Game show dilemma

[Libertine]

You have entered a game show, and have even made it to the final! Now, the game show host asks you to pick one of three doors. Behind one of those three doors is a big fucking prize, behind the others - nothing.

So, you make your pick. Now, the game show hosts says he'll make things even easier for you, and opens one of the doors you didn't pick with nothing behind it.
He tells you that you are still allowed to switch doors, if you so desire.

What do you do, and why do you do it? Do you switch or stick with your choice? In other words, what gives you the highest chances of getting the prize?

[BRISK]

I would switch because the probability of getting the big prize increases if you switch.

[Sarah_Jayne]

been watching reruns of lets make a deal?

[SpaceAce]

Always switch doors. Every time they take one away, your odds will improve by switching.

SpaceAce

[Libertine]

Originally posted by sarah_webinc
been watching reruns of lets make a deal?

Reruns of what?

[BRISK]

Originally posted by sarah_webinc
been watching reruns of lets make a deal?

It's more likely that he's an economics student studying game theory.

[Libertine]

Originally posted by BRISK


It's more likely that he's an economics student studying game theory.

Actually, I'm only studying your reactions. I already know the solution

[SpaceAce]

Originally posted by punkworld


Actually, I'm only studying your reactions. I already know the solution

So does BRISK. So do I, for that matter

SpaceAce

[BRISK]

Originally posted by punkworld


Actually, I'm only studying your reactions. I already know the solution

So you're not an economics student?

[Libertine]

Originally posted by SpaceAce


So does BRISK. So do I, for that matter

SpaceAce

Ehm... no you don't. You talked about the multi-stage variant, and in the multi-stage variant, your answer is wrong

[Sarah_Jayne]

Originally posted by punkworld


Reruns of what?

let's make a deal..a rather popular game show in the 80's that was still getting rerun last time I was home. Hosted by Monte Hall. Contestants would get three curtains, boxes, etc and they had to pick one. One would have a grand prize and the rest would have a junk prize like a goat. He would always make people think again when it got down to the last two.

see..same question Let's Make a Deal Demonstration - it's cool, was just having Monte flashbacks from afternoons at the babysitter. She watched it all the time.

[Libertine]

Originally posted by sarah_webinc


let's make a deal..a rather popular game show in the 80's that was still getting rerun last time I was home. Hosted by Monte Hall. Contestants would get three curtains, boxes, etc and they had to pick one. One would have a grand prize and the rest would have a junk prize like a goat. He would always make people think again when it got down to the last two.

see..same question Let's Make a Deal Demonstration - it's cool, was just having Monte flashbacks from afternoons at the babysitter. She watched it all the time.

Ah, so that's where they got the name (Monty Hall Dilemma) from Never heard of the show though.

[SpaceAce]

Originally posted by punkworld


Ehm... no you don't. You talked about the multi-stage variant, and in the multi-stage variant, your answer is wrong

Unles you misstated the question, the answer is to switch your choice of doors if you want the best odds of getting the prize.

When you make a selection of one door out of three, your chance of getting the prize are 1:3. It doesn't matter if they take one door away or blow it up with plastique or leave all three there. Your chance of being correct is 1:3. Then, when the host takes one away you can switch your choice at 50/50 odds or keep your original choice which is still at 1:3. The odds of your first choice don't change when a door is taken away. It seems like you now have a 50/50 chance by not changing your selection but that is false.

SpaceAce

[SpaceAce]

Originally posted by punkworld


Ah, so that's where they got the name (Monty Hall Dilemma) from Never heard of the show though.

Link, please? I'd like to see a website that claims switching does not improve your odds. I could use a hearty chuckle.

SpaceAce

[BRISK]

Originally posted by punkworld


Ehm... no you don't. You talked about the multi-stage variant, and in the multi-stage variant, your answer is wrong

Who's talking about the multi-stage variant?

http://www.comedia.com/hot/monty-answer.html

[Libertine]

Originally posted by SpaceAce


Link, please? I'd like to see a website that claims switching does not improve your odds. I could use a hearty chuckle.

