Game show dilemma

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

been watching reruns of lets make a deal?

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

SpaceAce

Reruns of what?

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

So does BRISK. So do I, for that matter

SpaceAce

So you're not an economics student?

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

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.

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

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

SpaceAce

Who's talking about the multi-stage variant?

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

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

Seems to me SpaceAce was:

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

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.

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

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...

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

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.

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

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

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

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.

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

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

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

summed up my thoughts EXACTLY

<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

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.

WFT

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.

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.

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.

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

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

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