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Monte Hall
So, there's this mathematical puzzle called the Monte Hall problem. Here's how it lays out.
There's three doors, and behind one is a car. Monte asks you to choose a door, so you do. Then he says, "Okay, before we reveal the door you've chosen, let me show you a door." He shows you a door that you didn't choose, of course with no car behind it. Then he asks, "Do you want to switch doors?"
What are the odds of your door being the door with the car behind it?
Naturally, your initial reaction is 50/50. But naturally (or else, this would be a boring puzzle), you'd be wrong. The odds are actually 1 in three that you've picked the door with the car. That is, you're actually better off switching.
I know, I know, it makes my brain hurt too. But here's how it breaks down.
You choose a door. If you choose the door with the car, then Monte reveals an empty door. Here if you stay, you win; if you switch you lose.
Now, let's say you choose one of the non-car doors. Monte shows the other non-car door. If you switch, you win; if you stay you lose. But since you are twice as likely to choose a non-car door, it's twice as likely for you to win by switching.
Here's the thing. Let's say you're standing there trying to decide whether or not to switch, and I walk onto the set. You turn and say to me, "Which door should I pick?" At this point, I have a 50/50 chance of choosing the right door.
Yes, we are faced with the same decision, and yet your choosing one door is 2/3rds correct and mine is only 50% correct.
That's just weird.
Obviously, there's some sort of historical information in the entire Monte Hall transaction that gives you different odds than me. I've been struggling to figure out what that is.
Then I realized, there's a problem with the assumptions laid out above. It assumes that at your first choice, choosing either of the two empty doors is a distinct event, giving three possible events. Yet, when Monte shows you a door, if you've already chosen the door with the car, his showing you either of the empty doors is a single event. That is, if it matters for you, it should matter for him. Or vice versa, if it doesn't matter for him, it shouldn't matter for you.
Looking at it this way, we can say: if you chose the door with the car, then there are two paths that Monte can take. He can show you empty door A or empty door B. For each of these, switching loses and staying wins.
That makes it two for and two against, bring the odds back to 50/50. I'd figured it out!
So, I wrote a Java program that runs through the scenario 1,000,000 times. And of course, I found the odds were: 67% for switching.
Yeah. It's actually true. Switching is better for you.
Okay, so I spotted the fault in my logic. I forgot that while there are 2 different events for Monte to take, each of those is actually half as important, because, remember, each is 50% of the original 1/3rd chance of me hitting the car in the first place. So, yeah. My dreams of being the smartest boy in the world were shattered. Again.
We're still left with the paradox that your decision is split 1/3-2/3 and mine is 50/50. What's even weirder is, if you explain the whole thing to me, then ask me if I want to switch or not, I'm in the same place you are, with a 2/3 chance of winning by switching.
So what's the difference? What's the information I have that changes the odds like that?
I still haven't figured it out. If I do, I'll let you know.
Addendum 7/2/05: In doing some reading I came across another way to look at it. When you make your initial choice, the odds were 1 in 3 that the door you chose had the car and 2 in 3 that one of the other doors did. When Monte opens one of those doors, it doesn't change your initial chances. The odds still have to be (have been?) 1 in 3 that your door has the car. Since that's the case, and since all the odds must "add up to 1", the remaining door must have a chance of 2 in 3.
So, if there were 1000 doors and Monte opened 998, the odds of the car being behind the last door that you didn't choose are 999 in 1000.
It makes a little more sense mathematically that way, but it's still freaky-weird.
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Some comments:
This thing will never be intuitive for me. NEVER.
I've been thinking about this a lot since your original post.
I think it helps to look at it like he's not really asking you to pick the door with the car. He's asking you which door is ineligible to be eliminated.
So the "piece" that is added there, is that the remaining door you could switch to had a chance to be eliminated but WASN'T allowed to be.
Your original door could have been eliminated, but wasn't eligible since you had reserved it.
Does that make sense? Obviously, the mathematics and the empirical explanations work to explain 2/3, etc. but the crux as you said is, what is that "extra thing" being gained to change it from 50/50 off the street to being a good thing to switch if you know the history. I think it's eligibility for elimination.
ARGH.
one of your assumptions, that refines the probability, is that Monte always shows you an empty door. This means that two random variables are no longer independent. Forget a friend walking on the set, if Monte picked a door at random, he'd have a 50% probability of selecting the door with the prize.
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