Re: math question verrrrrry off topic
Sorry, the value I quoted was 150! too large.
It is a bit academic though as we are only interested in m! results
where m is the number of entries from Montana State University faculty,
which could be as low as 3! leaving the remaining
1277916851111034312174368668471297860116606752864715959921012222958747794594
5942798968735988572227507979644437906069421788093063476741931161574090
combinations to fight it out between themselves.
Yves Gauvreau wrote:
Ian,
may be it's just a typo but C(n,m) =
600!
-------------------
150!(600-150)!
=
1277916851111034312174368668471297860116606752864715959921012222958747794594
5942798968735988572227507979644437906069421788093063476741931161574096
I t is however way much larger then what usual software can endle and this
is why I said it's infinite. Still remains the problem that somewhere in
these gazillion sets of 150 are our 3 people from institution A.
Yves
----- Original Message -----
From: "Iain" <iain.coghill@eiflex.com>
To: <alt-photo-process-l@usask.ca>
Sent: Saturday, January 19, 2008 12:56 PM
Subject: Re: math question verrrrrry off topic
There are actually 600!/(600 - 150)! ways to randomly pick 150 items
from a bag of 600, where n! (pronounced n-factorial) is defined as 1 x 2
x 3 x ... x (n-1) x n. If you don't care about which order those 150 are
picked you can divide that number by 150! - giving roughly 57 followed
by 261 zeroes
Yves Gauvreau wrote:
Hi,
if one consider the number of ways to pick randomly 150 items in a bag
with
600 items in it, then the answer is infinity and whatever you do after
that
like multiplying it by a tiny tiny tiny probability, even then your
stock
with infinity.
One would need to approach this problem another way and in our case the
150
people are not choosen randomly in practice. The only other way around
to
give an answer to Chris, is to find out what's in the bag. For example,
say
we fill the bag with 10 oranges and ask someone to pick 3 items in that
bag,
what is the probability of getting 3 oranges. 100% of course but if the
bag
contains 3 oranges, 4 bananas and 3 pear, the answer is différent of
course.
All this to say we need to know what's in the bag.
Regards,
Yves
----- Original Message -----
From: "Katharine Thayer" <kthayer@pacifier.com>
To: <alt-photo-process-l@usask.ca>
Sent: Saturday, January 19, 2008 9:51 AM
Subject: Re: math question verrrrrry off topic
Okay, look.
If this were a simple probability experiment, let's say there were
600 objects in a big jar, all exactly the same size and shape and
differing only in color: 450 red ones (for not-accepted) and 150
green ones (for "accepted") all mixed up really well, and the
question was, "if some people from College X reached into the jar
blindfolded and three of them pulled out green objects, what would
the probability of that result be, would it be 1/4x1/4x1/4?" then
the answer would be "only if just three people from College X
reached into the jar. If more than three people from college x
reached into the jar, let's say five people from college x reached
into the jar, then the probability of three of those five people
pulling out a green marble would be 1/4 x 1/4 x 1/4 x 3/4 x 3/4."
But of course this isn't anything like that. In this case, the 600
objects in the jar are made by the people who are being selected, and
the objects are all different, all different sizes and shapes and
colors, and made of different materials. People might contribute
different numbers of objects (though it's never been clear to me
whether that's the case or not) in which case the objects made by the
same person presumably aren't as different from each other as the
objects made by different people. And the selection is made not by
the people themselves reaching into the jar blindfolded, but by a
third party, a judge, who also doesn't reach into the jar blindfolded
and pull out objects randomly to make a selection, but pours them all
out on in a big tray and looks at all 600 of them closely before
deciding which ones he/she wants to include. This particular judge
might be drawn to metal objects, or even especially to platinum over
silver, or maybe he or she particularly dislikes street scenes and
prefers pictures of old mills, or is looking for a certain level of
craftsmanship in the work, or maybe the criterion is something even
more difficult to articulate, whether the judge "likes" something or
not. Whatever the criteria, by the end of the day, the objects are
separated into two piles, "accepted" and "not accepted," and there
are 150 objects in the first pile and 450 in the second pile.
I hope it should be clear to everyone at this point how different
this is from the problem above, and why you can't treat this as a
problem of simple probability and say that the probability of any
entry being accepted is the same as the probability of any other
entry being accepted, and that this equal probability for each entry
is 1/4, since 1/4 of all the entries were accepted. It's just not
that kind of problem, and it doesn't work to treat it that way.
Thank you.
Katharine
On Jan 18, 2008, at 6:49 PM, Katharine Thayer wrote:
:--)
On Jan 18, 2008, at 6:23 PM, Diana Bloomfield wrote:
Hey Katharine,
I don't know-- maybe. I honestly didn't read the other answers. :)
On Jan 18, 2008, at 8:43 PM, Katharine Thayer wrote:
Hmm, I thought that's what we all already have said, isn't it?
That that theoretical probability (1/4x1/4x1/4) would hold only
if assumptions were met, and since assumptions are obviously not
met (for example, judging is not a random lottery of course but
is done on the basis of criteria, arbitrary or otherwise but
certainly not random). Also, no one has said whether the 600
entries are 600 works or 600 people; I was assuming that they are
600 works representing fewer than 600 people, in other words
people could submit more than one work, in which case, as I said,
the number of works submitted per person would also have to be
figured into the equation somehow. Besides, if one person
submits ten pieces and another person submits one, the ten pieces
by the one person couldn't be considered independent entries in
the same way one of those ten could be considered independent of
the one from the other person, and independence is also an
assumption that must be met in order to consider the probability
of acceptance to be the same for all entries.
Katharine
On Jan 18, 2008, at 4:25 PM, Diana Bloomfield wrote:
Okay, Chris. Here is it-- straight from my resident
statistician here:
If they were the only 3 people from that institution who
applied, AND if judging was completely random, then the
probability of this is roughly 1 in 64 (key word: roughly). If
more than that applied from this same institution, and only 3
got in, then the calculation will be more complex.
Hope that helps. :)
On Jan 17, 2008, at 12:00 PM, Christina Z. Anderson wrote:
Where else but this list can I ask these weird questions about
chemistry and math and computers and alt???
OK for you math people (Yves?): If there is a show and 600
entries, and 150 are accepted, there is a 1 in 4 chance of
acceptance. If 3 people from the same institution are accepted
what percent chance is that--is it 1/4 x 1/4 x 1/4 or a 1.5%
chance or is it a more complex formula?
Forgive the off topic request but it does relate to photo as 3
of our program got into a photo show and I want to be able to
mathematically brag about it to the dept. head/dean.
Chris
Christina Z. Anderson
Assistant Professor
Photo Option Coordinator
Montana State University
CZAphotography.com
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