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Monty Hall problem - SP edition - Printable Version

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Monty Hall problem - SP edition - mrcowcow - 2009-03-16

Wait... Did the host guy know behind which door the car was located? Like what if the host picked one with the car? o___o *iRSuxz0rzattehmath(coughChemplzcough)*

Monty Hall problem - SP edition - mrcowcow - 2009-03-16

mrcowcow Wrote:Wait... Did the host guy know behind which door the car was located? Like what if the host picked one with the car? o___o *iRSuxz0rzattehmath(coughChemplzcough)*

Edit: Ohcrap I'm confusing myself now. ;_;

Monty Hall problem - SP edition - mrcowcow - 2009-03-16

Argh, delete please. Realized I accidently pressed 'quote' instead of 'edit'. >.<

Edit: Okay what the crap is wrong with me. ;___; Please, delete this post and the one before it. *bashes head into wall*

Monty Hall problem - SP edition - Kaasoljoyyx - 2009-03-17

TehMatt Wrote: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.

This is flawed logic which is the source of this famous problem.

Look at it from this perspective:

Let us say we had 100 doors. Only 1 door has a good prize. That means the chance is 1%.

Now, you are right in saying "Okay, lets eliminate 98 doors, if we were free to choose the two remaining doors at the beginning, the chance is 50% each"

But the fact is, we have stayed with our choice, which was, is, and will always be 1%.

Also look at it from this perspective, lets say we add 100 more doors. Does that mean the chance is now 0.5%? Of course not, and anyone will call you crazy for thinking that. But if we had to choose a door to start with from the 200 initial doors, then yes it would be 0.5%

Now, lets go back to 100 initial doors, picking one, and eliminating 98 incorrect doors. This leaves one door correct, and one door incorrect. Like above, if we started with 2 doors to start with, then the chance was 50%. But we had to make our initial guess from 100 doors, so that door still has a 1% chance of being correct. However, the other door has a very high increased probability of being correct because that door did undergo changing influences to affect its probability. While we kept with our original choice, the probability for that door kept increasing because technically, you can see it as "a free choice from an initial set of doors" as we have never chosen it.

I hope that is enough information to help you understand why it is better to switch and it is not a 50-50 chance to win. I'll be more than happy to explain anything you don't understand as long as it doesn't turn into trolling.

Monty Hall problem - SP edition - sicnarf - 2009-03-17

Alloy Wrote: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. >Sad

Huh, I'm revealing eliminating a bad door for you. I'm not that big an pimento host. D:

mrcowcow Wrote:Wait... Did the host guy know behind which door the car was located? Like what if the host picked one with the car? o___o *iRSuxz0rzattehmath(coughChemplzcough)*

We have no cars here. Just chinchillas.

Kaasoljoyyx, I think that's a great explanation; I should remember that if I need to try explaining it to someone. xD

Monty Hall problem - SP edition - GummyBear - 2009-03-17

Every time a door opens out to a chinchilla, a hunter will shoot it. now WOULD YOU STILL WANNA SWITCH??

*loading rifle!!*

Monty Hall problem - SP edition - Cancambo - 2009-03-17

sicnarf Wrote: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?

Kaasoljoyyx Wrote:This is flawed logic which is the source of this famous problem.

Look at it from this perspective:

Let us say we had 100 doors. Only 1 door has a good prize. That means the chance is 1%.

Now, you are right in saying "Okay, lets eliminate 98 doors, if we were free to choose the two remaining doors at the beginning, the chance is 50% each"

But the fact is, we have stayed with our choice, which was, is, and will always be 1%.

Also look at it from this perspective, lets say we add 100 more doors. Does that mean the chance is now 0.5%? Of course not, and anyone will call you crazy for thinking that. But if we had to choose a door to start with from the 200 initial doors, then yes it would be 0.5%

Now, lets go back to 100 initial doors, picking one, and eliminating 98 incorrect doors. This leaves one door correct, and one door incorrect. Like above, if we started with 2 doors to start with, then the chance was 50%. But we had to make our initial guess from 100 doors, so that door still has a 1% chance of being correct. However, the other door has a very high increased probability of being correct because that door did undergo changing influences to affect its probability. While we kept with our original choice, the probability for that door kept increasing because technically, you can see it as "a free choice from an initial set of doors" as we have never chosen it.

