Wei, the solution that makes immediate sense is the one proposed by Uzi Segal in Economic Letters, Vol 67, 2000 titled "Don't Fool Yourself to Believe You Won't Fool Yourself Again".

You find yourself at an intersection, and you have no idea whether it is X or Y. You believe that you are at intersection X with probability α. Denote by q the probability of continuing to go straight rather than taking a left. What lottery do you face? Well, if you are at Y, then EXITING will yield a payoff of 4, and CONTinuing will yield a payoff of 1. If you are at X, then EXITING will yield a payoff of 0, and CONTinuing on straight will take you to Y. However, when you do in fact reach Y you will not realize you have reached Y and not X. In other words, you realize that going straight if you are at X will mean that you will return to the same intersection dilemma. So, suppose we denote the lottery you currently face at the intersection by Z. Then the description of the lottery is Z = ( (Z, q; 0, (1-q)), α; (1, q; 4, (1-q)), 1 - α). This is a recursive lottery. The recursive structure might seem paradoxical since there are a finite number of intersections, but is simply a result of the absent-mindedness. When you are at an intersection, you realize that any action you take at the moment that will put you at an intersection will not be distinguishable from the present moment in time, and that when you contemplate your choice at this subsequent intersection, you will not have the knowledge that you had already visited an intersection, inducing this "future" you to account for the belief that the second intersection is yet to be reached.
The next crucial step is to recognize that beliefs about where you are ought to be consistent with your strategy. You only ever have to make a choice when you are at an intersection, and since all choice nodes are indistinguishable from each other (from your absent-minded perspective), your strategy can only be a mixture over EXIT and CONT. Note that the only way to be at intersection Y is by having chosen CONT at intersection X. Thus, if you believe that you are at intersection X with probability α, then the probability of being at Y is the probability that you ascribed to being at X when you previously made the choice to CONT multiplied by the probability on CONTinuing, as described by your optimal strategy. That is, Prob(Y) = Prob(X) Prob(CONT). So, consistency of beliefs with your strategy requires that (1 - α) = α q.
Then, the value of the lottery Z, V(Z) as a function of q is described implicitly by:
V(Z) = αqV(Z) + α(1-q)(0) + (1-α)q(1) + (1-α)(1-q)4. Solving for V(Z), and using the consistent beliefs condition yields V(Z) = (1 + 3q)(1-q), and so the optimal q is 2/3, which is the same mixed strategy as you chose when you were at START.

Of course, Aumann et. al. obtain that optimality of the planning-stage choice, but the argument they make is distinct from the one presented here.

Wei, the solution that makes immediate sense is the one proposed by Uzi Segal in Economic Letters, Vol 67, 2000 titled "Don't Fool Yourself to Believe You Won't Fool Yourself Again".

You find yourself at an intersection, and you have no idea whether it is X or Y. You believe that you are at intersection X with probability α. Denote by q the probability of continuing to go straight rather than taking a left. What lottery do you face? Well, if you are at Y, then EXITING will yield a payoff of 4, and CONTinuing will yield a payoff of 1. If you are at X, then EXITING will yield a payoff of 0, and CONTinuing on straight will take you to Y. However, when you do in fact reach Y you will not realize you have reached Y and not X. In other words, you realize that going straight if you are at X will mean that you will return to the same intersection dilemma. So, suppose we denote the lottery you currently face at the intersection by Z. Then the description of the lottery is Z = ( (Z, q; 0, (1-q)), α; (1, q; 4, (1-q)), 1 - α). This is a recursive lottery. The recursive structure might seem paradoxical since there are a finite number of intersections, but is simply a result of the absent-mindedness. When you are at an intersection, you realize that any action you take at the moment that will put you at an intersection will not be distinguishable from the present moment in time, and that when you contemplate your choice at this subsequent intersection, you will not have the knowledge that you had already visited an intersection, inducing this "future" you to account for the belief that the second intersection is yet to be reached. The next crucial step is to recognize that beliefs about where you are ought to be consistent with your strategy. You only ever have to make a choice when you are at an intersection, and since all choice nodes are indistinguishable from each other (from your absent-minded perspective), your strategy can only be a mixture over EXIT and CONT. Note that the only way to be at intersection Y is by having chosen CONT at intersection X. Thus, if you believe that you are at intersection X with probability α, then the probability of being at Y is the probability that you ascribed to being at X when you previously made the choice to CONT multiplied by the probability on CONTinuing, as described by your optimal strategy. That is, Prob(Y) = Prob(X)

Prob(CONT). So, consistency of beliefs with your strategy requires that (1 - α) = αq. Then, the value of the lottery Z, V(Z) as a function of q is described implicitly by: V(Z) = αqV(Z) + α(1-q)(0) + (1-α)q(1) + (1-α)(1-q)4. Solving for V(Z), and using the consistent beliefs condition yields V(Z) = (1 + 3q)(1-q), and so the optimal q is 2/3, which is the same mixed strategy as you chose when you were at START.Of course, Aumann et. al. obtain that optimality of the planning-stage choice, but the argument they make is distinct from the one presented here.