Naivojj, that would work, but all three options dont have the same chance of accuring... A) would be 1(you) divided by the amount of people in the lottery, B) would be 1(your wife) / the amount of people in the lottery, however C) would be the remaining percentage, which would be 1-A-B counted in decimal.
Okay who can explain this to me? How can he have 66% chance pf winning when theres 2 doors and the goat can be behind any of them. It must be 50%, numbers cant tell where the car is!
You might understand this better if you scale it up, imagine you had 10 000 doors, behind 9999 doors there are goats, and behind 1 is a car. Lets say you choose door number 10, the host then opens 9998 doors so you are stuck with the one you chose, door number 10, and another door of the host's choosing. Now its really obvious that the car must be behind the door the host did not open, and not behind door number 10 that you picked out which had a 1/10 000 chance of being correct.
The chance of winning + chance of losing must add up to 1. You have a 1/3 chance of winning with the initial pick. You have a 2/3 chance of losing.
Once given the option to switch, ask yourself, How could I lose if I do switch? The answer is, only if I had been lucky enough to have guessed the right door. (1/3)
I will WIN the car with the switch if I had guessed INCORRECTLY at first. (2/3) Thus the odds of winning by making the switch are equal to the odds that your initial choice was wrong.
I would like to go ahead and say it's 50% chance of getting the car, given that with the host opening 1 door it's virtually the exact same as simply giving the contestant 2 doors.
Suppose you choose door 1. Now imagine that instead of showing you a goat behind one of the other two doors, the host says to you, if you like you can stick with door 1, or you can choose doors (2 + 3)simultaenously. If the car is behind EITHER door 2 or 3, you will win. Do you agree that making this choice improves your odds of winning the car?
Do you see the connection? After the host's reveal, the power of your choice is still the same: there is a 2/3 chance of getting the car.
@BiggySn1p3r Yes it is bad luck. The question involves which choice gives you BEST odds of winning. It does not necessarily mean that it gives you a 100% chance of winning.
Let me clarify my statement somewhat: 100% of the ideologies out of Hollywood are wrong because Hollywood is of the devil. The only truth out of Hollywood is truth that is meant to establish a false sense of credibility in an attempt to fool you in the long run.
2. The host ALWAYS HAS TO OPEN A DOOR AND REVEAL A GOAT AND OFFER THE SWITCH. (This was not explicitely stated.)
When the second condition is not met, one can for example not rule out the possibility that the host will only offer the switch if you have picked the car, in which case the probability of winning the car by switching is zero.
If you play it out with cards, you will find that by "switching", your chances of choosing a goat become your chances of winning a car. if you cant comprehend the theory, do it with cards and keep track. it will be at or around 2/3 every time.
LISTEN. it basically boils down to having a 1/3 chance of getting a car, and the other two doors have a value of 2/3. by eliminating one door the value remains. you are now faced with a door with 1/3 chance of a car and a door of 2/3 chance of a car. which would you choose? NOT 50/50.
The key thing is that the host KNOWS what door holds the car. When the host decides to reveal a door, he can't pick the door that Ben has chosen and he can't pick the door that contains the car. This means that there is a 2/3 chance that the remaining door contains the car. If the host had picked any door randomly, including Ben's choice and the one with the car, then yes it would be a 50% chance Ben's choice was correct.
when there is 3 doors, you are more likely to choose wrong than right, so then if you switch, you will switch to the car as you chose the goat... (assuming)
its a hard concept to get but if you have 2 things you dont want like we'll say an old sock behind 2 doors and behind the 3rd one theres a car your more likely to pick a door with the old sock so the gameshow host will open one of the doors with a sock and he will ask you if you want to change because since one of the socks is gone and you morethan likely picked a sock then you should change because it will increase your odds from the 33.3% you started with to a new 66.7% due to variable change
The correct explanation is the t1320 one: it basically boils down to having a 1/3 chance of getting the car, and the 'other' two doors have a chance of 2/3. By eliminating one 'other' door the 2/3 chance stays on the single remaining 'other' door.
@Naivojj look,i tell you that you have to swap before you pick one card..if you pick the car 33,3% u lose cause u will swap it for a goat but if you pick a goat then the host has to reveal the other goat either u pick goat 1 or 2 as u say so u swap and take the car...possibility to pick a goat at first is 66.7%
@Naivojj You pick at random. Before he opens the door, one of three things happen, all equally probable:
1. You picked the car.
2. You picked goat 1.
3. You picked goat 2.
The host always opens a goat you did not choose. That means, if you picked the car and you switch, you lose. If you picked goat 1 and you switch, you win. If you picked goat 2 and you switch, you win. You have no way of knowing which one you picked.
