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[Criptfeind]

Okay, let me try to explain what I am thinking.


You pick door A You have a 1/3 chance to win.

A door is opened.

Now you can switch: Each door has a 50% chance. You can stay for a 50% chance or can switch for a 50% chance.

This ignores your door opening and the car being there, or a door opening and the car being there, but that is not the issue here.

[Nivim]

 HehahahahehuhehehahaahahaHAA! (At this argument that happens every time this riddle is brought up...Armok needs to come fail at this.)
 Following wikipedia, here's a good source on the "pigeon fails less" fact. Our incredible ability to reason shoots us in the foot! Again!

 Hm, now I know what it feels like to be "ninja'd".

[Leafsnail]

No.

You are given a choice between one out of 100 doors.  Each door has a 1/100 chance of being the right one, as there are 100 doors and only one of them can be right.

The host opens 98 doors, revealing all those doors to be wrong.  There are now only two doors left.  The one you originally chose, and one other.

There are two doors, one of them can be right, and you have to choose between the two.  What are the chances for each door to be correct?

Your door does not have a 1/100 chance of being correct, because it is one of two choices.  The other door does not have a 99/100 chance of being correct, because it is one of two choices.


Your door is not a 1/100 chance!  At the start, you picked one door out of 100 equal doors.  Then you had a 1% chance of being right.  But then the host eliminated 98 choices by opening them and showing them to be false.  Now you are presented with an entirely new problem where you must pick between one of two doors.

Your statistical chances do not carry over from the first problem.  There's no magic in this.  You are given a choice between two doors.  It's faulty logic that follows from being tricked by word play.

Hey, I know.  How about we actually play out this scenario a bunch of times via pm, and see what happens?  We'll do it with 1 million doors.  It should still be 50/50, right?

For "worlds" I'm just meaning "possibilities".  Nothing more complicated than that.

This ignores your door opening and the car being there, or a door opening and the car being there, but that is not the issue here.

The host knows where the car is, and always picks a door without the car.

[Criptfeind]

Okay, I am reading Wikipedia. Still do not get it as it comes down to a cumulative argument.

Hey, I know.  How about we actually play out this scenario a bunch of times via pm, and see what happens?  We'll do it with 1 million doors.  It should still be 50/50, right?

Okay, that sounds good to me, maybe I can understand it better that way.

But how do I know you will not cheat? (Not that I think you will, just asking.)

[Leafsnail]

But how do I know you will not cheat? (Not that I think you will, just asking.)

If you seriously, seriously think I'm cheating, we can also run the game with you as the host.

[dragnar]

Hey, I know.  How about we actually play out this scenario a bunch of times via pm, and see what happens?  We'll do it with 1 million doors.  It should still be 50/50, right?

Nope. The trick is, you are essentially betting against yourself if you switch. Think about it, it you pick the wrong door(which is almost guaranteed) then all incorrect choices are eliminated. Only if you picked the right door to start with(highly unlikely) will switching mean you lose.

[Criptfeind]

Okay.

I have one question though.

If you have a game where there is two doors, no switch, and no host, how is that different from the second choice.

Hey, I know.  How about we actually play out this scenario a bunch of times via pm, and see what happens?  We'll do it with 1 million doors.  It should still be 50/50, right?

Nope. The trick is, you are essentially betting against yourself if you switch. Think about it, it you pick the wrong door(which is almost guaranteed) then all incorrect choices are eliminated. Only if you picked the right door to start with(highly unlikely) will switching mean you lose.

But the chance you picked correctly goes up to 50% after a door is opened.

[Criptfeind]

Huh.

Well Leafsnail. That makes sense.

But I am having trouble connecting it to three doors.

[Criptfeind]

No wait.

Fuck.

[Leafsnail]

You probably realised while trying to pick a number from 1 to 1000,000 that the odds would be horribly against you if you stuck.  It's exactly the same for 3 doors, just... less so.

[Kagus]

Hey, I know.  How about we actually play out this scenario a bunch of times via pm, and see what happens?  We'll do it with 1 million doors.  It should still be 50/50, right?

Nope. The trick is, you are essentially betting against yourself if you switch. Think about it, it you pick the wrong door(which is almost guaranteed) then all incorrect choices are eliminated. Only if you picked the right door to start with(highly unlikely) will switching mean you lose.

Okay, let's start from there.

100 doors, pick the wrong one.  Host eliminates all other doors but the right one.  You are now allowed to pick between the right door and the wrong door.  1/2.

100 doors, pick the right one.  Host eliminates all doors except for yours and one empty door.  You are now allowed to pick between the right door and the wrong door.  1/2.


Explain to me how this is wrong.  And yes, I would very much like to see the results of a statistical program working out this problem.

[dragnar]

But the chance you picked correctly goes up to 50% after a door is opened.

No it doesn't. Why would it? You have a million possible choices, ONE of which is correct. If you picked that one, then switching will result in loss. If you picked ANYTHING else, then switching will result in victory.

If the second choice was 50/50 then that would mean you had a 50% chance of being right anyway. If you don't change your choice why should your chance of victory change?

Statistics is weird, but very fun once you've figured it out. I actually thought about becoming an actuary for a while because of how interesting they are.

Hey, I know.  How about we actually play out this scenario a bunch of times via pm, and see what happens?  We'll do it with 1 million doors.  It should still be 50/50, right?

Nope. The trick is, you are essentially betting against yourself if you switch. Think about it, it you pick the wrong door(which is almost guaranteed) then all incorrect choices are eliminated. Only if you picked the right door to start with(highly unlikely) will switching mean you lose.

Okay, let's start from there.

100 doors, pick the wrong one.  Host eliminates all other doors but the right one.  You are now allowed to pick between the right door and the wrong door.  1/2.

100 doors, pick the right one.  Host eliminates all doors except for yours and one empty door.  You are now allowed to pick between the right door and the wrong door.  1/2.


Explain to me how this is wrong.  And yes, I would very much like to see the results of a statistical program working out this problem.

Yes... but the first situation, in which the other door is correct, is 99 times more likely than the second. The trick to statistics is this: it doesn't matter how many possible states there are, it matters how many ways there are to reach each state.

[Leafsnail]

You're saying the chance of picking 1 door in 100 is 1/2?

[Criptfeind]

Kagus, let me do to you what Leafsnail did to me.

I have a random number between 1 and 1,000,000. Chose one and I will eliminate 999,998 other numbers.

What is your number.


Edit: Leaf is much better at this then me.

I have chosen a random number between 1 and 1,000,000.

Please choose a number within that range, and I will reveal 999,998 incorrect answers.  You then have the choice to stick or switch.

[Nivim]

Statistics is weird, but very fun once you've figured it out. I actually thought about becoming an actuary for a while because of how interesting they are.

It would be better to find more riddles like this that exploit holes in human logic.