Let's Make a Deal Puzzler

[Graham Smith]

I ran across this the other night and thought that some people here might find this interesting.

Imagine you are on "Let's Make A Deal". You are shown three doors and told that behind one is a real car and behind the other two are toy cars. You are asked to pick a door and you pick #1.

The host (who knows what's behind what door), walks over and opens door #3 to reveal a toy car. He now offers you a chance to change your mind.

Should you stick with door #1, switch to door #2, or doesn't it matter?

Explain your answer. I'll get back later with the right answer.

------

And no fair looking up the answer on the internet!!!

Edited July 1, 2010 by Graham Smith

[spanky]

Change your answer gives you better odds. I remember reading that somewhere but I don't remember the reasoning.

Edited July 1, 2010 by spanky

[g34]

Switch to door number 2. When you picked door number 1 there was a 2/3rds chance you were wrong. When the host opens one of the doors with the fake car there is still a 2/3rds chance you were wrong, which gives door number 2 a 2/3rds chance of being right.

[AlamoShooter]

I ran across this the other night and thought that some people here might find this interesting.

Imagine you are on "Let's Make A Deal". You are shown three doors and told that behind one is a real car and behind the other two are toy cars. You are asked to pick a door and you pick #1.

The host (who knows what's behind what door), walks over and opens door #3 to reveal a toy car. He now offers you a chance to change your mind.

Should you stick with door #1, switch to door #2, or doesn't it matter?

Explain your answer. I'll get back later with the right answer.

------

And no fair looking up the answer on the internet!!!

He just said behind 'One' is a real car

[Corey]

I've seen the movie "21" so yea i know the answer and the reasoning.

"i'll switch doors and thanks for the extra 33%"

[sperman]

I've heard the explanation, but I think it's all statistical double speak. Once we've eliminated door #3, it's still a 50/50 chance of which door it's behind. I'm not buying that switching doors increases the odds.

[Corey]

DISCLAIMER: DO NOT CLICK THIS LINK UNLESS YOU WANT TO KNOW HOW THE MATH IS DONE

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

yea, i know its wikipedia, but the math on it is right. its probability at its best.

[Graham Smith]

It would seem that several people have already seen this. I ran across it quite a while back but was reminded it when I caught part of a rerun of the movie "21" the other night.

Here's the logic. When you are first asked to choose, there are three doors and you have a 1 in 3 chance of picking the right one. That also means that there is a 2 in 3 chance of it being behind one of the other two doors.

Now, the host is going to open one of the doors you did not choose. This leads a lot of people to the assumption that you now have a 50/50 chance of being right. But that ignores the original odds which have not changed. There is still a 1 in 3 chance that your original choice was right and a 2 in 3 choice that one of the other two doors was right. By opening one of those two doors, the host has essentially transferred that 2 in 3 chance to the door he did not open. So you double your chances of being right by switching to the door he didn't open.

Clear as mud?

[-JQ-]

That kind of makes sense...but like Jamie said. He told you there is a car behind "one".

[badchad]

I've heard the explanation, but I think it's all statistical double speak. Once we've eliminated door #3, it's still a 50/50 chance of which door it's behind. I'm not buying that switching doors increases the odds.

Try it, you'll see. G34 nailed it.

Edited July 2, 2010 by badchad

[Team Amish 1]

Once the third door is identified as "toy car only", that first round of the game is over.

One door is taken out of the equation when the show host offers to let the competitor play the second time around.

The "switching" strategy is based on the assumption that the third door is still of relevance for the second round of the game.

It is not because it was taken out of the equation - new cards were dealt, so to speak.

Apple Game - Round One

Mr. Right and Mr. Wrong and Mr. Indifferent fight over an apple.

Each has a 33% chance of winning.

Apple Game - Round Two

Mr. Indifferent has lost interest and settled for an orange, i.e. his odds are at 0% to win the apple. He practically returned his 33% chance for the other two to split.

Now it's Mr. Right and Mr. Wrong fighting over one apple.

Each has a 50% chance of winning.

I read the wiki-entries but I could not agree. Nice brain-teaser, though.

Edited July 2, 2010 by Team Amish 1

[spanky]

Better explanation for Those still wondering

http://en.m.wikipedia.org/wiki/Monty_Hall_problem?wasRedirected=true

[g34]

Imagine that there were one million doors. There is still only one with a real car. If the host opened all of the doors except for the one you picked and one other one, do you still think it is a 50/50 shot? There is a 1/1,000,000 chance that you picked the right door the first time. Now that all the other doors are opened, there is a 999,999 /1,000,000 chance that it is in the remaining door. I know this is on a much larger scale, but it is the same concept.