If you doubt it, run a 100 simulations and see what happens.
@Naivojj Your scenario is wrong because winning the lottery and not winning are not equally likely. In the Monty Hall problem, picking any of the three doors are equally likely.
@TuningFreak23 That doesn't matter. Whats important here, is: The host WILL NOT reveal the car. So if you picked a goat to begin with, then he will reveal the other goat. If you picked the car originally, is the only way you will lose.
You pick car, he reveals, you switch: you lose.
Which gives you a 66.6% of taking out the first goat. The host will take out the second goat. (He won't reveal the car). You switch. You win.
@shadowgrail No. If originally you picked a goat. You have a 66.66% chance of that happening. The host will reveal the other goat. Now you switch, to the other door, which is the car.
If you don't switch. You have a 33% chance of winning the car. The only way you can win, if you don't switch, is if you picked the correct door in the first place. Which is a 1/3 chance. But if you picked a goat (2/3), and you switch, you win. The host won't reveal the car, thats the key point.
Naivojj, that would work, but all three options dont have the same chance of accuring... A) would be 1(you) divided by the amount of people in the lottery, B) would be 1(your wife) / the amount of people in the lottery, however C) would be the remaining percentage, which would be 1-A-B counted in decimal.
awsomereborn 3 months ago
Fun Fact: Birds are better at this problem than humans.
kirby4d 4 months ago
I call shenanigans. No way the professor would know a students name.
SonOfNye 7 months ago 2
@SonOfNye hahaha
kirby4d 4 months ago
All this thinking because i had to look up paradox after watching Doctor Who. DAMN YOU MASTER!!!
BiggySn1p3r 7 months ago
Okay who can explain this to me? How can he have 66% chance pf winning when theres 2 doors and the goat can be behind any of them. It must be 50%, numbers cant tell where the car is!
MrAssassinful 9 months ago
@MrAssassinful
You might understand this better if you scale it up, imagine you had 10 000 doors, behind 9999 doors there are goats, and behind 1 is a car. Lets say you choose door number 10, the host then opens 9998 doors so you are stuck with the one you chose, door number 10, and another door of the host's choosing. Now its really obvious that the car must be behind the door the host did not open, and not behind door number 10 that you picked out which had a 1/10 000 chance of being correct.
Zakyrion 7 months ago 2
The chance of winning + chance of losing must add up to 1. You have a 1/3 chance of winning with the initial pick. You have a 2/3 chance of losing.
Once given the option to switch, ask yourself, How could I lose if I do switch? The answer is, only if I had been lucky enough to have guessed the right door. (1/3)
I will WIN the car with the switch if I had guessed INCORRECTLY at first. (2/3) Thus the odds of winning by making the switch are equal to the odds that your initial choice was wrong.
derfunkhaus 7 months ago 3
The guy at the end traveled.
POBOXFU 9 months ago
lets try this way. you walk in and theres 3 doors but one is already open with a goat and monty says pick one of the 3. now what are the odds?
exactly.
jubileeshine 9 months ago
@jubileeshine Wrong.
DGrant1010 2 months ago
This has been flagged as spam show
so you are sayin if it was 5 doors and monty open 3 then you got a 80% chance to be right if you switch to the other door.
so if its 100 doors monty open 98 then you are 99 times in 100 winning the car if you change your choice from between 2 options.
shoot tell monty i want a zillion doors i cant lose.
damn inert mathematicians cant relate to new conditions.
2 choices = 50/50. what came before has no bearing on now mathwise.
jubileeshine 9 months ago
I would like to go ahead and say it's 50% chance of getting the car, given that with the host opening 1 door it's virtually the exact same as simply giving the contestant 2 doors.
FocalPoint1575 11 months ago
Suppose you choose door 1. Now imagine that instead of showing you a goat behind one of the other two doors, the host says to you, if you like you can stick with door 1, or you can choose doors (2 + 3)simultaenously. If the car is behind EITHER door 2 or 3, you will win. Do you agree that making this choice improves your odds of winning the car?
Do you see the connection? After the host's reveal, the power of your choice is still the same: there is a 2/3 chance of getting the car.
derfunkhaus 10 months ago
@derfunkhaus so what happens when you pick 2+3 and it turns out to be # 1? bad luck?
