math probability question for seven American nights - CORRECTION

Gerry Quinn · 07 Sep 2014, 20:20 · from 2014-September.txt.gz

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From: gerry at bindweed.com (Gerry Quinn)
Date: Sun, 07 Sep 2014 21:20:29 +0100
Subject: (urth) math probability question for seven American nights -
 CORRECTION
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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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