I hope that is enough information to help you understand why it is better to switch and it is not a 50-50 chance to win. I'll be more than happy to explain anything you don't understand as long as it doesn't turn into trolling.

You are given the option to select again, thus changing the problem. Now you aren't allowed to choose the third door. This makes you have a 50-50 chance of selecting the Chin. Or, as I see it. You chance from the beginning was 33%, or 1/3. Now, however, your odds have changed since you only have two options to choose from. That's how I see it. Even if you couldn't choose it you could say, in a way, that your chances are 50-50 of having the Chin. This is since you know that 1/3 doors contains it and that 1/3 was already said to not contain it. This leaves 2 doors and NOW you have 50% chance of having it. I agree, initially, that you had a 1/3 chance.

Monty Hall problem - SP edition - sicnarf - 2009-03-17

Yes, but I'm not asking for the probability of that part. I'm asking it of the entire problem.

If you switch blindly, you're more likely to get a chinchilla, as shown above.

Kaasoljoyyx Wrote:Choose A + Stay = Win
Choose A + Switch = Lose
Choose B + Stay = Lose
Choose B + Switch = Win
Choose C + Stay = Lose
Choose C + Switch = Win


Monty Hall problem - SP edition - Cancambo - 2009-03-17

sicnarf Wrote:Yes, but I'm not asking for the probability of that part. I'm asking it of the entire problem.

If you switch blindly, you're more likely to get a chinchilla, as shown above.

I guess, it just is confusing since at that instant after a door has been eliminated you have the same chance with just the 2 doors left. At the instant if you switch blindly it doesn't change your chances since there is no third door available to switch to. You are left with 2 doors and 2 decisions.

Sorry if I am sounding disagreeable.

Monty Hall problem - SP edition - y0y0y0y0shi0 - 2009-03-17

It doesn't matter. Either way you have the same probability, a 1/2 chance that one of the two doors holds the chinchilla. You eliminate the first door because it is revealed, and is not part of the problem, giving a 1/2 chance for a chinchilla or a pineapple.

BUT, chances are if the host picked the wrong one, he purposely did it, and knowing you would think that it the one with the chinchilla was the one that you originally picked, he would make the other door the pineapple door, leaving you with a pineapple if you stuck with your first choice. Though, that probably wouldn't happen anyway because people aren't THAT cruel (Except Nexon).
In which case, if it was Nexon that was hosting it, then you would have a 0% chance to get the chinchilla because they screwed up and put 3 pineapples. BUT, the chinchilla would want to destroy the pineapple so he would go and eat it, making him the 3rd position, in which case it would be a 1/3 chance because the chinchilla might have gone and eaten the pineapple from the door you already chose, giving you a 0% chance of getting the chinchilla, but a 1/3 chance right after you eliminated one.

tl;dr I like making problems more complicated than they really are and making heinous things that have nothing to do with the problem.

Monty Hall problem - SP edition - Alloy - 2009-03-17

sicnarf Wrote:Huh, I'm revealing eliminating a bad door for you. I'm not that big an pimento host. D:

I just copy/pasted from your post D:

Really, I wasn't sure what particular adjective you used...

Monty Hall problem - SP edition - sicnarf - 2009-03-17

Well, you already have a door selected. There are two doors you can pick initially, then switch, which will yield a chinchilla. However, there's only one door you can select and stick with to yield a cute chinchilla. That's the main factor in this, not the fact that there are two doors remaining.

Think of the chinchillas.

Alloy Wrote:I just copy/pasted from your post D:

Really, I wasn't sure what particular adjective you used...

Yeah I know, just going along with me being an ass.