BiggySn1p3r 7 months ago
This has been flagged as spam show
@BiggySn1p3r Yes it is bad luck. The question involves which choice gives you BEST odds of winning. It does not necessarily mean that it gives you a 100% chance of winning.
derfunkhaus 7 months ago
This has been flagged as spam show
@FocalPoint1575
A) You pick the car, host shows a goat, you switch to a goat and lose
(B) You pick goat #1, host shows goat #2, you switch to the car and win
(C) You pick goat #2, host shows goat #1, you switch to the car and win
that gives you a 2/3 or 66.6666.. % chance of winning
cotarna 8 months ago
@Naivojj but arent your initial chances of winning the lottery 1 in how many million?
not 1 in 3....
K3BANG 11 months ago
Smart guy!
MrKallemannen96 1 year ago
Comment removed
itcanbecheezcaketime 1 year ago
Let me clarify my statement somewhat: 100% of the ideologies out of Hollywood are wrong because Hollywood is of the devil. The only truth out of Hollywood is truth that is meant to establish a false sense of credibility in an attempt to fool you in the long run.
objectivethirdparty 1 year ago
Oh, my, what a load of NONSENSE!
Read "Refuting Marilyn Vos Savant's false math theory" on facebook for some SENSIBLE reading!
Thank you!!!!!!!!
One tipoff: If Hollywood promotes it, it's wrong AT LEAST 99.9999999999% of the time, because behind it IS a goat, and I'm referring to the devil.
objectivethirdparty 1 year ago
This has been flagged as spam show
1.The host knows where the car is. (OK)
2. The host ALWAYS HAS TO OPEN A DOOR AND REVEAL A GOAT AND OFFER THE SWITCH. (This was not explicitely stated.)
When the second condition is not met, one can for example not rule out the possibility that the host will only offer the switch if you have picked the car, in which case the probability of winning the car by switching is zero.
snoehvit 2 years ago
Comment removed
snoehvit 2 years ago
This shows that the movie makers have not really understood the problem.
Two conditions have to be met before one can conclude that the probability of winning the car by switching is 2/3:
snoehvit 2 years ago
this just answered my homework problem
xxkevinxx323 2 years ago
Comment removed
sepehrsareskani 2 years ago
If you play it out with cards, you will find that by "switching", your chances of choosing a goat become your chances of winning a car. if you cant comprehend the theory, do it with cards and keep track. it will be at or around 2/3 every time.
t1320 2 years ago
LISTEN. it basically boils down to having a 1/3 chance of getting a car, and the other two doors have a value of 2/3. by eliminating one door the value remains. you are now faced with a door with 1/3 chance of a car and a door of 2/3 chance of a car. which would you choose? NOT 50/50.
t1320 2 years ago 3
The key thing is that the host KNOWS what door holds the car. When the host decides to reveal a door, he can't pick the door that Ben has chosen and he can't pick the door that contains the car. This means that there is a 2/3 chance that the remaining door contains the car. If the host had picked any door randomly, including Ben's choice and the one with the car, then yes it would be a 50% chance Ben's choice was correct.
coldplasma 2 years ago 3
This comment has received too many negative votes show
The movie is wrong. There is a 50/50 chance not a 66% chance.
Who is the chick behind Ben. She's hot!
MsWiggleworm 2 years ago
dude u must be really stupid. go search wikepedia for variable change and learn something.
sushi1993InSAnE 2 years ago
wow ur a dumb fuck. that's why its called variable change. read some books.
dh144498 2 years ago
there are 3 doors to start off, you idiot
anhqtran 2 years ago
nope, is not wrong.
pedroetcr 2 years ago
i tried to explain this to my dad but it was a lost cause
invaderaid 2 years ago
O_O he made a 97% on that calculus 3 exam,, hes smart
Pharatos666 2 years ago
i think the easiest way to put this is....
when there is 3 doors, you are more likely to choose wrong than right, so then if you switch, you will switch to the car as you chose the goat... (assuming)
aZn0vvNz56 2 years ago
fix the volume ?
aplayaz2000 2 years ago
its vairble change, if you are offered the swtich you always take it becuase you gain 33.3 percent more of a better chance of winning. simple math
jure87 3 years ago
it took me 2 weeks ot figure it out... and then i googled it and was like what a stupid waste of time...
arman1008 3 years ago
But this concept has nothing to do with Variable Change... it's bayes and total probabilities
Givicencio 3 years ago
ignorance is bliss XD
arman1008 3 years ago
its a hard concept to get but if you have 2 things you dont want like we'll say an old sock behind 2 doors and behind the 3rd one theres a car your more likely to pick a door with the old sock so the gameshow host will open one of the doors with a sock and he will ask you if you want to change because since one of the socks is gone and you morethan likely picked a sock then you should change because it will increase your odds from the 33.3% you started with to a new 66.7% due to variable change
schnur13 3 years ago 2
yellownotebook10 (6 days ago)
A) You pick the car, host shows a goat, you switch to a goat and lose
(B) You pick goat #1, host shows goat #2, you switch to the car and win
(C) You pick goat #2, host shows goat #1, you switch to the car and win
that gives you a 2/3 or 66.6666.. % chance of winning
stockola 3 years ago 13
stockola and schnur12, great explanations. i understand it now. before i wasn't too sure. thanks.
