Monty Hall, dead at 96

F

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youtube.com/watch?v=_zAU7tZnCcQ
betterexplained.com/articles/understanding-the-monty-hall-problem/
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Literally who

who cares

Just a reminder, you've got a 66~67% chance at picking the correct door if you switch to the other one after the first reveal.

Never saw the original show but the Wayne Brady one is pretty comfy, rest in peace.

whom??

okay so door 1

Just to elaborate, because I didn't understand the concept at first either, and I think this is pretty neat

Say theres 1,000,000 doors, and only one of them has a car behind it, the others are junk.
You pick a door.
999,998 doors are opened, revealing junk.
There are now only 2 doors remaining.
Did you really pick that 1 in a million door your first try, or do you wanna switch?
The odds are in your favor if you change to the second door.

I guess he picked the wrong door. Serves you right asshole.

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Is Carlos our guy?

In the end, the real Monty Hall problem was heart failure.

F YOU BEAUTIFUL BASTARD

youtube.com/watch?v=_zAU7tZnCcQ

anyone notice the Monty Hall Problem in IT ?

Remember, when Monty opens one of the three doors to show it was not the right one, and there are only two left, you have a 50-50 chance of getting the prize with your first choice. That's how probability works - if the problem changes, then you start the calculations all over again.

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The problem with the analysis on wikipedia is that it has a flawed premise - that the initial choice is still relevant to the final choice. This is like assuming a coin on a second flip is more likely to be tail if the coin was a head on the first flip. Once a non-winning door is opened, the three-door problem is reduced to a two-door problem. Adding more doors doesn't change that - you always end up with a two door problem.

betterexplained.com/articles/understanding-the-monty-hall-problem/

Apologies if you're the 66% guy and not the 50-50 idiot I forgot to remove your post number

Alright, let's look at it this way.

Suppose you have a choice between two closed doors and an open door with a goat. Obviously the car is behind one of the closed doors, so the choice is 50-50.

Now you have three closed doors and you choose one. One of the two other doors is opened, revealing a goat. You now have two closed doors and one open door with a goat, and a choice of one closed door or the other.

Why would the two situations have different odds when at this point the situation is exactly the same? Remember, in probability prior results have no effect on the current choice.

>when Monty opens one of the three doors to show it was not the right one, and there are only two left, you have a 50-50 chance

That's the reason the problem is famous. It's not 50/50. It's not intuitive.

You can literally do this in any classroom, hence it is done, and it's always the same. It's used to teach people that:

> That's how probability works

How you think it works is wrong.

Because Monty knows where the car is, and he will always show you a goat.

If you picked the car first time, he will open one of two remaining goat doors.

If you picked a goat first time, he will open the only remaining goat door, leaving the car behind the last door.

66% of the time, you will have picked a goat. Once Monty reveals the other goat, switching doors will get you the car. This will only fail when you pick the car first, which will only be 33% of the time.

Toasting in sticky