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Monty Hall Problem

DaveW said:
This post sums up my thinking.
http://www.cut-the-knot.org/terry.shtml
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The problem is that it doesn't take into account that when I go into the 2nd choice, my probability of having picked the right door in the 1st choice remains the same. It's still a 1/3 probability that I chose the car the first time. That means that it's still a 2/3 probability that the remaining 2 doors contain the car.
 
It depends, as has been said before, on whether Monty will always open another door and offer to switch. If that's a given, you're better off switching, because there is a 2/3 chance that one of the remaining two, instead of the door you picked, has the car.

If you don't assume Monty will always open another door, it's back to 50-50.

Questions?
 
DaveW said:
Sure, but I only ever get one door at a time. I never "have" 2 doors.
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If you swap then you DO have two doors. What Monte is doing is saying "You can keep the first door you chose OR you can have either of the other two, and look here I've just opened one of them and it's not the right one, so if you switch you'll know which of the two doors to pick"
 
DaveW said:
Just to let everyone know... I'm not trying to bust anyone's chops, I really just don't get it, and I know I don't, but hashing this out like this will hopefully help me learn it. Thanks!
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Well, we're all trying to get at your intuition. The demonstration is much easier.

Demonstration: Start 100 identical games in 100 identical studios, all with the car behind door 57. In each game the contestant picks a different door.
Now Monty is going to go from studio to studio. 99 of these people didn't pick door 57. For those he leaves them only the door they have, and door 57. In the other studio he leaves one other door unopened.

Now all contestants switch to door 57. 99 of them win.

If none of the contestants switch, only 1 wins.

So in 99 out of 100 identical games, switching led to a win. Thus, the probability of a win from a switch is 99%.

Now the intuition. Let's try this a different way. You pick a door. You're pretty sure you picked wrong, correct? So Monty is going to tease you by opening doors. How certain are you that the car is among the doors he's going to open? Pretty sure. You should be, there's a 99% chance of that.

But Monty is feeling cruel today and opens goat doors first. Now he's down to 10 doors. You know he controls the game, and you knew before he opened anything that he had a 99% chance of having the car, right? He's just teasing you, opening that one last.

And now he's down to one door, and he starts to open it... what are the chances that you're going to see a car there? Don't you think it's pretty likely? You thought it was 99% sure he had the car before opening doors, he's just being mean about the order of opening. So why if he waits to open the car last, is there suddenly a 50% chance that he didn't have the car at all? You KNOW he has the car.

Switch places. Monty is guessing, you know where the car is. Monty picks a (probably wrong door) and you're going to play the cruel open-the-car-last game. But you know where the car is. Don't you think it's pretty likely that in nearly every game, you're going to get to one last door that has the car behind it?

Play it out with cards or coins.
 
rppa said:
Well, we're all trying to get at your intuition. The demonstration is much easier.

Demonstration: Start 100 identical games in 100 identical studios, all with the car behind door 57. In each game the contestant picks a different door.
Now Monty is going to go from studio to studio. 99 of these people didn't pick door 57. For those he leaves them only the door they have, and door 57. In the other studio he leaves one other door unopened.

Now all contestants switch to door 57. 99 of them win.

If none of the contestants switch, only 1 wins.

So in 99 out of 100 identical games, switching led to a win. Thus, the probability of a win from a switch is 99%.

Now the intuition. Let's try this a different way. You pick a door. You're pretty sure you picked wrong, correct? So Monty is going to tease you by opening doors. How certain are you that the car is among the doors he's going to open? Pretty sure. You should be, there's a 99% chance of that.

But Monty is feeling cruel today and opens goat doors first. Now he's down to 10 doors. You know he controls the game, and you knew before he opened anything that he had a 99% chance of having the car, right? He's just teasing you, opening that one last.

And now he's down to one door, and he starts to open it... what are the chances that you're going to see a car there? Don't you think it's pretty likely? You thought it was 99% sure he had the car before opening doors, he's just being mean about the order of opening. So why if he waits to open the car last, is there suddenly a 50% chance that he didn't have the car at all? You KNOW he has the car.

Switch places. Monty is guessing, you know where the car is. Monty picks a (probably wrong door) and you're going to play the cruel open-the-car-last game. But you know where the car is. Don't you think it's pretty likely that in nearly every game, you're going to get to one last door that has the car behind it?

Play it out with cards or coins.
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I played it out with Excel, 2000+ runs of the three door gag. According to what I have seen so far, there is not a doubling of probability by switching. There is a slight advantage, on average to switching, though (probably due to the whole effect of Monty having to pick a bad one for you, tipping the scales somewhat). But, it really appears to not support the doubling mantra. I can't post it here from work, so I'll try to do it from home a bit later.
 
Dragon said:
Then think again, because that post is wrong, wrong, wrong.

Try this link - and have a go at the simulator.
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I played with those simulators. Problem is, it would take a long time to get a good sample size.
 
DaveW said:
Ah, but he is! He lets me know one of the two doors that was wrong!
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Believe me, I went through the same arguments you did when I first heard of this.

It doesn't matter which of the two doors is wrong. You already know that at least one of them is wrong. Whether it's B or C is irrelevant to the odds.

If he offered you the choice to pick A or B & C, you would pick B & C, right? And you would do so KNOWING that one of the two, B or C, must have a goat. So when Monty reveals that the goat is actually behind B, that doesn't change the odds one bit.
 
I played the Excel sim I made out to about 8000 runs. I'll try to post it by tomorrow, but, if the theory is actually right, something is wrong with my spreadsheet (yeah, more likely, I know, but I want to know what!)
 
Assuming the programming is correct, this simulator: Montysim demonstrates the odds very effectively. Over 1000 tests, and the odds came out exactly as they should. Changing your door wins the car twice as much as keeping your door.

Try it!
 
Look at it this way. Someone who consistently adopts the switching strategy will always be rewarded if they chose wrongly in the first place.

You have a 2/3 chance of being wrong in the first place so your odds of getting the prize are 2/3 by changing your choice.

If you try this game 100 times and always switch you should win approximately 66 times (not 50 times).

If you try this game 100 times and always stick you will win approximately 33 times.
 
Wait a minute.

Your first choice with all 3 doors closed is not relevant, because the first pick was only for dramatic effect - it has no bearing on the outcome. Your option to choose a second time is the only real choice, so your odds are 50/50. The real choice is between the 2 closed doors and a goat. I guess there is a slight chance some kind of survivalist or pervert would pick the goat, leaving each door with a chance of 49.99 %.

Am I missing something?
 
fishbob said:
Wait a minute.

Your first choice with all 3 doors closed is not relevant, because the first pick was only for dramatic effect - it has no bearing on the outcome. Your option to choose a second time is the only real choice, so your odds are 50/50. The real choice is between the 2 closed doors and a goat. I guess there is a slight chance some kind of survivalist or pervert would pick the goat, leaving each door with a chance of 49.99 %.

Am I missing something?
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Yes.

Go run the simulator posted above. It will demonstrate the results very effectively.
 
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