The absent-minded driver’s paradox


Originally published at:


Call a taxi. Because if you need this much planning you have had too many drinks to be driving in the first place.


Complain to the city that they need better signage.

The real answer, as I see it, is that there are two scenarios: one where you are willing to deviate from your plan when you come to an intersection, and one where you’re not.

In the scenario where you’re not willing to deviate from your plan, the answer is to go straight, because if you apply the same calculation at each exit, “Straight” will get you to safety, where “exit” will not.

In the scenario where you are willing to deviate from the plan, the answer is still to go straight.

Your expected value from turning is still 2, but the expected value from going forward is

E = 1/2 (1) + 1/2 (4) = 2.5 (because you are willing to deviate from your original plan, so you have to include both possibilities at the next intersection).

In both cases, the greater value is to go straight.

Or, you know, to take off a glove after passing each exit so that you can keep track of which exit you’re approaching.


a recalculation with no new significant information shows that exiting yields twice the expectation of going straight

Haven’t you actually lost information along the way? Because while planning, you know that exiting will get you on the wrong way since you’re looking at the map and know which exit comes up first. But as soon as you’re on the road, you (implausibly) have no idea where you are and how many exits you’ve passed. So you’re wrongfully assigning a payoff of 2 to an action which has a 100% likelihood of delivering zero.


Flip a coin? Then your average payout is 0.5 * 0 + 0.5 * (0.5 * (4 + 1)) = 1.25, which is better on average than “always straight” which gives you 1. Maybe there’s a weighted coin that gives you a better outcome than that, but I’m lazy.


Lemme guess, the driver’s day job is driving a trolley.


I got “go straight 2/3 of the time”. But yeah, this is just an illustration that deterministic strategies are often inferior in game theory than taking an action with some probability. Once the driver started he forgot the past so he was “forced” into a probabilistic situation where the better solution became apparent.


At the pub, he knows for certain that both intersections are still ahead of him. Once he’s on the road, he doesn’t know how many intersections he’s passed, so it’s a different calculation.

But what he should think is: “OK, if this is the second exit, then I must have passed the first; but then I would have had the same idea, and turned off back there; therefore this cannot be the second exit”. So he’d be back to the strategy he came up with in the pub.


The only reason that it seems like there is a paradox is because you have described two different scenarios: one in which “planning to turn” means he is guaranteed to take the first exit, and the other where “planning to turn” gives a 50% chance that he has already passed an exit.

In the first scenario he knows he’s at the pub, in the second scenario he might be passed the first exit. Obviously this results in different strategies. It’s not a paradox at all.

My answer (which looks to be the same as @doop’s) is to plan from the start to flip a coin.

1/3 of the time you get 0. 2/9 you get 4. 4/9 you get 1. Your expected outcome is 1.33, which is indeed higher than 1.25. Is that the highest? Turning more often than 50% definitely seems to tilt your expected outcome lower.


Perhaps he sells ice cream on the beach.


P = Probability of Going Straight
(1-P)(0)+P((1-P)(4)+P(1)) => expected return
4P-4P^2+P^2 => Simplified
-3P^2+4P => Put into quadratic form, where -b/2a is the maximum value
P = 4/6 = 2/3 => Best Outcome


If i were in that scenario i would continue and sleep at the hotel rather than take a turn. If the guy really is that absent-minded then who’s to say that he didn’t notice that the number of intersections is correct or not from what he thinks they are. The only known fact is that if one were to keep driving there’s a hotel, for it’s the simplest choice to make.


What’s the answer. Steal a beer mug from the pub and when you pass the first intersection, throw it out the window. Then when you come to the second exit, you will know it’s the second because you no longer have the beer mug.


Yeah, that’s the answer alright - don’t be driving home from the pub.


So the takeaway is that he has no functional memory but a PhD in mathematics…?


I didn’t think there’d be math on the test.


Yeah, this is not thought through in the article. If he knew he could make choices based on impartial information then this could be planned in advance. So the whole dichotomy of planning versus execution being different is a fallacy. Further, the payout function based on probability of branching f§ = 4p(1-p)+(1-p)^2 = -3p^2+2p+1, which is maximized at 1/3, with a value of 1 and a 1/3.


Look for the second intersection. Both have taco trucks but the second one sells breakfast tacos all day.


Sounds like a mathematician to me.


Yeah, I’m with @tobinl. If you’re so drunk “absent minded” you can’t count to 1, don’t drive.