An interesting math riddle

[QUOTE="chessmaster1989"]

But, the probabilities shift when you open the door. After that, there are two possible doors, and the third is irrelevant. Since there is one of the two objects behind each, there is a 50-50 chance of one of the objects being behind the door.

Your explanation just uses some convoluted mathematical logic that, quite frankly, does not seem to work...

GabuEx

Out of curiosity, and partly to prove it to myself, I wrote a computer simulation of 2,000,000 trial runs with both the door choice and correct door picked at random. In the first 1,000,000, the simulation stays with the door it initially picked, and in the second 1,000,000, it switches to the other unopened door. The results:

Conclusion: probability is indeed 2/3 if you switch.

Again, this hinges on the fact that the host will always open a wrong door. If the host opened the door completely at random, you would be correct, as in that case you could have the situation where if you picked wrong the first time then it'd be impossible for you to win.

it has to be 50 % regardless if you switch doors because when 1 door is revealed, you know #3 is bogus, so you have door 1 and 2 .. 50% for the win, regardless if you stay with #1.. if you stay with #1 when #1 and 2 are options, theirs 50% chance. because door 3 isnt a factor anymore.. even tho it was 1/3 when you first picked, it narrows down because the 3rd is taken out of the equasion. think about it, you have 3 cups and theirs a quarter under one. you pick one cup. you put your hand on top of it. then lets say the 3rd cup is picked up and it is shown to you that theirs nothing under it.. now your selected cup and the 2nd one are left.. theirs a 50% chance the quarter is under your cup. not a 33% chance because cup 3 is eleminated.. my example is the exact same thing except i wanted to use a different example to try and make it more obvious, if the doors thing would be a little strange.nimatoad2000

-Right door----Host Removes a Door---Swap to wrong door

-Wrong door--Host Removes a Door---Swap to right door

-Wrong door--Host Removes a Door---Swap to right door

Just because the host removes a door does not change the fact that in 2 of the three outcomes will end in you swapping to the correct door.

If you were to repick then yes the chances would be 50/50.. but your not repicking from two, your assuming that you were originally wrong and swapping.Zoomin

There is a higher chance of you winning, but choosing a different door would be the same as sticking with the first door...since both doors=50%DrSponge

[QUOTE="DrSponge"]There is a higher chance of you winning, but choosing a different door would be the same as sticking with the first door...since both doors=50%gobo212

Wasn't this in the movie 21? And according to the movie, it is an advantage to switch.II-FBIsniper-II

The answer is yes, if you switch the doors you will have 50% chance of winning, if you dont, you will have 33% chance still because when you chose that first door the first time you had 33% (three doors)Lokantis

^He's wrong, switching has a 66% chance of winning. You're wrong, not switching has a 33% chance of winning.luccaface

http://www.youtube.com/watch?v=mhlc7peGlGgRed-Ravens

[QUOTE="Zoomin"] If you were to repick then yes the chances would be 50/50.. but your not repicking from two, your assuming that you were originally wrong and swapping.DrSponge

omg, anybody who believes this is dumb.

I like my analogy in another post.

What your saying, is if I have 3 doors, and say I don't go and pick one to begin with, and the host eliminates the last door. Now that I can't technically switch doors, because I haven't "imagined" one to begin with, I don't get my bonus 17%? Your implying there's some phsycic link between you and the door when you pick one at the start, which sounds just as stupid as it is.

FACT: IMAGINING DOORS AND THEN SWITCHING THEM DOES NOT GIVE YOU A SPECIAL 17% BONUS AHEAD OF SOMEBODY WHO DID NOT IMAGINE A DOOR TO BEGIN WITH

[QUOTE="Famiking"][QUOTE="luccaface"]^He's wrong, switching has a 66% chance of winning. You're wrong, not switching has a 33% chance of winning. luccaface

1)33% of the time, door #1 is correct. You shouldn't switch
2)33% of the time, door #2 is correct. You should switch.
3)33% of the time, door #3 is correct. You should switch.

But if the host doesn't know which door is correct, then the choice is 50/50 because...

1)33% of the time, door #1 is correct. You shouldn't switch.
2)16.6% of the time, door #2 is correct AND he offers you door #3. You should switch.
3)16.6% of the time, door #3 is correct AND he offers you door #2. You should switch.

The remaining 33% is comprised of the times when he accidentally opens the correct door. Thus he can't even offer you a choice.

well, the description most people are using about it becomming 50% after a door is eliminated and therefore you should switch is incorrect, thats not how this works. THe solution has nothing to do with probability, it's all just a trick.

I wrote a program to disprove this whole thing, as I assumed it would be 33% no matter what, however thats not the case, but it's also doesn't give you 50/50 odds as people seem to be suspecting.

Instead, switching the door gives you a 2/3rds chance of winning 66%. Which doesn't make sense with either logic explanation. This "game" is a trick. Basically, it reverses the odds. If the user chooses the correct door to start with, they end up wrong, as they switch away from the correct door to an incorrect one (obvously), making them wrong 1/3rd of the time (as they would choose the right door one time out of three).

However, when the user chooses the wrong door, the host eliminates another wrong door, leaving a right door, user switches, and grabs the correct door. It's simply reversing probabilities. Because of what the host does, you can win the game as long as you choose the wrong door first. It's a trick, and has nothing to do with scamming an extra 50% probability.

well, the description most people are using about it becomming 50% after a door is eliminated and therefore you should switch is incorrect, thats not how this works. THe solution has nothing to do with probability, it's all just a trick.

I wrote a program to disprove this whole thing, as I assumed it would be 33% no matter what, however thats not the case, but it's also doesn't give you 50/50 odds as people seem to be suspecting.

