Evil and Switching Doors
So about that Buick Skylark. After considerable discussion, my father and I have concluded that the question is deceptively posed.
I'd put it this way: What I didn't mention is that I know where the car is. If I were to open a door at random, you'd have a one in three chance of winning the car on your first guess. But I'm not going to open a door at random. If you guess the right door, I won't open it. The missing part of the problem is that I'll always make sure it's in your interest to switch. The game always has two moves, not one.
This is the way he puts it:
Now before we get to the problem of evil, there is the Monty Hall problem, an irresistible irrelevance.
There are three doors labeled A, B, and C. A prize is behind one door, nothing behind the other two. The prize is labeled 1, and nothing labeled 0. Thus the obvious notation: A = 1, B = 0, and Pr(A = 1), for the probability that A =1.
The assumptions:
1 Pr(A) = Pr(B) = Pr(C).
2 A represents the initial choice.
3 C = 0.
The question:
4 Is Pr(B = 1) > Pr(A = 1)?
If so, then switch, if not, then stay put.
Stay right where you are is the correct answer for the problem as it has been posed.
It is also the answer uniformly rejected by mathematicians, and while this is not evidence in its favor, it is evidence that it is hard to ignore.
Why switch? A non-sequitur is often introduced: If Pr(B = 1 or C = 1) = 2/3 and C = 0, then Pr(B = 1) must be 2/3.
On the contrary: Pr(B = 1) = 1/2.
Consider those three doors again, and with them all possibilities of gain or loss.
A B C
1 0 0
0 1 0
0 0 1
By Assumption 3, the last row may be eliminated: C = 0. Thus the true possibilities are
A B C
1 0 0
0 1 0
Whereupon by an obvious counting argument, Pr(B = 1) = 1/2. But ditto for Pr(A =1).
Stay where you are. You have nothing to gain.
My Pop and I have spent a lot of time thinking this through. Makes sense, doesn't it? But just to be sure, I spent all morning playing the game with my cats. I'm afraid one of them won the Skylark.
The Glory Badges will be awarded to human members of Ricochet presently.
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Comments :
May '10
Re: Evil and Switching Doors
I look forward to this discussion. I read it Mamet's book, and still can't figure out why the "official answer" is valid.
Mar '11
Re: Evil and Switching Doors
Here's a lengthy discussion of the puzzle which arrives at a different conclusion.
Jun '10
Re: Evil and Switching Doors
I'm so fascinated by this problem. Here's a place to play.
Apr '11
Re: Evil and Switching Doors
You are misstating the game. That C=0 is an initial assumption only for this explanation. I do not have this information until after I choose A. After it is revealed that C=0, the offer to swap is not "Find out what is behind A OR find out what is behind B", it is "(1) Find out what is behind A OR (2) find out what is behind B and C". Thus, that Pr(C=1 or B=1) =2/3 is not irrelevant.
Why isn't (1) "Find out what is behind A and C"? Because once you choose A, what you learn about the subset [B C] does not tell you anything more about [A]. If you opened B or C at random the truth table looks like this:
A B C Open Switching get you
1 0 0 B=0 C=0
1 0 0 C=0 B=0
0 1 0 B=1 B=1
0 1 0 C=0 B=1
0 0 1 B=0 C=1
0 0 1 C=1 C=1
But showing the car 1/3 of the time would have gotten the show canceled the first season. Notice that Claire has to hide information only about B and C to ensure that she always shows a badge. Thus, showing that C=0 doesn't tell me anything about A.
The "Switching gets you" column shows you win 2/3 of the time by switching, whether the door is opened randomly or by Claire so as always to show a badge. BTW, that Claire knows where the car is is irrelevant, but is assumed in the Price Is Right game.
Edited on Oct 12, 2011 at 8:19amMar '11
Re: Evil and Switching Doors
Lost me.
May '10
Re: Evil and Switching Doors
In the Monty Hall problem Assumption 3 does not exist. Randomization occurs before it is revealed whether door B or C has probability 0 of containing the prize. The order of operations is such:
Assumptions: Pr(A) = Pr(B) = Pr(C) = 1/3.
Let the contestant choose door A.
At this point Monty looks behind the doors and knows where the prize is. There 2 outcomes:
1: The prize is behind door A, probability 1/3. Monty opens either door B or C.
2: The prize is not behind door A, probability 2/3. Monty opens either door B or C.
The opening of the door provides no new information to the contestant whether the prize was behind door A or not. Door A still has a 1/3 chance to have the prize.
Sep '10
Re: Evil and Switching Doors
Claire Berlinski: Stay right where you are is the correct answer for the problem as it has been posed.
So I ended up making the apparently correct choice just by thinking: Why change? It's got to be either the one I've picked or the other one. And in my experience, whenever I have ended up changing my mind in such a situation, much more often than not I have regretted it. No knowledge of math whatsoever! The really interesting thing to me, though, is how there are so many different good answers to this problem based on a language I've always thought was all about eiither/or but now seems to be more about either/and. Talk about your paradigm shift!
