Puzzle - which door?

[Paul Adams]

Okay, for those that do not know this puzzle, the answer is below, but you'll need to highlight between the smiley faces to see it...




It may surprise you to hear that you are more likely to get the PC if you switch doors.

As for the reasoning behind it, I'll put that in a follow-up post at some point <gd&r>

[silent ben]

Switch, because i saw it on "how2" once

ben

[Paul Adams]

Okay, here's the reasoning behind the answer, hidden again so you'll need to highlight between the smilies (unless this just arrived in an email and you can read it anyway )...



This is based on the "Monty Hall" problem. (Google it for full details, there are loads of sites about it.)

Basically, the best way I can come to terms with it is to look at it this way:
At the start you have a 1/3 chance of picking the right door, thus a 2/3 chance of having picked the wrong one.
Whether you pick the right one or one of the two wrong ones, a wrong one from the other set you did not pick is resolved - it is guaranteed that probability of the other door you did not pick increases in the probability of being correct.
You have gained knowledge about the pair you did not pick, and one is resolved to be definitely incorrect.

It's enough to make your head hurt

[Stewart]

Yeah, but now I have that telling me the answer, and it makes sense, but the fact that with one door removed, its a 50% chance if you switch and a 50% chance if you don't.

Thats what my brain keeps telling me. A the end of the day, no matter how you dress it up, with one door removed, there are two doors left. With two doors left and one choice, you have, and always will have, a 50% chance of getting the right door.

2 doors, 1 choice = 50% chance. My ears hurt, from thinking about how something going on before this, no matter what it may be, could change this, so longas it doesn't invovle you bveing able to see whats behind more than one door, which has the item you don't want to win behind it.

[Iain]

Here's an alternative explanation that may help


You are on a game show and are asked to pick between three doors. Behind one of the doors is a million-dollar prize, behind the other two is either a date with Linda Tripp (for the boys) or Ken Starr (for the girls). I am the host and I know what is behind each door and will obviously try to keep you from winning.
I ask you to choose a door and, let’s say, you choose Door #2. Now, I open Door #1, revealing none other than Ken Starr. At this point, I ask if you would like to switch your choice. Here is the question:

1. Is it better to switch?
2. Is it better to stay with your original choice?
3. Does it make a difference in your ability to win?

If you are similar to 99.99% of the sapient human populace, you’ve chosen either the second or third option. So dust off the microphone and dark glasses, baby — you’re off on a date with Linda Tripp.

Yes, it’s hard to take, but it’s true — the best choice is to switch. Think about it. Do the math:

1. The prize is behind Door #3. You first choose Door #2 and so I open Door #1 (I will certainly never open the door with the prize behind it). Out pops Ken Starr. If you stay with Door #2, you lose, if you switch to Door #3, you win.
2. The prize is behind Door #1. You first choose Door #2. I know where the prize is, so this time I open Door #3. Out pops Linda Tripp. If you stay with Door #2, you lose, if you switch to Door #1, you win.
3. The prize is behind Door #2. You first choose Door #2. I know where the prize is, so I open either Door #1 or #3. If you stay with Door #2, you win, if you switch, you lose.

Result? If you switch, you win two out of three times. If you stay, you win one out of three times.

[Paul Adams]

Az are betterer at explainerizing than what me is.

Meanwhile, I'm still trying to figure out that March 1st 1980 twin thing LOL.
(I'm thinking something to do with leap years but I can't think why!)

[Stewart]

Right Az, coat off, outside.

Last one standing is correct. We'll settle this in the right and proper way.

[Devilbod]

The twins one could be : Different time zones i.e. born on the opposite side of the globe. Technicaly the same day by individual time zones but differernt day from the orginal time zone
Dont know why I put that in white....

As to the other thing - when you tie it into a third person who's trying to influence then it may be better to switch, but at a basic maths level I have to agree with vaul - it's 33.33% (1 in 3) with three doors with 2 it 50% (1 in 2).

This is old logic twisting by Mr. McTwistys from Twistyland, Confusevile. Thats the problem with logic there always some silly bugger who figures out how to prove the correct answer wrong by logical means.

