Posted on 03/18/2009 4:04:29 AM PDT by mattstat
It’s about time we did the Monty Hall problem.
Many of you will already know the answer, but read on anyway because it turns out to be an excellent example to demonstrate fundamental ideas in probability.
Incidentally, I just did this yesterday to a group of surgical residents: you might be happy to know that none of them got the right answer. One even insisted—for a while—that I was wrong.
Here’s the problem.
Setup: Monty Hall shows you three doors, behind one of which is a prize, behind the other two is nothing. Monty knows which door hides the prize. You want that prize.
Rule: You pick a door that Monty will open. You are free to change your mind as often as you like.
The Pick: I’ll suppose, since I can’t quite hear you, that you picked A (the resident yesterday took that one) and did not change your mind.
Question...
(Excerpt) Read more at wmbriggs.com ...
Oh Oh Oh I GET IT!!!!
That’s kind of a cool way of looking at it.
It easy to see that your strategy should be switching if you look at this way. If you don’t get the prize on the first pick (with a probability of 2/3) you automatically get it after you switch, but you don’t get after you switch if your first pick was the prize (with a probability of 1/3). Thus by switching, your conditional probability of getting the prize is being weighted towards the greater probability on the first pick, that is the probability of not picking the prize in the first place.
Err...
Allowing that Monty is constrained to NOT opening the Door with the prize.
If the Prize were behind “A” - he could have opened either “B” or “C”. Thus, he was not “forced” to open “B” - “B” is simply a valid choice of available options.
The author concedes this when he states:
“Now suppose you had first picked C. Which doors could Monty have opened? Right: A or B. And in this case you should stay.”
The problem with the hypothesis is that the author, in his conclusions includes assumptions as evidence!
He asks - assume the prize is behind “C”, then “A” = 1/3, “C” = 2/3.
The problem is - at the time you need to make the decision, you have no evidence that the prize is behind “C” - you only know that it “could” be behind it, but that was a given from the beginning.
That makes more sense to me that than the authors explanation!
However,
Given A = prize
Pick A - Open either B or C
Pick B - Open C
Pick C - Open B
What is not clear is why switch.
After having Picked A - I can not know whether Monty was forced to open either B or C, or whether he had a choice.
So, if B is eliminated after the first round - I have two choices (Stay / Switch) and two arguments, Monty:
1. Was forced to open B as C contains the prize, or
2. had a choice to open either B or C as A contains the prize.
I'm likely missing something - but I still don't see the reason for switching.
I think whats missing is not knowing the number of times Monty is going to give you a choice of switching.
But then again mind games like this start to make my mind spin.......I never like this Monty guy anyway.
Right. You the contestant has no idea if Monty’s choice is constrained or not. Just that there is a 2/3 chance that it is. Based on this lack of evidence, there is never any reason to switch.
Why is there a 2/3 chance his choice is constrained?
If I made the correct (first) choice, he is not constrained,
If I made the wrong (first) choice, he is constrained
From “my” perspective there is a 1 in 2 chance, AFTER my selection, that Monty was “forced” to open a specific door. BEFORE my selection there is a 2 in 3 chance of “forcing” Monty - which is simply the inverse of my probablity of having chosen correctly.
The author’s assumption that the prize is behind C invalidates his argument. Given that the prize is behind a closed door, there is no change in the equality of distribution of probability that the prize is behind a specific closed door, no matter how many other doors are opened.
If there are 5 doors, the chance that the prize is behind the first door is 1 in 5.
Open the fifth door, and the probability that the prize hides behind door number one is 1 in 4.
Open the fourth, 1 in 3.
The equal distribution of probability does not change, although the actual value of the probability does rise as each door is opened.
To take the argument presented here, one must assume that the initial pick was incorrect, with no evidence to support that assumption.
Why?
I was either right when I made made pick, or wrong. The number of doors I had to choose from is immaterial at that point.
With every door Monty then opens, my chance of being correct INCREASES - all the way to the last choice when I have a 1 in 2 chance of being correct.
Your initial chance of being correct does not change no matter how many doors he opened. It stays at essentially zero. I pick the other door because its now a certainty.
Can you explain that?
As I understand it, my initial chance of being correct 1 to n.
My chance of "still" being correct as doors open is 1 to n - doors opened.
What I am not undertstanding is your statment (and the contention of the author) that I have a HIGHER chance of being correct if I switch on the last "choice"
My chance of "still" being correct as doors open is 1 to n - doors opened.
Your initial chance of being correct is 1/n. Since n is infinity you have zero chance of making the correct first pick. No matter how many doors Monty opens your pick is still wrong. With two doors left you have your wrong pick and the winning pick remaining.
Since the game you imply has infinate doors, the game would never end. One the other hand, with n being a finite number - my contentions stands.
Further, even if n were infinite - my chance of being correct is NOT zero. It is infitismally small, but not zero.
Your final statement though is (almost) correct. "With two doors left you have your wrong pick and the winning pick remaining."
My bad.
Meant to say that, with two doors left there is one right pick and one wrong pick. In other words a chance of 1/2.
I remember this was a question somebody asked Marilyn and her (correct) answer was a topic of debate in Parade Magazine for weeks. This was many, mnay years ago.
Excellent summary!
By the way, back in the “Ask Marilyn” days when this was a hot topic, several classrooms of kids physically tested, through a classroom game show demonstration, the assertion that you should always switch and sure enough that answer was supported by the class’s empyrical data.
Rather than zero lets call the tiny little number 1/n represents as "@". You only had @ chance of being right when the door was picked.
Why should that change just because you are shown some other doors? Thinking that the odds somehow change is like thinking we have a "living" constitution subject to arbitrary change. If it was right the first time it's still right!. (Or in my example it's still [1-@]*100% wrong door )
If Monty went and opened the door to the right choice you'd surely switch, in my example he's done nearly the same thing. (As close to 1 as 1-@ gets you!)
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