A devil’s offer: Tossing a magic coin to get out of hell

2013-01-14 in MIRI/LW | 7 comments

(The following is adapted from a scenario by Graham Priest, depicted in his book ‘Logic: A Very Short Introduction‘.)

Suppose that at some point you find yourself in a posthuman hell. But you have one chance to get out of it. You can toss a coin; if it comes down heads, you are out and go to heaven. If it comes down tails, you stay in hell forever. The coin is not a fair one, however, and the posthuman entity that simulates the hell has control of the odds. If you toss it today, the chance of heads is 1/2 (i.e. 1-1/2). If you wait till tomorrow, the chances go up to 3/4 (i.e. 1-1/2^2). If you wait n days, the chance of going to heaven goes up to 1-1/2^n. How long are you going to wait before tossing the coin?

The associated values of remaining in hell or escaping are constant over time.


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  • Lukasz Stafiniak

    Is there ongoing personality change and memory decay while I’m in hell or is it just “idealized unpleasantness”?

  • http://kruel.co/ Alexander Kruel

    The value of remaining in hell is constant, does not change over time. No personality change or memory decay.

  • Lukasz Stafiniak

    If I could precommit, I would stay in hell for two-three weeks.

  • dmytryl

    The value of remaining in hell forever is x, the value of going to heaven is y, so when x*(1-p)+y*p >0, you toss the coin.

    Also, observe: if you are in a brain simulator, and at the moment of maximum pleasure the simulator does not swap it’s arrays (keeps re-computing same happy state from the previous state), you aren’t experiencing any pleasure – you are frozen in time. Same goes for pain in hell; number of states is eventually exhausted. Negative value of remaining in hell forever is finite even if all chunks of this value are equal. The value of a collection of items is not equal to the sum of values of components, likewise, the value of 10 days in hell is not the sum of values of each day in hell.

  • http://profiles.google.com/katsaris Aris Katsaris

    > The value of remaining in hell forever is x, the value of going to heaven is y, so when x*(1-p)+y*p >0, you toss the coin.

    So if the two values are exactly opposite to each other (x=-y where y is a positive number), you toss the coin on just the second day? That seems bad reasoning.

    Let’s just forget all the stupid paradises and hells for the moment which just muddle the brain, and imagine it’s just about winning 10,000 dollars vs losing 10,000 dollars.

    Imagine that the utility of winning and losing these money is just exactly the opposite of each other, and (not being in any particularly rush for the money) you don’t significantly care in comparison how long you’ll wait before tossing it. Tossing the coin on only the 2nd day seems a really bad idea, when that will give you 1/4 chances of losing 10,000 dollars, but if you wait a mere week the chances of losing the money will be only 1/128 and winning them will be 127/128.

  • http://www.facebook.com/peterhurford Peter Hurford

    I assume every additional day spent in Hell is at a specific finite marginal cost, and every additional day spent in Heaven comes at a specific marginal benefit. You should be able to slope this and find a point where benefit – cost times probability is maximized.

  • dmytryl

    Fair point. Need to count the utility of day in hell and day in heaven and I guess toss when disutility of 1 extra day is less than utility of extra chance. I.e. in the case of $ and if you don’t particularly need money for anything now, you toss it the moment improvements in your risk fall below the interest rate.