probability to find a car

[jian]

It is an old question, it says:

In a gambling game, there are three door numbered #1,#2 and #3. There are a car behind one of the doors. If you guess the right door number, you can get the car. Then you made a guess, say door #1, then the host opened the door3, it turned out to be empty. Then he offered you a change to guess again.

The question is " should you change your choice to #2?"

Edited January 18, 2009 by jian
There is a similar post on the forum,plz delete it

[Bignose]

It is an old one, but a very well studied one: http://en.wikipedia.org/wiki/Monty_Hall_problem

 

Short story, for all problems like this (for any number of doors), you stick with your original choice as the host opens "goat" prizes until the very last time you are offered a chance to switch (i.e. only two doors left) and then always switch. The reason you always change has to do with the conditional probabilities and with the extra information you have received by the host revealing what doors the prize definitely is not behind.

[swansont]

And if you have trouble visualizing it, consider the case where there are ten doors. You pick one, and the host opens eight doors. Should you switch?

[NIN]

I really don't get this problem (isn't it from the movie 21, lol?)

 

What I don't get is how the # of the door affects anything...Idk, just doesn't make sense to me. Maybe I'll get it after some more posts.

[mrburns2012]

I really don't get this problem (isn't it from the movie 21, lol?)

 

What I don't get is how the # of the door affects anything...Idk, just doesn't make sense to me. Maybe I'll get it after some more posts.

 

The problem:

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice? (Whitaker 1990)

 

The fact that the host always reveal the wrong door affects the outcome. Let me elaborate:

 

There are 3 doors,

 

Case 1: You DO NOT switch. You have two scenarios:

 

1) The probability of choosing the correct door will be 1/3

 

2) The probability of choosing the incorrect door will be 2/3

 

Case 2: You DO want to switch:

 

1) If you had initially picked the incorrect door (probability = 2/3), the chance of changing to the correct door will be 100% since that'll be the only door available after the other incorrect door is revealed.

 

2) If you had initially picked the correct door (probability = 1/3), the chance of changing to the incorrect door will be also 100% since that's the only door available.

 

---

 

So which method yields the best outcome in these conditions? If you DO NOT switch, you'll have 33% chance of winning (case 1, scenario 1). If you DO switch, you have a 66% chance of winning (case 2, scenario 1). Why? Because you're more likely to choose the incorrect door the first time.

Edited January 19, 2009 by mrburns2012

[NIN]

snip

 

Wow, I guess so. That's a hard one to think about, though. To me, the thing that's confusing is that it doesn't make sense that the door that YOU chose should change the probability, since the chooser is an external force, and shouldn't affect the outcome of the problem itself. It seems as though it should go from 33% 33% 33% into 50% 50%. I guess your right though, it's just confusing to me, for the reasons I just said. I'm sure I'll understand it more thoroughly with thought, though.

Edited January 24, 2009 by NIN

[ironizer]

This one is extremely easy to figure out. I figured it out on my own when i was 15. If you switch, you get twice as better chance of getting the car.

 

This problem is a quick IQ test I offer people when i meet them to see if they're smart or dumb.

[devrimci_kürt]

original question

Suppose you're on a game show, and you're given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say #1, and the host, who knows what's behind the doors, opens another door, say #3, which has a goat. He says to you, "Do you want to pick door #2?" Is it to your advantage to switch your choice of doors?

Craig F. Whitaker

Columbia, Maryland

 

answer

Yes; you should switch. The first door has a 1/3 chance of winning, but the second door has a 2/3 chance. Here's a good way to visualize what happened. Suppose there are a million doors, and you pick door #1. Then the host, who knows what's behind the doors and will always avoid the one with the prize, opens them all except door #777,777. You'd switch to that door pretty fast, wouldn't you?

 

 

 

http://www.marilynvossavant.com/articles/gameshow.html

[NIN]

original question

Suppose you're on a game show, and you're given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say #1, and the host, who knows what's behind the doors, opens another door, say #3, which has a goat. He says to you, "Do you want to pick door #2?" Is it to your advantage to switch your choice of doors?

Craig F. Whitaker

Columbia, Maryland

 

answer

Yes; you should switch. The first door has a 1/3 chance of winning, but the second door has a 2/3 chance. Here's a good way to visualize what happened. Suppose there are a million doors, and you pick door #1. Then the host, who knows what's behind the doors and will always avoid the one with the prize, opens them all except door #777,777. You'd switch to that door pretty fast, wouldn't you?

 

 

 

http://www.marilynvossavant.com/articles/gameshow.html

 

He'd avoid all except #777,777, but he'd also avoid the one you chose in this particular situation. So I still don't quite understand what the difference is...

[devrimci_kürt]

clue:host

 

host knows what's behind the doors ...

[DJBruce]

He'd avoid all except #777,777, but he'd also avoid the one you chose in this particular situation. So I still don't quite understand what the difference is...

 

Since your are struggling to visualize the problem it might help if you actually simulate the problem. I do not guarantee the complete accuracy of this simulation but I ran through it 20 times staying and 29 times switching and the results were close to the predicted values. For 20 trials of staying I got a probability of winning of 30% and out of 20 trials for switching I received a probability of winning of 60%. Both of which seem to be reasonable for such a small sample.

[Bignose]

He'd avoid all except #777,777, but he'd also avoid the one you chose in this particular situation. So I still don't quite understand what the difference is...

 

The difference is that every time a door without a prize is opened, you get a new piece of information you didn't have before. And that extra information changes the probabilities.

 

It isn't simply:

 

P(door picked = car) it is P(door picked = car | door x_1 does NOT have the prize) where the "|" denotes "given that" or "on condition that". The vertical bar is used for conditional probabilities.

 

Sometimes the conditions won't always add extra information. For a random variable that is truly random, e.g the probability of heads coming up when flipping a fair coin is completely independent of any of the previous flips, then the conditions don't mean anything. But, many variables are not independent. For example, if on Saturday there is a 30% chance of rain on Wednesday, and it rains on Tuesday, the condition that it rained on Tuesday probably changes the probability of precipitation on Wednesday.

[NIN]

I finally got it. I feel like an idiot now, it's simple as Hell. I was just over-thinking it and trying to come up with super complex ideas to how it works, when it's actually really simple.

 

What I didn't take into consideration is the fact that the host will not chose the door that you've already chosen.