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# The best explanation to Monty Hall

Brief problem description:

You're on a game show. There are three doors. Behind two there are goats. Behind one there's a car. You're asked to open one door. Then the host tells you one of the doors in which there's a goat, and gives you the opportunity to switch doors.

Question: Do you switch the door?

Answer: Yes. You have greater odds if you do so.

Problem: But what if I picked the right door? The car is only in one of the doors, and what difference it makes if the host shows me one of the doors? What if I have the car?

Usual Solution: maths.

No complex maths, no fancy thinking. Here's the best explanation I've found on the web:

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Case closed.

2 years, 8 months ago

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yeah this question has become popular among the world it seems · 2 years, 7 months ago

Well, those who do not get the 1 in 3 solution are not supposed to get the 1 in 50 solutions, either. · 2 years, 8 months ago

Yea they do this is far more obvious with 50 than with 3. Not everyone thinks math. · 2 years, 8 months ago

I have a much better way (in my opinion):

There are two cases:

Case 1:

If you pick the door with a goat behind it, he is forced to reveal the other goat. Thus, switching will give you the car. This has a probability of $$2/3$$.

Case 2:

You pick the car on your first try, and switching gives you a goat. The probability of this is $$1/3$$.

Thus it is more likely that Case 1 will occur and give you the car, as long as you switch. · 2 years, 8 months ago

Um actually I don't get it. Why in Case 1 the probability is 2/3? And why in 2nd one it's 1/3?

Plus, to anyone who hates math, I think the example with 1,000 doors does it. It's like lottery: you choose your numbers, and we'll tell you another combination. You can keep your Powerball combination or you could switch. Now, honestly, which one would you do? xD · 2 years, 8 months ago

Pretty much, as long as YOU pick a goat the first time, HE will HAVE to show you the other goat (right, because he can't show you the car). So, as long as you pick the goat (with probability 2/3), switching will always get you the car. Do you kind of get it? · 2 years, 8 months ago

Ohh Hulse lol... OBVIOUSLY I GET IT. But I know for a fact if you explained this to a non math person they wouldn't. And actually there's a bit more to it, with change of variables and stuff, y'know.

But a commoner always has this question floating in his head (and so did I): But what if I have the car? I mean, he will always show me a goat. That doesn't mean I don't have the car. In physics the car won't teleport all of a sudden to the other room because the guy just showed me a goat (possible, but I digress)... · 2 years, 8 months ago

No... The car doesn't need to teleport. It's already behind a door, but once Monty shows the other goat, switching will give you the car. · 2 years, 8 months ago

"Will give you the car"? Now that's an ambiguity. What if the door I chose has the car? Then it won't be. And back we go to probabilities and numbers and jumbo. · 2 years, 8 months ago

The probability that you initially get the car is smaller than of you getting a goat. Goat=win, car=lose. · 2 years, 8 months ago

Yes. And 2+2 is 4. And roses are red. And oranges are green. But what does this have to do with anything? lol

it seems like every time I reply you simplify the problem more and more until you completely deconstruct it - like now. what's next - car is bigger than a goat?

hehe and no offense here. not that it should be, but some people take it that way I dunno. none intended. · 2 years, 8 months ago

I'm confused as to whether you understand my solution or not... I mean, I've explained Monty hall to classrooms full of people multiple times... And most of them have been completely math-illiterate... :P · 2 years, 8 months ago

You? They let you do that at your age? Good school · 2 years, 8 months ago

Can someone prove that $$\lim_{N\rightarrow \infty}{P}=100%$$?, where N is the number of doors and P is the probability of getting the car upon switching. That'd be nice. · 2 years, 8 months ago

Umm... Yes.

$P=\frac{n-1}{n}$

So the limit is

$\frac{\infty-1}{\infty}$

and since these are the same $$\infty$$s, the limit is simply $$1$$. · 2 years, 8 months ago

Yeh, by L'Hospital's. This shows that the likelyhood of you picking a goat if you don't switch is 100%. So if you switch, the likelyhood becomes 0% to pick a goat, and thus 100% for the car :D Nice job Finn! · 2 years, 8 months ago

LOL, L'Hospital's. · 2 years, 8 months ago

Well... Kind of. You have to verify that you're talking about the same kind of infinity. So if I did the limit

$\frac{2n}{n}$

and plugged in $$\infty$$, I would get

$\frac{\infty}{\infty}$,

which you might think would be $$=1$$. Clearly, that's not the case, though. :P · 2 years, 8 months ago

Um that's what I said. By L'Hospital's you'd differentiate the top and bottom to reveal it to be 2. Am I missing something?

Btw, is it just me or there's such thing as "kinds of infinities"? Lol just thinking about them drove me nuts, but I've convinced myself that $$\infty^2>\infty$$, even though mathematically you cannot show that. How I think of it is the squared infinity is an infinite field of corn, whereas the just infinity is just corn stacked in a line.

And obviously if we cube it we get a corn space, but mathematically if you try to calculate using these your infinities all hold same strength... But eeh what the heck I'm sure this can be explained but just I can't. I have all the visuals about infinities, how they have their "rates of change", some greater than others, but I can't prove it :O · 2 years, 8 months ago

Hospital's??? ;) · 2 years, 8 months ago

Link · 2 years, 8 months ago