When you think about it it they come up in real life as well, just you aren't necessarily aware of them. Wouldn't it be strange, for example, if you measured the distance between two atoms and it turned out to be exactly 1 metre? Or an exact fraction like 1.4 metres? Or even a long fraction like 1.453162343055682m? It seems much more likely that the number would be an infinite series of random-looking digits, which would make the distance an irrational number.

That assumes a naive model for the universe in which we live. The distance between two atoms is not a number that can be determined with arbitrary precision, so it doesn't make a lot of sense to ask whether it is rational or irrational. I am not a physicist, but my understanding is that this is not just a limitation of our instruments, but a feature of nature.

The way I think of the world these days, everything that matters is discrete. Real numbers are a convenient approximation in situations where the numbers are large enough. This is often the case in physics; but there is nothing "real" about real numbers.

Yeah, fair point, I was assuming a simple Newtonian / Euclidian universe - just trying to build up some kind of intuition about what irrational numbers are.

Mathematics tell you that the likelihood of a goat being behind Monty's doors changes when he opens another one, and you're twice as likely to win if you change mind. Unless Monty's assistant is swapping prizes behind the scene (which of course we cannot know for sure), the likelihood doesn't change. Again, while the reasoning is correct, the assumptions and conclusions are wrong (the mathematican coming up with the proof has probably never seen the show).

I agree with what you said about mathematics being based on assumptions, but I think you've got this one confused. The Monty Hall problem isn't about absolute probabilities (which are simply 1 for the door with the prize behind it, 0 for the other two doors) it's about conditional probabilities given the information available to you. Initial that is nothing so they begin at 1/3, 1/3, 1/3, but when the presenter opens one of the doors you did not choose this affects your probabilities in an asymmetric way. If you aren't convinced, draw up all nine possibilities (3 positions of the prize and 3 possible guesses) and see how often you do better by changing!

Do the Monty Hall problem with 1,000,000 doors instead. Pick a door at random. Then the host opens up 999,998 doors and they all have goats behind them. Do you change your pick? If yes, how many doors do you think you need to have to make it a better choice to change? Correct answer: 3, as in the original problem.

Paul Erdos also disagreed with the Monty Hall problem, until he was shown a computer simulation ;) And Erdos used probabilistic arguments to prove results in combinatorics!

Most of samoth's examples indicate lack of understanding. In particular, I really don't get why people get stuck on the Monty Hall problem. When you learn more information about a situation, the probabilities do change.

I really don't get why people get stuck on the Monty Hall problem. When you learn more information about a situation, the probabilities do change.

But we don't generally encounter evolving situations like this much in education, so most people aren't really equipped to accept the counterintuitive result (I know I wasn't, the first time I saw it). This is exactly why it's such a good problem, because it teaches young mathematicians that their intuition can be wrong. If you got it right the first time, congratulations, but maybe one day something else will come along to keep your ego in check. :)

I can give you another probability problem that I like even better than the Monty Hall problem. Two people play Russian roulette the following way: A neutral referee goes into a room and secretly tosses a fair coin, either puts a bullet in a revolver or doesn't --depending on the result of the coin toss--, spins the drum, comes out of the room and hands the gun to one of the participants. Then the participants will take turns pointing the revolver at their own head and pulling the trigger. The revolver's drum has 6 chambers for bullets.

Imagine you are participating in such a game, your opponent goes first, pulls the trigger and no bullet comes out, then you go, then the opponent again, then you, then the opponent once more. Still no bullet. Now you are at the sixth position, so if there is a bullet in the revolver, you are going to get it. What is the probability that you will get your brains blown out when you pull the trigger?

Zero of course. I point the gun at the opponent's head, pull the trigger, and if there is no bang start running very fast.

Do the Monty Hall problem with 1,000,000 doors instead.

Yes, yes... Vos Savant's well-known explanation. Very entertaining, very smart. Except it's based on a wrong assumption. Monty isn't forced to, and indeed doesn't always open a door with a goat. And that's the point.

I watched that show a few times when I was around 10 years old, and sure enough, they'd sometimes (admittedly rarely) bring out the prize on the first door. If the host doesn't like you and you choose the goat, they just give you the goat. They'd occasionally have only one "zonk" (that's what the "goat", which is a fluff animal, is called here), too. In that case, you'd have a car in one vault, a home trainer or some equivalent value in another, and a "zonk".

You thus have a 1/3 (or 2/3, if there's a second prize) chance of "surviving" round 1. But of course you always "survive" since your vault isn't opened. Insofar the 1/3 chance is entirely irrelevant. You could just as well not make any kind of choice and only decide in round two.

The second round is an independent event. Which leaves you with two vaults, or a 50/50 chance.

The only "gained information" that you have between rounds 1 and 2 is that there are now 2 doors instead of 3. So the chance has "gone up" to 1/2 from 1/3, but it's none higher for either door. And it sure isn't magically 2/3 for one door because you didn't decide to change your mind.

That left aside, you don't know if the host is cheating and assistants move prizes around after you made your final choice. There's no way of knowing.

Zero of course. I point the gun at the opponent's head, pull the trigger, and if there is no bang start running very fast.

Di di mau! Di di mau!

Or asked differently, why would you give a prisoner a loaded gun...

Of course, a good description of the Monty Hall problem makes the assumptions explicit. The applicability to the show that inspired the problem is something I don't particularly care about.

Will anyone give my problem a try? You can write a simulation, if you can't think of the math...