Give yourself a minute or two to solve this : You have 2 boxes and 20 balls, 10 black and 10 white. How would you distribute the balls in the boxes (any number in each box so long as the total is 20) to maximize the odds of picking 2 whites on a random selection of 1 ball from each box ? i.e: how many blacks and whites in each box ?
Well to get the largets chance you want to maximize the numerator and minimize thedenominator of the fraction. So I guess 5 white and 10 black in one box and 5 white in the other box.
I think so too... There's a common trap that a lot of people fall into. Here's the argument I've heard most commonly (including out of my own head). All balls in Box 1 or all balls in Box 2 gives you a zero chance of success. The problem is symmetric w.r.t the boxes, and starting from one of the above extremum cases, and moving balls into the empty box can only improve your odds. So the best odds will the found at the midpoint, i.e : 5 black and 5 white balls in each box. The "apparent" similarity with problems like maximising entropy in a 2-chamber system, or finding the rectangle of largest area for given perimeter leads to the taking of the bait quite happily