Random dominoes

Difficulty: 3 of 10

This puzzle wouldn't be called a "classic", but it is nonetheless a good demonstration of non-intuitive probability.

Say I have a complete set of dominoes. Each domino has 0 to 6 dots on each side, and no two dominoes are the same. I take pick out a random domino with my eyes closed. While my eyes are still closed, I pick a random side of the domino and show it to you. You see six dots. What is the probability that the other side also has six dots?

There are 7 possible dominoes that I could have picked out: 6-0, 6-1, 6-2, 6-3, 6-4, 6-5, and 6-6. Only one of those dominoes has six dots on the other side. The standard way of calculating probabilities is to divide the number of "hits" by the total number of possibilities. Therefore, the probability must be 1/7, right?

Wrong. The standard way of calculating probabilities assumes that each possibility is equally likely. In this puzzle, that is not the case.

In a complete set of dominoes, there are 28 dominoes, and each domino has two sides. Therefore, when I showed you a random side of a random domino, there were 56 equally likely possibilities. Not 7, but 8 of these possibilities had six dots. Those possibilities were the following: 6-0, 6-1, 6-2, 6-3, 6-4, 6-5, 6-6, and 6-6. That's right, I was twice as likely to have picked the 6-6 domino, because I could have picked either side of the domino and still gotten a six. Before I showed you the six dots, I was just as likely to pick 6-0 as 6-6, but with 6-0 there was a 1/2 probability of picking the zero instead of the six.

Therefore, the probability of finding a six on the other side is not 1/7, but 2/8 = 1/4.

If you don't believe me, consider this. There are 28 dominoes, and 7 of them have the same number of dots on either side. Therefore, whenever I pick out a random side of a random domino, the other side has a 7/28=1/4 probability of being the same as the side I showed you.