Instead, switching the door gives you a 2/3rds chance of winning 66%. Which doesn't make sense with either logic explanation. This "game" is a trick. Basically, it reverses the odds. If the user chooses the correct door to start with, they end up wrong, as they switch away from the correct door to an incorrect one (obvously), making them wrong 1/3rd of the time (as they would choose the right door one time out of three).

However, when the user chooses the wrong door, the host eliminates another wrong door, leaving a right door, user switches, and grabs the correct door. It's simply reversing probabilities. Because of what the host does, you can win the game as long as you choose the wrong door first. It's a trick, and has nothing to do with scamming an extra 50% probability.

DriftRS

[QUOTE="lazzordude"]you switch and heres why....when you orignally pick you have a 1/3 chance of picking it right.2/3 times your wrong,now when he narrows it down to 2 most people think the odds improve to 1/2 but they are mistaken.when you switch you are saying that your first guess is wrong which is true 2/3 of the time.Lokantis

No, if you switch, your chances become 1/2, not 2/3.

[QUOTE="Lokantis"]

[QUOTE="lazzordude"]you switch and heres why....when you orignally pick you have a 1/3 chance of picking it right.2/3 times your wrong,now when he narrows it down to 2 most people think the odds improve to 1/2 but they are mistaken.when you switch you are saying that your first guess is wrong which is true 2/3 of the time.Dr_Brocoli

No, if you switch, your chances become 1/2, not 2/3.

[QUOTE="lazzordude"]you switch and heres why....when you orignally pick you have a 1/3 chance of picking it right.2/3 times your wrong,now when he narrows it down to 2 most people think the odds improve to 1/2 but they are mistaken.when you switch you are saying that your first guess is wrong which is true 2/3 of the time.Lokantis

No, if you switch, your chances become 1/2, not 2/3.

[QUOTE="Lokantis"]

[QUOTE="lazzordude"]you switch and heres why....when you orignally pick you have a 1/3 chance of picking it right.2/3 times your wrong,now when he narrows it down to 2 most people think the odds improve to 1/2 but they are mistaken.when you switch you are saying that your first guess is wrong which is true 2/3 of the time.Ontain

No, if you switch, your chances become 1/2, not 2/3.

haha, got a great way of explaining it which doesn't use any stats, explained it to one of my friends like this over the phone and he finally got it. The trick used reverses the probability, nothing else, it has nothing to do with 50% or anything, so, basically:

If you were right at the start, you end up wrong (due to switching) and if you were wrong at the start, you end up right (as the host eliminates the other wrong door, which only leaves the right one to switch to).

So yeah, basically in the situation described by the problem, if you were to switch, you'll win if you chose the wrong door, and lose if you chose the right door, hence reversing the probability and giving you a 66% chance of a win. Hope that explains it XD

ive heard this one. its better to stick with your choicefreshgman

Lol wut?

Dude... you have two doors left... the car is behind one of the two.

Your first guess is irrelevant:|

chessmaster1989

This is the way I always saw it too. I didn't like the logic for the other answer, It implies that choosing to stay with one of two doors is equivilant to picking from one of three.

Ah, the Monty Hall problem. It's so counterintuitive O_o

Anyway, you should switch.

Ah, the Monty Hall problem. It's so counterintuitive O_o

Anyway, you should switch.

Funky_Llama

I don't know. I have heard the reasoning but I don't buy it. I think the second choice is independant from the first. Say someone has flipped 9 heads in a row, what are the chances they will flip 10 heads in a row? 50% It isn't like saying what are the chances of flipping 10 heads in a row from scratch. I think this is the same logic.

[QUOTE="Funky_Llama"]

Ah, the Monty Hall problem. It's so counterintuitive O_o

Anyway, you should switch.

Sitri_

I don't know. I have heard the reasoning but I don't buy it. I think the second choice is independant from the first. Say someone has flipped 9 heads in a row, what are the chances they will flip 10 heads in a row? 50% It isn't like saying what are the chances of flipping 10 heads in a row from scratch. I think this is the same logic.

[QUOTE="Sitri_"][QUOTE="Funky_Llama"]

Ah, the Monty Hall problem. It's so counterintuitive O_o

Anyway, you should switch.

Funky_Llama

I don't know. I have heard the reasoning but I don't buy it. I think the second choice is independant from the first. Say someone has flipped 9 heads in a row, what are the chances they will flip 10 heads in a row? 50% It isn't like saying what are the chances of flipping 10 heads in a row from scratch. I think this is the same logic.

So I made a program and it confirmed that switching would be an advantage.dave123321

[QUOTE="dave123321"]So I made a program and it confirmed that switching would be an advantage.Jandurin

ive heard this one. its better to stick with your choicefreshgman

You should've paid closer attention when you heard it. :P

[QUOTE="freshgman"]ive heard this one. its better to stick with your choiceOleg_Huzwog

You should've paid closer attention when you heard it. :P

[QUOTE="Oleg_Huzwog"]

[QUOTE="freshgman"]ive heard this one. its better to stick with your choiceFunky_Llama

You should've paid closer attention when you heard it. :P

[QUOTE="Sitri_"][QUOTE="Funky_Llama"]

Ah, the Monty Hall problem. It's so counterintuitive O_o

Anyway, you should switch.

Funky_Llama

I don't know. I have heard the reasoning but I don't buy it. I think the second choice is independant from the first. Say someone has flipped 9 heads in a row, what are the chances they will flip 10 heads in a row? 50% It isn't like saying what are the chances of flipping 10 heads in a row from scratch. I think this is the same logic.

What is the article?

And how are the doors dependant on the previous events? Would that first door somehow be more valid if the person refused to pick before a door was opened?