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re: Marilyn vos Savant and the history of the Montel Hall question
Posted on 2/23/15 at 3:31 pm to CockHolliday
Posted on 2/23/15 at 3:31 pm to CockHolliday
quote:Your error is assuming that all 4 of these scenarios have an equal chance of occurring. I think the problem is better to frame from the original choice. Since the scenarios would play themselves out the same way for all three doors chosen, we will just assume door 1 is chosen.
If Door 1 contains the car:
1) Guest picks Door 1, host opens Door 2. If guest switches, lose. If guest stays, win.
2) Guest picks Door 1, host opens Door 3. Switch: lose. Stay: win.
3) Guest picks Door 2, host opens Door 3. Switch: win. Stay: lose.
4) Guest picks Door 3, host opens Door 2. Switch: win. Stay: lose.
Choose Door 1 and Switch: 2 wins; 1 loss
1.) 1/3 chance car is behind door one (either door can be open; it doesn't matter). Lose
2.) Car is behind door two (door three is open). Win
3.) Car is behind door 3 (door two is open). Win
Choose door 1 and Stay: 1 win; 2 loses
1.) 1/3 chance car is behind door one (either door can be open; it doesn't matter). Win
2.) Car is behind door two (door three is open). Lose
3.) Car is behind door 3 (door two is open). Lose
Basically, choosing the correct door on one's first choice is still 1/3 and it being behind one of the other two doors is 2/3. When you switch, you are basically choosing the other 2 doors (even though we know one doesn't have the car, as shown).
I think a common error is that there are technically 4 possibilities in the above scenarios, 2 of those possibilities occur when you choose the correct door immediately; however, those 2 possibilities can only add up to 1/3. For example, if you chose the correct door (1/3 chance), and each door then has a is a 1/2 chance of being opened, then choosing the correct door and door 2 being open is actually 1/6 (1/3*1/2); same for door 3. In other words, the odds either door 2 or door 3 is chosen must add up to 1/3.
Posted on 2/23/15 at 3:36 pm to buckeye_vol
(no message)
This post was edited on 2/23/15 at 3:37 pm
Posted on 2/23/15 at 3:56 pm to buckeye_vol
No matter which door you initially pick, there is always at least one door the host can open to show a goat. So why are your odds always better if you switch your original pick? In the case where the host opens door 3 and shows a goat, why does switching from door 1 to door 2 double your odds? Wouldn't saying that the same would be true had you picked door 2 originally and switched your pick to door 1 be contradictory?
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