jiminybeam 3 years ago 11
damn... i thought of a completely different and uselessly complicated version of that ... i wasted my time into what Google could do -.-
arman1008 3 years ago
@stockola Stockola did you know my family also have 66.666% change of winning the top prize at the lottery:
A) I bought a ticket and win the millions $ top prize
B) My wife bought a ticket and win the millions $ top prize
C) My wife and I don't win
That gives 2/3 change of winning!
Naivojj 1 year ago
Stockola, your B) and C) are basically the same possibility where you have swapped goat #1 and #2. Why don't you break down A) the same way like this:
A1) You pick the car, host shows goat #1, you switch to goat #2 and lose
A2) You pick the car, host shows goat #2, you switch to goat #1 and lose
Then you have 4 possibilities and according to your logic a 2/4 or 50% chance of winning.
Naivojj 1 year ago 2
The correct explanation is the t1320 one: it basically boils down to having a 1/3 chance of getting the car, and the 'other' two doors have a chance of 2/3. By eliminating one 'other' door the 2/3 chance stays on the single remaining 'other' door.
Naivojj 1 year ago
@Naivojj you cant pick the car twice.you start with all the possible paths.1)car swithct to a goat 2)goat1 switch to car 3) goat2 switch to car.
21Hellin 1 year ago
@Naivojj look,i tell you that you have to swap before you pick one card..if you pick the car 33,3% u lose cause u will swap it for a goat but if you pick a goat then the host has to reveal the other goat either u pick goat 1 or 2 as u say so u swap and take the car...possibility to pick a goat at first is 66.7%
supersayajim23 1 year ago
@Naivojj You pick at random. Before he opens the door, one of three things happen, all equally probable:
1. You picked the car.
2. You picked goat 1.
3. You picked goat 2.
The host always opens a goat you did not choose. That means, if you picked the car and you switch, you lose. If you picked goat 1 and you switch, you win. If you picked goat 2 and you switch, you win. You have no way of knowing which one you picked.
If you doubt it, run a 100 simulations and see what happens.
AngryScissor 1 year ago
@Naivojj
Interesting. I want you to believe :-)
stockola 1 year ago
@Naivojj wrong about the lottery,you put all other tickets in C...
supersayajim23 1 year ago
@supersayajim23 haha yea that's like saying that life existing on mars is 50%.
1) life lives on mars
2) life doesn't live on mars.
cherryquf 10 months ago
@Naivojj Your scenario is wrong because winning the lottery and not winning are not equally likely. In the Monty Hall problem, picking any of the three doors are equally likely.
magician13134 4 months ago
@stockola well isnt there like 2 (A) ?
(A1) You pick the car , host shows goat Nr.1 , you switch and lose
(A2) You pick the car , host shows goat Nr.2 you switch and lose
that would make a new 50/50 chance right?
TuningFreak23 3 months ago
@TuningFreak23 That doesn't matter. Whats important here, is: The host WILL NOT reveal the car. So if you picked a goat to begin with, then he will reveal the other goat. If you picked the car originally, is the only way you will lose.
You pick car, he reveals, you switch: you lose.
Which gives you a 66.6% of taking out the first goat. The host will take out the second goat. (He won't reveal the car). You switch. You win.
I'm horrible at explaining.
DGrant1010 2 months ago
@DGrant1010 yeah , I got it :D
TuningFreak23 2 months ago
@DGrant1010 What if you don't switch? Isn't it supposed to make a 50/50 chance?
shadowgrail 1 month ago
@shadowgrail No. If originally you picked a goat. You have a 66.66% chance of that happening. The host will reveal the other goat. Now you switch, to the other door, which is the car.
If you don't switch. You have a 33% chance of winning the car. The only way you can win, if you don't switch, is if you picked the correct door in the first place. Which is a 1/3 chance. But if you picked a goat (2/3), and you switch, you win. The host won't reveal the car, thats the key point.
DGrant1010 1 month ago
@DGrant1010 Oh! Now I see. Thanks for enlightening me.
shadowgrail 1 month ago
@shadowgrail
I am so bad at explaining. I'm not even sure how you understood that, lol.
Cheers
DGrant1010 1 month ago