Would you pay more to empty both loaded chambers of a six-shooter holding two bullets, or to empty one of...

No solution was found in the collected postings.

[ Editor: the poster promised "a nice demonstration which I'll give
next week"; it is not in the collected material. ]

[ Editor: further material on this puzzle, edited separately from other collected postings (batch 19); it may repeat some of the above. ]

[ Editor: problem as stated there: ]
You are compelled to play Russian Roulette with a six-shooter holding
two bullets: how much would you pay to empty both chambers?  With four
bullets: how much would you pay to empty one of the four?  Show that,
on certain assumptions, expected utility theory says you should pay the
same amount in both cases (Zeckhauser's problem).


First, suppose that you are compelled to play Russian Roulette, using a
six-shooter with two loaded chambers. How much would you pay to empty both
chambers before pulling the trigger?

Second, suppose that you are compelled to play Russian Roulette, with four
loaded chambers. How much would you pay to empty one of the four before
pulling the trigger?

On certain assumptions, it turns out that expected utility theory insists
that you be willing to pay the same amount in both cases (though of course
it doesn't say how much).

Most of the assumptions are obvious. You take the chambers to be
equiprobable (each has the same chance of lining up with the barrel, or
however those things work, when you spin the chambers). You prefer life to
death, and more money to less. Not so obvious, and probably not realistic:
assume that you are indifferent between being dead and poor, and being
dead and rich (that is, on the condition that you get unlucky in the game,
it doesn't matter to you whether you paid the sum of money or not).

Almost everyone pre-theoretically says that they'd pay a lot more in the
first case to empty both chambers (and be perfectly safe) than they would
to empty one of the four loaded chambers in the second case. But, under
the assumptions, this will violate the very same axiom that the Allais
preferences violate. (ObPuzzle: prove. Not terribly easy. There's a nice
demonstration which I'll give next week.) If you have the common
preferences, then you don't have any expectational utility function. You
cannot possibly be maximizing your expected utility.

Zeckhauser (who is, I think, a Michigan economist) did not publish this
problem, but it is mentioned in Kahneman and Tversky, "Prospect Theory: An
analysis of decision under risk", _Econometrica_ 47 (1979), 263-91; and
also in Jeffrey, R., "Biting the Bayesian Bullet", _Theory and Decision_
25 (1988) 117-122. Both articles attribute the problem to Zeckhauser.

-- Jamie Dreier <pl436000@brownvm.brown.edu>

I'll assume that in the second example, removing *one* bullet is the
only option, paying more money won't allow you to empty more chambers.

Then the second example is equal to first having a 50% chance of death,
then having to play the first game. The first 50% chance is riding on
faith and no amount payed will change its outcome, but removing
the subsequent 33% chance of death should be worth the same as in the
first game.

If however each bullet has a fixed price so that you could
remove all bullets in the second game as well, then this reasoning
doesn't apply, and I think that an analysis would be quite difficult.

-- Stein Kulseth <stein@hal.nta.no>