A parent in one of my math classes passed this along to me. The National Museum of Mathematics (MoMath) has partnered with the Wall Street Journal to create Varsity Math. There are three levels: freshman, JV, and varsity. I was impressed with the JV problem, it wasn’t obvious and it really made you think. It is posted below.

Your friend tells you: This is a remarkable urn. It contains only red and black balls. If you reach in and take one ball at random, there’s an equal chance of drawing red or black. So you might think that if you instead take two at random, it’s 50/50 that your balls will match. Well, you’d be wrong—but if they do match, and then you reach your other hand in and take two more, your chances of matching a second time are 50/50!

*How many balls in your friend’s urn (before you take any out)?*

Comment below if you think you’ve figured it out. Beware, it is tricky. I had to read it about 5 times until I finally fully understood the situation.

ChrisThe urn has 6 balls to start.

Since your very first selection has 50/50 probability of red vs. black, that means the number of red and black balls are equal. r = b

Once you’ve picked out two balls of the same color (assume they are black), then the probability of picking consecutive red balls in your next two draws = (r/(r+b-2))*(r-1)/(r+b-3) = 0.5

Substitute b in for r and do the multiplication: (r^2 – r)/(4r^2 – 10r +6) = 0.5

Do some algebra to isolate everything on one side: 0 = r^2 -4r + 3

0 = (r-3)(r-1) –> r=3 or r=1. But r cannot = 1 because you must be able to pick out 2 black balls at the onset. Therefore r=b=3, so the urn has 6 balls.

Tom ORourkeKathryn, kindly help this old geezer with double-faced dice quiz in Sept 2-3 2017 issue of Wall Street journal? Thank you so much.