My friend was asked the following problem in an interview a while back, and it has a nice answer, leading me to believe that there is an equally nice solution.


  Suppose that there are 42 bags, labeled $0$ though $41$.  Bag $i$ contains $i$ red balls and $42-(i+1)$ blue balls.  Suppose that you pick a bag, then pull out three balls without replacement.  What is the probability that all 3 balls are the same color?


The problem can be solved easily by using some basic identities with binomial coefficients, and the answer is $1/2$.  Moreover, if $42$ is replaced by $n$, the answer does not change, assuming $n>3$.  However, this computational approach obscures any hidden structure there might be.  Ideally, I would like a simple and direct proof that the probability of getting RRR is the same as the probability of getting RBB.

So, is there a nice solution to the problem, one that could be explained fully to someone without the use of paper?  Or is there no good way to explain this beyond computational coincidence?

The game can be reformulated in the following way: There is one urn with 42 balls numbered 0 through 41. You start by drawing and keeping one ball (this corresponds to picking a bag in your version). The urn is then taken away while an assistant paints all balls with a lower number than your first draw red and the rest of them blue. Then you draw three more balls and see whether they have the same color.

Now, this is equivalent to first drawing four of the numbered balls, then among those four pick a random one to be the "bag" ball and coloring the other three according to that choice. But doing it that way, we can see that the initial drawing of four balls is entirely superfluous -- only the order relation between them matters, and any ordered set of four balls have the same structure. So we might as well forgo the initial draw and just start out with four balls numbered 1, 2, 3, and 4. Then you win if the one you pick in the second draw is either 1 or 4, and the probability of that is, naturally, 1/2.
