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'[OT]: Friday Poser Challenge'
2001\10\26@051218 by Kevin Blain

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If it is company policy for an employee to bring in cakes on their
birthday.......

What is the probability of there being two people bringing cakes in on one
day?


Example case study:

Size of staff: 45 people
Working days: (5/7 x 365) - 8 days misc (bank holidays etc) = 252 working
days.

Don't forget also that there is 22 days annual leave to account for!


starter for 10....

the probability of someone bringing cakes in today is about 18% (45
people/252 days), However the probability of today being a birthday is only
about 12%.




Regards, Kevin

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2001\10\26@092524 by solar

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> What is the probability of there being two people bringing cakes in on one
> day?

0% :)

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2001\10\26@110438 by Dennis Hoskins

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Just "Let them eat more cake"

----- Original Message -----
From: Kevin Blain <spam_OUTkevinbTakeThisOuTspamWOODANDDOUGLAS.CO.UK>
To: <.....PICLISTKILLspamspam@spam@MITVMA.MIT.EDU>
Sent: Friday, October 26, 2001 5:10 AM
Subject: [OT]: Friday Poser Challenge


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2001\10\26@222005 by Russell McMahon

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Methinks the odds are MUCH higher than the starter suggests.
Ignoring leave etc for now.
On a given day the chance that it is NOT a given employees birthday is
364/365 (ignore leap years).
For all but two employees to NOT have this as their birthday is
(364/365)^(N-2)
364/365 = 0.997261
0.997 ... ^ 43 = 0.1178
This is the chance that all of 43 people won't have a birthday today.

ie there is a 88% chance on any given day that two people will bring cakes
in.

I can see holes in this (and the answer appears stupid) but it suggests that
the odds for cakes are much higher than may be expected. Someone point out
what refinements are required.


           RM


{Original Message removed}

2001\10\27@170454 by uter van ooijen & floortje hanneman

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> What is the probability of there being two people bringing cakes in on one
> day?

Note that you can not simply square the average chance that someone brings
cake on a day, because the monday and friday have a higher change (and the
mean of squares is not the same as the square of the mean).

Wouter

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2001\10\28@133123 by Gerhard Fiedler

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My statistics and combinatorics are a bit rusty, but isn't the probability
of this something like


          / 45 \
pNCakes = |    | * p^N * (1-p)^(45-N)
          \  N /


pNCakes: the probability of N cakes at any given day
N: the number of cakes expected
45: the number of employees
p: the probability of one employee bringing in a cake on a certain day

(You need probably a fixed pitch font to see what I mean -- I'm not sure
how this operation is called in English, it's probably "combinations of N
from 45" or so...)

This calculates to a probability of 10.9% for 1 cake and 0.66% for 2 cakes
on a given day -- if we consider a 7 day workweek (as we have it here
:)  Otherwise we'd have to come up with rules how the Saturday and Sunday
people bring their cakes in. In any case, the probabilities then are
probably not anymore equal for all days of the week, and the whole thing
gets probably too complex for my free time :)

A careful analysis of this formula also reveals what was wrong with the
thinking below -- even though the outcome below could intrigue me to go
back to work in companies where a considerable number of employees bring in
cakes, and 88% of the days two of them... :)

ge

At 00:53 10/27/2001 +1300, Russell McMahon wrote:
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