Riddle Me This: A Statistic Problem

[9_Lives]

I don't think switching doors affects your 1 in 3 chances. Both are still 33.3% likely to have the car, whether you change your mind or not I would think. And if each are equally likely, then it should be a 50/50. Like to the guy talking about the 100 doors thing; they would just both be 1/100. There is not enough information provided to conclude you would raise your chances of getting the car at all let alone by that significant of a margin. The logic has been presented, but not justified to the point where I could agree just yet. Can anyone elaborate on how there is any difference between the two remaining doors that does not involve "It's not your first option?" Or am I just doomed not to get this? '

 

[Killphil]

I don't think switching doors affects your 1 in 3 chances. Both are still 33.3% likely to have the car, whether you change your mind or not I would think. And if each are equally likely, then it should be a 50/50. Like to the guy talking about the 100 doors thing; they would just both be 1/100. There is not enough information provided to conclude you would raise your chances of getting the car at all let alone by that significant of a margin. The logic has been presented, but not justified to the point where I could agree just yet. Can anyone elaborate on how there is any difference between the two remaining doors that does not involve "It's not your first option?" Or am I just doomed not to get this? '

He says it pretty well.

But the point is it does matter. In switching doors you effectively gain an extra 33% chance of picking the right door, the odds become stacked in your favour.


EDIT:

Okay, let's walk through this completely. Lets just say (for simplicity sake) that doors 1 and 2 have a goat behind them and door 3 has the car (it doesn't really matter which door has which just using this to show the scenarios)


Not switching
You choose door 1; you lose
You choose door 2; you lose
You choose door 3; you win

Switching
You choose door 1, host opens door 2 and you switch to door 3; you win
You choose door 2, host opens door 1 and you switch to door 3, you win
You choose door 3, host opens either door 1 or 2 and you switch to the other door; you lose

Each of those scenarios has an equal probability of occurring. As you can see, if you switch 2/3 times you will be successful while if you stay only one out of 3 times you are successful.

 

[CrimsonShadow]

I don't think switching doors affects your 1 in 3 chances. Both are still 33.3% likely to have the car, whether you change your mind or not I would think. And if each are equally likely, then it should be a 50/50. Like to the guy talking about the 100 doors thing; they would just both be 1/100. There is not enough information provided to conclude you would raise your chances of getting the car at all let alone by that significant of a margin. The logic has been presented, but not justified to the point where I could agree just yet. Can anyone elaborate on how there is any difference between the two remaining doors that does not involve "It's not your first option?" Or am I just doomed not to get this? '

I went through this in detail -- the host's door will only be wrong if your first guess was right. In all other cases his door will be the winning door.

 

[9_Lives]

I went through this in detail -- the host's door will only be wrong if your first guess was right. In all other cases his door will be the winning door.

Well, reading that, maybe my problem is that I just don't agree. It looks 50-50 to me. The op said he reveals a goat, then it is not safe to assume my first guess is right or wrong.

Assuming that door 3 has the car and the door the host opens is always a goat:

Scenario 1:
Choose door 1, host opens door 2 and I switch to door 3; win
Choose door 1, host opens door 2 and I stay; lose

Scenario 2:
Choose door 2, host opens door 1 and I switch to door 3; win
Choose door 2, host opens door 1 and I stay; lose

Scenario 3:
Choose door 3, host opens door 1 and I switch to door 2; lose
Choose door 3, host opens door 1 and I stay; win
Choose door 3, host opens door 2 and I switch to door 1; lose
Choose door 3, host opens door 2 and I stay; win

There are two outcomes in which I'm right if stay, two outcomes in which I'm wrong. Same for switching. It looks like an equal shot all the way through.

 

[CrimsonShadow]

Well, reading that, maybe my problem is that I just don't agree. It looks 50-50 to me. The op said he reveals a goat, then it is not safe to assume my first guess is right or wrong.

Assuming that door 3 has the car and the door the host opens is always a goat:

Scenario 1:
Choose door 1, host opens door 2 and I switch to door 3; win
Choose door 1, host opens door 2 and I stay; lose

Scenario 2:
Choose door 2, host opens door 1 and I switch to door 3; win
Choose door 2, host opens door 1 and I stay; lose

Scenario 3:
Choose door 3, host opens door 1 and I switch to door 2; lose
Choose door 3, host opens door 1 and I stay; win
Choose door 3, host opens door 2 and I switch to door 1; lose
Choose door 3, host opens door 2 and I stay; win

There are two outcomes in which I'm right if stay, two outcomes in which I'm wrong. Same for switching. It looks like an equal shot all the way through.

