We’ve seen several cases of puzzles in which the answer can be found by simply plodding forward, or the answer can be guessed and then verified. If the guess if correct, this is a great labor and time saving method. (Note: we also discussed how the guessing method can lead to biased results and selectively using the evidence to support the guessed solution, so use it carefully. Also guessing and then verifying does not ensure you have all the possible solutions.) But we have also seen examples of puzzles in which the answer can be more quickly derived by considering something other than finding the answer directly. That is, by finding something that was not asked for, the actual target can be more easily found. This method often is used to great benefit in solving probability problems. When asked to find the probability of an event, finding the probability of it not happening is sometimes more direct. You cannot always tell in advance what the best method is to use, but the more puzzles you work, the more likely it is that you will start down the optimum path.

To illustrate these concepts, here are a couple of puzzles. Neither requires more than an introductory understanding of mathematics. The solutions can be found in several ways, but for the fun of it, try to solve them with the non-straightforward methods first.

I show you a stack of 18 cards. 6 of them are black and 12 are red. All of the black ones are face cards, and half of the red ones are face cards. I allow you to draw one card from the shuffled deck. You cannot see the face of the card or in any other way bias the outcome. What is the probability that you will draw a card that is either black, or a face card, or both?

This kind of mixing apples and oranges in the probability distribution is more common in the practical world than you might think, and so being able to rapidly see the underlying scheme is valuable.

With a nod toward Jeff Partridge, who is a stickler for stating puzzles such that they are unambiguous, consider the following along with my repeated warning about making unwarranted assumptions. That is not a clue. If it were a clue, I would probably say so.

Assume you have two tables. One is round, and the other is square with distinctly different four sides. You are hosting a party and a total of eight people will attend. You want to seat four of them (A, B, C, D) at one table and four (G, H, I, J) at the other. But you are not sure of how you want them to be organized. How many distinct seating arrangements can you have for each table?

Note: although I did not explicitly say that all of the guests are distinctly different, since they have unique names, even if they are identical twins, they are all distinct - no tricks there!

The answers for both cases are straightforward - or not.

For those who wish to delve further into decision theory without wading through a lot of equations, I have posted a tutorial on elementary decision theory. It shows examples of faulty physicians’ diagnoses (important for those considering surgery) and how to evaluate anti-terrorist activities (important for everyone). That tutorial can be found here.