Monty Hall problem

[CAT13]

I want to argue with people about this, so lets debate whether switching doors really does matter or not. If you don't know what the Monty hall problem is, then look it up. I don't think it really matters if you switch or not, because either way, it is still a 50/50 chance because there are 2 doors left and one of them has the car, one of them has the goat.

 

[Escher]

I want to argue with people about this, so lets debate whether switching doors really does matter or not. If you don't know what the Monty hall problem is, then look it up. I don't think it really matters if you switch or not, because either way, it is still a 50/50 chance because there are 2 doors left and one of them has the car, one of them has the goat.

wrong...

http://math.ucr.edu/~jdp/Monty_Hall/Monty_Hall_a.GIF

I think thats a good way of putting it, even though its not using a car and goat or whatever.

 

[Bryan]

Well, you're wrong.

http://en.wikipedia.org/wiki/Monte_Hall_Problem

The big issue is that the host has to reveal a door with a goat, he can't just reveal any door.

 

[Kian]

There is no debate. It's a mathematical certainty.

 

[Sa967St]

it's explained pretty well here:

It would be best to switch. The door you began still has a 1/3 chance of having the prize even when one of the other doors are opened. So if one of the other doors with a goat was opened, the remaining door which you didn't choose at the beginning will have a 2/3 chance of winning.

 

[JBCM627]

Here's a simulator:
http://www.grand-illusions.com/simulator/montysim.htm
...That only works in IE. Here's a better automated one (hit 'play now'):
http://www.theproblemsite.com/games/monty_hall_game.asp

 

[CAT13]

I want to argue with people about this, so lets debate whether switching doors really does matter or not. If you don't know what the Monty hall problem is, then look it up. I don't think it really matters if you switch or not, because either way, it is still a 50/50 chance because there are 2 doors left and one of them has the car, one of them has the goat.

wrong...

http://math.ucr.edu/~jdp/Monty_Hall/Monty_Hall_a.GIF

I think thats a good way of putting it, even though its not using a car and goat or whatever.

I'm sorry, but isn't it true that when the guy picks cup one (the one with the prize), there are 2 different cups that the host could've eliminated. Which then shows that two out of 4 times, switching is wrong. I'm not sure about it, though.

 

[beingforitself]

I want to argue with people about this, so lets debate whether switching doors really does matter or not. If you don't know what the Monty hall problem is, then look it up. I don't think it really matters if you switch or not, because either way, it is still a 50/50 chance because there are 2 doors left and one of them has the car, one of them has the goat.

PYTHAGOREAN THEOREM:

I want to argue with people about this, so let's debate whether it is really true for all right triangles or not. If you don't know what the Pythagorean Theorem is, then look it up. I don't think it really matters if you write up a proof or not, because doesn't that only prove it for THAT particular triangle? Who's to say that you can't find some right triangle for which the Pythagorean Theorem does not hold?

 

[PatrickJameson]

I like the increasing the amount of doors explaination. That's the one that made me believe that switching was the right answer.

Let's say there are 1000 doors instead of 3. You pick one. The host knows where the car is. He eliminates all doors except the one you picked and another.

Should you switch then? Of course. The chances of picking the correct door the first time is 1/1000. The chances that the door the host left is correct is 999/1000.

 

[CAT13]

I want to argue with people about this, so lets debate whether switching doors really does matter or not. If you don't know what the Monty hall problem is, then look it up. I don't think it really matters if you switch or not, because either way, it is still a 50/50 chance because there are 2 doors left and one of them has the car, one of them has the goat.

PYTHAGOREAN THEOREM:

I want to argue with people about this, so let's debate whether it is really true for all right triangles or not. If you don't know what the Pythagorean Theorem is, then look it up. I don't think it really matters if you write up a proof or not, because doesn't that only prove it for THAT particular triangle? Who's to say that you can't find some right triangle for which the Pythagorean Theorem does not hold?

sorry, geez.
And I just realized something anyway. I wrote out all of the 4 possibilities and half of them led to winning, half of them led to losing. Then after I looked at it a little longer, I noticed that 1 of the situations was not significant in the game, which is the thing that I pointed out earlier, so I get it now.

