An urn contains one red ball and one blue ball. A box of extra red and blue balls lie nearby. George performs the following operation four times: he draws a ball from the urn at random and then takes a ball of the same color from the box and returns those two matching balls to the urn. After the four iterations the urn contains six balls. What is the probability that the urn contains three balls of each color?
I solved this problem with brute force. I want to try and understand the solutions in the website.
Since we start with a red (R) and blue (B), we would need the permutations of RRBB in the next 4 draws. I have trouble understanding why the 6 permutations would have the same probability of $1/30$ (we can find 3 pairs that are equal because of symmetry.) , I do not really get the explanation in the AOPS solution (solution 2).
Source: https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_18
Best Answer
The gist of solution 2 is this:
Imagine you're brute forcing this instead. There are 6 different orders the Rs and the Bs can come in. So we take one, say RRBB and calculate the probability for that particular order: $$ \frac12\cdot\frac23\cdot\frac14\cdot\frac25 $$ Now take another order, say BRBR: $$ \frac12\cdot\frac13\cdot\frac24\cdot\frac25 $$ Do you see the similarities? We always get $2\cdot3\cdot4\cdot5$ in the denominator, because the number of balls in the urn changes the same way regardless of what we draw. And we always get, in some order, two $1$'s (from the first red ball and the first blue ball) and two $2$'s (from the second red ball and the second blue ball) in the numerator.
Solution 2 from AoPS merely points out this exact pattern, but without going through the motions of the brute force method first.