**From:** William Pearson (*wil.pearson@gmail.com*)

**Date:** Mon Mar 10 2008 - 10:22:48 MDT

**Next message:**Mike Dougherty: "Re: Separate Copies Contribute Separately to One's Runtime"**Previous message:**John K Clark: "Re: Is a Person One or Many?"**In reply to:**Lee Corbin: "Re: Eliezer's Coin Flipping Duplicates Paradox"**Messages sorted by:**[ date ] [ thread ] [ subject ] [ author ] [ attachment ]

On 10/03/2008, Lee Corbin <lcorbin@rawbw.com> wrote:

*> William writes
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*>
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*> > [Lee wrote]
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*>
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*> >
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*> >> >> Consider the case now after five days have passed. We compute
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*> >> >> that the expectation is that just one of you will still be alive
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*>
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*> >> >> abecause every day 100/101 are eliminated, whether or
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*>
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*> >> >> not they saw an H or a T.
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*> >> >>
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*> >> >> What will this one remember? It's possible that he will remember
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*> >> >> TTTTT, but that is very unlikely. That would only occur if each
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*> >> >> day the 100/101 death toll struck only those who had received
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*> >> >> "heads". The chances are (100/101)^5, which is close to .95,
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*> >> >> that he would remember HHHHH.
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*> >> >>
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*> >> >> And if this continues, then a "T" will crop up in a long sequence
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*> >> >> of mostly H's about one time in one hundred and one.
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*> >> >>
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*> >> >> Therefore, as before, the subjective probability is 100/101
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*> >> >> that on each trial you'll see an H.
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*> >> >
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*> >> William writes
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*> >>
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*> >> > Isn't there only a 64% chance anyone will be alive after one
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*> >> > iteration? And after 5 iterations only a 10.2% chance that
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*> >> > anyone will be alive?
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*> >>
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*> >> Sorry, I don't follow your reasoning and arithmetic. Can you
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*> >> explain?
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*> >
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*> > Chance one person will die 100/101
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*> > Chance that everyone will die on one day (100/101)^101 = 0.366
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*> > So the chance that at least one person will survive a single day =
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*> > 1-.366 = 0.634
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*> > So the chance that at least one person will be alive after 5 days =
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*> > 0.634^5 = 0.102
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*>
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*>
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*> Ah, I was computing the expectation, i.e., the expected number of
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*> instances alive. You are computing the chance that at least one
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*> instance of the person will be alive.
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*>
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My point was that 90% of the time in this experiment everyone dies, so

0 people should be the expectation after 5 rounds. You calculated the

expectation after one round and extrapolated, which I think was wrong.

I did the math for the probability of exactly one person surviving*

and it is 37 % chance, the same as the chance that everyone dies.

However the chance of exactly one person surviving has to be less than

10% after 5 rounds.

Will Pearson

* 1/101*((100/101)^100)*101 = 0.37

**Next message:**Mike Dougherty: "Re: Separate Copies Contribute Separately to One's Runtime"**Previous message:**John K Clark: "Re: Is a Person One or Many?"**In reply to:**Lee Corbin: "Re: Eliezer's Coin Flipping Duplicates Paradox"**Messages sorted by:**[ date ] [ thread ] [ subject ] [ author ] [ attachment ]

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