Subj : Re: efficient way of processing m combinations of n numbers in p sets
To   : comp.programming
From : Mike
Date : Wed Aug 17 2005 12:08 pm

In article <1124188190.499826.88190@g44g2000cwa.googlegroups.com>, mykyp@pearcey2001.freeserve.co.uk says...
> Hi,
> 
> 
> I am trying to process all combinations of 6 sets of numbers (using
> VB).  The
> minimum number in each set will be 2 and the maximum number is
> definable but will be the same for all 6 sets.  The increment in each
> set is also definable and is also the same for each set.
> 
If I understand the problem correctly - you don't need to do any significant computation to get the answer.

So I assume the problem is as follows:

You have n horses H1...Hn, each is at odds O1..On, you have a total of B to bid on the horses, and (finally) you want 
to minimise the worst case loss (or maximise the worst case gain) resulting from any particular horse winning, by 
choosing a series of bids B1..Bn, on each horse.

I am assuming your odds Oi, for horse Hi, means that if you pay x dollars and the horse wins, you get Oi*x dollars 
back. This means that if horse Hi wins you pay B1+B2+B3+....+Bn, and get back Oi*Bi.


So, the defining equations are:

Sum(Bi) = B
Wi = Oi*Bi-Sum(Bi) = Oi*Bi - B

Now, you want to maximise the minimum value of Wi, this will occur when all the Wi are identical, i.e equal to some 
value Wmim.

So, Wi = Oi*Bi - B = Wmin, thus

Bi=(Wmin+B)/Oi

Still want to know what Wmin is, so if we sum the above for all i we get...

sum(Bi) = sum((wmin+B)/Oi) = (Wmin+B)*Sum(1/Oi)

so, if we define P = sum(1/Oi) and do a little more algebra we finally end up with the solution...

Your bet Bi on each horse should be Bi = B/(P*Oi), where P = Sum(1/Oi),
and your expected return regardless of which horse wins
will be Wmin = B*(1/P-1).   QED and all that.

Mike

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