One of the things that was throwing me off was how if one of the cups were eliminated and someone else just entered the room and picked a cup without knowing what you picked, the chances 'magically' became 50/50 again.

 

[shelley]

It's not about what you think is right. It's a simple exercise in probability. Probability doesn't care about your opinion.

If the math still doesn't convince you, run the simulator and see if the results are consistent with what you think is correct.

 

[bamman1108]

The reason people have trouble understanding this problem (like I used to) is because people think, "If I pick one door, he will open another door. Now, there are two left. From here, there is a 50/50 chance that the door has a car in it now."

The reason this is wrong is because either the switched door has a car or has a goat. 1/3 of the time, you pick the car, 2/3 of the time, you pick a goat. That means that 2/3 of the time, the host reveals the other goat, and switching wins, and staying loses. 1/3 of the time, it's the other way around.

Very similar to this is gambler's fallacy. In the case of flipping a coin, people think, "Since it landed heads before, the odds that it lands heads again is 1/4." This is only true if you haven't flipped the coin yet, just like how switching doors would have no effect if the host hadn't revealed a goat yet.

 

[Lotsofsloths]

Its 50/50, because you have to constantly apply what you know about the situation into the formula. It WAS 1/3, but he removed one, so now its 1/2.

 

[Sa967St]

Its 50/50, because you have to constantly apply what you know about the situation into the formula. It WAS 1/3, but he removed one, so now its 1/2.

Although one door was removed, the door that was choosen still has a 1/3 chance.

 

[Derrick Eide17]

i remember i came across this in a book we just read last year in english or so and it was pretty interesting and i thought for sure it doesnt matter but it really does.

because lets look at all the options.

1. You pick the door with a lama and then the other lama is revealed.
if you stay with that door you lose but if you switch you win the car

2. you pick the door with another lama, basically same situation, you switch you win the car you stay you dont

3. you pick the car and one lama is revealed, you switch you dont win, you stay you do win

so 2/3 times if you switch you win the car

 

[teller]

Its 50/50, because you have to constantly apply what you know about the situation into the formula. It WAS 1/3, but he removed one, so now its 1/2.

That's the fallacy...you know more about the situation than if you had come across 2 doors cold with no other information. Monty has revealed a little bit extra to you. It's not 50/50 at all.

 

[Karthik]

By opening one false door for you, the host has effectively given you the choice of selecting two doors and you win if either one of them has the car. You lose only if the car is in the door you didn't select.
So your winning probability is 2/3.

 

[Kian]

Its 50/50, because you have to constantly apply what you know about the situation into the formula. It WAS 1/3, but he removed one, so now its 1/2.

Your probability stays at 1/3 if you do nothing, 2/3 if you change.

 

[Lucas Garron]

You are blindfolded, holding a solvable cube. I ask you to take out a corner, toss and catch it randomly, and insert it back in.

I pick direction A as either clockwise or counterclockwise, such that the following statement is true, and say it to you:
"If you twist a corner in direction A, the cube will be in an unsolvable state (where ).

If you'd like the cube to be solvable again, should you leave the corner alone, or twist it in direction A'?

 

[qqwref]

I ask you to take out a corner, toss and catch it randomly, and insert it back in.

...and I reply, "Sorry, my cube's not loose enough."

So the probability that the corner is solved is obviously 100%, assuming i don't twist it in direction A'.


About Monty Hall: the problem statement itslef is often ambiguous if it isn't worded correctly. If the host *randomly* opens a door and it happens to have a goat, yes, you actually do have a 1/2 chance of having the car, since you're actually looking at the probability that you picked the car given that the host didn't. But the deal is that the host knows where the car is and will always open a door with the goat, so he'll open a goat door every time, which means that the probability that you picked the car at the beginning (1/3) stays the same.