SpaceAce

Switching improves your odds in the single stage version, but in the multi-stage version you should only switch the last time you get the chance to switch.

http://en.wikipedia.org/wiki/Monty_Hall_problem

There is a generalization of the original problem to n doors: in the first step, you choose a door. Monty then opens some other door that's a loser. If you want, you may then switch your allegiance to another door. Monty will then open an as yet unopened losing door, different from your current preference. Then you may switch again, and so on. This continues until there are only two unopened doors left: your current choice and another one. How many times should you switch, and when, if at all?

The answer is: stick all the way through with your first choice but then switch at the very end. This was proved by Bapeswara Rao and Rao.

[Libertine]

Originally posted by BRISK


Who's talking about the multi-stage variant?

http://www.comedia.com/hot/monty-answer.html

Seems to me SpaceAce was:

Originally posted by SpaceAce
Every time they take one away, your odds will improve by switching.

[SpaceAce]

Originally posted by punkworld

Switching improves your odds in the single stage version, but in the multi-stage version you should only switch the last time you get the chance to switch.

http://en.wikipedia.org/wiki/Monty_Hall_problem

I'll read that page in a second but I have to point out that the paragraph you posted didn't say anything about getting multiple chances. Just three doors, take one away, switch or stay.

SpaceAce

[BRISK]

Originally posted by punkworld

Switching improves your odds in the single stage version

The single stage version is what you described in your first post, and that is why I chose to switch. You never said it was a multi stage.

[SpaceAce]

I just saw your other reply. I didn't mean to imply that he would keep taking doors away. By "every time" I meant every time you are faced with the situation in the paragraph you posted.

SpaceAce

[GFED]

001

pick1

0x1

switch=win;stay=lose


001

pick2

x01

switch=win;stay=lose


001

pick3

xy1

switch=lose;stay=win


hmmm... ur right... there is a 2/3 chance of winning by switching and 1/3 chance of winning by staying...

[SpaceAce]

Haha, from the web page punkworld linked to:

"In spite of being an elementary problem, it is notorious for being the subject of controversy about both the statement of the problem and the correct answer."

That seems to be true.

SpaceAce

[Libertine]

Originally posted by BRISK


The single stage version is what you described in your first post, and that is why I chose to switch. You never said it was a multi stage.

Yes, that's what I was originally talking about, and your answer was indeed correct. However, from what SpaceAce posted I thought he was talking about a situation which in fact was the multi-stage variant.
The "Every time they take one away" looked to me like an implication of <i>n</i> doors, of which a number are taken away one at a time.

[GFED]

That's some cool shit I haven't thought about before... I didn't learn that in statistics...

[Libertine]

Originally posted by SpaceAce
I just saw your other reply. I didn't mean to imply that he would keep taking doors away. By "every time" I meant every time you are faced with the situation in the paragraph you posted.

SpaceAce

Yeah, just realized that. A little bit of miscommunication, I guess

[SpaceAce]

Originally posted by punkworld


Yeah, just realized that. A little bit of miscommunication, I guess

Eh, no biggie. I'm going to look into the math behind the multiple-removal scenario.

SpaceAce

[Libertine]

Originally posted by SpaceAce


Eh, no biggie. I'm going to look into the math behind the multiple-removal scenario.

SpaceAce

Oh, that's fairly simple. Since by switching, you actually also switch probabilities, it is best to have as small a chance as possible with your last switch, since that will become the chance of losing instead of winning.

With 4 doors, 2 doors being removed and 2 chances to switch, picking one and only taking the last switch would give you a chance of .75 to get the prize (the door you just left had a .25 chance). However, when you take the first switch, your chances increase to .375, and since you'll be switching again, your next switch will only give you a .625 chance.

