What would you do?

[AlexGTV]

A little question for your intelligence:

A game-show host has placed a car behind one of three doors. There is a goat behind each of the other doors.

"First you point toward a door," he says. "Then I'll open one of the other doors to reveal a goat. After I've shown you the goat, you make your final choice whether to stick with your initial choice of doors, or to switch to the remaining door. You win whatever is behind the door."

You begin by pointing to door number 1. The host shows you that door number 3 has a goat. What should you do (assuming you want the best odds of winning the car)?

 

[TB]

Switch. Your chances jump to 2/3.

 

[Famine]

Switch. Your chances jump to 2/3.

This - and I recall we already have a thread for the Monty Hall problem.

 

[lbsf1]

Yep swap doors.

I worked out why in my mind however I'm gonna struggle to get it down on paper, so I won't bother.

 

[ASH32]

I'll keep what I have and I'll win the same amount of times as the people who switch. You effectively have a 2/3 chance anyway, since you know it can't be one of the doors. This isn't really a Maths issue and thinking of it that way just confuses things.

 

[Famine]

I'll keep what I have and I'll win the same amount of times as the people who switch. You effectively have a 2/3 chance anyway, since you know it can't be one of the doors. This isn't really a Maths issue and thinking of it that way just confuses things.

Not really. If you switch you'll win twice as often as if you don't.

When you select initially, you have a 1/3 chance of picking the right one door from three - and a 2/3 chance of being wrong. If one of the doors is opened and shown to be wrong, you still have a 1/3 chance of being right and a 2/3 chance of being wrong, but now there's only two doors. The one you picked is half as likely to be right as the one you didn't.

It's called the Monty Hall problem. There's a good Wiki article about it.

 

[Moglet]

I don't understand how the odds change from 1/3 to 2/3 when you know what's behind door 3. Surely they should changed to 1/2?

What if the 3rd door was removed altogether? It's exactly the same, assuming you don't want a goat.

 

[Famine]

I don't understand how the odds change from 1/3 to 2/3 when you know what's behind door 3. Surely they should changed to 1/2?

They don't change at all - your original choice always has a 1 in (original field size) chance of being right and everything else is a (original field size - 1) in (original field size) of being wrong.

If you had a million doors and picked one, you have a 1 in 1,000,000 chance of being right and a 999,999 in 1,000,000 chance of being wrong. If all but one wrong door and your original choice were opened, you'd still have the same odds - your original choice is still 1 in 1,000,000 to be right and the remaining door is still 999,999 in 1,000,000 to be right.

The odds are set when you make your initial choice. They don't change at the reveal.

 

[Moglet]

Ah, so the fact that those others have been 'revealed' wouldn't change it, but removing them completely would? I guess that's the difference between revealing and removing. Or I'm totally wrong and confused already.

 

[Famine]

You wouldn't need to see that the "wrong" doors are wrong - just that they are - so they could be removed.

In either case the odds are set the moment you make your guess - you have a 1 in (the number of original options) chance of being right and a (the number of original options -1) in (the number of original options) chance of being wrong. Even with all the alternate wrong options removed, those are still the odds - it just becomes more obvious with a million choices than with three.


Say I have £10 for you if you can guess the number I'm thinking of between one and four hundred. You get one guess. Once you've guessed I'll remove all but one of the other numbers and leave you with a choice between the number you originally picked and the number I've left. Stick or switch?

 

[Moglet]

Well since I don't understand how it works, it would make no difference to me. As far as I'm concerned in that situation I have a 50/50 chance of getting it right as there are now only 2 choices!

Why is there any difference between starting with 2 and starting with 400 but narrowing it down to 2?

My head hurts. This goes against everything I was ever taught or have read.....It's annoying me because I genuinely want to understand it

 

[ASH32]

I am well aware of what the Maths is, however it doesn't quite work like that. When you give me a choice between the two remaining choices, whether I keep what I have or change I am effectively picking again and I'll win the same amount of times as I'll lose in, regardless of the maths in the first place. That's why I said to ignore the maths as it confuses things. I know Maths tell me that I should always change, but I don't agree with it, the same way Maths also tells me that if I times your £10 by 0 you suddenly have £0, which is obviously wrong.

 

[Moglet]

Never mind, I read about it on Wiki and just got it. I was thinking about it the wrong way, this is what made me understand the theory:

On average, in 999,999 out of 1,000,000 times the other door will contain the prize, as 999,999 out of 1,000,000 times the player first picked a door with a goat.

 

[Famine]

I am well aware of what the Maths is, however it doesn't quite work like that.

Err, yes it does.

When you give me a choice between the two remaining choices, whether I keep what I have or change I am effectively picking again and I'll win the same amount of times as I'll lose in, regardless of the maths in the first place.

