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Topic: Probability Pill
Replies: 20   Last Post: Jan 31, 2013 7:23 PM

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quasi

Posts: 10,396
Registered: 7/15/05
Re: Probability Pill
Posted: Dec 28, 2012 12:28 AM
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quasi wrote:
>quasi wrote:
>>quasi wrote:
>>>quasi wrote:
>>>>William Elliot wrote:
>>>>

>>>>>Each day I take 1/2 an aspirin tablet. I bought a bottle
>>>>>of 100 tablets; each day I take out one, if it's whole I
>>>>>break it half and eat a half and put the other half back:
>>>>>if I pull out a half tablet I eat it. I was wondering
>>>>>after I break the last whole one what the expected number
>>>>>of halves are in the bottle? I assume that any piece I
>>>>>pull out has uniform probability.

>>>>
>>>>I suspect that the above question is not actually yours.
>>>>
>>>>If that's the case, what is the actual source?
>>>>
>>>>Is it from a poster in another forum? If so, why do you omit
>>>>mention of the poster and the forum?
>>>>
>>>>Is it from a book or math contest?
>>>>
>>>>Why do you repeatedly post questions that are not your own
>>>>without giving credit to the source?
>>>>
>>>>In any case, the expected number of halves left when the last
>>>>whole pill is split is
>>>>
>>>> (199!) / ((4^99)*((99!)^2))

>>>
>>>Which is slightly more than 11 half pills.

>>
>>Oops -- ignore my answer -- it's blatantly wrong.
>>
>>I'll rethink it.
>>
>>In the meantime, can you identify the source of the
>>problem?

>
>Ok, the correct answer is x/y where x,y are given by
>
> x = 14466636279520351160221518043104131447711
>
> y = 2788815009188499086581352357412492142272
>
>As a decimal, x/y is approximately 5.18737751763962
>
>Thus, on average, about 5 half pills.


As a point of interest, when starting with n pills, the
expected number of half pills remaining at the end (after
the last whole pill is gone), as confirmed by the data, is

1 + 1/2 + 1/3 + ... + 1/n

However, while I'm sure the above result is correct, I don't
have a proof.

Let f(n) denote the expected number of half pills remaining at
the end, starting with n whole pills at the outset. It's
immediate that f(1) = 1, hence a natural approach would be to
try to prove the recursive relation f(n) = f(n-1) + 1/n, but
I don't see how to prove it. Can someone provide a proof?

quasi



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