Increasing your chances when those chances will later turn into the chances of NOT getting the prize is generally a bad idea

[Triple 6]

i would not switch, and think the host asking if i want to switch my answer is a ploy to make me choose the wrong one, cuz maybe he knows i chose the right one and his producers told him in his earpiece to make me an offer.


ya never know.,

and im not a math genius but i dont see how switching gives you better odds. You are now at a 50:50 chance either way, since you are given a choice to either stay with your original pick or move. The 1:3 odds dont matter because there is no longer three doors... and you have a 50/50 chance of staying with your pick or moving.

[SpaceAce]

Originally posted by punkworld


Oh, that's fairly simple. Since by switching, you actually also switch probabilities, it is best to have as small a chance as possible with your last chance, since that will become the chance of losing instead of winning.

With 4 doors, 2 doors being removed and 2 chances to switch, picking one and only taking the last switch would give you a chance of .75 to get the prize (the door you just left had a .25 chance). However, when you take the first switch, your chances increase to .375, and since you'll be switching again, your next switch will only give you a .625 chance.

Increasing your chances when those chances will later turn into the chances of NOT getting the prize is generally a bad idea

Do you know of any sites with a discussion or explanation that expands on your post? I don't feel like digging books out of boxes, right now. I understand the principle but I would like to see a more in-depth analysis and I can't find anything but by Bapeswara Rao except binary search algorythms.

SpaceAce

[SpaceAce]

Originally posted by Triple 6
i would not switch, and think the host asking if i want to switch my answer is a ploy to make me choose the wrong one, cuz maybe he knows i chose the right one and his producers told him in his earpiece to make me an offer.


ya never know.,

Except that the contestant <I>always</I> gets a chance to switch so it wouldn't be a very good conspiracy

SpaceAce

[Libertine]

Originally posted by SpaceAce


Do you know of any sites with a discussion or explanation that expands on your post? I don't feel like digging books out of boxes, right now. I understand the principle but I would like to see a more in-depth analysis and I can't find anything but by Bapeswara Rao except binary search algorythms.

SpaceAce

Don't really know about any sites giving a more detailed explanation. If you want me to, though, I'll write a more detailed one myself after I get some cigarettes

[myjah]

Originally posted by Triple 6
i would not switch, and think the host asking if i want to switch my answer is a ploy to make me choose the wrong one, cuz maybe he knows i chose the right one and his producers told him in his earpiece to make me an offer.


ya never know.,

and im not a math genius but i dont see how switching gives you better odds. You are now at a 50:50 chance either way, since you are given a choice to either stay with your original pick or move. The 1:3 odds dont matter because there is no longer three doors... and you have a 50/50 chance of staying with your pick or moving.

summed up my thoughts EXACTLY

[Libertine]

<b>The Game Show Dilemma - Multiple Stages</b>

There are 4 doors, behind one of them lies a prize, behind the others nothing. After your initial pick and your first chance to switch, a door with nothing behind it that you didn't pick will be revealed. After your last chance to switch, the prize will be revealed... if you got the right door, you will get it.
This little piece assumes you fully understand the original one-stage dilemma and it's solution.

<b>First scenario: Pick and stay</b>
(your choice is indicated by a *, after each door follows it's chance to contain the price)

You make a choice, namely door A. Here are the chances of each door:
Door A: .25 *
Door B: .25
Door C: .25
Door D: .25

The host removes a door, but you stay with your initial choice, door A:
Door A: .25 *
Door B: .375
Door C: .375
Door D: 0

The host removes another door, but you still stay with your initial choice:
Door A: .25 *
Door B: .75
Door C: 0
Door D: 0

Clearly, this doesn't work well. How about another strategy?

