Interesting strategic puzzle

[Cascade]

For what its worth, this types of practices create some interesting disclosure issues. From my perspective, player who use these types of mixed strategies can be required to describe the existence of a mixed strategy to the opponents and register their "keys" with the Tournament Directors. However, their partner can't have any information regarding the particular key that is being used on any given day.

Why?

 

If I use a mixed strategy and sometimes I do why does the director need to know. Not only that I am free to deviate from my own mixed strategy on any particular hand.

 

I don't see any advantage in telling the director my "key" and it might be a disadvantage if on another day that director becomes my opponent.

[hrothgar]

For what its worth, this types of practices create some interesting disclosure issues.  From my perspective, player who use these types of mixed strategies can be required to describe the existence of a mixed strategy to the opponents and register their "keys" with the Tournament Directors.  However, their partner can't have any information regarding the particular key that is being used on any given day.

Why?

 

If I use a mixed strategy and sometimes I do why does the director need to know. Not only that I am free to deviate from my own mixed strategy on any particular hand.

 

I don't see any advantage in telling the director my "key" and it might be a disadvantage if on another day that director becomes my opponent.

The primary problem with mixed strategies is one of disclosure: How can players accurately describe bidding agreements that are inherently random? In an ideal world, players would have enough experience with their opponent's methods that they could make accurate judgements regarding the frequencies that they randomize. In practice, its necessary to trust the opponents when they describe their methods.

 

Unfortunately, this creates some very clear incentive problems. Why should I as a player accurately describe the Probability Density Function that I use when I randomize? Equally significant, assume that I did provide an inaccurate PDF... How would anyone ever know? (Please note: I am not claiming that any one pair would necessarily cheat. Rather, that the rules set should not be designed in such a way that people have easy ways to cheat). You may WANT to be able to based your randomization on inherently subjective considerations. However, I think that its best for everyone if you aren't allowed to.

 

The most logical way to eliminate this possibility is to force players to tie their hands. Players have the right to randomize/adopt mixed strategies. However, if they do so the TDs need some mechanism to determine whether they are accurately disclosing their methods. In turn, this requires that players describe the "keys" when they randomize their methods.

 

Please note: There is no reason that keys need to remain static between sessions. Indeed, I'd argue that players should probably be required to vary their keys for fear that their partner's might "key in" on whats going on... In all honesty, its not particularly difficult to devise a set of different keys that could be used for any given mixed strategy.

 

(As usual, this posting is only half serious. I severely doubt that anyone would ever bother going to this much trouble. Equally significant, I doubt that the existing Zonal authorities would ever want to deal with all the issues associated with opening this can of worms. With this said and done, if your sole concern is coming up with a fair way to allow so-called "random" bids I suspect that you'll need to implement something similar to this.

 

For whgat its worth, I storngly beleive that anyone who claims that they use 1S as a "random" overcall of a strong club opening should be forced to adopt these types of restrictions. If you can't accurately describe your methods you don't get to use your methods.

[Echognome]

No problem at all Frances, you explained things much better than I did. (and I was way to lazy to compute the 91.5%)

 

Matt, it would be nice if you could explain your calculations without using sophisticated notation. I think that this is something that many bridge players are able to understand, but not if they are not familiar with the notation.

A fair point Han. I am also certain that many people will NOT be interested in it. So for those that are not, just ignore the rest of what I'm about to say. For those that are, read on.

 

The beginning of my thread was discussing simply the line that declarer would take. In effect, declarer has the option of playing for the drop or playing a two-way finesse. However, in playing for the finesse one way, he always succeeds, whereas playing for it the other way, he has a guess. In essence, if he guesses correctly the first time, he doesn't need to guess the second time.

 

Now, let us take the example given by Gerben and assume that declarer has:

 

AT98

 

K5432

 

If declarer guesses to play small to the A, he wins all the tricks. It doesn't matter what card East plays.

