size>=1, otherwise you should go all-in but the frequency change)

I don't have MoP, but after some thinking seem that using every possible bet-size between 0 and n should be optimal, but i've not worked out the exact function between the hand that you have and the action that you should do.

Should be something similar to:
bet to f(x) with any hand (f(x) should be 0, so checking, for a lot of middle value hand)
call with every hand between g(bet size) and 1
for same function f() and g() that depends on the stack-size.
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Old 03-12-2012, 02:37 PM   #8
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Re: Has The Basic Simplest NL Toy Head Up Game Been Solved

The solution to the constrained game carlop was talking about, ie if the player can either bet amount B or check, he can't choose how much to bet, is the following:

First player bluffs [0, B/D] and value bets [1-(B+1)/D, 1]
Second players calls [1-2(B+1)/D, 1]
Where D = (2+B)(1+2B) = 2B^2+5B+2

The solution to the unconstrained problem can be found in MoP where Gibert said
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Old 06-18-2012, 04:41 AM   #9
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Re: Has The Basic Simplest NL Toy Head Up Game Been Solved

if theres a 1 dollar freerol every hand, i will multitabling that game and check 1 M hands, having a infinite ROI and a 500k expected value
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Old 06-18-2012, 01:05 PM   #10
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Re: Has The Basic Simplest NL Toy Head Up Game Been Solved

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if theres a 1 dollar freerol every hand, i will multitabling that game and check 1 M hands, having a infinite ROI and a 500k expected value
Sure, because you're freerolling.

However, I think this problem could be altered in this way:
Instead of being a stranger putting the 1$ in the pot, it would be each player putting 1$ in the pot every other hand, like a blind. Now, each player would play the roles of 1st and 2nd players an equal amount of hands, be it switching every hand, every 100 hands, every 1k or whatever arbitrary number.

The goal is now to make more money than your opponent.

Edit: To avoid confusions and/or speed the game if played in real life, the blind could always be posted by the same character, either 1st player or 2nd player, if they were to switch roles every N hands with N <> 1.

Last edited by uberkuber; 06-18-2012 at 01:08 PM. Reason: Added thought
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Old 06-18-2012, 03:56 PM   #11
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Re: Has The Basic Simplest NL Toy Head Up Game Been Solved

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Sure, because you're freerolling.

However, I think this problem could be altered in this way:
Instead of being a stranger putting the 1$ in the pot, it would be each player putting 1$ in the pot every other hand, like a blind. Now, each player would play the roles of 1st and 2nd players an equal amount of hands, be it switching every hand, every 100 hands, every 1k or whatever arbitrary number.

The goal is now to make more money than your opponent.

Edit: To avoid confusions and/or speed the game if played in real life, the blind could always be posted by the same character, either 1st player or 2nd player, if they were to switch roles every N hands with N <> 1.
If you change the game like this it's just a zero-sum game with the players switching roles...if we're looking for an optimal strategy for each player, when they switch roles the strategies won't change so long term both players would have to have 0 ev
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Old 06-18-2012, 04:39 PM   #12
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Re: Has The Basic Simplest NL Toy Head Up Game Been Solved

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If you change the game like this it's just a zero-sum game with the players switching roles...if we're looking for an optimal strategy for each player, when they switch roles the strategies won't change so long term both players would have to have 0 ev
Agreed, granted they both played GTO.

My post was in response to the one just above. He seemed to think that since it was a freeroll, he could use the check line 100% of the time as the "1st player". Sure he would make money on the long run (because he's not the one putting the 1$ in the pot), but that is certainly not GTO.
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Old 06-18-2012, 04:52 PM   #13
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Re: Has The Basic Simplest NL Toy Head Up Game Been Solved

It's not, but it's actually not a terrible starting point if you're looking for an optimal strategy. I haven't seen any of the books that have been referenced in this thread so I don't know how explicit their solutions are, but since player one has ev of x just by checking he needs to be betting with a frequency/amount so that he has at least ev of x (or exactly if variance isn't a concern) for each x given player two's calling range.

(In thinking about it more, I think there might be a range of values that player 1 will always bet, so the above would only have to hold for values outside of that range)

Last edited by muuuuuuufasa; 06-18-2012 at 05:06 PM. Reason: clarification
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Old 06-18-2012, 05:53 PM   #14
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Re: Has The Basic Simplest NL Toy Head Up Game Been Solved

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Originally Posted by muuuuuuufasa View Post
It's not, but it's actually not a terrible starting point if you're looking for an optimal strategy. I haven't seen any of the books that have been referenced in this thread so I don't know how explicit their solutions are, but since player one has ev of x just by checking he needs to be betting with a frequency/amount so that he has at least ev of x (or exactly if variance isn't a concern) for each x given player two's calling range.

(In thinking about it more, I think there might be a range of values that player 1 will always bet, so the above would only have to hold for values outside of that range)
I'm on the same page, it's not a bad starting point, although it looked like it was also an ending point for the said poster. Time will tell us.

By always checking, it seems that we give up a positional advantage, but possibly a positional disadvantage as well. At first, I assumed that player one had a better position (because two cannot raise nor bet if checked to him), but I'm not 100% convinced anymore.

Edit: Checking a 1.000000 being a mistake, that alone proves that always checking isn't GTO.

Last edited by uberkuber; 06-18-2012 at 05:55 PM. Reason: Added thought
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Old 06-18-2012, 10:02 PM   #15
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Re: Has The Basic Simplest NL Toy Head Up Game Been Solved

Not only that, but P2 can't raise even if it's bet to him, so P1 definitely has a positional advantage. As a start, if we constrain P1's options to bet 1 or check and look at only 1 hand, P2's strategy has to be that he calls with any hand ≥ x where F(x)=.33 and F(•) is the conditional distribution of of player 1's hand values given that 1 has bet.

However, if 2 is employing a cutoff strategy like this, one isn't going to want to mix up his play once he receives his value; he's either going to want to bet or check depending on what has the highest ev. On the bottom of his range, he's better off checking hands that are close to x (since the only calls beat him and he folds out what he'd beat anyway). So he's going to check anything above y where y satisfies:

(1-x)*-1+x*1=y
2x-1=y

We have a similar situation on the top side of his range; one only wants to bet hands above z where z satisfies:

(z-x)/(1-x)*2+(1-z)/(1-x)*-1+z=z
(z-x)=2*(1-z)
3z-x=2
z=(2+x)/3

The problem I'm running into is that since x needs to satisfy F(x)=1/3, x either has to be less than y or greater than z, which violates 1's problem above. Obviously I might be making some mistakes here and there might be a more complex set of strategies that solves the initial game (taking into account multiple hands) but at the moment I can't see how to escape the fact that for any set of strategies on any given hand it seems like both players would want to deviate.

That's just where I'm at now, interested to see if anyone else has thoughts on this
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