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Trying to come up with an even simpler non-realistic toy example based on the classic Russian roulette example everybody uses. Imagine N people playing Russian roulette with six-shooter revolvers, repeatedly.
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1/ Been thinking about non-ergodicity and path dependency due to provocations from @doriantaylor and @TaylorPearsonMe among others. Trying to come up with the simplest toy example I think clarifies the basic question. Here’s what I came up with...
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Consider remainder-fatality statistics. After the first shot, 1/6 are dead. After the second, 1/5 of the remaining people are dead, after the third, 1/4, after the fourth, 1/3, after the fifth, 1/2, and with the 6th, everybody left dies. And then there were none.
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There's a very Monty Hall energy about this game. Obviously nobody would play this even once without incentives, and nobody would play the 6th round at all, since you're guaranteed death. What would make people play?
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Relative to the initial N, each round eliminates 1/6. So with N=6, 5 rounds cuts it down to 1 person left alive. With N=12, you get L=2 (last round population), etc.
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Example. There are 12 people trapped on a planet about to explode. Vogons offer to save 2 people in return for all the Altairian dollars all 12 have together. Iterated Russian roulette would be a good way to choose. Survivors take dead people's dollars. Last 2 alive take seats
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Actually, the primary candidate selection process we just went through is kinda like this. If you don't get 15% you're dead. Delegates get divided up among the rest.
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Here, Gabbar uses a revolver with 3 bullets, 3 empty chambers. All 3 escape the first round and everybody starts laughing that they all escaped. Then he turns around and shoots then all anyway.
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Though crucially, he empties the first 3 chambers by firing into the air, so the remaining three are in sequence, not randomly distributed. So when he randomizes by spinning the cylinder, there's only 1/6 chance of starting in a way that yields that particular outcome.
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Hmm... the first guy must estimate his survival chances at 1/2. After he survives, the second guy can actually estimate his survival chances at 2/3. After he survives too, the third guy will have estimate of 1/2 again. After all survive the first round, all 3 have 0% in 2nd.
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Ah thought of Monty Hall connection. In Sholay example, if he offers the first guy a chance to switch to second position, should he take it? Makes no difference *knowing what he knows then*. But if he offers it to 2nd guy after first one survives, he should *not* switch.
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So survival of 1st guy is non-trivial info. Survival of 2nd guy is as well, but in that case, makes no difference to the odds: value of information is balanced by increased closeness to the full sequence of barrels. So it's worth competing to be #2 *in the #1-survives scenario.*
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