21, The Movie: Variable Change

Explanation of the problem.
[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 66.67 % chance of winning.

Category:

Education

Tags:

• 21
• The
• Movie:
• Variable
• Change
• Maths
• Monty
• Hall
• Problem
• From
• Movie

[stockola]

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

[Naivojj]

@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!

[stockola]

@Naivojj

Interesting. I want you to believe :-)

[jiminybeam]

stockola and schnur12, great explanations. i understand it now. before i wasn't too sure. thanks.

[DGrant1010]

@shadowgrail

I am so bad at explaining. I'm not even sure how you understood that, lol.

Cheers

[shadowgrail]

@DGrant1010 Oh! Now I see. Thanks for enlightening me.

[DGrant1010]

@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.

[shadowgrail]

@DGrant1010 What if you don't switch? Isn't it supposed to make a 50/50 chance?

[TuningFreak23]

@DGrant1010 yeah , I got it :D

[DGrant1010]

@jubileeshine Wrong.

[DGrant1010]

@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.

[TuningFreak23]

@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?

[awsomereborn]

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.

[magician13134]

@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.