Many puzzles are difficult to solve from one perspective but easy from another. A challenge on stackexchange was to find an equivalent version of the Monty Hall problem where the correct solution of switching is obvious. Joshua B. Miller has an excellent answer. To recall, in the original there is a great prize hidden behind one of three doors. You choose a door. Monty Hall then reveals a lousy prize behind one of the other two doors (it’s always a lousy prize). Do you switch doors? Most people see no reason to switch. Even Paul Erdos was a no switcher! Moreover, most of those who do switch get to that conclusion with an unintuitive Bayesian calculation.
Here’s the intuitive version.
There are three boxers. Two of the boxers are evenly matched (no draws!); the other boxer will beat either them, always.
You blindly guess that Boxer A is the best and let the other two fight.
Boxer B beats Boxer C.
Do you want to stick with Boxer A in a match-up with Boxer B, or do you want to switch?
See also Miller’s new piece in the JEP which looks at the Monty Hall problem and the Hot hand puzzle.
6. How and when do people change their minds on-line. Does this article make any sense to you?
I’ve been reading Leonard Mlodinow’s The Drunkard’s Walk: How Randomness Rules our Lives. The book covers the Monty Hall problem, Bayes’s Theorem, availability bias, the illusion of control and so forth. If these are unfamiliar, look no further for an entertaining account.
On the other hand, I can’t say that I learned much I didn’t already know. Nevertheless, I still enjoyed reading the book – it’s well written and filled with interesting nuggets (Did you know that the great mathematician Paul Erdos refused to believe that you should switch doors?). If you teach probability theory or intro stats you will find lots of good examples to brighten up your lectures.
One problem did intrigue me. Suppose that a family has two children. What is the probability that both are girls? Ok, easy. Probability of a girl is one half, probabilities are independent thus probability of two girls is 1/2*1/2=1/4.
Now what is the probability of having two girls if at least one of the children is a girl? A little bit harder. Temptation is to say that if one is a girl the probability of the other being a girl is 1/2 so the answer is 1/2. That’s wrong because you are not told which of the two children is a girl and that makes a difference. Better approach is to note that without any additional information there are four possibilities of equal likelihood for the sex of two children (B,B), (G,B), (B,G), (G,G). If we know that at least one is a girl we can remove (B,B) so three equally likely possibilities, (G,B), (B,G), (G,G), remain and of these 1 has two girls so the answer is 1/3.
Ok, now here is the stumper. What is the probability of a family having two girls if one of the children is a girl named Florida?
At first it seems impossible that knowing the name should make a difference. Surely, the answer is 1/3 just as before? After all, every child has a name. But knowing the name does make a difference. Here’s a hint, Florida is a rare name.