Frictional Games Forum (read-only)

You are in a show and you can win a car. The car is located behind one of three doors. The other two doors are empty. You have one choice. After you told the showmaster which door he should open, he just opened an other door which is empty and makes you a suggestion: "Now, there are only two doors remaining. You can now stay at your first choice or switch to the other possible door."

Makes it sense to switch? Or doesn't it matter?

And then an other candidate comes to the show and has to choose between the two remaining doors. He doesn't know that there's already one door omitted and he doesn't know your choice. But he chose the same door like you. Does he has the same prospects to win the car?

I had this problem in statistics class but you could win a sheep,
So, there is higher probability that you'll get the car when you switch.

Doesn't matter, no matter which door you pick first, you always end up with %50 chance.
Because they always reveal an empty door, leaving only 2 to choose.

Yes, I heard about this "paradox" but it makes little sense to me. Even if your original pick only has a 33% chance, the fact they always remove one of the wrong choices makes that 33% a 50% because effectively there are only two choices.

(09-16-2015, 07:05 PM)Googolplex Wrote: [ -> ]You have one choice.

This is a lie. The showmaster clearly gives me another choice after picking 1 door out of 3.

If I stay at my first choice, the door that was revealed empty, then I have a 0% chance of getting the car.

If I choose another door I have a 50% chance to get the car.


Which would I pick, you ask? The one where I have a chance of course.



Then another candidate comes in. He has the same conditions as me when I picked a door among 3.

You ask whether he has the same chance than I had. But I had 2 picks where he only has 1.

If we compare it to my first pick, then yes. We both had 33% chance.
If we compare it to my second pick, then no. I have 50% chance, and he only has 33%.

(09-16-2015, 11:58 PM)Daemian Wrote: [ -> ]Doesn't matter, no matter which door you pick first, you always end up with %50 chance.
Because they always reveal an empty door, leaving only 2 to choose.

(09-17-2015, 07:49 AM)Mugbill Wrote: [ -> ]Yes, I heard about this "paradox" but it makes little sense to me. Even if your original pick only has a 33% chance, the fact they always remove one of the wrong choices makes that 33% a 50% because effectively there are only two choices.

No. Like brus said it makes sense to switch. Then you will have the double chance.

(09-17-2015, 04:14 PM)FlawlessHappiness Wrote: [ -> ]

(09-16-2015, 07:05 PM)Googolplex Wrote: [ -> ]You have one choice.

This is a lie. The showmaster clearly gives me another choice after picking 1 door out of 3.

You always have one final choice. Even when you switch, you have to choose one door at the end.

(09-17-2015, 04:14 PM)FlawlessHappiness Wrote: [ -> ]If I stay at my first choice, the door that was revealed empty, then I have a 0% chance of getting the car.

If I choose another door I have a 50% chance to get the car.


Which would I pick, you ask? The one where I have a chance of course.



Then another candidate comes in. He has the same conditions as me when I picked a door among 3.

You ask whether he has the same chance than I had. But I had 2 picks where he only has 1.

If we compare it to my first pick, then yes. We both had 33% chance.
If we compare it to my second pick, then no. I have 50% chance, and he only has 33%.

That's wrong. If you stay at your first pick, you have 1/3 chance, when you switch you have 2/3 chance.

How can you have 2/3 chance when there are two options?





This is the explanation for the puzzle, but I do not agree with it.

I made a sketch that may will clear it for you:


Does it help?

You have good handwriting, I'll give you that. However, I completely understand the situation so the explaination isn't necessary. I just don't agree with it. Supposedly it IS a paradox, but I just think people are overthinking and overcomplicating it. The end result is that there are two options, so the first pick can be disregarded from the conclusion.
1 / 3 = 0.33 = 33% = chance of getting right pick first round (but you have no idea whether you got it right)
2 - 1 = 1 = remove one of the wrong picks to remain with 1 right and 1 wrong.
1 / 2 = 0.5 = 50% = chance of getting right pick second round.

A guy from reddit said this

Guy from reddit Wrote:Door 1: Car
Door 2: Goat
Door 3: Goat

If we don't switch, the chance is 1/3 for all of them.
Here are the outcomes if you switch:

Chooses Door 1, host opens Door 2, you switch. You lose.
Chooses Door 2, host opens Door 3, you switch. You win.
Chooses Door 3, host opens Door 2, you switch. You win.

As you can see you win in 2 of 3 times if you switch.

But it makes no sense to me.
If the host always reveals a wrong door before the final choice, then that door shouldn't even exist.

If there are 3 doors, but my final choice is between 2 of them, why does the third exist, when it has nothing to do with my choice?


The percentages that is shown in the video are from when there are 3 doors to pick from. When there are 3 doors, there are 3 different results. That means the 66% of winning the car, are 66% of 3 different results.

But Mugbill and I don't see 3 doors. We see 2 doors. With 2 doors there are 2 results. That gives us 50% chance of 2 results.


Taking your picture into account



End result
You are only choosing between 2 doors. One with a prize and one without. The last wrong choice is non-existant.