# Conditional probability. Unable to prove the general result.

1. Oct 30, 2010

### Karlx

Hi everybody.
I keep on reading Rohatgi's book "An introduction to Probability and Statistics" and I have worked out the following problem:

"An urn contains r red marbles and g green marbles. A marble is drawn at random and its color noted. Then the marble drawn, together with c > 0 marbles of the same color, are returned to the urn. Suppose that n such draws are made from the urn. Prove that the probability of selecting a red marble at any draw is r/(r+g)."

I have obtained the following expression for the required probability:

$$P(n) = \frac{1}{\prod_{j=0}^{n-1}(r+g+jc)} \sum_{k=0}^{n-1}(\stackrel{n-1}{k}) \prod_{j=0}^{k}(r+jc) \prod_{j=0}^{n-k-2}(g+jc)$$

This expression gives the result r/(r+g) for values n=2,3.
I have been trying to prove it for all values of n>=2, by induction, but with no success.
Perhaps anyone of you could help me.
Thanks and Happy Halloween !!

2. Oct 30, 2010

### bpet

Hint: P(n) is the expected value of R(n)/(R(n)+G(n)) where R and G are the number of red and green balls after n draws.

3. Nov 3, 2010

### Karlx

Thanks bpet.
Calculating P(n) as the expected value of R(n)/(r+g+(n-1)c) I arrive to the same expression I wrote in my first post.

So it remains the problem of proving that

P(n)=r/(r+g) (1)

for all values of n>0.

Finally, I achieved it.
I have proved that expression (1) holds for n=1,2,3.
Then, from the general expression of P(n) I have proved that P(n+1)=P(n).
It's not difficult, but in my first attempt I had some trouble with the indexes of summations and products.

Once again, thanks for your hint.
It has been useful for me in order to look the problem from another point of view.

By the way, I think this problem is known as the Pólya's urn.
The Hungarian mathematician gave it as an example of a martingale, that is, a random variable whose expected value at any time t is equal to his expected
value at a previous time s < t.

Last edited: Nov 3, 2010
4. Nov 3, 2010

### bpet

That's interesting. Also the following identity might be useful:

E[R(n+1)|R(n),G(n)] = R(n)+c*R(n)/(R(n)+G(n))

5. Nov 4, 2010

### Karlx

Once again, thanks a lot, bpet.
Your last expression has been, indeed, very useful !!
In fact, after "fighting" with summations and products in my first approach to the problem, your expression is the clue to solve it without such a mess.
The new approach follows these steps:

1) Let R(n) and G(n) be respectively the number of red and green marbles at the n-th draw.

2) At the (n+1)th draw,
a) The probability that R(n+1)=R(n) is G(n)/(R(n)+G(n))
b) The probability that R(n+1)=R(n)+c is R(n)/(R(n)+G(n)

So, E[R(n+1)|R(n),G(n)] = R(n)+c*R(n)/(R(n)+G(n)), as you said.

3) And from this last expression,

P(n+1) = E[R(n+1)|R(n),G(n)]/(R(n+1)+G(n+1))
= [R(n)+c*R(n)/(R(n)+G(n))]/(r+g+nc)
= R(n)/(R(n)+G(n)) = P(n)

4) Thus P(n) = r/(r+g) for all n>0.

6. Nov 4, 2010

### bpet

Almost there. The statements in step 3 can be made more precise with the law of total expectation E[X] = E[E[X|Y]].

7. Nov 6, 2010

### Karlx

Yes, you are right.

P(n+1) = E[R(n+1)/(R(n+1)+G(n+1))]
=E{E[R(n+1)|R(n),G(n)]}/(R(n+1)+G(n+1))
=E{R(n)+c*R(n)/(R(n)+G(n))}/(r+g+nc)
=E{R(n)/(R(n)+G(n))}= P(n)

Thanks.