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:

Case closed.

Source: The Business Insider: *The Most Controversial Math Problems*

## Comments

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TopNewestyeah this question has become popular among the world it seems – Vishal Ch · 2 years, 10 months ago

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Well, those who do not get the 1 in 3 solution are not supposed to get the 1 in 50 solutions, either. – Agnishom Chattopadhyay · 2 years, 10 months ago

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– John Muradeli · 2 years, 10 months ago

Yea they do this is far more obvious with 50 than with 3. Not everyone thinks math.Log in to reply

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. – Finn Hulse · 2 years, 10 months ago

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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 – John Muradeli · 2 years, 10 months ago

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– Finn Hulse · 2 years, 10 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?Log in to reply

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)... – John Muradeli · 2 years, 10 months ago

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– Finn Hulse · 2 years, 10 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.Log in to reply

– John Muradeli · 2 years, 10 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.Log in to reply

– Finn Hulse · 2 years, 10 months ago

The probability that you initially get the car is smaller than of you getting a goat. Goat=win, car=lose.Log in to reply

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. – John Muradeli · 2 years, 10 months ago

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– Finn Hulse · 2 years, 10 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... :PLog in to reply

– John Muradeli · 2 years, 10 months ago

You? They let you do that at your age? Good schoolLog in to reply

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. – John Muradeli · 2 years, 10 months ago

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\[P=\frac{n-1}{n}\]

So the limit is

\[\frac{\infty-1}{\infty}\]

and since these are the same \(\infty\)s, the limit is simply \(1\). – Finn Hulse · 2 years, 10 months ago

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– John Muradeli · 2 years, 10 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!Log in to reply

– Finn Hulse · 2 years, 10 months ago

LOL, L'Hospital's.Log in to reply

\[\frac{2n}{n}\]

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 – Finn Hulse · 2 years, 10 months ago

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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 – John Muradeli · 2 years, 10 months ago

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– Anuj Shikarkhane · 2 years, 10 months ago

Hospital's??? ;)Log in to reply

Link – John Muradeli · 2 years, 10 months ago

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– Finn Hulse · 2 years, 10 months ago

He was making fun of you for saying Hospitals rather than L'Hopitals. I was very tempted to do the same, actually.Log in to reply