Monty Hall problem - SP edition - Alloy - 2009-03-17

sicnarf Wrote:Yeah I know, just going along with me being an ass.

Why you little...
[Image: al-hakim-shaking-fist.jpg]

Monty Hall problem - SP edition - Cancambo - 2009-03-17

sicnarf Wrote:Well, you already have a door selected. There are two doors you can pick initially, then switch, which will yield a chinchilla. However, there's only one door you can select and stick with to yield a cute chinchilla. That's the main factor in this, not the fact that there are two doors remaining.

Think of the chinchillas.



Yeah I know, just going along with me being an ass.
You don't know any of this though. Saying you are picking two doors is, well, not known to you. y0y0y0y0shi0 pretty much summed up my whole point of view. Minus his completely off-tangent story involving Nexon being the possible game show host.

Off-topic: What the hell is this-Poast? Meme I don't know about?

Monty Hall problem - SP edition - Russt - 2009-03-17

Poast Poast
Site Suggestions plx

Anyway, as Kaasoperson says, imagine there are 100 doors. You pick one. Then he opens 98 of the doors that you didn't pick, and you see that they all had pineapples behind them.

Switch or stay?

Monty Hall problem - SP edition - Cancambo - 2009-03-17

Russt Wrote:Poast Poast
Site Suggestions plx

Anyway, as Kaasoperson says, imagine there are 100 doors. You pick one. Then he opens 98 of the doors that you didn't pick, and you see that they all had pineapples behind them.

Switch or stay?

Well since now you have 2 doors left and 2 doors to pick from it doesn't really matter. You can't choose the other doors, you only have 2 choices now. This still makes the most sense to me.

Monty Hall problem - SP edition - Kaasoljoyyx - 2009-03-18

TehMatt Wrote:You are given the option to select again, thus changing the problem.

Wrong. You are not given the option to select a door from 2 doors, you are given the option to switch to the remaining door.

With that logic, look at the other situation I posted

You have 100 doors and you have a 1% chance of being right. I add 100 more doors. Does that make the door you chose 0.5% now? No.

Monty Hall problem - SP edition - Cancambo - 2009-03-18

Kaasoljoyyx Wrote:Wrong. You are not given the option to select a door from 2 doors, you are given the option to switch to the remaining door.

With that logic, look at the other situation I posted

You have 100 doors and you have a 1% chance of being right. I add 100 more doors. Does that make the door you chose 0.5% now? No.

I looked at it this way, you have 2 choices. Stay or switch. Effectively, you are given the option to choose from the remaining doors.

So you have 3 doors. 2/3 chance of getting a Pineapple, 1/3 chance getting the Chin.

Now, minus a pineapple since he eliminated a door. So now you have two choices. Stay or switch. 2 choices, 2 doors.

If you choose to stay you have a 50% chance of getting the Chin and if you switch yo have 50% chance of getting the Chin. The last door was eliminated, it doesn't count any longer since you can't choose it.

With your other situation if you stay you have a 1% chance still if you switch then your chance will be .5%. A little different. Except if you knew where the first 100 doors were, one containing the Chin, you could effectively make yourself still have a 1% chance of winning. Only if you knew which the new doors were, of course.

Again, sorry if I am seeming disagreeable. I hope I am not as if you are talking with a brick wall.

Monty Hall problem - SP edition - Tir - 2009-03-18

TehMatt Wrote:With your other situation if you stay you have a 1% chance still if you switch then your chance will be .5%. A little different. Except if you knew where the first 100 doors were, one containing the Chin, you could effectively make yourself still have a 1% chance of winning. Only if you knew which the new doors were, of course.

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.

TehMatt Wrote:Now, minus a pineapple since he eliminated a door. So now you have two choices. Stay or switch. 2 choices, 2 doors. [...] The last door was eliminated, it doesn't count any longer since you can't choose it.

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.

Monty Hall problem - SP edition - Kaasoljoyyx - 2009-03-18

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.