Wolfgang Schwarz

Absentmindedness and persistent instability

Hey there. I've been a bit busy moving house, sitting in the garden, watching the falling leaves, etc. I've also thought some more about the absentminded driver. Here's something odd: on a certain interpretation of this case, we get a an unstable decision problem that remains interestingly unstable even when mixing (randomization) is allowed.

Some background. A decision problem is unstable if a decision to do one thing inevitably makes another thing preferable. In a classic example, Death, who is very good at predicting people's whereabouts, has predicted where you will be tomorrow and awaits you there. Should you stay where you are (in Damascus) or flee to Aleppo?

Whatever you do, the fact that you do it is evidence that Death has predicted this and awaits you just where you decided to be. If you decide to stay in Damascus, you would have better gone to Aleppo; if you go to Aleppo, you would have better stayed in Damascus. As soon as you're inclined to do one thing, you will find the other thing more attractive.

Suppose you are, say, 60% certain that you will go to Aleppo. Then you should give slightly higher credence to Death being in Aleppo than to him being in Damascus. And that's a reason to stay in Damascus; so you should lower your confidence that you will leave. Let's say you lower it to 50%. Then you have to adjust your belief about where to expect Death: you should now give equal credence to him being in Damascus and him being in Aleppo. Given these updated expectations, you have no more reason to go to Aleppo than to stay in Damascus; so you can remain 50% confident that you will do either. Your beliefs have reached an equilibrium.

It turns out that for every decision problem, there exists such an equilibrium. How does this help you if you want to know what to do, rather than what to believe about what you will do? Well, suppose you have access to some sort of chance device that yields different outputs with probabilities corresponding to your probabilities in the equilibrium state. For instance, in the Death example, you could delegate your choice to a fair coin toss. Unless Death can predict the outcome of your coin toss, this gives you a 50% chance of survival. More importantly, the decision to toss a coin doesn't undermine itself in the way the two 'pure' decisions do.

Of course, this solution won't always be available. Sometimes, you may not have a suitable chance device at your disposal. And sometimes, using the chance device may interfere with the expected outcome -- as when Death can predict the result of your coin toss, or when you'll be killed by the Anti-Gambling Police if you toss a coin.

OK, now for the much more peculiar case.

An absentminded driver has to take the second exit off the highway to get home. The first exit leads to a disastrous area. If she takes neither exit, she has to stay at a motel at the end of the highway. Due to her absentmindedness, she is unable to tell whether she is at the first or the second exit when she reaches an intersection.

In fact, let's assume that if the driver reaches both intersections, then she will be in exactly the same subjective state at both of them.

When she reaches an intersection, she faces an unstable decision problem. For if she decides to exit, she can be confident that she is at the first intersection (if she was at the second, she would have made a different decision at the first, despite being in the exact same state she is in now). And if she is at the first intersection, she ought to continue. On the other hand, if she decides to continue, then she knows she will reach both intersections, and there is a reasonable chance that she is now at the second. Since she much prefers getting home over the two other outcomes, she then ought to exit. Given that she decides to do one thing, the other thing is always better.

Let's try to break the instability by giving the driver a coin, so that she can toss it and exit iff it lands heads. To figure out the optimal coin bias, assume that the utility of reaching home is 1 and that of the two other outcomes is 0. (This makes for easier calculations than the original numbers in Piccione and Rubinstein 1997.) It then appears that a fair coin will do best: for a chosen bias b towards tails, the chance of reaching home is b(1-b); and this has its maximum (1/4) at b = 1/2.

Now consider again what happens when the driver reaches an intersection and has decided to toss our fair coin. Even though she can't be certain where she is, she can be (reasonably) certain that a fair coin is tossed at the first intersection, and perhaps also at the second. Following the Principal Principle, she might assign equal probability to heads and tails for each coin. (Our driver is a halfer in the Sleeping Beauty problem.) Moreover, let's assume that conditional on the first coin landing tails (so that she reaches both intersections), she gives equal credence to being at the first intersection and being at the second.

More generally, if she has decided to toss a coin with bias b towards tails, her beliefs will be approximately distributed as follows (with T1 = the first coin lands tails, N1 = I am at the first intersection, etc.):

To break the instability, we're looking for a value of b (ideally, 1/2) for which selecting a coin with this bias does not make a different bias preferable. That is, we need to calculate the expected utility of tossing a coin with one bias B while assuming that a coin with a possibly different bias b is actually tossed.

We consider three dependency hypotheses: i) the driver is at the first intersection and the coin at the second intersection (if reached) lands tails, ii) the driver is at the first intersection and the coin at the second intersection (if reached) lands heads, and iii) the driver is at the second intersection. I have included the outcome of the second coin toss because even though the driver has influence over whether or not she reaches the second intersection, she has no influence over the result of the coin toss there.

If hypothesis i is true, the driver will get payoff 0 no matter what she does. The expected utility of tossing a coin with bias B is therefore

From the above table, we get P(ii) = P(N1 & H2) = (1-(b/2))(1-b), and P(i) = P(N2) = b/2. Notice that what matters for the probability of ii and iii is the bias b the driver assumes she will use, not the counterfactual bias B whose utility is under consideration.

V(iii & B-toss) equals the probability that the driver reaches home given that she is at the second intersection and tosses a coin with bias B. Since the driver exits iff her present coin lands heads, this equals 1-B.

The tricky case is V(ii & B-toss). Conditional on ii, the driver reaches home iff the B-toss results in tails. V(ii & B-toss) therefore equals the conditional probability of the B-coin landing tails given ii. From our table, we get P(T1/N1) = b/(2-b), where b is the bias of the coin tossed at the first intersection. Conditional on ii & B-toss, the coin tossed at the first intersection has bias B. So V(ii & B-toss) = P(T1/ii & B-toss) = B/(2-B).

Putting all this together, we have

This is the expected utility of tossing a coin with bias B, assuming that the coin actually tossed has bias b. Setting B=b, we get the 'self-conditional expected utility' of selecting a certain bias; U(B-toss) then reduces to B-B^2. This is the utility the driver will expect after actually carrying out a B-toss. It is just what we would expect: if she tosses a B-coin at one intersection, she will do the same at any intersection she reaches, hence the probability that she gets home is B(1-B) = B-B^2 (see above).

So far, so good. Now let's see if we can find an equilibrium at which a certain value of b does not make a different value of B preferable. We might try b = 1/2. Then

For B = 1/2, that's 1/4. But for B = 1, it is 1/2. That is, on the assumption that the driver tosses a fair coin, it is better for her to toss a coin with bias 1, which amounts to the same as continuing without tossing a coin at all. -- Of course, if the driver actually decides to continue, her expected payoff is no longer 1/2, but 0. And given this assumption, every bias other than 1 has higher expected payoff.

This holds in general: for any value of b, the optimal B is either 0 or 1.

That is, if the driver has decided to delegate her decision to a (non-trivial) coin, she will prefer to not have delegated her decision! Her decision problem remains unstable even if she is allowed to toss a coin. Strange!