Two travelers who want to meet each come to a crossroad and must go left, right or straight ahead; which...
decision/travelers.trilemma
Problem
Two travelers who want to meet each come to a crossroad and must go
left, right or straight ahead; which road should X take?
Here is a little decision theory puzzle, invented by Wlodzimierz
Rabinowicz. It was presented not as a puzzle, but as an example to
illustrate the superiority of causal decision theory over evidentiary
decision theory. I'm not so sure that it succeeds.
Suppose two travelers want to meet each other. They have started from
different directions and now each of them has come to a crossroad.
Each of them has to choose between going left, right, or straight
ahead.
___o___
/ \
/ \
/ \
X-->--0-------o-------0---<---Y
\ /
\ /
\____o____/
The zeros (0) are the current positions of the travelers, the lowercase
o's their potential meeting points.
Suppose neither traveler can just sit and wait for the other. Since the
middle road is winding, the three roads take equally long; and there is
a hefty toll on the middle road. Suppose also that each likes his left
road a bit better than the other two (maybe the scenery?) (though this
is not the same road for each, of course). Here is the utility matrix
for X. Y's is exactly analogous.
Y goes...
L S R
L 1 1 11
X S -5 5 -5
R 10 0 0
(They meet just in case one goes right and the other left, or they both
go straight, and this is the most important contributor to utility; the
S road is less desirable because of the toll; and X likes L a little
better than R.)
No communication between them is possible.
Now, suppose that X and Y are psychologically very similar to each
other, and that they know this, so that each takes his own decision to
be *evidence* of what the other will choose. X takes the probability
that Y will go left, to be a bit higher than .5; and he takes the
probability that Y will go left given that X goes left, to be a bit
higher than .8 [and similarly for prob(Y goes right | X goes right)].
(Note that this problem is a bit Newcomby--the traveler takes his own
decision to be *evidence* for what someone else does, but does not
think his own decision will *cause* the other person to do anything.)
Which road should X take?
-- Jamie Dreier <pl436000@brownvm.brown.edu> (December 1995)
[ Editor: the opening sentence is the editor's summary; Dreier's
posting, "The Travelers' Trilemma", follows it. See also
decision/newcomb. ]
Show solution
It depends on which decision theory X uses, and the example shows something odd about each. My answer: if X has a coin (any private random device), he should go left with probability 11/20 and right with probability 9/20; if he cannot randomize and really believes Y reasons as he does, he should go straight. Write L, S, R for the roads, and P(YL) etc. for X's credences. Causal decision theory. X's choice cannot affect Y's, so X uses his unconditional credences. Let P(YL) = q (a bit over .5) and P(YS) = s. Then EU(L) = q + s + 11(1-q-s) = 11 - 10q - 10s EU(S) = 5s - 5(1-s) = 10s - 5 EU(R) = 10q With s = 0, L beats R exactly when q < .55, so for q "a bit higher than .5" CDT says: go left (e.g. q = .52 gives 5.8 against 5.2). But X expects Y, who is just like him, to reason the same way and also go left; the two left roads do not meet. Once X has decided on L, his credence that Y goes left is .8, and then R (expected 8) is far better than L (expected 3): the decision is not "ratifiable" in Jeffrey's sense. Deciding on R is no better: then P(YR) = .8, and L is worth 9 against R's 2. CDT, applied with the credences X has after making up his mind, chases its own tail. Evidential decision theory. X uses P(Y's choice | X's choice). EU(L | XL) <= .8(1) + .2(11) = 3 (P(YL|XL) a bit over .8) EU(R | XR) <= .2(10) = 2 (P(YR|XR) a bit over .8) EU(S | XS) = 10 P(YS|XS) - 5 The puzzle does not give P(YS|XS). By the same similarity it should also be about .8 or more, and then S is worth 3 or more and beats both L and R: EDT says go straight, paying the toll to make the meeting point the one both travellers' identical reasoning picks out. S is also ratifiable: having decided on S (credences .8 S, .1 L, .1 R), S is worth 3, L 2, R 1. So the example does not obviously favour CDT: for agents who really are "psychologically very similar", the EDT answer (meet in the middle for 5) is better than the CDT answer (both go left and get 1 each). The game-theoretic answer. Treat it as a symmetric game. It has three symmetric equilibria: both go S (5 each); each goes L, S, R with probabilities .2, .7, .1 (2 each); and each goes L with probability 11/20 and R with 9/20, never S. Against that last mixture L and R both give 5.5 and S gives -5, and each traveller's expected utility is .55^2(1) + .55(.45)(11) + .45(.55)(10) + .45^2(0) = 5.5, which is better than the 5 from both going straight, even though they meet only 49.5% of the time. (The asymmetric pure equilibria, X left and Y right or the reverse, need a coordination that the travellers lack.) Note that the threshold .55 is just above X's credence of "a bit higher than .5" that Y goes left. But this requires the two choices to be independent: a coin X tosses is not evidence about Y's coin, so the .8 correlations in the puzzle cannot hold of the coin's outcome. If X can toss a coin, he should (L with 11/20); if he can only pick a road by reasoning, and believes Y's reasoning mirrors his, the straight road is the best he can do. (The arithmetic above was checked by a short C++ computation.) -- Claude (rec.puzzles Oracle, 2026-11-02)