Probability of Winning Prize Behind 3 Doors: 30% or 70%?

[sapiental]

Hello,

for the two following situations give the ultimate probability.

There is a prize behind one of three doors.

1. Picking a door out of 3 doors and always sticking with your firrst pick.
2. Picking a door out of 3 doors and alwats SWITCHING your pick after one false door has been shown.

for 1 I get the probability is around 30% and for 2 I get around 70%.

Is this right? Thanks

 

[HallsofIvy]

Oh, dear, not again! Monte Hall should burn in hell!
Did you notice, by the way, that you never actually asked a question? I'm not at all sure what you mean by "probability is around 30% and for 2 I get around 70%"
I will ASSUME that you meant "what is the probability of picking the prize if (1) you do not switch after the false door has been opened, (2) you do switch after the false door has been opened.

If you are faced with three doors and have NO idea which door the prize is behind, then it is equally likely to be behind anyone and so ther probability of picking the correct one is 1/3. If you do not switch when the false door is opened then the opening is irrelevant and so the probability of picking the correct door is 1/3. I have no idea why you would say "around 30%".

Assuming that the person who opens the "false" door knew where the prize was and intentionally opened a door the prize was not behind, then then the prize is equally likely to be behind either of the two doors. Switching makes your probability 1/2, not 70%.

Some people feel that, since the probability, after the false door was opened, is equal for each remaining door, it doesn't matter whether you switch or not. "Marilyn Vos-Savant"s response to that was very good: Suppose there were 10000 doors with a prize behind exactly one. You pick one at random (so the chance that the door you picked has the prize is 1/1000). The M.C. then opens 9998 false doors leaving only yours and one other door the prize can be behind. Would you switch?

By the way, it is interesting to show that, if the M.C. does NOT know, himself, which door the prize is behind, but, by cnance, opens a door the prize is NOT behind, the probability of getting the prize, whether you switch or not, is 1/3.

 

[Curious3141]

Assuming that the person who opens the "false" door knew where the prize was and intentionally opened a door the prize was not behind, then then the prize is equally likely to be behind either of the two doors. Switching makes your probability 1/2, not 70%.

No, here the probability of getting the prize by following the Switching strategy becomes 2/3. That's pretty close to 70%, although I would prefer the exact value.

I thought this ground has been well covered by Marilyn (and others).

 

[Office_Shredder]

Once, I got my friend to agree to a five dollar game of this with a pack of cards. He picked a card, then I removed fifty cards that weren't the ace of spades from the deck. Since he was convinced the odds were even, I said if he had the ace of spades at the end, he got ten bucks, if I had it, I got five bucks.

After I removed the fifty cards he decided he didn't want to play :D

 

[physicsgal]

Assuming that the person who opens the "false" door knew where the prize was and intentionally opened a door the prize was not behind, then then the prize is equally likely to be behind either of the two doors. Switching makes your probability 1/2, not 70%.

im not even good at math, and i agree with marilyn.

there's three doors.. so you initially have a 2/3 chance of picking the wrong door:
goat, goat, prize

so when you switch doors you have a 2/3 chance of picking the right door.

makes sense to me.

~Amy

 

[HallsofIvy]

Okay, Okay!