Probability of the last ball picked from a box is blue

conditional probabilityprobabilityprobability theory

In a box there are $R$ red balls and $B$ blue balls. $R$ and $B$ are even numbers, and $R>B$.

Initially a random ball is drawn from the box, we take a look on its color and keep the ball out of the box. After, we keep drawning balls from the box until we get a ball where its color is different from the first one. When it happens, we place this ball(the one with different color) back in the box and the process is restarted until there is no balls left.

The question is, what is the probability of the last ball picked from the ball is blue?

EDIT:
What I tried to do was, solve the problem backwards, that means that for the blue ball be the last one the drawns in the box must be specific. But I got nowhere with this idea.

The other thing that I tried was to write all possibilities with $R=4$ and $B=2$, but I don't know how to deal with the restart of the process.

Someone has any ideia how to model this problem?

EDIT 2:
Here is an example:
Let's say that we have $50$ red balls and $14$ blue balls.

Suppose that the firts drawn ball is red, in a such a way that now there are $49$ balls and $14$ blue balls inside the box. The second drawn is a red ball too, so now we have $48$ red balls and $14$ blue balls inside the box. Let's say that in the third drawn the ball is blue, as the last ball is a different color from the others taken out of the box, the blue ball is put back in the box and the process restarts with $48$ red balls and 14 blue balls.
This keep going until there is no balls left.

Best Answer

Let the desired probability be $P_{R,B}$ for $R$ red balls and $B$ blue balls. We characterize the process into 2 steps:

  1. A ball is chosen at random and discarded.
  2. A 2nd ball is chosen. If it has the same colour as the first ball, we discard it. If it has a different colour as the first ball, we replace it into the box and return to step 1.

Let's complete this cycle once, so that we do both steps and return to step 1. There are 3 mutually exclusive and exhaustive events (one of them HAS to occur):

  1. All red balls have been discarded. Let's call this event $\mathcal{A}$. Clearly $P(\mathcal{A})=\frac{R!B!}{(R+B)!}$.
  2. All blue balls have been discarded. Let's call this event $\mathcal{B}$. Clearly $P(\mathcal{B})=\frac{R!B!}{(R+B)!}$.
  3. The box still contains at least 1 red and 1 blue ball. Let's call this event $\mathcal{C}$. Due to mutual exclusivity and exhaustivity, $P(\mathcal{C})=1-\frac{2R!B!}{(R+B)!}$

Our first line of attack in these types of problems should always be inductive. It turns out, in this case, we prepare strong induction on $k=R+B$, the total number of balls. We have $R,B>0$. Let our base case be $k=2$, so $R=B=1$. Clearly, due to symmetry, the probability $P_{1,1}=\frac{1}{2}$. To simplify the notation, let $P_{R,B}=P_{k}$ when $k=R+B$.

Inductive hypothesis: Let's assume $P_{k'}=\frac{1}{2}$ for all $2\leq k'\leq k$. Then, what is $P_{k+1}$? We can do both steps once and return to step 1. Then, we can condition on the exhaustive events described previously: If the 3rd event is the case, our inductive hypothesis guarantees the probability to be $1/2$. In the first 2 cases the probabilities of a blue ball being last are $1$ and $0$, respectively.

$$P_{k+1}=\frac{1}{2}\cdot P(\mathcal{C})+0\cdot P(\mathcal{B})+ 1\cdot P(\mathcal{A})$$

$$=\frac{1}{2}\cdot \left(1-\frac{2R!B!}{(R+B)!}\right)+ 1\cdot \frac{R!B!}{(R+B)!}=\frac{1}{2}$$

This completes our induction. The probability is ALWAYS $1/2$ for the blue ball to end up last. It doesn't matter how many total balls there are, nor what the relative proportion of red and blue balls are. It also doesn't matter whether $R,B$ are even or odd.

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