Start with a bag containing 5 white beans. Randomly draw one at a time employing the following rule:
If the bean is white, color it black and put it back in the bag;
If the bean is black, keep it out.
What is the probability that at some point there will be a single white bean in the bag?
Generalize to start with N beans.
Does the probability converge, and if so, to what value?
for n=5 :
We mark each specific bean with a number. After 10 draws the bag is empty as each bean has been drawn twice. If we list the possible sequences of draws, we'll get a list of 10 digits comprised of digits 1-5 , each appearing twice. All possible permutations of this list are 10!, but here the first and the second appearance of each digit is counted as a seperate case, so we have to divide the total number of possible sequences by 2^5, and we get :
10!/2^5
The asked for incidents are all those ending by :11, 22, 33, 44, 55
The number of permutations answering to this, will be the number of permutations of a sequence of the first 8 draws containing only the other 4 digits, and that should be multiplied by 5 for each of the 5 endings, so we get:
(8!/2^4)*5
and the required probability :
P = [ (8!/2^4)*5 ] / (10!/2^5) =
1/9
Analytical solution
