Mosca
Mosca
Joined: Dec 14, 2009
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February 17th, 2020 at 8:00:40 AM permalink
If you pick a door, and then a losing door is revealed, and you are NOT offered a switch... does that mean that you are less likely to win? My intuition says that the solution then turns into the misconception of the answer for the original problem, because the order in which the doors are revealed is irrelevant if there is no choice. The choice is what creates the problem.
NO KILL I
unJon
unJon
Joined: Jul 1, 2018
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Thanks for this post from:
Mosca
February 17th, 2020 at 8:03:35 AM permalink
Quote: Mosca

If you pick a door, and then a losing door is revealed, and you are NOT offered a switch... does that mean that you are less likely to win? My intuition says that the solution then turns into the misconception of the answer for the original problem, because the order in which the doors are revealed is irrelevant if there is no choice.

No. You had a 1/3 chance originally. After the losing door is opened, you still have a 1/3 chance. Itís just that the other door now has a 2/3 chance.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
Ayecarumba
Ayecarumba
Joined: Nov 17, 2009
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February 17th, 2020 at 11:41:42 AM permalink
Quote: unJon

No. You had a 1/3 chance originally. After the losing door is opened, you still have a 1/3 chance. Itís just that the other door now has a 2/3 chance.


Why isnít the updated probability divided between the remaining doors/curtains/envelopes? With one of the three revealed, and no option to switch, the updated odds are 50/50 between the two that remain, no?
Simplicity is the ultimate sophistication - Leonardo da Vinci
ThatDonGuy
ThatDonGuy
Joined: Jun 22, 2011
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February 17th, 2020 at 11:50:10 AM permalink
Quote: Ayecarumba

Why isnít the updated probability divided between the remaining doors/curtains/envelopes? With one of the three revealed, and no option to switch, the updated odds are 50/50 between the two that remain, no?


This has been discussed to death in any number of locations. It comes down to whether or not Monty knew in advance which door was the winning door.

If the first (losing) door that was opened was selected randomly from the two unchosen doors (so, if the contestant's door was a losing door, there was a chance that the revealed door was the winning door), then it is now a 50/50 chance.

However, it is presumed that "everybody knows" that, since at least one of the two unchosen doors was a losing door, Monty, who knew in advance which door was the winning door, would always show a losing door. Since the probability of showing a losing door is 1, the probability that the contestant chose the winning door does not change because of it.
helpmespock
helpmespock
Joined: Mar 6, 2010
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February 17th, 2020 at 12:09:44 PM permalink
One variation of the Monty Hall problem is what if there were 1000 doors?

You choose 1 door, and then Monty opens 998 doors to reveal a goat behind each of them. Do you switch to the other unopened door? Yes because effectively you're getting to choose all the other doors by switching.

I saw an academic study that posed the Monty Hall style question to people, but just varied the number of doors. I can't remember all the details, but once you get to 10 doors a large majority of folks switch doors. Also 3 doors caused maximal indecision among those in the study.

Probability is all about perspective.

--helpmespock

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