 

If declarer guesses to play that 8 (or 9 or 10) from dummy and sees the J while winning the K, he must guess whether to finesse West for the missing Q or to play for the drop. He leads another small card from dummy and sees the last small card from West. He now has to make the decision. So there is only one card outstanding and he is guessing whether the suit is laid out as:

 

Q76 opposite J or

76 opposite QJ

 

Now we get into the discussion of the often cited "restricted choice". The idea is simple enough in saying that IF East has J singleton, then East will have to play the J on the first round. If East has QJ doubleton, then East can play the J or the Q on the first round. Now before we get into East making a choice (and thus altering the odds), let us consider the odds ex ante. That is to say, what are the odds of the various holdings of the EW cards before we observe E and W's plays to the first couple tricks. Here I used suitplay to speed up the calculations. If you are interested in those combinatorics, that can be discussed separately.

 

West-----East----Prob

Void-----QJxx----4.78%

x---------QJx-----12.43%

xx--------QJ------6.78%

J----------Qxx----6.22%

Jx--------Qx------13.57%

Jxx-------Q-------6.22%

Q---------Jxx-----6.22%

Qx-------Jx-------13.57%

Qxx------J--------6.22%

QJ-------xx-------6.78%

QJx------x--------12.43%

QJxx----Void-----4.78%

 

So these are the odds of the various holdings of East and West. However, we can rule out several of the possibilities given that we have seen three cards. Namely, we now know that East holds the J and West holds the two x's. So we know that it is one of the following two possibilities:

 

West-----East----Prob

xx--------QJ-----6.78%

Qxx-------J------6.22%

 

Note that it is more likely ex ante that East holds QJ doubleton than J singleton. On that premise it may seem better to play for the drop than the finesse. And actually that premise would be true IF East always played the J from QJ doubleton. As an exercise, we can consider the ex ante conditional probabilities that East holds J or QJ. That is to say, what if we asked the following question: "Now that we know East holds either QJ or J, what are the odds he holds QJ compared to J?" We "condition" on the argument that it must be one of QJ or J and then look at the relative probabilities on these. Now we note that East holds QJ or J some 13% of the time. So the "conditional" probability that East holds QJ doubleton is 6.78/13 = 52.2% and the "conditional" probability that East holds J singleton is 6.22/13 = 47.8%. So if this were the only consideration, then we would be better off playing for the drop rather than the finesse.

 

However, we now add the premise that when East holds QJ doubleton, East might sometimes play the J and sometimes play the Q. So we now need to define a variable that I called p to be the probability (or if you like proportion of time) that East plays the J when he holds QJ doubleton. Another way of saying this is that when East holds QJ doubleton, he is only playing the J a certain percentage of the time and we are calling that percentage p. If you want, think of p as the percentage of time you "true card" and 1 - p as the percentage of time you "false card". However, when East holds J singleton, East has no choice but to play the Jack. Thus when East holds J singleton, the J is always a true card. So we can adjust our table to now think of the ex ante probabilities (given that East falsecards with percentage p) as:

 

West------East-----Prob

xx---------QJ-------6.78p%

Qxx--------J--------6.22%

 

We now make the exact same calculations as before, with the added wrinkle that the total percentage is no longer 13% but is 6.78p% + 6.22%.

 

So the odds that East holds QJ doubleton are now:

 

Conditional Probability of QJ = 6.78p/(6.78p+6.22)

Conditional Probability of J = 6.22/(6.78p + 6.22)

 

Which means that whether we play for the finesse or the drop depends entirely on the value of p! Let us consider the two extreme cases.

 

Suppose that p = 1. This means that East always plays the J from a holding of QJ doubleton. In this case, we can use the odds calculated above and play for the drop.

 

Now suppose that p = 0. This means that East never plays the J from a holding of QJ doubleton. The consequence of this is remarkable. We can now RULE OUT QJ doubleton as a possibility and play for the finesse.

