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Bean coloring (Posted on 2015-07-13) Difficulty: 4 of 5
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?

No Solution Yet Submitted by Jer    
Rating: 4.0000 (1 votes)

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Solution Final correction of analytical solution | Comment 9 of 11 |
  Charlie - Thanks for your comments !

  I have corrected accordingly and recalculated the probability which results now as yours  !!!
 
 here is the corrected solution :

 
 We denote the situation of having completed "d" draws (including returns of blackened beans) ending up  with "w" white beans and "b" black beans in the bag, as : (d,w,b), and the probability of such occurrence, as:
   P(d,w,b).
 Dealing with the case of 5 beans, a single white bean in the bag can only result at the end of the 8th drawing. This can easily be shown by attributing a value of 2 to each white bean and a value of 1 to each black bean. Clearly the initial value of the bag is 10, and each draw reduces this value by 1, so the final value of 2 will be reached after the 8th draw. We are therefore looking for P(8,1,0).
We shall find it by writing the relevant equation backwards till arriving at the known  P(1,4,1)=1.

 (8,1,0) can be produced only from a previous situation of (7,1,1), by drawing a black bean, the probability of which is 0.5 . Therefore we get :  P(8,1,0)=P(7,1,1)*0.5
 Going one step back-  P(7,1,1) can be produced only by drawing a black bean from a previous situation (6,1,2), or from drawing a white bean from (6,2,0), therefore :-
    P(7,1,1) = P(6,1,2)*2/3 + P(6,2,0)*1
where 2/3 and 1 are the probabilities of drawing the respective beans from the bag.
Going further back by the same method we get the following equations :-
 P(6,1,2)=P(5,2,1)*2/3 + P(5,1,3)*3/4
 P(6,2,0)=P(5,2,1)*1/3
further backward steps yield :-
 P(5,2,1)=P(4,3,0)*1 + P(4,2,2)*0.5
 P(5,1,3)=P(4,2,2)*0.5 + P(4,1,4)*4/5

 P(4,3,0)=P(3,3,1)*1/4
 P(4,2,2)=P(3,3,1)*3/4 + P(3,2,3)*3/5
 P(4,1,4)=P(3,2,3)*2/5 + P(3,1,5)*5/6

 P(3,1,5)=P(2,2,4)*1/3 + P(2,1,6)*6/7
 P(3,2,3)=P(2,3,2)*3/5 + P(2,2,4)*2/3
 P(3,3,1)=P(2,4,0)*1 + P(2,3,2)*2/5

 P(2,1,6)=0
 P(2,2,4)=0
 P(2,3,2)=P(1,4,1)*4/5
 P(2,4,0)=P(1,4,1)*1/5

  But P(1,4,1) is known to be :     P(1,4,1) =1

   So now we substitute back through the above equations till we get the asked for probability :

                                      P(8,1,0) = 0.3055388888


  Posted by Dan Rosen on 2015-07-16 00:57:58
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