This is a fun brain teaser in probability. I think it's been known for a long time, but it became prominent in the early 90s when it was asked in a Marilyn vos Savant column in Parade Magazine. Many people didn't believe her answer (which was correct), including some pompous mathematicians.
It's highly unintuitive, so if you don't see the solution at first, you're in good company!
There are many web sites that have explanations, simulations, and so forth, so if you're interested, do some searching.
Suppose you're on a game show, something like "Let's Make a Deal," which was popular in the 70s. As I remember, in that show, after you won a game, the host, Monty Hall, would offer to let you choose from doors 1, 2 or 3, one of which contained a nice prize, but the others were lame. Sometimes, after you'd chosen a door, Monty Hall would offer you money instead. Or he'd allow you to change your choice. It was all very silly, even though valuable stuff was at stake.
Okay, that's the setup, now for the actual puzzle:
Since Monty Hall can always open a door to reveal a goat, he has given you no information. So, there are now two doors and the chance is now 1/2 instead of 1/3, but the chances are equally likely. You might as well stick, since there's no gain in switching.
Your original probability of winning was 1/3, and if you stick, that probability doesn't change. Therefore, the probability of winning if you switch must be 2/3.
Think about this for a while and see which argument you like. I'll wait.
hum de dum
la de dah
Thought about it enough? Good!
The answer is that you should switch. We can draw out decision trees, we can simulate, or we can present mathematical arguments, but the answer is the same. Strange, but true.
Suppose that, at the very end, someone walks into the room and is offer the same choice. For them, there are two doors, so the chances of winning are 1/2. For you, the chances of winning are 2/3, since you know something they don't: you know your original choice. Strange, isn't it? To determine probabilities, we can't just look at the number of choices and divide. Sometimes, it can be more subtle!
Here's why the probability is 1/2 for the newcomer. Your door has a probability of 2/3 and the other door has a probability of 1/3. If the newcomer flips a coin to choose between the two doors, the probability of winning is
P = (0.5)(1/3) + (0.5)(2/3) = 1/6+1/3 = 1/2
Here's a model in Extend. We'll explore it some.
The St. Petersburg game is another delightful puzzle. It's moderately important in itself, but it will also help to understand another fascinating puzzle that we'll get to next.
The basic idea is that a casino offers the following gambling game:
The question is, how much would you be willing to pay to play this game?
The Wikipedia has an excellent article on the St. Petersburg Paradox. We'll look at that after some discussion.
Here's another interesting puzzle:
A related question: does it matter if you open the envelope you have?
Let X be the amount of money in your current envelope. If you switch, you'll either get 2X or X/2, and those are equally likely. The expected value of switching, then, is:
V = (0.5)2X + (0.5)X/2 = (0.5)(2X+X/2) = (0.5)(5/2)X = (5/4)X = 1.25 X
So you gain 25 percent by switching!
This is bogus! Nothing has changed so how could your expected value go up? In fact, by that argument, if you switch a second time, back to your original envelope, you could gain even more money!
I have an idea, but I'm interested in your ideas, too.
Again, the Wikipedia has an excellent article on the Two Envelopes Problem. We'll look at that after some discussion.
Any of the following might be right; I'm not yet sure. All of these people are undoubtedly more knowledgeable than I am, but I remain unconvinced.