Scenario 3, it doesn't matter what the host does. It's not 2 scenarios, it's one: You picked the winning door, he will leave behind the wrong door. Which wrong door he leaves does not affect the outcome at all.

Remember, the only choice you are given is stay vs. switch. So you don't get extra choices in scenario 3. You either switch and lose or stay and win.

There are only 2 possible outcomes in this entire situation:
1) You picked the right door, and he will leave a wrong door -- so switching loses
2) You picked the wrong door, and he will leave the right door -- so switching wins

Your odds of 1) are 1/3.
Your odds of 2) are 2/3.

Statistically that is all that matters.

 

[9_Lives]

Scenario 3, it doesn't matter what the host does. It's not 2 scenarios, it's one: You picked the winning door, he will leave behind the wrong door. Which wrong door he leaves does not affect the outcome at all.

Scenario 3 does matter.

I should just break it down into two scenarios:

I pick one of the two wrong doors and he reveals the other wrong door. I can either be wrong for staying or right for switching.
I pick the one right door and he reveals one of the two wrong doors. I can either be right for staying or wrong for switching.

Whether I have picked the winning door first or not, there will always be a guess because there are two wrong doors. If I pick the right one, it will eliminate one wrong door; if I pick the wrong one, it will eliminate one wrong door. I'm outlining every possibility to show that no matter what, I will still have one wrong or right door to switch to or stay on.

 

[CrimsonShadow]

Scenario 3 does matter.

I should just break it down into two scenarios:

I pick one of the two wrong doors and he reveals the other wrong door. I can either be wrong for staying or right for switching.
I pick the one right door and he reveals one of the two wrong doors. I can either be right for staying or wrong for switching.

Whether I have picked the winning door first or not, there will always be a guess because there are two wrong doors. If I pick the right one, it will eliminate one wrong door; if I pick the wrong one, it will eliminate one wrong door. I'm outlining every possibility to show that no matter what, I will still have one wrong or right door to switch to or stay on.

Yeah, but you didn't calculate the odds correctly.

If you pick a right door first, switching means you lose.
If you pick a wrong door first (odds of this are 2/3), switching means you win.

Nothing else affects the odds, and as you can see, they are in favor switching overall, because statistically it is more likely that you will pick a wrong door.

 

[CrimsonShadow]

Scenario 3 does matter.

I should just break it down into two scenarios:

I pick one of the two wrong doors and he reveals the other wrong door. I can either be wrong for staying or right for switching.
I pick the one right door and he reveals one of the two wrong doors. I can either be right for staying or wrong for switching.

Whether I have picked the winning door first or not, there will always be a guess because there are two wrong doors. If I pick the right one, it will eliminate one wrong door; if I pick the wrong one, it will eliminate one wrong door. I'm outlining every possibility to show that no matter what, I will still have one wrong or right door to switch to or stay on.

Yeah, but you didn't calculate the odds correctly.

If you pick a right door first, switching means you lose, no matter what (nothing will change this).
If you pick a wrong door first (odds of this are 2/3), switching means you win, no matter what (and nothing will change this).

Nothing else affects the odds, and as you can see, they are in favor switching overall -- because statistically it is more likely that you will pick a wrong door at the start.

After you pick the first door, the options are set in stone. There's a 1/3 chance that your first pick was right and switching will lose.. A 2/3 chance that your first pick was wrong and switching will win.

 

[9_Lives]

Yeah, but you didn't calculate the odds correctly.

If you pick a right door first, switching means you lose.
If you pick a wrong door first (odds of this are 2/3), switching means you win.

Nothing else affects the odds, and as you can see, they are in favor switching overall, because statistically it is more likely that you will pick a wrong door.

Oh, I see what you're saying. There goes half an hour of my life arguing an invalid point. XD

 

[CrimsonShadow]

Oh, I see what you're saying. There goes half an hour of my life arguing an invalid point. XD

Haha don't worry -- I was going to argue the same point before I realized why it didn't matter.

 

[9_Lives]

Haha don't worry -- I was going to argue the same point before I realized why it didn't matter.

I'm just...so used to being right...defeat is like a slap in the face. D: I'll get over it though. I'm just sorry it took me so long to get it, now reading back through and seeing the right answer all the way in the first post. XP

 

[astronout]

this veridical paradox is old hat sir.

 

[NoobHunter420]

you chose door 1 when your chance of getting the right door was 33%. when you take door 3 from the equation, is only logic to change your door or at least consider it since you now have a 50% chance of getting the right door.

 

[Pan1cMode]

you chose door 1 when your chance of getting the right door was 33%. when you take door 3 from the equation, is only logic to change your door or at least consider it since you now have a 50% chance of getting the right door.

You now have a 66% chance if you switch not a 50% chance.