… that this puzzle is still making the rounds, and is still often missolved. (OP’s slight variation doesn’t change the problem importantly, just makes it less ambiguous.)
Unless I’ve had a mental lapse (not unheard of these days :smack:), correct answers are in a minority, and even some of those with correct answer compute odds wrong.
The chance that the car is behind the 3rd curtain was 1/3 to begin with and remains exactly 1/3 after the plays. Since probabilities sum to 1, and one has dropped to zero, your probability must be 2/3 exactly.
This is loosely related to a principle called “restrict choice,” which gives the odds in certain finesse situations in when playing contract bridge.
[spoiler]Yes, he shows you a losing curtain 100% of the time. 2/3 of those times, however, it’s the only losing curtain he can show. Because 2/3 of the time, one of the contestants has a winner and one a loser.
Just like the ‘classic’ Monty Hall problem, 2/3 of the time his hands are tied (assuming we know he’s going to go through the behaviors described). It’s just he’s showing one of the *chosen *curtains instead of one of the unchosen curtains.[/spoiler]
No, I’m only assuming a 50-50 chance in the instance that neither contestant chose the right door. In the event that one of them chose the right door, I’m assuming that Monty opened the door selected by the contestant that got it wrong.
[spoiler] “2/3 of those times, however, it’s the only losing curtain he can show. Because 2/3 of the time, one of the contestants has a winner and one a loser.”
He knows this, you don’t. He has provided you no useful information.
" It’s just he’s showing one of the *chosen *curtains instead of one of the unchosen curtains."
just?
[/spoiler]
quotes and spoilers at the same time is annoying. I know I could have done it better.
[spoiler]
In a scenario with no switching, 2 out of every 3 times one of the contestants should be a winner. After one non-winning contestant is removed, the expected winning % of the remaining contestant should still be 66.66%
So for example, if this was run 100 times, with the remaining contestant not given the option to switch, he should be expected to win 67 times. If you take those same 100 runs, but this time have the remaining contestant switch every time, then he would only win 33. He shouldn’t switch.[/spoiler]
From thinking about it and reading the thread, I have two answers to the riddle:
1:
This is still the same puzzle as the original Monty Hall problem. The final contestant had a 1/3 chance of picking the right door at the beginning, and should therefore switch.
2:
It’s a different puzzle, because it’s teamwork. The two contestants together had a 2/3 chance of picking the right door between them at the start. The final contestant should therefore keep his door.
Clearly, they are both right. As a result, the universe will now implode from the sheer ridiculousness of this in ten seconds. 10, 9, 8…
Actually, though, I think
option 2 is the right one. In other words, what **RTFirefly **said.
Since I’m wedging comments in, in between trying to get stuff done at work, I’m less than certain that I’ve got the right probability: it could be 2/3 chance, rather than 4/5, that you’ll win by standing pat. But I’m sure about the standing pat part.
Let me put it this way. You are only around to wonder if you won a car because one of two things happened. You and the other contestant had won the car between the two of you and you must be the one to remain standing because you have the car. The car was in the third door and you just happened to be the one that was allowed to remain standing.
Monty has given you no further information. Two options 50/50 chance. Otherwise you wouldn’t still be playing to ask the question. Sort of like a monty hall anthropic principle.
[SPOILER]This puzzle is unambiguous. Monty picks a selected curtain that does not conceal a car. Some people misconstrue the original puzzle to think Monty might reveal a car. (Even without that, Monty did have the option of opening no curtain at all – that he didn’t refuse to open a curtain may provide info.)
OP’s problem is indeed different BUT has the same solution AND avoids the ambiguities in the usual phrasing.
As for the new information Monty has given you: He has given you the information that he DIDN’T open YOUR curtain!
[/SPOILER]
The original Monty hall problem was counter-intuitive because initially it seemed like you couldn’t carry over information from before Monty did stuff to after Monty did stuff but you could.
Interestingly, this version is counter-intuitive because it seems like you should be able to carry information from before Monty did stuff to after Monty did stuff but you can’t. You don’t have enough information. You and the other contestant are interchangeable.
I don’t see why you can’t. What Monty does doesn’t change the probability of the two contestants together picking the set of doors that contains the big prize at the beginning.
Think in terms of running through this scenario many times from the point of view of one of the contestants. You won’t always be around to pick.
Also remember what the special element was in the original Monty hall problem that leads to the correct answer. The key is that Monty has to be able to eliminate any of the doors and he knows which door has the goat.
Here, he’s restricted from choosing one of the doors.
