RE: Shooting room paradox

Sean Hastings (
Mon, 09 Dec 1996 00:54:21 -0600

rom: Hal Finney <>
Date: Fri, 6 Dec 1996 09:23:45 -0800
Subject: RE: Shooting room paradox

>David Musick and Richard Brodie have said that, if they found themselves
>in the shooting room, they would be willing to bet that the dice won't
>roll double sixes. In fact David wants odds in his favor(!).

>I wonder if they and others will agree, though, that this is a losing
>strategy for most people who follow it.

No matter what the odds, betting against getting shot is always a
stratagy, as you don't have to pay out if you take the bullet.

P(Death)*Zero + P(Life)*Bet Amount = Expected Payout

That aside, the "shooting room paradox" and the "doomsday paradox"
are both based on faulty mathamatics. When you examine the
experiment at its point of completion and deduce that 90% of
those involved are in the terminal group, you are not calculating
the odds of a random person who is known to be in *any* experiment
being in the terminal group, but rather, the odds of a person known to
be in a *chosen* experiment being in that final group. By picking the
experiment first, and the choosing the person as if each is equally
probable, you lose any claim that the person is a random selction from
the set of People X Experiments.

Try it this way (the opposite and equally incorrect paradox):

Choose the person first (yourself), then, given the information
that you have been selected for the experiment, add up all
the possible experiments that it could be. The one that ends
on this itteration, the next, the next, and so on... Since we
have an infinite number of possible choices, the chance that
the experiment will end this time drops to zero. So we're
garunteed never to get shot or experience doomsday. Hurrah!

--Sean H.
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