Subject: (urth) math probability question for seven American nights

From: Marc Aramini <marcaramini at gmail.com>
Date: Sun, 07 Sep 2014 12:34:02 +0000

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?
-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://lists.urth.net/pipermail/urth-urth.net/attachments/20140907/b64d7344/attachment.htm>

---

From: Gerry Quinn <gerry at bindweed.com>
Date: Sun, 07 Sep 2014 13:19:40 +0000

On 07/09/2014 13:34, Marc Aramini wrote:
> 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

---

From: Gerry Quinn <gerry at bindweed.com>
Date: Sun, 07 Sep 2014 13:26:41 +0000

On 07/09/2014 14:19, Gerry Quinn wrote:
>> 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 
> 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.
>
> 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

---

From: Marc Aramini <marcaramini at gmail.com>
Date: Sun, 07 Sep 2014 14:13:33 +0000

Thanks.

(Almost every Middle Eastern name in the text is from the Adventures of
Hajji Baba, and interestingly enough, the detective sent to look for Nadan
at the start of the book, Hassan Kerbalai, is the name of Hajji Baba's
father.  I may have to read at least some of that book beyond the chapters
dealing with Nadan and Mirza and Osman Aga (a writer in Seven American
Nights and Baba's master in the older text, who sends him to purchase
lamb-skins to start the novel).  Baba means father or grandfather, and
when (in Wolfe's story) Nadan sees the old American man who has the writing
machine at the musuem he calls him grandfather. Pretty sure Hajji Baba
tries to defraud Nadan in Morier's book)

The play Visit to a Small Planet is by (EuGENE) Gore Vidal and the cat in
the play is named Rosemary - the second play, Mary Rose, with the two
disappearances of the titular character for periods of time, also seems to
play on Rosemary's name besides the allusions to Nadan's disappearance.

Even though The Adventures of Hajji Baba was very critical of the Middle
Eastern/Persian characters, it was popular there as kind of a satire that
revealed their societal shortcomings.  The same is true of Visit to a Small
Planet and its take on McCarthyism for America. Ironically these Americans
see it as a kind of  testament to their old greatness.
On Sun, Sep 7, 2014 at 6:26 AM, Gerry Quinn <gerry at bindweed.com> wrote:

>>> 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?
>>>
>>>
>> 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
>>
>
> 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.
>
> 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.
>
> _______________________________________________
> Urth Mailing List
> To post, write urth at urth.net
> Subscription/information: http://www.urth.net
>
-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://lists.urth.net/pipermail/urth-urth.net/attachments/20140907/58487c40/attachment-0001.htm>

---

From: Gwern Branwen <gwern at gwern.net>
Date: Sun, 07 Sep 2014 14:16:58 +0000

On Sun, Sep 7, 2014 at 9:19 AM, Gerry Quinn <gerry at bindweed.com> wrote:
> 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.
>
> 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!

Exactly. On the other hand, if he has some source of information about
the egg, then conditional on his new information each day, he may be
able to update his beliefs as he goes along and do better than
guessing just 1/6th each day.

As it happens, there's some literary precedent here: Graham Green's
novel https://en.wikipedia.org/wiki/Doctor_Fischer_of_Geneva_or_The_bomb_party

And if one doesn't know the egg is there, then it becomes an
interesting statistical problem:
http://gwern.net/docs/statistics/1994-falk

-- 
gwern

---

From: Dimitar Nikolov <dimitar.g.nikolov at gmail.com>
Date: Sun, 07 Sep 2014 20:06:13 +0000

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

-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://lists.urth.net/pipermail/urth-urth.net/attachments/20140907/39e376a1/attachment.htm>

---

From: Marc Aramini <marcaramini at gmail.com>
Date: Sun, 07 Sep 2014 20:17:21 +0000

hmmm ... but all the probabilities add up to .665 then ... and if he eats
all six, he's GOING to eat the egg on one of those days
...  sometimes I feel like probabiity is the most nebulous and abstract
aspect of lower math.

On Sun, Sep 7, 2014 at 1:06 PM, Dimitar Nikolov <dimitar.g.nikolov at gmail.com
> 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.
>
>>> to eat them in advance, and laid all six in a row, each marked with its day
-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://lists.urth.net/pipermail/urth-urth.net/attachments/20140907/0153a942/attachment.htm>

---

From: Gerry Quinn <gerry at bindweed.com>
Date: Sun, 07 Sep 2014 20:20:29 +0000

On 07/09/2014 21:06, Dimitar Nikolov wrote:
> 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

---

From: Dimitar Nikolov <dimitar.g.nikolov at gmail.com>
Date: Sun, 07 Sep 2014 20:49:47 +0000

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:

>> 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.
>
> 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
-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://lists.urth.net/pipermail/urth-urth.net/attachments/20140907/f7a95d1d/attachment.htm>