Initially, almost any prisoner will do -- the light bulb is initially
off, and almost any prisoner will turn it on.  And then you have an
expected wait of 100 prisoners before it gets turned back off.
Simple...

As time goes on, though, you have to wait longer and longer for a
prisoner that's not turned the light on to take action.  And, then,
you have to wait another 100-ish prisoners to turn the light bulb off.
 But these are two independent events.  So I think at the end, we have
an expected value of approximately 100 prisoners to turn the light
bulb on for the last time, and then another expected value of
approximately 100 prisoners to turn it back off.  So that's 200 visits
at the end for the last light cycle.

If every case were like this, we would have 100 repeats of the on/off
sequence, for 20000 expected days.  So 2e4 should be an upper bound on
expected value.  I think the actual expected value is 100*(100+101%2)
or 15050 days (or slightly over 41 years).

Of course, if any of the prisoners get sick and die before this time
elapses, that could mean that they never get out.

Or, did I miss something?

-- 
Raul

On Mon, Sep 10, 2012 at 11:32 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?
>>
>>
>>
>> On 9/10/12, Jordan Tirrell <[email protected]> wrote:
>> > 99 of the prisoners switch the light if and only if it is currently off
>> and
>> > they have not turned it on in a prior round.
>> >
>> > The one other prisoner turns the light bulb off every time time he finds
>> it
>> > on (otherwise leaves it off). He asserts the claim after he has turned it
>> > off 99 times.
>> >
>> >
>> >
>> > On Mon, Sep 10, 2012 at 11:55 AM, Roger Hui
>> > <[email protected]>wrote:
>> >
>> >> 100 prisoners are in solitary cells, unable to see, speak or communicate
>> >> in
>> >> any way from those solitary cells with each other. There's a central
>> >> living
>> >> room with one light bulb; the bulb is initially off. No prisoner can see
>> >> the light bulb from his own cell. Everyday, the warden picks a prisoner
>> >> at
>> >> random, and that prisoner goes to the central living room. While there,
>> >> the
>> >> prisoner can toggle the bulb if he wishes. Also, the prisoner has the
>> >> option of asserting the claim that all 100 prisoners have been to the
>> >> living room. If this assertion is false (that is, some prisoners still
>> >> haven't been to the living room), all 100 prisoners will be shot for
>> >> their
>> >> stupidity. However, if it is indeed true, all prisoners are set free.
>> >> Thus,
>> >> the assertion should only be made if the prisoner is 100% certain of its
>> >> validity.
>> >>
>> >> Before the random picking begins, the prisoners are allowed to get
>> >> together
>> >> to discuss a plan. So -- what plan should they agree on, so that
>> >> eventually, someone will make a correct assertion?
>> >> ____________________
>> >>
>> >> Posed to me by Arthur Whitney at Iverson
>> >> College<https://sites.google.com/site/iversoncollege/>.
>> >> This puzzle resembles the 88
>> >> Hats<http://www.jsoftware.com/papers/88hats.htm>puzzle but is (IMO)
>> >> easier.
>> >> ----------------------------------------------------------------------
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>> >>
>> > ----------------------------------------------------------------------
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>> >
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