Your payoff for choosing CONTINUE with probability p becomes α[p^2+4(1-p)p] + (1-α)[p+4(1-p)], which doesn’t equal p^2+4(1-p)p unless α = 1.
No. This statement of the problem pretends to represent the computation performed by the driver at an intersection—but it really doesn’t. The trouble has to do with the semantics of alpha. Alpha is not the actual probability that the driver is at point X; it’s the driver’s estimate of that probability. The driver knows ahead of time that he’s going to make the same calculation again at intersection Y, using the same value of alpha, which will be wrong. Therefore, he can’t pretend that the actual payoff is alpha x (payoff if I am at X) + (1-alpha) x (payoff if I am at Y). Half the time, that payoff calculation will be wrong.
Perhaps a clearer way of stating this, is that the driver, being stateless, must believe P(I am at X) to be the same at both intersections. If you allow the driver to use alpha=.7 when at X, and alpha=.3 when at Y, then you’ve given the driver information, and it isn’t the same problem anymore. If you allow the driver to use alpha=.7 when at X, and alpha=.7 again when at Y, then the driver at X is going to make a decision using the information that he’s probably at X and should continue ahead, without taking into account that he’s going to make a bad decision at Y because he will be computing with faulty information. That’s not an alternative logic; it’s just wrong.
The “correct” value for alpha is actually 1/(p+1), for those of us outside the problem; the driver is at Y p times for every 1 time he’s at X, so he’s at X 1 out of p+1 times. But to the driver, who is inside the problem, there is no correct value of alpha to use at both X and Y. This means that if the driver introduces alpha into his calculations, he is knowingly introducing error. The driver will be using bad information at at least one of X and Y. This means that the answer arrived at must be wrong at at least either X or Y. Since the correct answer is the same in both places, the answer arrived at must be wrong in both places. Using alpha simply adds a source of guaranteed error, and prevents finding the optimal solution.
Does the driver at each intersection have the expected payoff alpha x [p^2+4(1-p)p] + (1-alpha)*[p+4(1-p)] at each intersection? No; at X he has the actual expected payoff p^2+4(1-p)p, and at Y he has the expected payoff p+4(1-p). But he only gets to Y p/1 of the time. The other 1-p of the time, he’s parked at A. So, if you want him to make a computation at each intersection, he maximizes the expected value averaging together the times he is at X and Y, knowing he is at Y p times for every one time he is at X:
p^2+4(1-p)p + p[p+4(1-p)] = 2p[p+4(1-p)]
And he comes up with p = 2⁄3.
There’s just no semantically meaningful way to inject the alpha into the equation.
Alpha is not the actual probability that the driver is at point X; it’s the driver’s estimate of that probability.
What do you mean by probability, if not “someone’s estimate of something-or-other”?
But to the driver, who is inside the problem, there is no correct value of alpha to use at both X and Y.
There’s also no correct value of p to use both when you’ll continue and when you won’t. But that doesn’t mean you should omit p from the calculation.
This means that the answer arrived at must be wrong at at least either X or Y. Since the correct answer is the same in both places, the answer arrived at must be wrong in both places.
The driver is computing an expectation. A value of an expectation can be wrong for X, and wrong for Y, and right for the union of X and Y.
(I agree, of course, that the formula involving alpha isn’t the right computation for the original problem. But that’s separate from whether it’s computing something interesting.)
That’s a very good explanation. I tried to generalize the problem to the case of partial additional knowledge about one’s intersection, and I invite you to take a look at it to see if it makes the same kind of error. For the case of “ignorance about one’s intersection”, my solution yields “continue with probability 2⁄3 at any intersection”, just the same as everyone else, and it does so by introducing the parameter r for “probability of guessing intersection correctly”. In the problem as stated, r=1/2.
No. This statement of the problem pretends to represent the computation performed by the driver at an intersection—but it really doesn’t. The trouble has to do with the semantics of alpha. Alpha is not the actual probability that the driver is at point X; it’s the driver’s estimate of that probability. The driver knows ahead of time that he’s going to make the same calculation again at intersection Y, using the same value of alpha, which will be wrong. Therefore, he can’t pretend that the actual payoff is alpha x (payoff if I am at X) + (1-alpha) x (payoff if I am at Y). Half the time, that payoff calculation will be wrong.
Perhaps a clearer way of stating this, is that the driver, being stateless, must believe P(I am at X) to be the same at both intersections. If you allow the driver to use alpha=.7 when at X, and alpha=.3 when at Y, then you’ve given the driver information, and it isn’t the same problem anymore. If you allow the driver to use alpha=.7 when at X, and alpha=.7 again when at Y, then the driver at X is going to make a decision using the information that he’s probably at X and should continue ahead, without taking into account that he’s going to make a bad decision at Y because he will be computing with faulty information. That’s not an alternative logic; it’s just wrong.
The “correct” value for alpha is actually 1/(p+1), for those of us outside the problem; the driver is at Y p times for every 1 time he’s at X, so he’s at X 1 out of p+1 times. But to the driver, who is inside the problem, there is no correct value of alpha to use at both X and Y. This means that if the driver introduces alpha into his calculations, he is knowingly introducing error. The driver will be using bad information at at least one of X and Y. This means that the answer arrived at must be wrong at at least either X or Y. Since the correct answer is the same in both places, the answer arrived at must be wrong in both places. Using alpha simply adds a source of guaranteed error, and prevents finding the optimal solution.
Does the driver at each intersection have the expected payoff alpha x [p^2+4(1-p)p] + (1-alpha)*[p+4(1-p)] at each intersection? No; at X he has the actual expected payoff p^2+4(1-p)p, and at Y he has the expected payoff p+4(1-p). But he only gets to Y p/1 of the time. The other 1-p of the time, he’s parked at A. So, if you want him to make a computation at each intersection, he maximizes the expected value averaging together the times he is at X and Y, knowing he is at Y p times for every one time he is at X:
p^2+4(1-p)p + p[p+4(1-p)] = 2p[p+4(1-p)]
And he comes up with p = 2⁄3.
There’s just no semantically meaningful way to inject the alpha into the equation.
What do you mean by probability, if not “someone’s estimate of something-or-other”?
There’s also no correct value of p to use both when you’ll continue and when you won’t. But that doesn’t mean you should omit p from the calculation.
The driver is computing an expectation. A value of an expectation can be wrong for X, and wrong for Y, and right for the union of X and Y.
(I agree, of course, that the formula involving alpha isn’t the right computation for the original problem. But that’s separate from whether it’s computing something interesting.)
That’s a very good explanation. I tried to generalize the problem to the case of partial additional knowledge about one’s intersection, and I invite you to take a look at it to see if it makes the same kind of error. For the case of “ignorance about one’s intersection”, my solution yields “continue with probability 2⁄3 at any intersection”, just the same as everyone else, and it does so by introducing the parameter r for “probability of guessing intersection correctly”. In the problem as stated, r=1/2.