The Monty Hall Problem

[Jeryhn]

God, you just don't get it.

When you are offered the 2nd chance at picking a door, you are not betting on getting the car. You're betting on whether or not your original choice was the door with the car.

[Seyrr]

If you guys think my math is wrong then tell me exactly how your math is correct. Here is what you guys are saying:

Odds at start:
A = 1/3
B = 1/3
C = 1/3

A + B + C = 1/3 + 1/3 + 1/3 = 3/3


Odds after Door B is revealed:
A = 1/3
B = 0/3
C = 1/3

A + B + C = 1/3 + 0/3 + 1/3 = 2/3 = WTF???

That's not what we're saying at all.

We're saying that before AND after,

A = 1/3
B + C = 2/3.

The host revealing a door shows that either B = 0 or C = 0, so the other has to be 2/3.[/quote]

[Byrd]

This is different because B is still unknown. In your case, of course it is smarter to switch. However, B is revealed. So the choice is now only between A and C.

No, actually, it's exactly the same question just phrased a little differently, which is my point. If you switch off of A, and B & C are both goats, you get a goat. But if EITHER ONE is a car, you get the car by switching.

[BarthelloSylph]

Diet, Millail on page 1 has it right.

I won't retype it here.

The key point to understand - and I know it's tough is - when you are shown the goat, the door which you choose is NO MORE LIKELY TO HAVE THE CAR THAN IT WAS BEFORE. Nothing has changed. Nothing has been moved. Thus the PERCENTAGE CANNOT CHANGE!

It was 1/3 before. The other 2 doors, combined, had a 2/3s chance. Now, since one of those two other doors is zero, the other door has to be 2/3.

It's tough to wrap your head around it, but it's true. There's only one right answer to this.

[Plow]

Just watch the video that was posted on the first page, http://www.youtube.com/watch?v=o2L_2psS9uI


I'll... try to make a graph I guess.

Here's the deal, O is where the actual car is, X is where the goats are.

3 possible scenarios:

O X X
X O X
X X O

Now, pretend you pick the first door. Z will be the door that gets taken away.

O Z X
X O Z
X Z O

In the first scenario, if you switch, you lose.

In the second *TWO* scenarios, if you switch, you win.

Therefore, in 2 out of 3 possible scenarios, if you switch, you win. In 1 out of 3 scenarios, if you stay, you win.

So, you're twice as likely to win if you switch.

[Xfaustx]

how about you all get a friend and test it out... like 5,000 times and then post the results. you cant tell him which door is the winning obviously. the answers will be skewed depending on what he chooses. telling the friend to switch from "stay or move" 50/50 will also skew the results. the guy explained the problem very linear which i didn't like. i don't think its that straight forward.

[Millael]

Ok, here's how we'll do it. Are you ready for this?

100 runs. Pick door A every single time and stick with it after seeing one incorrect door. How many times will you win?

If you say approximately 33, you are correct, and you win the prize. What is so hard to understand about this? If you say this isn't the proper scenario, or that this will produce improper results... you are beyond all help.

[Seyrr]

Yeah, if you want to do the test it out method, choose some particularly large number, then choose to always stay or always switch. I can pretty much guarantee that if you don't use something less than 1000, you'll end up with about a 33% success rate for Stay, and a 66% success rate for Switch. Creating Java code to run that simulation was like the second or third assignment in the introductory Comp Sci class my frosh year.

[Xfaustx]

the problem with this is Monty knows which door is the winning door. if Monty did not know which door was the winning it could be 2/3, but since he skews the results, and you can be fooled, the results drop down lower than 2/3. its luck and mind games just like poker. its 2/3 if the door changed linear like the video example, but Monty can make door 1 the prize twice and door 3 the prize once and never even choose door 2. 2/3 is the best you can possibly get and pretty much only if it switches in a linear fashion as shown.

and to Seyrr, the problem with that is, some of the options arent always there. you dont have a simulated Monty to switch up the doors in a completely random fashion? and not a 1/3 chance per each. just cuz it may never be door 2 since Monty likes to trick the contestants to choose certain doors.

thats my take on it. i personally dont think theres a 100% mathematical answer to this question, since theres error in human randomness.

[Millael]

the problem with this is Monty knows which door is the winning door. if Monty did not know which door was the winning it could be 2/3, but since he skews the results, and you can be fooled, the results drop down lower than 2/3. its luck and mind games just like poker. its 2/3 if the door changed linear like the video example, but Monty can make door 1 the prize twice and door 3 the prize once and never even choose door 2. 2/3 is the best you can possibly get and pretty much only if it switches in a linear fashion as shown.

Dude. Dude. Duuuuuuuude. What the hell are you talking about? We're talking utterly random here, not "Monty Hall is a dick, knows that the test is going to involve the player picking A every time, and thus decides to fuck up a completely scientific test by choosing B every single time." But if you really have to put dickery in there, fine. The door picked every time is random, but the player still sticks with it.

[Byrd]

the problem with this is Monty knows which door is the winning door. if Monty did not know which door was the winning it could be 2/3, but since he skews the results, and you can be fooled, the results drop down lower than 2/3.

I don't believe it's been stated anywhere in the problem as I always thought it was a given, but the basic premise of the problem is that the prize can't be moved from door to door, and is stationary. If it can be moved, then it comes down to 50/50 or whatever percentage Monty feels like making it. Presuming he can't switch it, your odds are 2/3 if you switch and 1/3 if you don't.

