Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others...

...

If i dont swap my pick its still at 1/3 chance to win.

If i swap i get 2/3'ds chance to win.

Its pretty easy math if you haven't spent your entire teen years watching anime, like you fucking twats have been.

your asking all the wrong questions

>not studying and watching anime at the same time

yeah senpai 2/3 more favourable to switch

...

wrong

tfw pigeons are more successful at this than humans

>if you haven't spent your entire teen years watching anime, like you fucking twats have been
God I regret that so much today. I even knew I would, but I refused to listen to me.

>he writes a ton of shit and still gets it super wrong while smugly giving a one word "wrong" post
kill yourself fucking faggot freshman

stayorswitch.com/
Switching have a higher probability of winning has been proven both empirically and theoretically, refusing to admit your initial beliefs were wrong is just immature and stubborn.

>tfw you know stochastic processes, measure theory and financial math but don't understand this

So first you choose a door, you have a 1/3 probability to get it right.

The host then opens one door, and reveals a goat.

You are left with two possible doors, behind which the car could be. Don't both doors have a probability of 1/2 then?

yes assuming you reset and shuffle the car around. But you don't. Don't think of it in the numbers and probabilities of each door, think of the probability of scenarios. There is a 1/3rd chance that ended up in the scenario where you picked the car. In this scenario, if you switch you will lose. But it is more likely that you ended up in the other scenario. It is a 2/3 chance that you didn't pick the car, that you are in the didn't initially pick the car scenario. In this scenario, which is twice as likely to occur as the other one, switching will result in your win.

Why is the 100% goat door no longer a door?

If:
A = 50% goat
B = 50% goat
C = Goat

Then either a or b are 50/50 as their has to be two goats

>Content on this page requires a newer version of Adobe Flash Player.
no

>not using google chrome
Kill yourself.

Switching gave you 2/3 possibilities of winning. The 2 possibile scenarios are:
A: you had picked the door with the car (1/3). If you switch , you'll lose.
B: you had picked the goath (2/3). Now switching get you a new car.

>not using netscape navigator

End your worthless """life"""

>3 stars
EL EM AYY O

you literally stole this from 21.

The Monty Hall problem dates back to 1975, you ignorant pleb.

Please lurk more, your stupidity makes me feel weird

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?"

Thinking that it is to your advantage, you switch, but then the host reveals that he didn't actually know which door the car was behind, and if he had picked that door you would have lost immediately. He then says to you, "Do you want to switch back to door No. 1?"

What if the host has a secret button in his hand which can switch the car with the goat if he decides to pick the door with car.

>Suppose you're on a game show, and you're given the choice of GRAHAM NUMBER doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens ALL DOORS which has a goat EXCEPT ONE (27584648754973). He then says to you, "Do you want to pick door No. 27584648754973?" Is it to your advantage to switch your choice?

just test it out, which will give you
2/3 winning for switching

if the host doesn't know, then the chance was always 50/50 and it makes no difference if you switch back or not

Wow. The odds that the number of the door the host picked could be displayed in a single post are basically 0. You sure got lucky, user

Do you think whoever came up with this game was aware of its complexity?
I mean, complexity is not the right term as it's not difficult to prove the right thing is to switch; but as in just how intrincate the reasoning behind it can get, and the amount of discussion it would eventually spawn.

The way the problem is set out for the player is just genius man.

>Do you think whoever came up with this game was aware of its complexity?

Even the original was never this simple (as compared to the recent iteration with Howie Mandel), there were often many many more elements within the game, not just three doors a car and two goats. And enough so that the knowledge of the problem would only very marginally provide advantage.

en.wikipedia.org/wiki/Bertrand's_box_paradox

Also this was a solved problem long before broadcast television.

This isn't true. Even if the host didn't know, you still initially had a 1/3 chance of picking the right door and by him revealing a goat, the switch will still have a 2/3 chance of containing the car. The only difference is that he could mess up and reveal the car on accident which would end the game prematurely.

...

If you picked the car, then he would pick a goat every time, but if you picked a goat he would only pick a goat 1/2 the time.

So knowing he picked a goat increases the likelihood that you picked the car initially. (Raising it from 1/3 to 1/2)

>all that "math"
>50/50: you win or you loose.
( R E T A R D E D )

I really want to see this simulated now. I think I can see the difference.

>I really want to see this simulated now
myth busted

Assume n doors.

P(Host picks (n-2) Goats(G)) = P(G|Picked Car)*P(Picked Car) + P(G|Picked Goat)*P(Picked Goat)
= 1 * 1/n + 1/(n-1) * (n-1)/n
=2/n

P(Picked Car|G) = P(G|Picked Car)*P(Picked Car)/P(G)
=(1*1/n)/(2/n)
=1/2