[Stewart]

As to the other thing - when you tie it into a third person who's trying to influence then it may be better to switch, but at a basic maths level I have to agree with vaul - it's 33.33% (1 in 3) with three doors with 2 it 50% (1 in 2).

Aha!

'Ave it Einstien! 'Ave it Hawkings! 'Ave it Archimedes! 'Ave it Babbage!

I rule you all.

And don't try to play the sympathy card Hawkings; it won't wash, my son.

[Zathras]

Tell you what Vaul, we'll settle this mano a mano. Want to put your money where your mouth is? Okay, one grand, let some third party such as Dave or Saracen play quizmaster so it's fair, we'll run 100 trials through this game. I'll switch each time, you stick each time. Whoever gets the most correct doors and prizes gets to keep the pooled money. Up for it?

[Stewart]

I still fancy the fighting option, to be honest. Or perhaps we could settle it over a game of Mouse Trap?

[Devilbod]

Right

Three doors, one prize so to make this easy you pick door 1 that gives three possible outcomes:

prize is behind door 1. You win.
prize is behind door 2. You lose.
prize is behind door 3. You lose.
chance of winning 1 in 3 = 33%

Reguardless of the choice the orignal chance of winning is one in 3.

And this bit's for AZ your maths is wrong your point (3.) is infact two seperate possible loosing situations if you switch and two possible chances to win if you stick.
I'll explain buy breaking this down into all possible outcomes:
So lets say your told to pick a door and and then one of the wrong doors will be opened and you'll have the chance to change.

You pick door 1.

Door 2 is opened, you stick to door 1, the prize is behind door 1. You win.
Door 2 is opened, you stick to door 1, the prize is behind door 3. You lose.
Door 3 is opened, you stick to door 1, the prize is behind door 1. you win.
Door 3 is opened, you stick to door 1, the prize is behind door 2. You lose.

Door 2 is opened, you switch to door 3, the prize is behind door 1. You lose.
Door 2 is opened, you switch to door 3, the prize is behind door 3. You win.
Door 3 is opened, you switch to door 2, the prize is behind door 2. You win.
Door 3 is opened, you switch to door 2, the prize is behind door 1. You lose.

So out of 4 possible outcomes with sticking you can win twice 2 in 4 chance of winning which is the same a 1 in 2 chance = 50%
Swtiching 4 outcomes result in the 2 win, 2 in 4 which again is the same as 1 in 2 = 50%.

Defy my logic I dare you. Muhahahahaha

[Stoo]

yup, I'd stick as well..

I couldn't give a crap about the probability, you either win or you don't - so it was only ever a 50/50 shot

[Stoo]

"There are two identical boys born on 1st March 1980 to the same biological parent and yet they are not twins ? How is this possible ?"

Human cloning?

[fondie]

Defy my logic I dare you. Muhahahahaha

you are changing the probability of where the prize was originally placed. in each 'paragraph' you have 2 examples of the prize being in box 1 to only 1 of each of 2 and 3.

first you must fix the prizes location, then work out the outcomes from the players choices.

for example's sake, the prize is in box 1

if you stick :-

you choose box 1 and stick, you win!
you choose box 2 and stick, you lose!
you choose box 3 and stick, you lose!

1/3

if you change :-

you choose box 1 and change, you lose!
you choose box 2 and change, you win!
you choose box 3 and change, you win!

2/3

obviously the same outcomes if the prize is in either box 2 or 3 as well.

[Devilbod]

I was trying to show that if you take all the possible outcomes provides a different answer and more accurate, with your example:

you choose box 1 and stick to box 1. you win!
you choose box 1 and change to box 2. you lose!
you choose box 1 and change to box 3. you lose!

you choose box 2 and change to box 1. you win!
you choose box 2 and stick to box 2. you lose!
you choose box 2 and change to box 3. you lose!

you choose box 3 and change to box 1. you win!
you choose box 3 and change to box 2. you lose!
you choose box 3 and stick to box 3. you lose!

Sticking gives you 1 in 3 chance.33%
Switching gives you 2 in 6 chance.33%
Overall gives you 3 in 9 chance. 33%
It's all the same.

In the door example the odds are 50% because a door is removed.