Err, no you won't.

Your new choice is between an answer you know has a 1 in 3 chance of being right against an answer you know has a 2 in 3 chance of being right. Or if there's 400 choices, between an answer you know has a 1 in 400 chance of being right against an answer you know has a 399 in 400 chance of being right. Or if there's a million choices, between an answer you know has a 1 in 1,000,000 chance of being right against an answer you know has a 999,999 in 1,000,000 chance of being right.


What are the chances of you guessing, correctly, a number I've picked between 1 and 400? They're 1 in 400. What are the chances of you guessing it incorrectly? They're 399 in 400.

Say you guess a number. You know you have a 1 in 400 chance of being right. I now remove 398 other answers, all of which are wrong, to leave you with just two numbers - your guess and another number. You still have a 1 in 400 (0.25%) chance of your original choice being right, which means the remaining odds of correctness - 399 in 400 (99.75%) - are with the other number.

That's why I said to ignore the maths as it confuses things.

It doesn't - it makes it quite simple.

In the Monty Hall problem, people who change their answer win twice as often as those who stick with their answer. The maths of the Monty Hall problem says that you are twice as likely to win by changing your answer as you are by sticking to it. Maths bears out reality.

I know Maths tell me that I should always change, but I don't agree with it, the same way Maths also tells me that if I times your £10 by 0 you suddenly have £0, which is obviously wrong.

What?

If you have no ten pound notes, your wealth of ten pound notes is zero.

 

[AlexGTV]

I know Maths tell me that I should always change, but I don't agree with it, the same way Maths also tells me that if I times your £10 by 0 you suddenly have £0, which is obviously wrong.

This wins the Razzie math award!

 

[GT_Prologue5]

My brain hurts but I still understand what's being said.

 

[Villain]

I understand why it is thought to be more logical to switch doors, but you are essentially going from a 1/3 decision to a 1/2 decision. I voted that it doesn't matter. I remember this from maybe 10 years back or so when my math teacher had us do it and we came out to a stalemate. Some people think you should always switch, and some people don't.

The fact is, you start out with a 1 in 3 chance of being right, and then once that third door is removed, you now have a new flip-of-the-coin 50-50 choice. The past decision is no longer relevant. You can stay or switch.

 

[SRV2LOW4ME]

I would stop watching the movie "21" and go do something productive.

 

[Famine]

I understand why it is thought to be more logical to switch doors, but you are essentially going from a 1/3 decision to a 1/2 decision. I voted that it doesn't matter. I remember this from maybe 10 years back or so when my math teacher had us do it and we came out to a stalemate. Some people think you should always switch, and some people don't.

The fact is, you start out with a 1 in 3 chance of being right, and then once that third door is removed, you now have a new flip-of-the-coin 50-50 choice. The past decision is no longer relevant.

You don't have a new flip - nothing changes. You have a 2/3 chance of being wrong.

Look, you can prove it to yourself in Excel (other calculation software packages are also available).

1. In column 1, generate a sequence of 100 random numbers between 1 and 3. This is your first choice.
2. In column 2, generate a sequence of 100 random numbers between 1 and 3. This is where the car is.
3. In column 3, type "Stick". This is your choice to stick with the original choice.
4. Tot up the number of times column 1 and column 2 are the same. You stuck with the original choice, the car was behind that door, so you win the car. This will be in the order of 33/100.
5. In column 3, type "Change" (regenerate the first two columns if you wish). This is your choice to change with from original choice to the unrevealed door you didn't choose.
6. Tot up the number of times column 1 and column 2 are the different. You changed your original choice, the car was behind the new door, so you win the car. This will be in the order of 66/100.

Changing will result in winning the car twice as often as sticking, because the chances of you being wrong initially are twice as high as the chances of you being right and the field of (doors you didn't pick) is twice as likely to contain the car.

Do it with ten numbers/doors instead. The result is the same, only you're now nine times as likely to win by switching. Do it with a hundred numbers/doors instead. The result is the same, only you're now ninety-nine times as likely to win by switching. Do it with a million numbers/doors instead. The result is the same, only you're now nine hundred and ninety-nine thousand, nine hundred and ninety-nine times as likely to win by switching.


Please read the Wiki article on the Monty Hall Problem, so I don't have to explain it again.

 

[lbsf1]

Famine. What he was saying was irrelivant of the switching doors.

At first you have 1/3 chance, now a door is removed from the equation completely so there are 2 left.

Now one is right and one is wrong so If you make a completely new choice now you have a 1/2 chance.

This is completely sepperate from the the whether it is better to switch or not.

 

[Villain]

I've read that Wiki before and I think it's fascinating. No, you don't have to explain yourself again (you never had to in the first place).

