21 (2008) movie Monty Hall Probability Problem [High IQcels GTFIH]

[human304]

Anyone think its just bullshit and the probability is 50/50

 

[ChadL1te]

no, the math is true it just takes a bit to conceptualize

Basically, when you chose the first door theres a 2/3 chance the car was behind the other two doors. The host opening one of the doors dosent change this fact. Thus, you'll have a 2/3 probability of winning if you switch. If you stay its only 1/3

 

[human304]

no, the math is true it just takes a bit to conceptualize

Basically, when you chose the first door theres a 2/3 chance the car was behind the other two doors. The host opening one of the doors dosent change this fact. Thus, you'll have a 2/3 probability of winning if you switch. If you stay its only 1/3

nope. the host opening the door doesn't have any validity now.

cause now it's not a door problem anymore, it becomes a coin toss problem.

the 66.6% probability is just an assumption.

the actual probability is 50/50

Your Opening choice has no link to it now.

 

[BrahminBoss]

nope. the host opening the door doesn't have any validity now.

cause now it's not a door problem anymore, it becomes a coin toss problem.

the 66.6% probability is just an assumption.

the actual probability is 50/50

Your Opening choice has no link to it now.

Probability is n abstract concept

 

[human304]

Probability is n abstract concept

still but in no way its gonna be 66.6%.

talked to many browns, blacks, asians on this. except the ones who are living in west everyone else thinks the explanation is wrong and its just 50/50

 

[BrahminBoss]

Probability is n abstract concept

Instead of 3 let’s keep 6 doors w only one having the car , so first when I choose it’s 1/6, if 2 doors a opened to reveal goats then probability of switching would increase odds to 1/4 . The monty hall problem is confusing because they keep it very close to each other like 1/3 and 2/3

 

[nobodylovesme]

Just think

Instead of 3 let’s keep 6 doors w only one having the car , so first when I choose it’s 1/6, if 2 doors a opened to reveal goats then probability of switching would increase odds to 1/4 . The monty hall problem is confusing because they keep it very close to each other like 1/3 and 2/3

That's incorrect, switching doors would give you 5/18 probability

 

[mrdouchebag]

Before looking at the table below, keep in mind that the game host always knows which door has the car so he will always open one that doesn't have it.

​

You Pick	Prize Door	Don’t Switch	Switch
1	1	Win	Lose
1	2	Lose	Win
1	3	Lose	Win
2	1	Lose	Win
2	2	Win	Lose
2	3	Lose	Win
3	1	Lose	Win
3	2	Lose	Win
3	3	Win	Lose
		3 Wins (33%)	6 Wins (66%)

Here I saved you the trouble. Of course the table is not mine. I found it on the internet. Smart people steal.

 

[ChadL1te]

nope. the host opening the door doesn't have any validity now.

cause now it's not a door problem anymore, it becomes a coin toss problem.

the 66.6% probability is just an assumption.

the actual probability is 50/50

Your Opening choice has no link to it now.

Youre conflating probability with chance. For example. lets say the best soccer team, team A, plays the worst soccer team, team B. There is a 50/50 chance team A will win because those are the only two outcomes. However, because of external factors, the expected probability for team A winning is higher than .5

You had the right idea but when using the word probability in monty hall its 2/3.

 

[Bushman]

just use bayes theorem n some conditional probability. You’ll get 2/3.


If an 85 IQ swamp nigger like me can solve this shieet, you can too.
I believe in you