Monty Hall problem - SP edition

It's better if you think about the other 100 doors example: There are 100 doors, you pick one, and 98 doors are opened revealing pineapples. So there's the door you picked, and the door that's left. We can both agree that before the other doors were opened, you had a 1% chance of picking the right door. According to you, afterward, you have a 50% of getting the chinchilla if you stay. That means that your probability went up from 1% to 50% of picking the door with the chinchilla by doing nothing. This means that no matter door you picked at the beginning, you have a 50% chance of it having the chinchilla.



That last door DOES matter, even if you can't choose it, because you could have chosen it. Although you have 2 choices, there are still 3 doors to consider. You either stay with one of the 3 doors you picked, or switch after one of the bad doors was taken away.

Taking the 100 door example even further, let's you have an infinite amount of doors, 1 with a chinchilla, the rest with pineapples. The probability that the door you pick has a chinchilla is 1/infinity, or effectively 0. All doors except for the door you picked and the door with the chinchilla are opened, revealing pineapples. Knowing that the first door you picked has a pineapple, wouldn't it be smart to switch?

What I'm basically trying to say is having the contents behind another door revealed doesn't change the probability of what you picked before the contents were revealed. Otherwise, each of the 3 doors would have a 50% chance of being either pineapple or chinchilla, since a door that you will not pick will always be revealed to be a pineapple.

It should if part of the problem was eliminated. I will apply my chart below to this as well using the the simpler example from the original post.

Maybe this example will make more sense.

Let us say you are playing roullette and there are 10 evenly divided sections. The host tells you that one of the sections will get you $1 while the other slots will take your money away.

Let's say the marble lands on section #1. This has a 10% chance of being the section that will get you $1

Now, the host tells you that slots #3~#10 are all wrong and will give you no money. He gives you the option to stay on slot #1 or move your marble to slot #2.

If you stay on slot #1, your chance is still 10% as the board has not changed. It is still 1 of 10 sections that you chose from the beginning

Now, if you switch to slot #2, the only remaining sections are slot #1 and slot #2 so the chance of #2 being correct is much higher. This does not apply to slot #1 as you chose it from an initial 10 slots while switching has a variable of 2 slots maximum, but you cannot apply the 2 slot maximum rule to slot #1 as you didn't choose it from 2 slots but rather 10 slots.

The chart I showed earlier should have been enough as it showed all 6 possibilities. You choose one door and another one is eliminated, and you switch or stay. Maybe you're unaware that the host ALWAYS eliminates a bad door/situation and never the good one.

I knew he will never eliminate a good door. If he did then the chance would be 0% AFTER he eliminates the door. AFTER the door is eliminated your chance goes up. You know there are 2 doors and ONE out of TWO contains the winning , the other is the . The same is for the roulette. Let me revise your chart to see how I am thinking this out.

Initially you have the three doors: A, B, C.

Let's just assume you chose B, it wouldn't matter if you chose A either, and that he eliminates C. A wins. You can substitute doors if you like. Just remember one of the remaining has to win and one has to lose.


So:
Choose A + Stay = Win
Choose A + Switch = Lose
Choose B + Stay = Lose
Choose B + Switch = Win
Choose C + Stay = Lose
Choose C + Switch = Win

C is struken since it was eliminated by the host, revealing it contained a . This makes the probabilty 50% chance of winning if you stay and 50% chance of winning if you switch.

Think of it this way:

You first choose one of the three doors.
Now, he eliminates one bad door.
That would equal to choosing two doors if you now choose the other door.

Think of it this way:

You first choose one of the three doors.
Now, he eliminates one bad door.
That would equal to choosing two doors if you now choose the other door.

I guess this helps, a little. It is cloudy since I can see it as you throwing away the other door since you never found out the result.

So:
Choose A + Stay = Win
Choose A + Switch = Lose
Choose B + Stay = Lose
Choose B + Switch = Win
Choose C + Stay = Lose
Choose C + Switch = Win

C is struken since it was eliminated by the host, revealing it contained a . This makes the probabilty 50% chance of winning if you stay and 50% chance of winning if you switch.

No.

Choose A + Eliminate B or C [does not matter] + Stay = Win
Choose A + Eliminate B or C [does not matter] + Switch = Lose
Choose B + Must eliminate C + Stay = Lose
Choose B + Must eliminate C + Switch = Win
Choose C + Must eliminate B + Stay = Lose
Choose C + Must eliminate B + Switch = Win

For choosing A, there are not two difference possibilities if he picks B or C. This makes it 1/3 if you stay and 2/3 if you switch.

If you really want more proof, wikipedia it or something. This problem is very very famous and used in tv shoes, movies, etc... It is proved that the above is true.

Pictures help a lot more, seeing the scenarios rather than just reading them. This is so counterintuitive. I more wanted to understand the problem then just take your word for it.

Not only that, I just realized how much I respect the SP community, for the most part. Instead of calling me a thick-headed idiot, or something of that sort, you worked with me.

It's 50%.

Because you already *know* that 1 incorrect door will be gone.

It's 50%.

Because you already *know* that 1 incorrect door will be gone.

But at the beginning, you don't know which door that is!

 chart listing out all the cases

If you count, you'll see that if you stay, there are two cases where you'll get a pineapple, and 1 case where you'll get a chinchilla, but if you switch, there are two cases where you'll get a chinchilla, and 1 case where you'll get a pineapple.

Choose A + Stay = Win
Choose A + Switch = Lose
Choose B + Stay = Lose
Choose B + Switch = Win
Choose C + Stay = Lose
Choose C + Switch = Win

C is struken since it was eliminated by the host, revealing it contained a . This makes the probabilty 50% chance of winning if you stay and 50% chance of winning if you switch.

You can't strike C because if you had chosen it, the host would have revealed what is behind door B.

edit: Ok, I re-read your post and have a clarification to make.

Yes, in the case you stated, the probability is 50%, but under the assumption that initially door B is always picked and you have the option of staying or switching, in which case you're only really picking between door A and door B (door C is irrelevant in this case). What you should do is repeat your steps assuming you pick door A first, and again assuming you pick door C first. Individually you'll still get 50% for each case, but when you combine them you'll find that you have a 2/3 chance of picking the chinchilla if you switch.

But at the beginning, you don't know which door that is!

No, but you know it's not the door you selected, and it's incorrect. So you have 2 doors anyway, 1 door could never be chosen.

Here is a similar problem.

You have 3 cards

White on both sides
Black on both sides
White on one side, black on the other

You only see one side of the card

You see the side is white

What is the probability that the other side is black

No, the answer is not 50% and you can go and test this yourself over and over and over and you'll see it converges to a different number.

Psh. Simple. Open the two doors, then select the one that has the chinchilla in it. You just said that you had to select the door - didn't matter if it was opened or closed.