Re: The Harem-problem
- From: Frank Cornelis <info@xxxxxxxxxxxxxxxx>
- Date: Wed, 31 Jan 2007 10:26:21 GMT
Hi,
Your formula seems to hold against the values I calculated with this simulator. Could you explain how you came up with it?
Regards,
Frank.
matt271829-news@xxxxxxxxxxx wrote:
.
On Jan 28, 8:52 am, Frank Cornelis <i...@xxxxxxxxxxxxxxxx> wrote:Hi,
When I was still a student I came up with a problem I called the
Harem-problem. The problem goes like this (sorry if I offend some
people, but it's an interesting mathematical quest): how many women does
a man need to be able to 'do it' every day of the month, and this with a
given probability?
I started solving this probability problem but never found a closed
formula for it. The document can be found at:http://www.frankcornelis.be/conqueror.pdf
Can anyone give any hint on how to continue the work?
Unfortunately I can't read Dutch(?), but your question started me thinking so I made some assumptions about what you're trying to do based on other people's replies and some of the diagrams in the paper. (Though looking at the length and complexity of your paper I wonder if my assumptions might be way off beam and way too simple).
Anyway, I assumed n women, each with a monthly cycle of t days, with each woman unavailable for a block of b consecutive days within that cycle, the blocks being uniformly randomly distributed. Provided t >= 2*b (which seems reasonable!) I make the probability that there will be at least one day in the cycle when all women are unavailable equal to (b^n - (b - 1)^n) / t^(n - 1). If t < 2*b then an adjustment is needed, but I have not tried to work out that case.
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