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?

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**