I've posted this before, but right now I can't find the answers that
I've gotten, and I'd like to repost to see if I can get a simpler
solution than what I've gotten before.
Let's say that n people agree to meet between 2:00 P.M. and 3:00 P.M..
If each person
is willing to wait the same amount of time d where 0<=d<=1 and the
amount of minutes
the person is willing to wait is d * 60 minute (if d=0.25, means that
the people will
wait 15 minutes), then the probability that ALL of the people will meet
is
n*d^(n-1)-(n-1)*d^n
OR
(1+k*n)*d^n where k=(1/d)-1
and the probability that NONE of them will meet is
(1-(n-1)*d)^n
Now, what if each person is willing to wait a different amount of time
d1, d2, d3, etc,
so that if d1 =.25 that's a 15 minute waiting time, or if d1 or d2 or d3
= .2, that is a
0.2*60 minute = 12 minute waiting time. What is the probability that ALL
of them meet?
The probability that NONE of them meet? if m<n what is the probability
that m people
will meet eachother, while the other n-m people are alone? If you can
express your
answers as a formula or algorithm rather than as an integral please do
so, as I have a
hard time deciphering integrals that are written in e-mail or newsgroups
and they are
slower to implent on computers than formulas (unless I find a super fast
integration
algorithm), but... I'll take whatever you come up with. TIA :-)
--
Patrick D. Rockwell
mailto:[EMAIL PROTECTED]
mailto:[EMAIL PROTECTED]
mailto:[EMAIL PROTECTED]