 

My main point here is that p matters. If East takes a strategy of ALWAYS playing the J or ALWAYS playing the Q, then it should affect our decision and we can use that knowledge to our advantage. But if East sometimes plays the J and sometimes the Q, we can think about how often East plays one card or another in making our decision.

 

The last step is to associate the probabilities on the holdings above with the relevant line. When East holds QJ doubleton we want to play for the drop. When East holds J singleton, we want to play for the finesse. So we compare them via an inequality. Playing for the finesse must be better than playing for the drop when:

 

Conditional Probability of J > Conditional Probability of QJ

 

or

 

6.22/(6.78p + 6.22) > 6.78p/(6.78p+6.22)

 

Since the denominators are the same, we can rewrite this as:

 

6.22 > 6.78p or

 

p < 6.22/6.78 or

 

p < 0.915 = 91.5%

 

And here is where we get our rule. When East is looking at QJ doubleton, he wants declarer to finesse. Thus he should play the J from QJ doubleton less than 91.5% of the time.

 

Finally, we can go through the entire exercise again when East instead plays the Q on the first round rather than the J. Luckily, our new problem is completely equivalent to our old one and we find that East should play the Q from QJ less than 91.5% of the time. Now we have our solution for East. When you hold QJ doubleton, it is best to randomize between playing the Q and the J. However, your randomization should not be too extreme (in this case).

[Echognome]

By the way, is it the case that anywhere up to 91%, but not 92% will find my prior post uninteresting?

[Guest Jlall]

By the way, is it the case that anywhere up to 91%, but not 92% will find my prior post uninteresting?

no I think that that number is over 92 matt :lol:

[pclayton]

Why do I have visuals of Richard and Matt standing on the 2nd floor of a dormitory in New. Engalnd in January scribbling these formulas on a window pane in crayon and talking to imaginary people? ;) :P

[han]

Why do I have visuals of Richard and Matt standing on the 2nd floor of a dormitory in New. Engalnd in January scribbling these formulas on a window pane in crayon and talking to imaginary people? tongue.gif tongue.gif

 

Hard to say without knowing you better Phil. Perhaps it is because you don't understand them and prefer to think of people you don't understand as being lunatics who have no idea what life is about? Perhaps because in most of the Western world intellectuals are often seen this way? I'm just guessing here, as I said, I can't read your mind from such a distance.

[hrothgar]

I suspect that you've been watching too many Russell Crowe movies...

(BTW, New Jersey is a mid Atlantic state)

[Guest Jlall]

Han, you should watch the movie a beautiful mind. I think Phil was just making a joke and was in reference to that.

[han]

I have seen the movie and enjoyed it. I understood the joke (and even thought it was funny, but don't tell that to Phil), but I made a serious reply.

[Guest Jlall]

If Clayton compared me to John Nash because of my intellect and knowledge, I would be flattered. Sadly he only compares me to Nash when he is discussing my mental state ;)

[adhoc3]

Game Theory Professor said: The best strategy is opponent's strategy --- but make sure you will do it better.

 

So for weaker team, don't try to perform such analysis because your opps are better than you, just do it randomly. For stronger team, better play the game/cards in normal way because you're the better one. -------In either case: no one need the intellegent % calculation. :)

 

May be this is useful for two teams who are the exactly same competitive... ;)

 

Good lucky and have fun!

 

Jack

[jdeegan]

:D When it comes to playing the Q from Q10x, isn't the correct answer always? Otherwise declarer will always go right since the nine is by far his percentage play if you dont 'split'. In short, you convert a sure loss into a possible win. Educate me, show me where I am wrong.

 

jdeegan, MA, PhD, PDQ, ASAP

[han]

B) When it comes to playing the Q from Q10x, isn't the correct answer always? Otherwise declarer will always go right since the nine is by far his percentage play if you dont 'split'. In short, you convert a sure loss into a possible win. Educate me, show me where I am wrong.