Jun '10
Re: Evil and Switching Doors
I disagree. Run some trials and you can't help but come up with wins 2/3 of the time.
Apr '11
Re: Evil and Switching Doors
FWIW, I chose the first door 10 times and switched every time. I got the car 8 times.
The summary of the explanation in the NYT is an excellent way to look at it:
"If your strategy is to always switch doors, you will lose only if your initial choice is the door with the car, which is a 33.3 percent chance. In the other two cases (66.7 percent of the time) you will switch to the car and walk away a winner."
What Monty/Claire knows is significant only if you always swap unopened doors. If your strategy is to play the odds (100% or 2/3), when the door is opened randomly, you still win 2/3 of the time, so, really, Monty/Claire aren't changing the odds, just affecting your strategy.
Jun '10
Re: Evil and Switching Doors
If we're going to reword questions after the fact, let's reword the question as follows: Under what circumstances does the probability in this game over a number of tries, say 100, approximate 2/3. Of course the game would require that the host show you the only other goat prize every time. Why would he do anything else? Seriously? Over multiple trials this only works out one way, and it's not 50-50, no matter how you parse the words. Leslie Watkins, are you up for a poker game any time soon?
Jun '10
Re: Evil and Switching Doors
I just tried 50 times, switched every time and won 70% of the time. A waste of time? Sure, but I've just spent an hour trying to access my gmail account to no avail.
Aug '11
Re: Evil and Switching Doors
I love Ask Marilyn. This read is fun, she gets the damnation of academia's highest - and ends up proving them wrong. Switching will get you a win 2/3 of the time.
May '10
Re: Evil and Switching Doors
Another way to look at it, if I understand the game right.
You pick A. There is a 1/3 chance of winning.
Later, you could switch to B for 1/3 chance to get the car, or to C for a 1/3 chance to get the car. Those probabilities never change. There is a 2/3 chance the car is in either B or C-- this fact does not change. If you could pick "B or C" as your choice, that is, have a chance to open both doors simultaneously, you'd have a 2/3 chance to win.
Then one of B or C are opened-- the one WITHOUT the car. (Correct? Is this how the game is played)? By doing so, it removes the losing choice between B or C, effectively giving you that chance to pick "B or C" as your option. Again, the probability that "B or C" is the winning choice remains at 2/3.
Jun '10
Re: Evil and Switching Doors
Really, you can try it for yourselves here.
Aug '10
Re: Evil and Switching Doors
FWIW, I chose the first door 10 times and switched every time. I got the car 8 times.
...
I just tried 50 times, switched every time and won 70% of the time.
At the risk of taking a diversion too seriously... I went to the NYT page, downloaded the Flash file, and hacked it a bit. I am fairly certain it is not using true random number generation, only pseudo RNG.
That may or may not have an effect on the average outcome of the game... but in any version of the game, physical or virtual, I doubt the initial distribution of the prizes and goats is truly random either.
May '10
Re: Evil and Switching Doors
BlueAnt
At the risk of taking a diversion too seriously... I went to the NYT page, downloaded the Flash file, and hacked it a bit. I am fairly certain it is not using true random number generation, only pseudo RNG.
They are always pseudorandom. If you can find a way to make a computer generate true random numbers you'll win a Nobel prize.
Apr '11
Re: Evil and Switching Doors
Humph. If a true random number generator talked with a pseudo-random number generator via teletype, how long would it take them to figure out which was which?
May '10
Re: Evil and Switching Doors
Just pick up a deck of cards and try this problem using 5, or 10, or 52 cards. Have someone lay out the cards face down and randomly place the Queen of Spades among them. Then you select one card. Have your friend turn up all the remaining cards except the Queen of Spades. The more cards you started with, the more obvious it should be that the remaining face down card is more likely than not the Queen of Spades. You should always switch.
The only case in which the remaining face down card is not the Queen of Spades is if you got lucky and picked it on your first try, in which case your friend would turn up the other cards, leaving behind any one card he wants while feigning that it might be the Queen of Spades. In this case, switching makes you lose, but it only happens 1 in n times, where n is the number of cards you're playing with.
Edited on Oct 12, 2011 at 1:19pmJun '10
Re: Evil and Switching Doors
Thanks Mark! Which is why when there are three doors, switching makes you lose only 1 in three times.
Oct '10
Re: Evil and Switching Doors
I think it's simpler than all the formulaic explanations and other demonstrations. Consider:
At the beginning of the game show I have a 67% chance of having selected a door with a goat. When the game show host opens one of the two goat doors, that doesn't change the fact that the door I originally selected has a 2-in-3 chance of having a goat behind it.
IOW, at the outset I have a 2-in-3 chance of being wrong. Thus, changing doors is the only rational decision to make, because it flips the 2-in-3 odds to favor the car, not the goat (and that's borne out by the approx. 70% wins that you see in the trials).
The problem seems more difficult than it actually is when we focus on winning (where's the car?) instead of not losing (where's the goats?).
Edited on Oct 12, 2011 at 11:42am