Go to the hotel. Turn around. Take the first intersection you see.
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If the driver starts the 50-50 strategy at the pub, then it yields the following results: 1/2 chance of the hazardous road, 1/4 chance of home, 1/4 chance of the hotel, making it better than the 100-0 straight line strategy.
If the driver starts with a 100-0 straight-line strategy at the pub, incrementally switching to a 0-100 turn strategy, then the driver gets further ahead.
(That’s the case you’re presenting, right? More likely to turn later on?)
If the driver starts with a 100-0 straight-line stratey, suddenly switching to a 0-100 turn strategy after the first intersection, that’s not allowed, but the incremental switch amounts to getting part of that from proxies such as a sense of having travelled far enough.
You win the internet for the day. Good job!
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Sweet! Thx!
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Be careful with it:
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Answer: Anybody with that bad of a memory should never be allowed to drive.
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Suppose the probability to go straight ahead is p and the probability to turn is 1-p
Then the expected pay-off is
E = 0*(1-p) + 4p(1-p) + p^2 = 4p - 3p^2
Differentiate with respect to p and set equal to zero, 4 - 6p = 0 -> p = 2/3
E = 4/3
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As a mathematician, i am insulted.
Edited* to add: Wait, not insulted. It’s another word. I’m sorry, it’s just that i’m so forgetful. … That’s it! : As a mathematician, i am forgetful.
*Not really 
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Sleeping beauty variant…
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Exactly!
Stay at the pub!
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His pub brain failed to consider the most valuable action, which is to put a pre-made sign on the dashboard in front of him as he passes the first turnoff, that tells him to take the next turnoff. He’d probably have to put a second pre-made sign on the dashboard as he left the pub, reminding him to put the other sign in front of him as he passed the first turnoff.
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I wanted to mention this when I saw the Futility Closet post, but they don’t have comments. There is a formally similar problem called the “Sleeping Beauty” problem that appeared on Quanta a little while ago:
and
and
I thought about this, too. But then I realized, there’s no mention that the pub is closing, so why bother going home at all?
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On the glove:
I think, the glove wouldn’t help much, since you don’t know, if you missed an exit.
If I see no glove: either I missed the first intersection, or I am at the first intersection (both 1/2). Turning then would give me 2 points, staying would give me 4, since I put down a glove now and turn at the next section. So if I see no glove and stay straight, E would be 2.5, turning would lead to E = 2. So no glove: drive on.
If I see a glove: turn left.
So it’s 1/2 for seeing the first intersection (4 points, since I definitely turn).
So the glove-putting strategy would lead to E = 1/2 (4) (saw first intersection) + 1/2 (1) (didn’T see first intersection), so still 2.5.
Right?
As far as I read it, you can see the intersections, you just can neither tell them apart nor remember how many you’ve seen. So, for every intersection you see, you take a glove off.
And, worst case scenario, you end up at the hotel if you forget to take your glove off at the first intersection.
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Reread the post and yeah, you are right. Well, nice brain-teaser anyway.
I can’t see why the payoff values are as he assigns them: taking the first exit is not zero benefit, it’s either negative (not reaching home, since presumably home isn’t along that road) or strongly negative (not reaching home and crashing the car in the attempt). Staying in the pub would be 0. Reaching the hotel might be 1, but it has a certain cost which is probably no lower than taking a taxi to the pub and back from home. Since that outcome is worth 4 times as much as the hotel, the solution would seem to be easy. When you reach the intersection, it isn’t a choice between a half chance of 4 and 1, but rather risking death for a 50% chance of sleeping in your own bed rather than taking a guaranteed night in a hotel.
Too drunk to remember which exit he’s passing, but sober enough to put this together on the fly:
\displaystyle E = \frac{1}{2}\left ( 0 \right ) + \frac{1}{2}\left ( 4 \right ) = 2.
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Sounds like my ex roommate.
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ask Monty Hall for help?
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