[Seyrr]

the problem with this is Monty knows which door is the winning door. if Monty did not know which door was the winning it could be 2/3, but since he skews the results, and you can be fooled, the results drop down lower than 2/3. its luck and mind games just like poker. its 2/3 if the door changed linear like the video example, but Monty can make door 1 the prize twice and door 3 the prize once and never even choose door 2. 2/3 is the best you can possibly get and pretty much only if it switches in a linear fashion as shown.

and to Seyrr, the problem with that is, some of the options arent always there. you dont have a simulated Monty to switch up the doors in a completely random fashion? and not a 1/3 chance per each. just cuz it may never be door 2 since Monty likes to trick the contestants to choose certain doors.

... how can Monty trick you? He can't move the car, and his actions are dictated by rules. He has to show you a door with a goat behind it, and he has to offer you the choice to switch.

And for the Java program, the simulation I ran did the following (paraphrased):

1. Select the number of runs and whether the strategy will be Switch or Stay.
2. Put the prize behind a random door. Call this door PRIZE.
3. Select a random door. Call this door CHOICE.
4. Determine which door Monty reveals (if CHOICE = PRIZE, randomly select one of the other two numbers; else, select whichever of the other two has the goat).
5. Apply the strategy. If Stay, CHOICE is unchanged. If Switch, set CHOICE to be the door that was not chosen nor revealed.
6. Check if CHOICE now equals PRIZE. If so, mark that as a win.
7. Repeat 2-6 for the number of runs you selected.

Notice that when you use the Stay strategy, you get a win only when CHOICE = PRIZE from the get go. That's a 1/3 chance.

[Xfaustx]

i never said monty was a dick o.o;; nor did i say he could move the prize. the only way that mathematical problem works is if he switches the prize (not during the same round) from door 1 to 2 to 3. but each round he does not hafta change where the prize is. so im saying round 1 is door 1 round 2 is door 2. and so on. but monty can make it so the prize is door 1 all 3 rounds. get what im saying this time around? i dont mean he switches the prize after you choose, cuz thats an illegal move.

[Tajin]

i never said monty was a dick o.o;; nor did i say he could move the prize. the only way that mathematical problem works is if he switches the prize (not during the same round) from door 1 to 2 to 3. but each round he does not hafta change where the prize is. so im saying round 1 is door 1 round 2 is door 2. and so on. but monty can make it so the prize is door 1 all 3 rounds. get what im saying this time around? i dont mean he switches the prize after you choose, cuz thats an illegal move.

-_- more senseless stuff

and each round your chances of winning are 1/3 if you dont switch door.

But i guess Monty is one mean mother fucker and will steal your prize on backstage ;;

[Byrd]

i never said monty was a dick o.o;; nor did i say he could move the prize. the only way that mathematical problem works is if he switches the prize (not during the same round) from door 1 to 2 to 3. but each round he does not hafta change where the prize is. so im saying round 1 is door 1 round 2 is door 2. and so on. but monty can make it so the prize is door 1 all 3 rounds. get what im saying this time around? i dont mean he switches the prize after you choose, cuz thats an illegal move.

Each run is independent of each other. One run does not affect the next, therefore it doesn't matter if he puts the prize behind door #1, 2, or 3, the odds work out exactly the same regardless.

[Millael]

i never said monty was a dick o.o;; nor did i say he could move the prize. the only way that mathematical problem works is if he switches the prize (not during the same round) from door 1 to 2 to 3. but each round he does not hafta change where the prize is. so im saying round 1 is door 1 round 2 is door 2. and so on. but monty can make it so the prize is door 1 all 3 rounds. get what im saying this time around? i dont mean he switches the prize after you choose, cuz thats an illegal move.

In this case, no matter what door Monty decides beforehand to be the "correct" one, you still have a 33% chance of winning if you stay, since the door you choose is chosen at random. As long as there is at least one random factor of the two factors given (Selected door, correct door), this test will be scientifically accurate. Sure, you'll get streaks, but over time it will average out to around 33%.

[Seyrr]

the only way that mathematical problem works is if he switches the prize (not during the same round) from door 1 to 2 to 3. but each round he does not hafta change where the prize is. so im saying round 1 is door 1 round 2 is door 2. and so on. but monty can make it so the prize is door 1 all 3 rounds.

Yes, if he did make it door 1 all 3 rounds, and you chose door 1 all 3 rounds, then staying would be best.

But you're generalizing it to say that the car is always behind door 1, and we always pick door 1. That's not the problem we stated. Either the car needs to be randomly placed, or you need to start with a random door, or both. Otherwise, it's a different problem.

[Xfaustx]

i never said monty was a dick o.o;; nor did i say he could move the prize. the only way that mathematical problem works is if he switches the prize (not during the same round) from door 1 to 2 to 3. but each round he does not hafta change where the prize is. so im saying round 1 is door 1 round 2 is door 2. and so on. but monty can make it so the prize is door 1 all 3 rounds. get what im saying this time around? i dont mean he switches the prize after you choose, cuz thats an illegal move.

-_- more senseless stuff

and each round your chances of winning are 1/3 if you dont switch door.

But i guess Monty is one mean mother fucker and will steal your prize on backstage ;;

lol yeah he is >.> he might add a bitchslap in too just for added insult.

but on the cereal side. its just like going to vegas. your odds of winning are very slim cuz the house effects the outcome. its pretty close to the same principle, but because youre dealing with smaller numbers does not make it completely solvable by math alone. there is some luck involved.

the math makes complete sense, im not trying to argue that. im just saying that its not a perfectly mathematical world and there is human error to be dealt with. if it were perfect then yes it would be 2/3

[Devek]

******************************* <-------- this is the line

If you don't understand why it is 2/3 after reading this far in the thread, you never will.

[Xfaustx]

******************************* <-------- this is the line

If you don't understand why it is 2/3 after reading this far in the thread, you never will.

i cant see the line, make it bigger.