Game Theory: Unusual Question #3 and #4

[Bozeman]

"The value-raise and bluff-raise criteria for B must be the same as in #3 because A has no additional options when B raises. So, I don't see how there can be any difference there."

This is false: value-raise criterion depend on A's option because now B will raise more hands because it preempts A's raise. And the bluff-raise criterion is set by the value-raise criterion.

I am willing to bet on the correctness of the proposed optimal strategies.

Especially in the absence of anything close to a viable alternative. This isn't rocket science,
Craig

[Aisthesis]

Expanding B's raising criteria to pre-empt A makes sense to me.

But the rest of what you say conflicts with your solution to the problem: You have B value-raising exactly the same hands in #3 and #4 (5/6 through 1).

But you have B bluff-raising more often in #4 (1/9) than in #3 (1/18). That makes no sense to me if "the bluff-raise criterion is set by the value-raise criterion."

But I definitely see your point on B making some additional pre-emptive raises. It's just a little surprizing to me that you don't then expand B's value-raises at all.

[PairTheBoard]

I thought this was the agreed upon correct solution. Posted by wells and agreed to by Bozman?

http://forumserver.twoplustwo.com/sh...;o=14&vc=1

It has B value raising from 3/4 up to 1 as opposed to the 5/6 up to 1 for problem #3.

Although the Raise by B meets similar conditions as in #3, the Calls do not. So it could make sense for the Calling Interval to change. Both on the low side to folds and the high side to raises. Assuming the Linked to Solution is Correct.

Also Aisthesis, I understand you to say that A's bluffing frequency should be greater than B's because of the pot odds. When A raises B has greater Pot Odds to call. Doesn't Game Theory say the raising frequency should be less in that case?

PairTheBoard

[Aisthesis]

On raising frequency, B seems to me to have exactly the same pot odds. But A is risking less to win more when he raises, since he has already put $1 in blind.

That's not Bozeman's view of optimum. He agrees with JA, who has B raising top 1/6. Cf. his comment to well's post on this: "Jerrod and I got the same strategies, and you and I got the same value, Jerrod apparently making an arithmetic mistake."

So, Bozeman thinks Jerrod's criteria are correct but that the EV for B is 17/72 on that scenario rather than 1/4.

[PairTheBoard]

When B raises A is getting 3-1 pot odds to call. So B should bluff raise 1 time for every 3 value raises.

When A raises B is getting 5-1 pot odds to call. So A should bluff raise 1 time for every 5 value raises.

Do I have this right? Game Theory is new to me as well.

PairTheBoard

[Aisthesis]

Well, you should probably take this with a grain of salt pending an answer from a non-newbie to this area. But it sounds to me like the proportions between bluff- and value-raises are correct GIVEN the pot odds you menion.

However, when A is making a raising decision in problem #4, B is still just getting 3:1 on the call, not 5:1. A bet $1 blind, then B put in a $1 limp, and now A has added one additional dollar with the raise. So, there is still just $3 in the pot, and it will cost B an additional $1 to show down.

[PairTheBoard]

Yes of course. I don't know what I was thinking. 3-1 either way isn't it. oh well. nevermind.

Did we ever get it settled on what's suppose to be the correct strategies?

PairTheBoard

[Bozeman]

Sorry, it looks like I was not careful enough in saying that there was agreement. I looked at most of the points in well's result, found them to agree, and then just scanned the rest to make sure no strange #'s appeared. I found JA's results exactly, and apparently well made a typographical error in that one of his 3/4's should have been 5/6.

As for your supposed improvement, I can only guess that you made an arithmetic error (easy to do here), or miscalculated the EV or size of one of the areas.

As for how I reached the solution, sop: define several decision points, and then note that at each decision point a player is indifferent to 2 possibilities. For example, if b is Player B's minimal value raise, then EVraise(b)=EVcall(b). These 7 eqs. then let you solve for the 7 decision points. To get game value, I look at the square [0,1] x [0,1]. The diagonal denotes a tie if showdown is reached, and each region divided by the diagonal and the lines corresponding to decision pts. has a particular expectation value and a probability of occuring equal to its area, so I just add them up. Many terms, so I occasionally make arithmetic mistakes here since I am not using Mathematica or Maple.

Craig

[Jerrod Ankenman]

[ QUOTE ]
I thought this was the agreed upon correct solution. Posted by wells and agreed to by Bozman?

http://forumserver.twoplustwo.com/sh...;o=14&vc=1

It has B value raising from 3/4 up to 1 as opposed to the 5/6 up to 1 for problem #3.

Although the Raise by B meets similar conditions as in #3, the Calls do not. So it could make sense for the Calling Interval to change. Both on the low side to folds and the high side to raises. Assuming the Linked to Solution is Correct.

Also Aisthesis, I understand you to say that A's bluffing frequency should be greater than B's because of the pot odds. When A raises B has greater Pot Odds to call. Doesn't Game Theory say the raising frequency should be less in that case?

PairTheBoard

[/ QUOTE ]

I think the one posted by well has at least one error. The solution I posted on 6/5 (which has a lightbulb next to it) is correct except that the value of the game is wrong.

B raises the same amount in both cases, because there's no third bet to consider. If a third bet were included (ie, B could limp-reraise), then the raising amounts would be different. As it stands, A calls to prevent B from bluff-raising, and B value raises half of the times that A will call.

Jerrod