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?

As Steve correctly commented, my previous solution is mistaken, so here is a correct one :

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.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

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)82/3

P(3,3,1)=P(2,4,0)*1

P(2,1,6)=0

P(2,2,4)=0

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

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) = 0.256077776**