(urth) math probability question for seven American nights - CORRECTION

[Dimitar Nikolov]

You are right! This would only work for an infinite number of eggs. Sorry
for bringing my own confusion into this.

On Sun, Sep 7, 2014 at 4:20 PM, Gerry Quinn <gerry at bindweed.com> wrote:

>
> On 07/09/2014 21:06, Dimitar Nikolov wrote:
>
>> It does depend what question exactly you're asking. Indeed, the
>> probability starts at 1/6, then assuming you know that wasn't the special
>> egg, the probability for the next day is 1/5 and so on.
>>
>> You could however, ask the question, on which day is he most likely to
>> eat the egg?
>>
>> Then, you start with 1/6 for the first day once again. But the
>> probability the egg is eaten on the second day, is 1/6 times the
>> probability it wasn't eaten on the first day (=5/6), so .~.139. The
>> probability it was eaten on the third day is 1/6 times the probability it
>> wasn't eaten on either previous day, and so on. This is given by the
>> geometric distribution that you can read about on Wikipedia and results in
>> the following probabilities:
>>
>> eaten on day 1: 0.167
>> eaten on day 2: 0.139
>> eaten on day 3: 0.116
>> eaten on day 4: 0.096
>> eaten on day 5: 0.080
>> eaten on day 6: 0.067
>>
> Those probabilities don't add to 1!  You're forgetting that tghere is one
> fewer egg every day.
>
> Your figures work for a scenario in which he adds an ordinary egg every
> day, so he always has six, and chooses at random from them. Of course this
> could go on forever, as the special egg might never be chosen.  You'll find
> that the limit of the sum of probabilities goes to 1.0 after infinitely
> many days.
>
>
> - Gerry Quinn
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