# D_Alex comments on Zeckhauser's roulette - Less Wrong Discussion

11 19 January 2012 07:22PM

You are viewing a comment permalink. View the original post to see all comments and the full post content.

Sort By: Best

You are viewing a single comment's thread.

Comment author: 20 January 2012 04:58:00AM -1 points [-]

I think the argument is wrong. Proof by counterexample: say you have \$10 and you value your life at \$5. Note that according to the terms of the question, the value of the money is lost as well as the value of the life if you get shot. Then:

Case 1: expected value of playing the game with 4 bullets = -4/6(\$10+\$5) = -\$10, 3 bullets: -3/6(\$10+\$5) = -\$7.5, delta = \$2.5

Case 2: ... 2 bullets: -2/6*(\$10+\$5) = -\$5, 0 bullets = \$0, delta = \$5

So you should pay up to \$2.5 to take the option in Case 1, but up to \$5 in case 2.

I'm pretty sure this generalises, and only in some cases is the expected amount the same.

Comment author: 20 January 2012 07:43:54PM 0 points [-]

Case 1: expected value of playing the game with 4 bullets = -4/6(\$10+\$5) = -\$10, 3 bullets: -3/6(\$10+\$5) = -\$7.5, delta = \$2.5

[...]

So you should pay up to \$2.5 to take the option in Case 1

This does not follow. Rather, you should be willing to pay an amount \$X such that -3/6(\$10+\$5) - 3/6 \$X > -\$10. That means that you are willing to pay as much as \$5 in Case 1.

Comment author: 23 January 2012 06:59:41AM *  1 point [-]

I hate to admit this, but you seem to be right. My mistake was that I did not allow for the fact that when you pay, the stake you are risking gets reduced by the amount you have paid.

This leads to bizarre follow-ons... for example, you should pay only half as much to remove the last bullet (1/6 --> 0/6) as you would to remove 1 bullet from 4 (4/6 --> 3/6)... and you'd pay 6 times as much to remove the first bullet (6/6 --> 5/6) as you would to remove the last bullet (1/6 --> 0/6). Which incidentally contradicts the original version of the paradox as stated here:

That paper states that most people would pay MORE to remove the last bullet than the first, but the right number should be the SAME in both cases. However, it does not account for the the loss of utility when you pay the money, and the subsequent lowering of the stake if you lose. So, the paradox is even more extreme than originally envisaged...!

Edit: and furthermore, under the assumptions made, you'd pay the same amount exactly to remove one bullet out of 6 (6/6 --> 5/6) as to remove all six bullets (6/6 --> 0/6), this amount being all the money you have plus debt to the point where life would hardly be worth living, so your utility is pretty much zero in any event... I guess this shows that weird assumptions lead to bizarre conclusions.

Comment author: 23 January 2012 10:22:06PM *  0 points [-]

Which incidentally contradicts the original version of the paradox as stated here:

That paper states that most people would pay MORE to remove the last bullet than the first, but the right number should be the SAME in both cases.

I don't think that there is any contradiction here. The scenario in Schick's book is different from the one in the OP. Schick is considering the case where the decision to pay ends up costing you the same amount whether or not you end up getting shot.

It's not clear to me that you were claiming otherwise, but I just wanted to emphasize that you weren't contradicting Schick in the sense that at least one of you had to be making a mistake.