<b>Second scenario: Pick, switch and stay</b>
(your choice is indicated by a *, after each door follows it's chance to contain the price)

You make a choice, namely door A. Here are the chances of each door:
Door A: .25 *
Door B: .25
Door C: .25
Door D: .25

The host removes a door, and you switch to C:
Door A: .25
Door B: .375
Door C: .375 *
Door D: 0

The host removes another door, and you stay with C:
Door A: 0
Door B: .625
Door C: .375 *
Door D: 0

Looking better, but still not great. How about yet another strategy?

<b>Third scenario: Pick, switch and switch again</b>
(your choice is indicated by a *, after each door follows it's chance to contain the price)

You make a choice, namely door A. Here are the chances of each door:
Door A: .25 *
Door B: .25
Door C: .25
Door D: .25

The host removes a door, and you switch to C:
Door A: .25
Door B: .375
Door C: .375 *
Door D: 0

The host removes another door, and you switch again, to B this time:
Door A: 0
Door B: .625 *
Door C: .375
Door D: 0

Now we're getting somewhere. But wait, what if we stay first, and then switch?

<b>Fourth scenario: Pick, stay and switch</b>
(your choice is indicated by a *, after each door follows it's chance to contain the price)

You make a choice, namely door A. Here are the chances of each door:
Door A: .25 *
Door B: .25
Door C: .25
Door D: .25

The host removes a door, and you stay with A:
Door A: .25 *
Door B: .375
Door C: .375
Door D: 0

The host removes another door, and you switch to B this time:
Door A: .25
Door B: .75 *
Door C: 0
Door D: 0

[- Jesus Christ -]

The logic used to justify switching is inccorect.
The only way the odds change is if the door revealed is the one you had picked.


Just cuz you read it on tbhe internet doesnt make it true.

[EZRhino]

WFT

[Libertine]

Originally posted by Triple 6

and im not a math genius but i dont see how switching gives you better odds. You are now at a 50:50 chance either way, since you are given a choice to either stay with your original pick or move. The 1:3 odds dont matter because there is no longer three doors... and you have a 50/50 chance of staying with your pick or moving.

Actually, the 1/3 odds do matter.

There are three options:

1. A has the prize, B doesn't, C doesn't
2. A doesn't, B has the prize, C doesn't
3. A doesn't, B doesn't, C has the prize

Each scenario has a 1/3 chance of being the case.

Now, if you pick the wrong door, there will always be another one that's wrong. The host will always pick that one if that's the case.

Say you have picked A, and scenario 2 is the case... the host will open door C. However, if you have picked A and scenario 3 is the case, the host will open door B. Your initial choice has a chance of 1/3 of being correct. What the host does is not increasing that chance. How could it be? Whether you are right or wrong, he always has a door to open.

What he does is like saying <i>"<b>IF</b> it's scenario 2 or 3 (2/3 chance), it <b>will</b> be scenario 2"</i>, or the opposite ofcourse.
So, he doesn't give you any information on if you are wrong or not, but on what the case would be assuming you're wrong. Rather than altering your 1/3 chance, he eliminates the 50/50 split between the choices you didn't pick.

[fiveyes]

The easiest way to see this is to realize that your initial guess has a 1/3 chance of being correct and a 2/3 chance of it being another door. Switching doors at this point will not affect the odds at all, each door has an equal 1/3 chance.

But, when the host opens a door to reveal what would've been a bad choice, the odds that you chose correctly have not been affected. And, if it is still 1/3, then the other closed door logically has to have a 2/3 chance of being correct.

[Libertine]

Originally posted by - Jesus Christ -
The logic used to justify switching is inccorect.
The only way the odds change is if the door revealed is the one you had picked.


Just cuz you read it on tbhe internet doesnt make it true.

The whole point of switching is that the odds on the door you already have picked do not change, since that door won't be revealed no matter what, and there is always another door to open for the game show host.

However, the odds on the INDIVIDUAL doors that you have not picked DO change, because possibilities are taken out, and they could be revealed. However, the combined odds on the GROUP of doors that you have not picked DO NOT change.... they are just redistributed among less individual doors.

