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

[Dimitar Nikolov]

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

So, in this phrasing of the question, which I think was the original
question, he was most likely to eat the egg on the first day. Perhaps
someone else can verify or challenge this, as I am not familiar with this
particular story.

Best,
Dimitar

On Sun, Sep 7, 2014 at 9:26 AM, Gerry Quinn <gerry at bindweed.com> wrote:

>
> On 07/09/2014 14:19, Gerry Quinn wrote:
>
>>
>> On 07/09/2014 13:34, Marc Aramini wrote:
>>
>>> I had a quick question but trying to articulate it to look it up
>>> independently is hard.
>>>
>>> Is there a statistically most probable day for Nadan to eat the special
>>> one of the six eggs on any given day given that the first day is 1/6 and
>>> then from there the probability that the egg is there begins to be
>>> something like 1/5  on the next day ( but the chance that it isn't there
>>> should be factored in somehow, but I wasn't sure if it was 1/5 - 1/6 ...
>>> Then 1/4-1/5-1/6 etc. ) Or is the system set up in such a way that the
>>> chance of getting the egg is always 1/6 regardless on any given morning
>>> assuming no eggs are stolen or disappear?
>>>
>>>
>> I don't have the book to hand, but if he is given six eggs, one special,
>> and eats one every day at random, the chance of getting the special egg on
>> any day is indeed 1/6, as you reckoned in your other post.
>>
>> The easiest way to see it is to imagine he decided at random the order to
>> eat them in advance, and laid all six in a row, each marked with its day
>> for eating.  Clearly the chance is the same for each day!
>>
>> - Gerry Quinn
>>
>
> Actually, that is the chance before the experiment,  If the first egg is
> to be eaten on Sunday, then before he eats it he knows that the chance of
> eating the special egg on any given day (say Tuesday) is 1/6.
>
> But that presupposes he has no clue as to when he has eaten the special
> egg.  If he can tell immediately, then after he eats Sunday's egg. he knows
> that either it was the special one, or it wasn't.  Depending on which, the
> chance that he will eat the special egg on Tuesday becomes either zero, or
> 1/5.
>
> If he has some information that might help him decide which egg is
> special, but cannot be certain, the answer is somewhere in between. I
> cannot remember the story well, but I suspect this is most likely the
> situation!
>
> - Gerry Quinn
>
>
>
>
>
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