 

jdeegan, MA, PhD, PDQ, ASAP

You are not exactly wrong, but I think that you are missing something. If you always play high then the percentage play for declarer is to come back to hand and play low to the nine. So you also always lose with this hand.

[kfgauss]

B) When it comes to playing the Q from Q10x, isn't the correct answer always?  Otherwise declarer will always go right since the nine is by far his percentage play if you dont 'split'.  In short, you convert a sure loss into a possible win.  Educate me, show me where I am wrong.

 

jdeegan, MA, PhD, PDQ, ASAP

You are not exactly wrong, but I think that you are missing something. If you always play high then the percentage play for declarer is to come back to hand and play low to the nine. So you also always lose with this hand.

There's something to be said for this practical approach. We presume everyone will put in the 9 if you play low (assuming they don't know something about your habits), but if you play high, there will be some who won't expect that from H10x(x..) and will play for KQx(x..).

 

Andy

[awm]

There are some interesting observations about all this. If declarer basically ignores the defense and determinedly plays to the nine (or beats an honor with the ace and then plays to the nine the next round) then he will win precisely when the ten and one of the K or Q is onside. This is 37.5% or thereabouts.

 

If we always play low from KQx, KTx, QTx... then this is basically the best declarer can do. The same applies if we always play high the first time and duck the second.

 

However, there is a psychological ploy here. Most weaker players would never consider to play the king or queen from KTx or QTx. However, they might play an honor from KQx. Of course, playing this way is exactly wrong; declarer can then play to the jack on the second round every time you play an honor and pick up KQx as well as KTx/QTx for a roughly 50% success rate (if you always split from KQx) or in any case better than 37.5% (assuming sometimes split and sometimes not). On the other hand, this assumes declarer is aware of this tendency.

 

So the psychological ploy, against someone who doesn't think you're capable of such things, is to play an honor from KTx/QTx and play low from KQx. Of course this doesn't really work either, in the sense that if declarer knows you do this he can succeed roughly 50% of the time. But declarer is more likely to assume that you wouldn't think of this -- if he plays you to do exactly the opposite (split KQx, low from KTx/QTx) then he will succeed only when all three honors are onside (something like 12.50%)! If declarer assumes you play randomly or ignores you honor plays, he's still only 37.5% likely to succeed. Only if he guesses that you are playing "backwards" can he improve his success rate.

[pclayton]

Why do I have visuals of Richard and Matt standing on the 2nd floor of a dormitory in New. Engalnd in January scribbling these formulas on a window pane in crayon and talking to imaginary people? tongue.gif tongue.gif

 

Hard to say without knowing you better Phil. Perhaps it is because you don't understand them and prefer to think of people you don't understand as being lunatics who have no idea what life is about? Perhaps because in most of the Western world intellectuals are often seen this way? I'm just guessing here, as I said, I can't read your mind from such a distance.

Blame it on Einstein I guess; great minds are often portrayed as eggheads having poor hygiene and dress strangely.

 

But Han, I totally respect and can barely comprehend the inner workings of a lot of the posters around here, especially Matt's and Richard and even yours. At least when it comes to non-bridge issues :o

 

Just kidding :) .

[han]

At least when it comes to non-bridge issues tongue.gif

 

Now that was funny!

[jdeegan]

So the psychological ploy, against someone who doesn't think you're capable of such things, is to play an honor from KTx/QTx and play low from KQx. Of course this doesn't really work either, in the sense that if declarer knows you do this he can succeed roughly 50% of the time. But declarer is more likely to assume that you wouldn't think of this -- if he plays you to do exactly the opposite (split KQx, low from KTx/QTx) then he will succeed only when all three honors are onside (something like 12.50%)! If declarer assumes you play randomly or ignores you honor plays, he's still only 37.5% likely to succeed. Only if he guesses that you are playing "backwards" can he improve his success rate.

:P In a game where RHO knows my habits, I'm going to have to split from KQx at least part of the time. What are the odds now, Prof. Bayes?