[Team Amish 1]

Imagine that there were one million doors. There is still only one with a real car. If the host opened all of the doors except for the one you picked and one other one, do you still think it is a 50/50 shot? There is a 1/1,000,000 chance that you picked the right door the first time. Now that all the other doors are opened, there is a 999,999 /1,000,000 chance that it is in the remaining door. I know this is on a much larger scale, but it is the same concept.

Round One

You pick one door in 1,000,000.

That's a 1/1,000,000 chance.

Round Two

You have two doors left. All others have been eliminated and have odds of 0.00% for the second round. They are irrelevant.

You can make the decision to

A ) stay with door #1

B ) switch to door #2.

In other words, there are two doors. Pick which one you bet on. 50/50 chance

Edited July 2, 2010 by Team Amish 1

[Graham Smith]

It is not because it was taken out of the equation - new cards were dealt, so to speak.

If you were given the option of choosing one door or two, clearly it would be to your advantage to choose two doors. That's essentially what happens here. You chose one door and the host gets the remaining two doors. The host then magnanimously shows you one of his doors and in essence asks you if you now want to trade your one door for his two. The odds of it being behind one of the two doors you didn't choose are still exactly the same as they were the first time because nothing has changed.

What makes this difficult to follow is that in most circumstances random probability is not based on prior events. The revealing of door #3 leads to the belief that the prior choice is a prior event and that the "reveal" somehow "resets" the odds. It doesn't. The only way that this would become a new event is if once door #3 was opened, the contents of doors 1 and 2 were shuffled. Otherwise, they retain the same probability as they did originally.

[Team Amish 1]

I just did some testing. I think I figured it out.

It's the OPTION to make a different decision on the second go-around that increases your odds.

It seems the math for the Wiki entry is based on the assumption that one WILL stick with his first choice. In fact, if one were to always stick with the first decision or always reverse the first decision, that person would always be stuck with a 33% chance of winning.

Since you have the option to either stay with door #1 or switch to door #2, you can either WIN or LOSE. Pick one of two = 50% chance for the second go.

Edited July 2, 2010 by Team Amish 1

[-JQ-]

This is why I hated my stats classes...

[g34]

IF you don't believe the logic, do some testing. Have someone play as the host and you as the contestant with an object hidden under a card or something like that and try it about twenty times. I think you will be surprised at your results.

[bbbean]

Round One

You pick one door in 1,000,000.

That's a 1/1,000,000 chance.

Round Two

You have two doors left. All others have been eliminated and have odds of 0.00% for the second round. They are irrelevant.

You can make the decision to

A ) stay with door #1

B ) switch to door #2.

In other words, there are two doors. Pick which one you bet on. 50/50 chance

I'm with you.

[rvb]

A couple years ago I couldn't get my head around this, and also was convinced it became 50/50. So I wrote a python script with randomize functions to take on the roll of the contestant and host. It then played the came a few thousand times with the contestant keeping his original choice and then a few thousand times w/ the contestant changing his choice. I was surprised when, in fact, the contestant won 1/3 of the time staying with the original, and 2/3 of the time switching...

What I figured out, thanks to coding it up making me look at it closer, was that by switching you only lost if you picked the car the on your first guess, which is a 1/3 chance... hence the 2/3 chance of winning....

-rvb

[Graham Smith]

by switching you only lost if you picked the car the on your first guess, which is a 1/3 chance... hence the 2/3 chance of winning....

That's a good way of putting it. Here's another:

There are three possible situations:

If the car is behind door #1, then the host can reveal either door #2 or #3. If you stick with #1 you win and if you switch you loose.

If the car is behind door #2, then the host can only reveal door #3. If you stick with #1 you loose but if you switch to door #2 you win.

If the car is behind door #3, then the host can only reveal door #2. If you stick with #1 you loose but if you switch to door #3 you win.

In two out of three of these situations, you win if you switch and lose if you don't.

[Team Amish 1]

Thanks for clarifying.

This means that you either pick A(33%) or you pick the winner of B/C (66%).

And I guess if you randomly switch between sticking with your first choice and switching to the other, you'd end up around 50%.

Feels like a lightbulb came on.

Thanks again.

Edited July 2, 2010 by Team Amish 1

[DWFAN]

For anyone still in disagreement, I'll play the game with you, you put $100 bill behind one of the doors. I'll pick one, then you show an empty one and I'll switch my guess EVERY time.

We'll go for 10 rounds and I bet I win more than $500.