You need the random draw to pick a prisoner who hasn't been, and then for
the random draw to pick the designated "counter" prisoner.



On Tue, Sep 11, 2012 at 7:52 AM, Skip Cave <[email protected]> wrote:

> Jordan
>
> I think 1e6 is much too high. The problem seems to be equivalent to drawing
> from a bag of 100 uniquely-numbered balls (0-99) with replacement. Then the
> question is: how many draws on average does it take to see every number?
>
> A quick modelling of this process:
>
>     $~.?100$100  NB. how many unique balls in 100 random draws, with
> replacement ?
> 65
>     $~.?100$100  NB. Try a couple more times.
> 63
>     $~.?100$100
> 62
>     $~.?300$100   NB. Try 300 draws with replacement.
> 94
>     $~.?300$100
> 99
>     $~.?300$100
> 92
>     $~.?500$100   NB. Try 500 draws with replacement.
> 98
>     $~.?500$100
> 100
>     $~.?500$100
> 97
>     $~.?500$100
> 100
>     $~.?700$100    NB. Try 700 draws with replacement.
> 100
>     $~.?700$100
> 100
>     $~.?700$100
> 100
>
> So it looks like making somewhere around 500 random draws on average, will
> let the drawer see all 100 numbers. So it will typically be a couple of
> years before the prisoners can be sure of their fate.
>
> Skip
>
>
> On Mon, Sep 10, 2012 at 10:32 PM, Jordan Tirrell <[email protected]
> >wrote:
>
> > I got an expected number of rounds around 1e6. It seems a bit high but I
> > think it might be right. So if they want to be 100% they'll die of
> natural
> > causes while waiting.
> >
> > I wasn't on the forums 5 years ago but its certainly a good problem.
> >
> >
> >
> > On Mon, Sep 10, 2012 at 6:06 PM, Roger Hui <[email protected]
> > >wrote:
> >
> > > V. good.
> > >
> > > A related question for the 100 Prisoners problem, is the expected
> > > number of rounds before the prisoners go free, if each prisoner has
> > > the same chance of being picked in each round.
> > >
> > > Did you solve the 88 Hats problem when it was posed here 5 years ago?
> > >
> >
> --
> Skip Cave
> Cave Consulting LLC
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>
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