What is the probability of drawing a red ball in the fourth trial? Find a simpler solution. A bag has $r$ red balls and $b$ black balls. All balls are identical except for their colours. In a trial, a ball is randomly drawn from the bag, its colour is noted and the ball is placed back into the bag along with another ball of the same colour. Note that the number of balls in the bag will increase by one, after the trial. A sequence of four such trials is conducted. What is the probability of drawing a red ball in the fourth trial?
My approach
We can make one of two choices at each of the first three trials. There are eight possible series of choices. (RRR), (RRB),...,(BBB).
Probability of choosing red in fourth trial in each of these eight cases is $\frac{r+3}{r+b+3}, \frac{r+2}{r+b+3}, \cdots, \frac{r}{r+b+3}$.
Using this I get the following expression for the probability of drawing a red ball at the fourth trial.
$P(Red) = \frac{r}{(r+b)(r+b+1)(r+b+2)(r+b+3)}\left[(r+1)(r+2)(r+3) + 3b(r+1)(r+2) + 3b(r+1)(b+1) 
+ b(b+1)(b+2) \right]$
Which simplifies to $\frac{r}{r+b}$.
My Question
My approach has been a tedious one.
Now that the answer looks so simple I am wondering if there is a more elegant solution to this. Could you please provide a simpler solution?
 A: Let's do the experiment in a slightly different way.
After each draw, instead of adding another ball of the same color, instead add a white ball with the trial number written on it.  So after the first draw, you replace the ball and add a white ball with the number "1".  After the second trial you add a white ball with the number "2".  And so on.
Now, when you are drawing a ball for the fourth time, there are all of the original balls plus the 3 white balls from the previous three trials.
So in fact we can skip the first three trials and skip ahead immediately to the fourth trial.  You draw out one of the balls.  If it is red or black, then you are done -- you do not have to worry about the first three trials -- the result would have been the same under the original rules, because you did not draw one of the added balls.
But if you draw a white ball, then you have to do a bit more work to discover what color (red or black) it should have been under the original rules.  For example, say you draw white ball #2.  Then to figure out its color, you have to recreate the second draw.  To do this, throw out white balls #2 and higher, leaving only the lower numbers (in this case just #1).  Now the bag is exactly how it looked on the second draw, so you can simulate the second draw now by drawing out a ball at random.
Again, if you draw a red or black ball, you are done.  But if you draw a white ball, you will have to do even more work to discover the correct color, again removing any white balls with higher numbers and trying again.
As you can see, repeating this process, we will eventually draw a red or black ball, and this will tell us the color that the 4th draw would have given under the original rules.
Since we are drawing balls until the first time we draw one of the original red or black ones instead of one of the white added ones, the probability of it being red vs. black is simply given by the original number of red vs. black balls.
A: We claim that the probability of drawing a red ball on the $n$th trial is $r/(b+r)$ for any $n \ge 1$.  We give a proof by induction over $n$.
The case $n=1$ is clear.  So suppose the claim holds for some $n \ge 1$ and for all $b,r \ge 0$.  We compute the probability of drawing a red ball on the $(n+1)$th trial by conditioning on the color of the first ball drawn.

*

*With probability $b/(b+r)$ the first ball drawn is blue, and the urn
now contains $b+1$ blue balls and $r$ red balls.  By the inductive
hypothesis, the probability of drawing a red ball after $n$ more trials, i.e. on the $(n+1)$th
trial, is $r /(b+r+1)$.

*With probability $r/(b+r)$ the first ball drawn is red, and the urn
now contains $b$ blue balls and $r+1$ red balls.  Again by the
inductive hypothesis, the probability of drawing a red ball on the
$(n+1)$th trial is $(r+1)/(b+r+1)$.

So the probability of drawing a red ball on the $(n+1)$th trial is
$$\frac{b}{b+r} \cdot \frac{r}{b+r+1} + \frac{r}{b+r} \cdot \frac{r+1}{b+r+1} = \frac{r}{b+r}$$
This completes the proof by induction.
