Monty Hall problem - SP edition

Since we're having probability fun...

Imagine you are on a game show. You are offered three doors. Behind two of these three doors are pineapples. Behind the third door is a chinchilla. You don't like pineapples. Pineapple pineapples. You want one of those damn cute chinchillas.

You are allowed to select one of the three doors. Upon selecting a door, your pimento host (me) opens a door that you did not select, revealing a pineapple. You now have a selection of two doors; the one you chose and the third one that's unchosen and unrevealed. You can either stay with your choice or you can switch doors.

What action will lead to a higher chance of getting the cute chincilla?

switch. =)

Always switch. If you choose a pineapple and switch, you get a chinchilla, if you choose a chinchilla and switch, you get a pineapple. Since there's a 2/3 chance you choose a pineapple in the beginning, odds are better if you switch.

My magical will enable me to see which door the is hiding behind.

Depends. If I chose a pineapple door, switch. If I chose the chinchilla door, stay. ^^

You never mentioned anything about random chance. And in the case that the chinchilla is behind the door I chose, it's obviously better not to switch.

You never mentioned anything about random chance. And in the case that the chinchilla is behind the door I chose, it's obviously better not to switch.

But what's the probability that it is?

But what's the probability that it is?

1/3. But that doesn't mean that if I hit that 1/3 chance on my original pick, it's still a 'good' choice to switch from an omniscient point of view.

Stay. pineapple probability, I'm 100% sure I got it right on my first guess

This question was more fun when certain death was two of the three choices.

Wait... the way I see it, there are three doors. You pick a door. The host opens one that you didn't pick, which contains a pineapple. There are two pineapples total and one chinchilla. Therefore, one pineapple is eliminated; there is one pineapple behind a door and one chinchilla behind a door, with two doors still to open. So the way I see it, you now have a 50/50 chance of getting a chinchilla.

Switch your door.

You have a 1/3 chance of getting the Chinchilla. Not good odds. Since the host opens a door revealing a pineapple, you are down to choosing two doors. Now since your door has a 1/3 chance of getting the Chinchilla, and the host already eliminated one door, it stands that the only door left has the Chinchilla 2/3 of the time, since you know the door you picked has the Chinchilla only 1/3 of the time.

This problem has sparked a huge raging debate amongst probability mathe-wannabes.

Look at Fiel in the audience, and see what door he's staring at and sweating over.
Since he has chin-radar, fallow his gaze.

I'd say it truly doesn't matter what you do. It all ends up being a mind game.

"Is he trying to get me to switch doors so I don't get the ?"

or

"Is he trying to get me to stay so I don't get the ."

In a case like this, I'd just do. Don't think, just do. He is trying to get you to rip your own mind apart.

Wow, doesn't this just make SP seem so smart?

I'd say most of the people that know the answer have seen this before

You (pimento host) could ALWAYS open a pineapple door. Even if it was our correct guess or not.

THEREFORE, I vote for it doesn't matter, you are playing with us. >

Wow, doesn't this just make SP seem so smart?

I'd say most of the people that know the answer have seen this before

I know I have. 's why I didn't vote.

Switch your door.

You have a 1/3 chance of getting the Chinchilla. Not good odds. Since the host opens a door revealing a pineapple, you are down to choosing two doors. Now since your door has a 1/3 chance of getting the Chinchilla, and the host already eliminated one door, it stands that the only door left has the Chinchilla 2/3 of the time, since you know the door you picked has the Chinchilla only 1/3 of the time.

This problem has sparked a huge raging debate amongst probability mathe-wannabes.

I don't see why it doesn't split into a 50-50 chance. Couldn't you say either door now has a 2/3 chance of having a ? Also, You could possibly say that you knew your door had a 1/3 chance of not having a and now, since a door was eliminated, that your door has a 2/3 chance of not having a . The numbers can be manipulated too much; just choose a door.

EDIT: Another thing I can point out, you can say the other door had 1/3 chance of not getting the chinchilla. Therefore, once a door is eliminated, using that logic, you could effectivly say the other door now has a 2/3 chance of not getting the . The numbers can go anyway you want them to, as I see it.

You could make some other possible scenarios, both for or against you getting the .

I don't see why it doesn't split into a 50-50 chance. Couldn't you say either door now has a 2/3 chance of having a ? Also, You could possibly say that you knew your door had a 1/3 chance of not having a and now, since a door was eliminated, that your door has a 2/3 chance of not having a . The numbers can be manipulated too much; just choose a door.

EDIT: Another thing I can point out, you can say the other door had 1/3 chance of not getting the chinchilla. Therefore, once a door is eliminated, using that logic, you could effectivly say the other door now has a 2/3 chance of not getting the . The numbers can go anyway you want them to, as I see it.

Door A: Good
Door B: Bad
Door C: Bad

You choose 1 door and can either switch or stay. That yields 6 possibilities

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

As you can see, switching gives you 2/3 winning while staying gives you 1/3

Also can be thought of variable change.

Door A: Good
Door B: Bad
Door C: Bad

You choose 1 door and can either switch or stay. That yields 6 possibilities

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

As you can see, switching gives you 2/3 winning while staying gives you 1/3

Also can be thought of variable change.

But, now a door is gone. Let's say you chose door A and he opened up door C. You still have no idea what is behind which door. He you have two choices. Stay or switch, 50-50 chance of selecting it right. EDIT: Once one door was opened a new factor was introduced completely changing the problem.

My magical will enable me to see which door the is hiding behind.

i like this one. we are using SP memes for this after all. and make the pinaeapples explode like in bloons TD 3. then people will be scrambling for the goggles. >=)