This all reminds me of watching the WSOP and seeing players get all furious when they lose out on those 70/30 rivers and ranting about how they were the smart ones when they made their decision to call.

By the way, I don't gamble.

Famine. What he was saying was irrelivant of the switching doors.

At first you have 1/3 chance, now a door is removed from the equation completely so there are 2 left.

Now one is right and one is wrong so If you make a completely new choice now you have a 1/2 chance.

This is completely sepperate from the the whether it is better to switch or not.

This 👍

 

[TB]

Give this a try.

I just got 2 out of 10 right by staying with my first door and 6 out of 10 by switching, very close to the expected results.

 

[Famine]

Famine. What he was saying was irrelivant of the switching doors.

At first you have 1/3 chance, now a door is removed from the equation completely so there are 2 left.

Now one is right and one is wrong so If you make a completely new choice now you have a 1/2 chance.

This is completely sepperate from the the whether it is better to switch or not.

This 👍

... is wrong. You are not generating a new independent field. You are playing the same one.

It's also nothing to do with Poker.


I'm thinking of a number. The number is between 1 and 5,000. You have one guess at that number. When you have made your guess, I will remove 4,998 wrong answers and tell you to choose between the number you have picked and the number that is left. If you stick to your choice, you're putting your faith in a 1 in 5,000 chance of having guessed it right first time. If you change your choice, you're putting your faith in a 4,999 in 5,000 chance of having guessed it wrong first time.

Make your guess.

 

[Foglight]

I understand what's happening here, but am a little confused on the process. If the host never opened up the other door does this still apply? Or does the host knowing where the car is change the maths?

Edit: This only works because the host and people behind the scenes know where the car is. The probability could even be 0% in my books, as there are variables outside of math involved. Edit #2 (Outside of math - In the sense that the show "could" control the outcome)

 

[Max_DC]

Ok I read through the wikipedia article... a little tricky. In reality you never know the intentions of the moderator or the game show producers and what not. Reduced to a a mathematical statistics game changing the door seems to make sense.
It is however a very artificial situation that you won't find outside of some game (-show), so you better don't add that logic to your everyday problem-solving mechanisms or you might end up with significantly less right decisions from now on

 

[TB]

Make your guess.

Forty. Two.

If the host never opened up the other door does this still apply? Or does the host knowing where the car is change the maths?

If a wrong door isn't revealed, the 1 out of 3 factor still applies and then it's just a crap shoot if you want to change or not - 1 out of 3 is 1 out of 3 no matter how you shake it.

Edit: This only works because the host and people behind the scenes know where the car is.

How does them knowing where the car is affect your decision, short of showing you where it isn't?

 

[Stevisiov]

Oh dear.

That's why I said to ignore the maths as it confuses things. I know Maths tell me that I should always change, but I don't agree with it, the same way Maths also tells me that if I times your £10 by 0 you suddenly have £0, which is obviously wrong.

Obviously wrong? You had better brush up on your maths.

0 x 10 = 0 That is mathematically AND logically correct.

Using the 10 pound note example:

If I have 4 examples of 10 pound notes in my possession (and no other currency), you would say I have 40 pounds worth of notes.

Expressed Mathematically it reads:

4 x 10 = 40

If I have 0 examples of 10 pound notes in my possession (no 10 pound notes and no other currency for that matter), you would say I have 0 pounds worth of notes.

Expressed Mathematically it reads:

0 x 10 = 0

------

It really is that simple.

The Monty Hall paradox is just as correct logically, and mathematically as the examples of maths and logic above. Although it has to be said that the Monty Hall Paradox is a little counter-intuitive.

 

[Foglight]

Forty. Two.

If a wrong door isn't revealed, the 1 out of 3 factor still applies and then it's just a crap shoot if you want to change or not - 1 out of 3 is 1 out of 3 no matter how you shake it.

How does them knowing where the car is affect your decision, short of showing you where it isn't?

Well not so much of them just knowing where the car is, but they could switch the car from one door to another. In essence they could make the odds of winning the car 0%.

But if you take all variables as a constant across the board including human behavior, yes switching doors will yield the better result.

 

[Villain]

It has everything to do with poker and probability. Poker just has waaay more variables.

You have 5,000 doors. You take away 4,998. There's a 50/50 chance of it being behind the first door.

 

[Stevisiov]

It has everything to do with poker and probability. Poker just has waaay more variables.

You have 5,000 doors. You take away 4,998. There's a 50/50 chance of it being behind the first door.

No!

You are re-setting the variables after the doors have been eliminated.

If once left with two doors, you have a blind fold put on you and the doors are shuffled, you would now have a 50/50 chance as you no longer can reference your initial choice.

However, that is no what is happening, if you stick, you have a 1/5000 chance of being right, if you switch, you have a 1/5000 chance of being wrong.