Just a short one today, all. I’m heading further North. So far from civilization that when I was a boy, it was a cause to celebrate when a cit thirty miles distant opened a Holiday Inn with an—get this— enclosed swimming pool. Such a marvel!
Good probability problems are those that affront the intuition, and the best are simple-sounding but cause fights or furious indignation. The Monty Hall dilemma is one such problem.
Most have heard of the stink made when the self-named Marilyn Vos Savant submitted her correct solution to Parade magazine. Not just civilians, but your actual college mathematics professors wrote in to say Wrong! But wrong they were.
We met something of the same symptom with the “One Boy Born Tuesday” problem. How the hell can knowing which day a kid was born give us any information about his sex?!
This kind of thing doesn’t fit into any of the ordinary slots we set aside for probability knowledge. And that’s what makes it such a wonderful example, because it forces us to carve out new slots once understanding arrives.
The “Three Cards: One Black, One Red, One Half & Half”, and the subject of today’s puzzler, is nothing more than a gentle push on the boundaries. The trick will come in trying to expand it by adding a twist, something I’ll invite you to do.
I have three cards, one of which is all black, one all red, and the last black one side and red the other. Before you on the table is a card whose face is red. Given this information—for we recall that all probability is conditional on stated premises—what is the chance that the other side is black?
Most intuitions suggest that the chance is 1/2, since you have either the all red card or the card that is half and half. But the actual chance is 1/3. Here’s why.
In all probability problems, the only thing to do is to lay out everything that can happen. This can be laborious, but once you do so you cannot go wrong. The mistake people make is to try and jump to the right answer using just their intuition.
What’s everything that can happen? Well, every card has two sides, 1 and 2. You could have seen B1 or B2 of the black card, R1 or R2 of the red card, or CR (combo red) or CB (combo black). You see red, and that can happen three ways: R1, R2, or CR.
If you see R1, then the other side must be R2. Likewise, if you see R2, the other side must be R1. The third and final possibility is you see CR, which makes the other side CB. Thus, the probability—give all the evidence we have—is 1/3.
Not so bad, right? Almost to the point of being dull.
So can we think of a clever way of tarting it up? My guess is that adding seemingly irrelevant, but actually pertinent, information can add the necessary amount of bogglement. Like in the “One Boy Born Tuesday” problem.
How about this? Same initial premise of three colored cards, but this time I say I deal you the card on Tuesday. Given this additional information, what is the probability the other side is black?
I haven’t worked this out—it’s a spur-of-the-moment thought. It might well be that the day of the week is irrelevant. But can we prove it is so?