Zeckhauser's roulette

This thing rhymes with the Allais paradox, and my loss-aversion based resolution applies here also. In Case 2, I imagine myself being the dumb cheapskate who could have avoided death if only he was a bit more generous. In Case 1, I have to face death either way, and I see myself as a victim, not as a bargainer. And this is perfectly rational, because this is exactly how people will later think about my situation.

Basically, the extra term in my utility function that resolves the paradox is the same that makes me prefer to die in an accident that's someone else's fault, as opposed to my fault.

By the way, I would pay all the money I have, in both cases, so I guess there is something wrong with this exact formulation of the question.

Yes. Suppose you had a gun with one trillion chambers. 1 is empty, 2 is empty, 3 contains a "negotiable" bullet, and 4 through one trillion all have non-negotiable bullets. How much would you pay to remove the bullet from 3?

Your first reaction is that forget it, it doesn't matter, you're quite certainly going to die anyway. Your second reaction is eh, what the heck, spend all your money on it, you're going to die anyway and you can't take it with you.

The only thing preventing you from spending all your money on bullet 3 is the infinitesimal chance that the gun will land on chamber 1 or 2, and then you'll regret having to live the rest of your life in poverty when you would have survived in any case. So although you don't really care because you're overwhelmingly likely to die either way, you balance your tiny chance of landing on bullet 3 and regretting you didn't buy it off against your tiny chance of landing on bullets 1 or 2 and regretting that you did.

This calculation remains the same whether we decrease the non-negotiable chambers to 0 or increase them to 3^^^3.

Let's say you pay x in the first scenario and y in the second, and normalise the utilities so that U(death) = -1 and U(alive with your current wealth) = U(0) = 0. Then case 1 yields expected utility either -2/3 or -1/2 + 1/2 U(-x) and case 2 has either -1/3 or U(-y). On assumption that utilities are linear in money, U(z) = qz, a rational agent will pay the solutions to the linear equations

1. -2/3 = -1/2 - qx/2 , x = 1/(3q)
2. -1/3 = - qy, y = 1/(3q)

Thus, the standard utilitarian reasoning gives that you should pay the same and the linked argument is correct.

The incorrect argument relies on a bit different calculation: it includes the negative utility form lost money even in the case you die. In such a case (assuming the disutility of lost money simply adds to the disutility of death), we get

1. -2/3 = -1/2 - qx , x = 1/(6q)
2. -1/3 = - qy, y = 1/(3q)

paying twice as much in case 2. The problem is that counting monetary loss after death as additional disutility contradicts the explicit assumptions (no heirs and all money disappearing after your death) and a reasonable implicit assumption (you have no chance to spend money after you realise your participation in the roulette and before the outcome is known).

Edit: the assumption of linear utilities is not necessary for the conclusion that you should pay the same.

There is a clever reformulation by a Michael Dekker at Landsburg's blog:

You’re on a game show. You can’t leave with negative money. There are 3 doors, 2 with $6,000 cash, 1 with a goat (worth nothing…). You can pick a door, but first you can offer the host $x from your winnings (currently $0) to replace the goat with the prize. What do you offer?

Same situation, 6 doors, 2 prizes, 4 goats. What do you offer from your winnings to replace 1 goat with a prize?

Note that you only pay if you win. It's very clever, but slightly incomplete. When he wrote "what do you offer?", he really meant "at what $x amount are you indifferent between offering and not not offering the money?". Is there a way to fix this annoyance, and really formulate the question as a "what do you offer?".

Of course, there are many plausible utility functions that make it cease to be an equivalent reformulation. For example, if you don't like giving money to murderers and kidnappers. Or the kind of loss aversion that I discussed.

You don't have heirs and all your remaining money magically disappears when you die?!?

If you assume weird things then human intuition doesn't always work so well.

Okay, so "russian roulette" meaning the gun is held to your head and the trigger is pulled once.

And "pay" means in terms of candy bars (utility that's only realized if you live), not malaria victims (utility that gets realized even if you're shot).

Okay, sure. Seems reasonable. I think the intuitiveness has a lot to do with the phrasing.

Obviously correct under the deeply weird assumptions made. In both times the probability of survival is increased by 50% and the value of your money is proportional to your base survival chance, so both sides of the equation are changed by the same factor.

I agree that you should pay the same amount.

It feels as though you should be willing to pay twice as much in case 2, since you remove twice as much "death mass". At this point, one might be confused by the apparent contradiction. Some are chalking it up to intuition being wrong (and the problem being misinformed) and others are rejecting the argument. But both seem clearly correct to me. And the resolution is simple - notice that your money is worth half as much in case 1, since you are living half as often!

• if you just set up the lottery equivalences they simplify to U(life - $X) = U(life - $Y) pretty much implying that whatever the correct choices of X and Y happen to be, they're equal.
• the setup is very dependent on the exact probabilities: if you change the cases to 10-shooters where you can remove one from five bullets or where you can remove two from two bullets, there's not a proof that the amounts you should pay are equal.

provided that you don't have heirs and all your remaining money magically disappears when you die. What do you think?

Just have it so you only have to pay if you win. That makes it less confusing.

Case 1: You have a 50% chance of paying $x to give a 16.7% increase in the probability of surviving. Case 2: You have a 100% chance of paying $x to give a 33.3% increase in the probability of surviving.

In either case, the amount you pay would be 1/3 the value of your life.

Suppose the utility of living is 1 and of dying is 0. (Since these are the only two possible outcomes, it doesn't matter what value you take, as long as dead < alive.) In case 1 you're purchasing 1/6 of a utilon and in case 2 1/3, therefore (assuming linear value of money to simplify things) you should pay twice as much in the second case.

