[helene_t]

One of the reasons why bridge is such an interesting game from a math point of view is that it involves problems that integrate decision theory and information theory. Optimizing information exchange, taking costs and other consequences of the messages into account.

Alice and Bob are in a forcing pass situation. Their respective desire to defend, A and B, are uniform on [-1;1] and independent. If A+B>0 they should defend so for defending they get a positive award A+B, for not defending a negative award -(A+B ). Alice chooses two thresholds, a1 and a2, Bob thresholds are b1 and b2. If A>a2, Alice will double for penalties and Bob will pull if B<b1. If a1<A<a2, Alice expresses doubt by passing, and Bob will double if B>b2.

It turns out that the optimal thresholds are -0.6 and +0.2 for both of them, and that the expected award is 0.64 which is quite close to the full-information optimum of 0.67 and much better than the 0.5 they would expect if either partner made a unilateral decision.

So far it doesn't matter if they implement the strategy by playing penalty doubles (as described) or optional doubles. But let's compute the information available to the partnership if they decide not to defend. That may be relevant if they still need to decide on strain and/or level, or to adjust the prior distribution of their defense desires in case they will face an fp situation again. The key to the problem is that if A<a1 (Alice doesn't want to defend) she can either pass-and-pull with probability q or bid directly with probability p(A<a1)-q, where q is another parameter to optimize.

Now it turns out that optional doubles are better than penalty doubles. The maximum information to exchange is 1.01 bit versus 0.93 bit.

You can download the S script if you want to try with other versions of the problem. In particular, things may change if there is asymmetry between the roles of Alice and Bob, such as when opps interfere in a relay sequence.

To make this realistic, the award function should be transformed to an IMP or matchpoint scale, and the distributions of A and B should be negatively correlated.

[FrancesHinden]

What do you mean by an 'optional' double?

I have a few other thoughts, but that's the first one...

[helene_t]

By "optional double" I mean that dbl and pass are reversed, i.e. dbl expresses doubt and pass is penalty.

[ArtK78]

I know that this is a serious discussion, but that last line sounds like an old joke.

How do you play (insert your favorite ethnic group here) Fishbein?

Double is for takeout and the next higher suit is for penalties.

[hrothgar]

I need to spend more a lot more time thinking about this one, however, here are a couple (minor) points.

1. As you already note, I don't think that its reasonable to assume that A and B are independant of one another. If A has a trump stack then B is more likely to have shortage and vice versa.

2. I suspect that a normal distribution is more likely than a uniform distribution. (To some extent this might depend on the level of the auction)

[helene_t]

Yes, your two points actually boil down to the same since the reason for chosing a normal distribution instead of the uniform is that it's easier to cope with correlation. For the rest it doesn't matter if you chose other distributions for the desire to defend, or if you achieve the same by using another award function.

I maximize the gross information exchange. If Alice choses to pass-and-pull she receives the information if Bob would have wanted to defend opposite the primary meaning of her pass. But that information is probably not so valuable.

[hrothgar]

helene_t, on Aug 2 2007, 03:41 PM, said:

Yes, your two points actually boil down to the same since the reason for chosing a normal distribution instead of the uniform is that it's easier to cope with correlation. For the rest it doesn't matter if you chose other distributions for the desire to defend, or if you achieve the same by using another award function.

Not sure that I agree:

Both assumptions (using a normal distribution versus a uniform distribution) and assuming that A and B are negatively correlated will cause A+B to fall closer to the mean. However, its unclear to me that one is a simple transform of the other.

Case in point:

Lets assume that

Distribution 1 is distributed uniformly with mean 0 and 0 covariance
Distribution 2 is distributed normally with mean 0, stdev X and 0 covariance

Can I tweak the convariance of distribution 1 and (precisely) duplicate the shape of distribution 2?

[helene_t]

hrothgar, on Aug 2 2007, 03:02 PM, said:

Can I tweak the convariance of distribution 1 and (precisely) duplicate the shape of distribution 2?

For standard normal distributed variables there is a canonical way of making them dependent. But for the uniform distribution it's not quite clear what the bivariate distribution would look like. Suppose we have [0;1] uniformly distributed variables A and B. Now define two normal distributed variables alpha and beta as the A- and B-quantiles of a standard normal distribution, alpha=Phi^(-1)(A), beta=Phi^(-1)(B ). You can turn this around by letting alpha and beta be correlated, and transforming the other way, i.e. A =Phi(alpha), B=Phi(beta). Now if you think a realistic award function is alpha+beta you can also compute it as Phi^(-1)(A)+Phi^(-1)(B ).

Back to bridge: Suppose we believe in the LOTT so A and B are simply the differences between the number of trumps and the number of enemy trumps. If they include all kind of LOTT-adjustments so that the scale becomes (almost) continous, a normal distribution does not seem unreasonable. What's the award of making the correct lawfull decision? Must depend on the kind of decision and the form of scoring. Suppose for example the differences between alternatives are very small, amounting to LOTT violations of 1/10 of a trick or such. Then the expected score impact of such a violation is propotional to the probability that this fraction of a trick actually materializes in a real trick. So the award becomes simply A+B.

Hmmmm.... sounds as if you're right, a normal distribution is probably better.

I chose uniform distributions because I deal with cut-offs, i.e. I have to compute expected awards under conditions like (A>a1)&(B>b2) etc. With the normal distribution that's more complicated.