[SpaceAce]

Originally posted by punkworld
<b>The Game Show Dilemma - Multiple Stages</b>

There are 4 doors, behind one of them lies a prize, behind the others nothing. After your initial pick and your first chance to switch, a door with nothing behind it that you didn't pick will be revealed. After your last chance to switch, the prize will be revealed... if you got the right door, you will get it.
This little piece assumes you fully understand the original one-stage dilemma and it's solution.

<b>First scenario: Pick and stay</b>
(your choice is indicated by a *, after each door follows it's chance to contain the price)

You make a choice, namely door A. Here are the chances of each door:
Door A: .25 *
Door B: .25
Door C: .25
Door D: .25

The host removes a door, but you stay with your initial choice, door A:
Door A: .25 *
Door B: .375
Door C: .375
Door D: 0

The host removes another door, but you still stay with your initial choice:
Door A: .25 *
Door B: .75
Door C: 0
Door D: 0

Clearly, this doesn't work well. How about another strategy?

<b>Second scenario: Pick, switch and stay</b>
(your choice is indicated by a *, after each door follows it's chance to contain the price)

You make a choice, namely door A. Here are the chances of each door:
Door A: .25 *
Door B: .25
Door C: .25
Door D: .25

The host removes a door, and you switch to C:
Door A: .25
Door B: .375
Door C: .375 *
Door D: 0

The host removes another door, and you stay with C:
Door A: 0
Door B: .625
Door C: .375 *
Door D: 0

Looking better, but still not great. How about yet another strategy?

<b>Third scenario: Pick, switch and switch again</b>
(your choice is indicated by a *, after each door follows it's chance to contain the price)

You make a choice, namely door A. Here are the chances of each door:
Door A: .25 *
Door B: .25
Door C: .25
Door D: .25

The host removes a door, and you switch to C:
Door A: .25
Door B: .375
Door C: .375 *
Door D: 0

The host removes another door, and you switch again, to B this time:
Door A: 0
Door B: .625 *
Door C: .375
Door D: 0

Now we're getting somewhere. But wait, what if we stay first, and then switch?

<b>Fourth scenario: Pick, stay and switch</b>
(your choice is indicated by a *, after each door follows it's chance to contain the price)

You make a choice, namely door A. Here are the chances of each door:
Door A: .25 *
Door B: .25
Door C: .25
Door D: .25

The host removes a door, and you stay with A:
Door A: .25 *
Door B: .375
Door C: .375
Door D: 0

The host removes another door, and you switch to B this time:
Door A: .25
Door B: .75 *
Door C: 0
Door D: 0

Hehe, thanks, but that's not quite what I meant. Like I said, I understand the principle of what you said. I was looking for more a discussion on the theory of the problem than the actual numbers.

It doesn't matter, anyway. It's an interesting dilemma.

SpaceAce

[SpaceAce]

Originally posted by - Jesus Christ -
The logic used to justify switching is inccorect.
The only way the odds change is if the door revealed is the one you had picked.


Just cuz you read it on tbhe internet doesnt make it true.

No, that's not correct. The percentage does change. If you have 1,000,000,000 doors, you have a 1:1,000,000,000 chance of picking the right door. That percentage already exists, you've picked the door. It doesn't matter what happens to the other incorrect doors, the chance that you picked the right one the first time is still 1:1,000,000,000, period. It was 1 in a billion when you picked it and it's still one in a billion once the other doors are eliminated. Switching improves your odds.

SpaceAce

[Libertine]

Originally posted by SpaceAce


Hehe, thanks, but that's not quite what I meant. Like I said, I understand the principle of what you said. I was looking for more a discussion on the theory of the problem than the actual numbers.

It doesn't matter, anyway. It's an interesting dilemma.

SpaceAce

Well, the numbers kinda sum up the discussion... for me, at least. Then again, I might be a bit freaky in that