Game Theory: Unusual Question #3 and #4

[David Sklansky]

By now some of you must realize that my "unusual questions" (that I don't profess to have an exact answer for) are meant to help you see why I believe excellent players should limp fairly often in very early position. Even when your opponents are not fish (but do play worse than you).

To complete this process I now ask two very precise game theory questions (again I have not calculated the solutions but I believe I will know the right answer when I see it.)

Two players, A and B receive one card with a real number between zero and one. High card wins. Player A bets one dollar in the blind. Player B can fold, call, or raise one dollar. If he raises, A can call or fold only. End of game.
What is the optimal strategy for each player and the value of the game to B?

To get you started notice that if there were no raises allowed, B would simply call whenever he had above .5. When he called he would win three quarters of the time (half of the time they were both above .5 and all of the time A was below.) Thus in eight tries he would fold four, win a dollar three times and lose a dollar once. The game is worth 25 cents to him. Since the blind is not live, the existence of an option to raise must help player B because if it didn't he could revert to the never raise strategy.

The second version of this question makes A's blind live. So if B just calls, A can raise if he chooses. Game over. Logic says that the value of this second game to B, if both players play optimally, is less than it would be if the blind wasn't live. Solving this second game for optimal strategy is probaly a lot harder than the first one. But I know we have some two plus twoers who can do it.

For the majority of you who know they are not familiar enough with game theory to tackle these proplems precisely, I ask a more general question. Should B raise more often in the first game or in the second game where's A's blind is live?

[ZeeJustin]

[ QUOTE ]
Two players, A and B receive one card with a real number between zero and one. High card wins. Player A bets one dollar in the blind. Player B can fold, call, or raise one dollar. If he raises, A can call or fold only. End of game.
What is the optimal strategy for each player and the value of the game to B?

[/ QUOTE ]

B clearly folds anything less than .5, and calls with a .5. The tough question is when is raising profitable? B will raise with #'s > X, where X > .5. A will then call with Y, where Y >X. If Y calls with # < X, he is gaurenteed to lose.

Let's say X = .75 (i like the ol' guess and check method). What would the optimal value for Y be?

Y is getting 3:1 pot odds on the call, so he has to win 1/4th of the time. Calling with .76 would obviously not win 1/4th of the time, and calling with .875 would win half the time. I'm assuming the average of .75 and .875 is what he calls with -> .8125.

So what's the EV of this strategy for B?

1/2 the time, he folds. EV = 0
1/4th of the time, he calls. When he calls, he wins 3/4 of the time. EV = $.75
1/4th of the time he raises, we have to split this up.

When he raises, A calls 18.75% of the time.

25x.1875 = 4.6875% of the time, B raises, and A calls.
25-4.6875=20.3125% of the time, B raises, and A folds. EV = $1

When A calls the raise, he wins 1/4th of the time. So in 4 tries B wins $2, wins $2, wins $2, and loses $2. He wins $1.50 on avg when he raises, and A calls.

50% of the time ev = 0.
25% of the time ev = $.75
20.3125% of the time ev = $1
4.6875% of the time, ev = $1.50

[(.5)(0)+(25)(.75)+(20.3125)(1)+(4.6875)(1.50)] / 100 = .4609375

I think this is player B's ev with this strategy, + $.4609375 per game.

So did I calculate the EV correctly? If I did, is this the highest EV attainable? I prolly screwed up somewhere. I'm still in highschool, so I haven't taken any game theory classes or anything.

[Bozeman]

Justin, B's optimal strategy will be to raise with some of his best hand but also to bluff raise with some bad hands.

[ZeeJustin]

[ QUOTE ]
Justin, B's optimal strategy will be to raise with some of his best hand but also to bluff raise with some bad hands.

[/ QUOTE ]

This is incorrect. Let's say he raises all hands .6 (the number is arbitrary) and above, and then bluffs with 10% of his hands. Wouldn't it make more sense to just change that to raise .5 and above instead? Instead of raising 10% of bad hands, just raise the next 10% of hands after the cutoff point.

[Louie Landale]

B should pretty clearly call with hand .51. So if he was to add bluffing hands he shouldn't do it with hands that have positive EV by calling; he should do it with hands that have no EV at all; in this case his worse hands.