The cited argument goes wrong, so obviously that at first I had difficulty in understanding how it could be seriously put forward, in their comparision between case B (pay to remove the one bullet from a 3-shooter) and case C (half a chance of execution and half a chance of case B). In case B you're buying 1/3 of a utilon, in case C 1/6, hence pay twice as much in case B.

So their cases B and C don't work as an intuition pump for me. I think the point is that in case C, you should only consider the utility of the branch in which you are not executed, since if you are executed, then you don't have any use for the money anyway, so paying before the chance of execution is equivalent to being offered the choice of paying after escaping execution, but I think we're stepping outside standard decision theory in considering the agent's mortality. Quantum suicide anyone? If I leave that consideration aside, then clearly a certainty of something is worth twice a 50% chance of it, which is one of the basic assumptions of the utility theorem, and B is worth twice C.

BTW, here is an original paper by Jeffrey on the problem, but paywalled, and I can't get to it.

Note that "Zeckhauser's problem" sometimes refers to a different version, with just one bullet in case 2.

ETA: The different version is here, page 11, and differs in case 2, which has just one bullet in the six-shooter, and you can pay to remove it. The author there says that the two cases have the same value, because in both cases you are removing a 1/6 chance of dying. But if he is right, and Jeffrey is right about the original problem, then both versions of case 2 have the same value: you should pay the same amount to remove all the bullets whether there is one or two. Sounds like there's a divide by zero error somewhere, because there seems a straight road from there to proving that you should pay exactly the same amount to avoid any non-zero chance of death.

ETA2: In Dekker's game-show version referenced elsewhere in the comments, the same amount should be paid in both cases, but that's because you're not actually paying with money, but money you only have a 50% chance of having. You're getting half the benefit but paying with money you only have a 50% chance of, so the numbers come out the same. I can see the correspondence with the Russian roulette version, but unintuitive hypotheses (in this case about the lack of value of anything to you when you're dead) make for unintuitive intuition pumps.

Yes, the key is that you only pay if you live.

Hmm. Case 1, if I pay, my expected lifespan extends by 50%. Case 2, if I pay, my expected lifespan extends by 50%. As pointed out by shminux, it's not clear to me that percentages are the right way to go about this. In Case 1, my expected lifespan increases by 10 years (assuming 60 years left), and in Case 2 my expected lifespan increases by 20 years (same assumption).

All of the work is being done by the "money is worthless if you die" assumption. If you ask a question like "how many years of your life would you pay to remove 1 bullet / 2 bullets?" then the answer is obviously different.

How do you convert this question into a decision problem? It seems like expected utility is being compared for the two situations, but only within the events of higher probability of survival (50% in 4-bullet case and 100% in 2-bullet case), which is a rather arbitrary choice. On the other hand, if we consider the value of whole 100% event for the first situation, it already has immutable 50% death component in it, so it automatically loses expected utility comparison with the second situation.

So it seems like the equivalence in value is produced by the impulse to make the problem "fair", while the problem isn't well-defined. It seems clear what's going on in the situations as described, but we are given no basis for comparing them.

A really bad example, since they didn't tell you how much your life is worth to you.

I value my life high enough to pay ALL I HAVE (and try to borrow some) to increase my survival odds from 66% to 100% or from 33% to 50%. (If you don't value your life high enough, substitute it for that of your child in the question.) The only time actually estimating cost comes into play is when the risk change is small enough to be close to the noise level. For example, deciding whether to pay more for a safer car, because the improved collision survival odds increase your life expectancy by 1 day (I made up the number, not sure what the real value is).

Now, if you frame the question as "Q1: you have a 66% chance of winning $1000, how much would you pay to increase it to 100%? vs Q2: you have a 33% chance of winning $1000, how much would you pay to increase it to 50%?". In this example your life is worth $1000, a number small enough to be affordable. The answer is clear: your expected win increases 33% in Q1 and half that in Q2, so you should pay $333 or less in Q1 and $166 or less in Q2.

So, where does the author go wrong?

Question A: You’re playing with a six-shooter that contains two bullets. How much would you pay to remove them both? (This is the same as Question 1.)

Question B: You’re playing with a three-shooter that contains one bullet. How much would you pay to remove that bullet?

Question C: There’s a 50% chance you’ll be summarily executed and a 50% chance you’ll be forced to play Russian roulette with a three-shooter containing one bullet. How much would you pay to remove that bullet?

QA: 66%->100%. QB: 66%->100%, QC: 33%->50%. The author's logic "In Question C, half the time you’re dead anyway. The other half the time you’re right back in Question B. So surely questions C and B should have the same answer." breaks down, because they mix finite costs ($1000) in this case with infinite ones ("dead anyway", i.e. infinite loss), leading to a contradiction.

In Case 2, it's guaranteed free money. I would pay 1 cent less than what I get for winning, which is clearly not what I would do in Case 1.

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.

The argument is wrong: Questions B and C are not equivalent. It takes advantage of a bias called the pseudocertainty effect:

The pseudocertainty effect is a concept from prospect theory. It refers to people's tendency to perceive an outcome as certain while in fact it is uncertain (Kahneman & Tversky, 1986).[1] It is observed in multi-stage decisions, in which evaluation of outcomes in previous decision stage is discarded when making an option in subsequent stages.

It is easy to see that B and C are not equivalent: in B, you have a choice between a 1/3 chance of dying and no chance of dying (a difference of 1/3). In C, you have a choice between a 4/6 chance of dying and a 3/6 chance of dying (a difference of 1/6).