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:
{Quote hidden}>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}