[Gandor]

[ QUOTE ]
[ QUOTE ]
Justin, B's optimal strategy will be to raise with some of his best hand but also to bluff raise with some bad hands.

[/ QUOTE ]

This is incorrect. Let's say he raises all hands .6 (the number is arbitrary) and above, and then bluffs with 10% of his hands. Wouldn't it make more sense to just change that to raise .5 and above instead? Instead of raising 10% of bad hands, just raise the next 10% of hands after the cutoff point.

[/ QUOTE ]

If you select just the next 10% of hands, player A can adjust his strategy knowing he needs a little less to call. Using the bottom 10% to bluff is much more effective. When you raise, player A must decide if you are bluffing or not. If you always raise with the top 40% in your example, then 40% of the time you have a strong hand and player A must be even stronger to call. But player A also knows that 10% of the time you raise, he has you beat. Then to call becomes a guessing game for him.

[Aisthesis]

Ok, we've seen that B's strategy of calling at above 1/2 and raising at over 3/4 is not optimal. But what is A's best counter to this strategy?

On the one hand, if A ALWAYS calls the raise when he has a hand 3/4 and above, it doesn't change B's EV at all over and above the non-raising scenario: A folds to the raise below 3/4 and loses only the $1 he would have lost anyway. But if both players have hands above 3/4, it's a wash with 0 EV for B. So, the value of the game is just .25 for B, as it is in the non-raising game.

This just seems a little strange to me, because A is only getting 3:1 pot odds on the call. So, it would in a sense seem that if he had the hand .75000000001, he doesn't have the odds to call. B will win his additional dollar 99.99999999% of the time?!

So, if A calls the raise on on the top 2/3 of the hands that B will raise (calls raise at 5/6), what happens? I get that he actually REDUCES B's EV from the original scenario by 1/48 to 1/4-1/48 = 11/48.

Is this correct?

[Bozeman]

I showed above that for (bad) strategies where B calls with > 1/2 and raises with >b, and A uses optimal response, the value is

"1/4-1/8*(1-b)^2"

This leads to 31/128 (=1/4 - 1/128) for b=3/4. The error you made is that A needs to call with 3/4 (not 2/3) as many hands as B raises getting 3:1 odds. If A calls with (less than optimal) top 1/6 of hands, B's EV is 1/4 -1/144. (the correction is 1/3 of what you computed).

Craig

[Aisthesis]

Yep, checked it through, and you're right. Thanks!

It's rather interesting that the ONLY thing making the raise profitable at all is the bluff. If B never bluffs, then A can always find a (tight) calling criterion to make it in fact unprofitable for B to raise (as opposed to the just calling with 1/2 or better and never raising). So, if B never bluffs, A can successfully punish B for raising.

On the other hand, it doesn't take very much bluffing to force A into dramatically looser calls, hence making the raise profitable for B. Moreover, the bluffing criteria and raising criteria for B always force A into an exact call point. It would be kind of interesting to figure out (given B's optimal value-raise threshold) just what the correlation is to A's call treshold if B bluffs more or less than is optimal.

On the other hand, while B's optimal values force A into an exact calling/folding strategy, it doesn't seem to me that this applies the other way around: If A didn't know that B was switching strategies and hence kept calling the raise on the top 1/3 of his hands, instead of bluffing, B could increase his winnings by value-betting only.

I guess what I'm basically saying is: B's optimal strategy forces A into exactly one optimal betting strategy. If B sticks with his optimal strategy and A changes his, A will just lose more money.

But A's optimal strategy by no means forces B into his true optimal strategy. It only does so under the assumption that A will adjust his strategy if B changes his. If A is incapable of adjusting to changes in B's strategy, then B can also capitalize on his inflexibility.

Or yet another way of putting it: A has no way to punish B for inflexibly sticking with the optimal (bluffing) strategy, but B can punish A for inflexibility.

[ProfessorJC]

I don't think there is any calling in version 1.
Bet top 1/2 and bluff bottom 1/6.
Fold remaining 1/3.

THe blind calls with his top 4/9 of hands

B wins .27 units per hand.