Helmuth - Ferguson Head Up Match Theory Question

[Karmic]

Sorry to reply to my own post but I thought about this a bit more on the drive home. You are right if the problem is meant to be structured in a way that you know your opponent so well and are so evenly matched that you have determined the perfect optimal strategy against him and applying that strategy gives you a 50% chance of winning. In that case the games are independant, except for the dependance of a game existing because you lost the last one, and there is no reason to vary your play because you cannot improve your chances of winning.

Also removing all other factors of it being in your advantage to end non terminal games quickly if you would like to play the first 4 as a one hand coin flip and revert back to strategic thinking on the last game, which incidentally is the same scenario if the question is structured that way.

Is this a poker question or a logic puzzle?

[WRX]

[ QUOTE ]
OK. I am now ready to hear explanations on your answers. And a guess as to why I see an analogy with the Snyder head up situation. And a theory about what is going on in the heads of the people who are screwing this up.

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The answer is actually simple.

The question was, "Say you are in a position where you win the 'championship' as soon as you win just one more out of a series of head up identical size freezouts against a player you have already beaten, know well, and who is an opponent who will not alter his own play based on the 'standings'. Let's say that you have five opportunities to win this one match. ...

"Given the above, in which of the following ways would you tend to alter your play as compared to how you would play if it was simply a one match championship?

...

"For the sake of this question we will assume that both you and your opponent are tough players of approximately equal skill and playing styles."

In a few words, the five freezeouts are independent events. The result of one does not depend upon the result of any other (although the championship tournament would end early as soon as you won one freezeout). Your probability of winning the championship is the sum of your probability of winning the first freezeout, plus your probability of winning the second conditioned on your not winning the first, and so forth. For each possible branch of the decision tree, which goes into this sum, your probability of winning is maximized when you maximize your probability of winning the individual freezeout.

Stated more formally, and using, as best I can, the notation from Epstein's The Theory of Gambling and Statistical Logic:

Let A_1, A_2,A_3,A_4,A_5 be the five events, the five freezeouts. Let P(A_i) be the probability of winning event i. Let P be the probability of winning the full tournament.

Then P = P(A_1) + P(A_2¬A_1) + P(A_3¬A_2¬A_1) + P(A_4¬A_3¬A_2¬A_1) + P(A_5 ¬ A_4¬A_3¬A_2¬A_1). "P(A_2¬A_1)" means the probability of the event A_1 and the event "not A_2" both occurring, etc.

Because the events A_1,A_2,A_3,A_4,A_5 are independent events:

P = P(A_1) + P(A_2)P(¬A_1) + P(A_3)P(¬A_2)P(¬A_1) + P(A_4)P(¬A_3)P(¬A_2)P(¬A_1) + P(A_5)P(¬A_4)P(¬A_3)P(¬A_2)P(¬A_1)

Because the five freezeouts are identical, and therefore P(A_1) = P(A_2) = P(A_3) = P(A_4) = P(A_5), this reduces to:

P = P(A_1)*(1 + P(¬A_1) + P(¬A_2)P(¬A_1) + P(¬A_3)P(¬A_2)P(¬A_1) + P(¬A_4)P(¬A_3)P(¬A_2)P(¬A_1))

Because the second factor on the right side of this equation is positive, P increases if P(A_1) increases, and P decreases if P(A_1) decreases.

In other words, if your probability of winning one freezeout goes up, your probability of winning the championship goes up, and if your probability of winning one freezeout goes down, your probability of winning the championship goes down. Therefore, you should always play the strategy that gives you the highest probability of winning the individual freezeout in which you are currently engaged. If you have already been playing your best game, you should not alter it in any way. Note that it makes absolutely no difference whether your skill is superior or inferior to your opponent's skill.

The Snyder heads-up situation is another example of the result of an overall tournament depending on the addition of the results of a series of independent events, in that case the results of individual hands. So there is a similarity to this example, although the analysis is not identical.

[gergery]

[ QUOTE ]
the announcer said something about how Phil can be a little more reckless with a one match lead.


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You don't need to know how to solve poker logic puzzles, you only need to know what he thinks about the value of announcers commentary -- play the player [img]/images/graemlins/smile.gif[/img]

and I don't alter my style.

-g

[Kimbell175113]

I've already said this, kinda, but I think it's one of the most interesting parts of this whole thing, so here we go...

Let's assume it really is correct to make no adjustments in both the Philips/Tomko tourney and the Hellmuth/Ferguson situation, which I think we mostly agree on anyway. Why do you suppose it is, then, that people's intuition tells them to be super cautious in one situation and to gamble it up in the other? I'm thinking that if we understand this, we understand the fallacy.

[DrVanNostrin]

-If you maximize your chances to win each individual match you'll also maximize your chance to win 1 of 5 matches.
-Since winning 1 of 5 matches is your goal you should maximize your chances of winning each individual match.
-You should not alter your strategy based on the fact that you must only win 1 more match to win the championship (unless your opponent alters his strategy)

[PiquetteAces]

I wouldn't change anything at all.

- jpp

[David Sklansky]

Who knows me well enough to answer this fellow for me?

[DrVanNostrin]

[img]/images/graemlins/confused.gif[/img] [img]/images/graemlins/frown.gif[/img]

(I realized I was wrong too)

[BobK]

I wrote this up before I read the replies. Seems logical to me.

Dave said the opponent will continue to play the same way and you know them very well. Since you know them well, you must already be using your best strategy against them.

It is unlikely for you to be less skilled than your opponent if you're 5 matches up.
Worst case is your about equal(+-very little). That makes each of the following matches a 50-50 coin flip.

.5 ^ 5 = .03125. So you have a 3.125% chance of losing the tournament. If you are actually the better player, that figure would be even smaller.

I don't think I'm going to alter my strategy in an attempt to gain part of about 3.125% or less. I'll stick with the one that got me there.

[Shandrax]

[ QUOTE ]
this means that variance is maximized when your winrate is 50%. so people who think they can lower variance by playing less optimal are mistaken. play as optimal as you can and you will maximize the probability that you win that leg of the match and the match itself.

[/ QUOTE ]

So this is the solution? I guess my idea of maximizing variance wasn't far off.

The question remains, how do you play close to optimal? Move all-in if you got the SC number for it? That may only be half of the answer, what if you are in the BB and he limps or moves all-in himself?