…and (rubbing his chin) that makes it all very interesting.
The Ryan is this what you were getting at?
I fully agree with your view, if you include Ryan and his horses.
Monty chooses randomly between the two wrong doors.
Is what what I was getting at?
In which case both players should switch to the other door giving them the prize 2/3 of the time, you are wrong in your assertion that B should be indifferent.
Your OP seemed to suggest that you felt there was something anomalous about your two-player game, I wasn’t sure as I couldn’t honestly follow what you intended the rules to be. Meanwhile Little Nemo, zut et al brought to my attention a counter-intuitive (but not anomalous) result of a different version of the two-player game, that’s when I thought “Is this what the OP was getting at?”
Then you read that, but presumably didn’t read the posts around it that would have given it a context.
Then you posted your post.
Then I wrote this.
Now we’re both up to speed, I believe your OP has been answered in the posts above.
But wait, I don’t think you’re correctly interpreting what The Ryan meant. I think the scenario is as follows:
- Three doors, only one prize.
- Contestant chooses a door.
- Monty randomly opens one “wrong” door regardless of whether or not it’s been chosen by the contenstant.
- Contestant gets a chance to “switch” or “stay”.
Different from the classic MH problem in that Monty is allowed to choose the contestant’s door.
So let’s make a table; without loss of generality assume the prize is in C. All the following possibilities are equally likely:
Case Orig-Choice ---Odds----
Num. Cont. Monty Switch Stay
1 A A 1/2 0
2 A B 1 0
3 B A 1 0
4 B B 1/2 0
5 C A 0 1
6 C B 0 1
(note Monty never chooses C, because that’s where the prize is)
Looking at the table, one sees that blindly choosing “switch” as a strategy gives an overall 1/2 odds (not 2/3), but which is still better than blindly choosing “stay” overall, which only gives 1/3 odds overall. Note, however, that the contestant has one additional piece of information: he knows, before the final switch/stay choice, whether the game is in subset (case 1 and 4) or subset (case 2, 3, 5, and 6), by whether or not Monty opened the contestant’s door. If you break up the possibilities like this, it’s easy to see you should switch in game subset (case 1 and 4), where “staying” means you keep what you know is a booby prize. This would be intuitively obvious to the contestant. However, in the other subset (case 2, 3, 5, and 6), you have odds of 1/2 whether you switch or stay.
Zut
I’m not sure I agree or not agree with you.
If Monty randomly opens either losing door, then the odds are 1/2. The switching/staying doesn’t matter. Of course if your door was opened you would switch.
If Monty randomly opens one of the losing doors, why even waste time with your first pick.
Indeed. The game may as well go something like:
“Okay Player, you have three doors to choose from.”
“I pick Door-”
“Before you choose, I’ll tell you that Door A is wrong.”
“Oh, right. I’ll pick B then.”
Or even get rid of one door entirely. If Monty’s choice is independant of the player’s choice, why not just eliminate one of the doors before the show even airs?
Exactly. The only difference is that, in my post, Monty’s “wrong door” is still considered a viable choice in a couple cases (when Monty chooses the contestant’s door, and the contestant “stays”). Of course, you could validly argue that choosing what the contestant knows is a booby prize isn’t really a valid choice, in which case the difference disappears.
Zut
We’re on the same line then.
If this is all Ryan meant by part of his original post, I’m at a loss why he posted it – “If you have two doors, what are you’re chances of winning”.
There’s been a lot of interesting mathematical discussion, but no one has bothered to answer the original question.
The answer is blue. Everyone’s favorite color is blue.
If your favorite color is something other than blue, you’re probably just an eccentric or a trouble-maker and the government should keep a close eye on you.