If you have two bags each containing 100 numbered balls and pick 1 ball at random from each bag, what are the odds that you will pick two similarly numbered balls?
I believe this is equivalent to the Birthday paradox equation:
T! 1 - ----------------- T^n . ( T - n )!
Which for T=100, n=2 comes out to:
use strict; use bignum; sub fact{ my($r,$n) = (1,shift); $r *= $n-- while $n; return $r; } print 1-((fact(100)/fact(98))/100**2);
gives 0.01 or 1% chance.
My question is, if I have 2 pairs of 2 bags and pick one ball from each, what are the odds of the first pair being identical to the second pair?
For my purposes, order does matter. So, the question is really what are the odds that ball 1 from each of pair1/bag1 & pair2/bag1 will be identical; and ball 2 from each of pair1/bag2 & pair2/bag2 will also be identical?
Is this a simple combination of odds? Eg. 0.01 * 0.01 == 0.0001?
In reply to Math help by Anonymous Monk
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