ok then, since we've all got our brains warmed up on stu's thread, lets see what you guys make of this one...

 

a contestant on a gameshow is asked to pick one of three mystery doors. behind one of the doors is a prize (don't know what but i'm sure its dead good!)

 

after the contestant picks one of the doors, the host then opens one of the doors that wasn't picked to reveal that there is nothing behind it. the contestant is then offered the choice to stay with his/her original choice or switch to the other one.

 

does the contestant have a better, worse or equal chance of winning if he/she chooses the other box? discuss.

Originally posted by Peter Hood

O.K. I'm deffinitely wrong, does this problem hinge around the fact

that the host knows which box is empty i.e. if he removes any box randomly (full or empty) the odds of picking the right one when there are two left are 1/2 and the advantage of changing your original pick is lost?

My head hurts now too! But I think that it would make no difference whether you change or not, but you'd win 1/3 of the time in each case.

Originally posted by james300

This is flawed in so many ways!

 

Each one has a 1/3 chance of being right.

A-1/3

B-1/3

C-1/3

You chose A, 2/3 chance that its B or C.

You are told its not B.......A and C are now left, just because you've chosen one, why does that make it 1/3 and the other 2/3 ?? Please tell me............please!

If it's B, the host takes C away and you switch to B. If it's C, the host takes B away and you switch to C. So if you were wrong in the first place (which happens 2/3 of the time), then you'll automatically be right by changing.

Originally posted by ste

ok then what if you put it this way.

 

three choices, A B C, one is right, two are wrong.

 

you choose A, it has a 1/3 probability of being write.

 

you are left with a 2/3 chance that it is B or C

 

you are told its not B so that means there is a 2/3 chance it is C

 

2/3 is better odds that 1/3 so you change your answer to C

 

its right, i promise :D

Totally agree with you but you didn't say does he have better odds you said "does the contestant have a better, worse or equal chance of winning if he/she chooses the other box" which means that he has the same CHANCE of winning a prize if he swaps.

Originally posted by ste

you are told its not B so that means there is a 2/3 chance it is C

 

You also have 2/3 chance that is it A :confused: .

 

WTF !!!!!

Originally posted by Dave Marley

OK, I'll spell it out! :p

 

There are only 9 possibilities:

 

Pick A. A is right. Host removes B or C. If we don't change, we win. If we change, we lose.

Pick A. B is right. Host removes C. If we don't change, we lose. If we change, we win.

Pick A. C is right. Host removes B. If we don't change, we lose. If we change, we win.

 

Pick B. A is right. Host removes C. If we don't change, we lose. If we change, we win.

Pick B. B is right. Host removes A or C. If we don't change, we win. If we change, we lose.

Pick B. C is right. Host removes A. If we don't change, we lose. If we change, we win.

 

Pick C. A is right. Host removes B. If we don't change, we lose. If we change, we win.

Pick C. B is right. Host removes A. If we don't change, we lose. If we change, we win.

Pick C. C is right. Host removes A or B. If we don't change, we win. If we change, we lose.

 

So, if we don't change, we win 3 out of 9.

If we do change, we win 6 out of 9.

 

Surely you can't argue with that, can you?

 

I sure can Argue with that.

There are not 9 possibilities because we have been told that B is empty, so why the hell would we pick it FFS !!!!

The question was if B was empty then what. so Take B out of the equation.

Originally posted by smw1

You also have 2/3 chance that is it A :confused: .

Why? You had a 1/3 chance of it being A before, so what changed to increase the chances of it being A?

Originally posted by smw1

I sure can Argue with that.

There are not 9 possibilities because we have been told that B is empty, so why the hell would we pick it FFS !!!!

The question was if B was empty then what. so Take B out of the equation.

I'm not with you.

 

If you pick A, B & C are left, and if you're told B is empty, then you choose C. The chance of it being A in the first place was 1/3, so the chance of it being B or C was 2/3.

 

If you are told that C is empty instead, then you'd pick B.

 

You're effectively being allowed to pick whichever of B or C isn't empty, so the chance of you getting it right is 2/3 if you swap, but only 1/3 if you don't.

Originally posted by Dave Marley

Why? You had a 1/3 chance of it being A before, so what changed to increase the chances of it being A?

 

You are the one who said 2/3 chance it's C right ?

 

Ok you had 1/3 chance it being A, 1/3 chance it being B and 1/3 chance it being C.

 

So you tell me what now gives C a 2/3 chance ?.

Originally posted by Dave Marley

If you pick A, B & C are left, and if you're told B is empty, then you choose C. The chance of it being A in the first place was 1/3, so the chance of it being B or C was 2/3.

The chance of it being C in the first place was also 1/3 so again what has changed ?. The fact that B is no longer an option ?. If this is the case then A's chances will also increse bringing us back to the 50/50 senario.

 

If you are told that C is empty instead, then you'd pick B.

no I wouldn't, I'd stick with A. Because the only thing that has changed when you were told that B is not correct is that you now know it's either A or C.

 

You're effectively being allowed to pick whichever of B or C isn't empty, so the chance of you getting it right is 2/3 if you swap, but only 1/3 if you don't.

No you are not, this is where you are getting confussed. You only get the option of swapping after you are told that B is empty. Therefore you only have 2 choices. If you were told you could swap before knowing B was empty then you would have the choice of B or C of which each would still be 1/3.

Originally posted by smw1

You are the one who said 2/3 chance it's C right ?

 

Ok you had 1/3 chance it being A, 1/3 chance it being B and 1/3 chance it being C.

 

So you tell me what now gives C a 2/3 chance ?.

You choose A. The host removes either B or C - whichever one was empty, leaving the other.

 

So, if A was right, and you change your mind, you'll lose. There was a 1/3 chance of A being right.

 

If *either* B or C was right, by changing you'll get the right one of the 2 as the other has been removed, so that's a 2/3 chance.

Dave's right. Here's another way of answering the question:

If you swap, you win when your initial guess was wrong, probability 2/3.

If you stick, you only win when your initial guess was correct, probability 1/3

 

And more explanation is here

 

You can even play the game here

 

My brain hurts - can I go down the pub now?

 

:D Gio

Originally posted by smw1

Therefore you only have 2 choices. If you were told you could swap before knowing B was empty then you would have the choice of B or C of which each would still be 1/3.

OK, take this point and assume that we don't change. The chances of being right are 1/3. Correct?

 

What if we change to one of the others? Still 1/3. Agreed?

 

What if we decide to change, and then the host removes whichever of B or C is empty before we have a chance to choose one of them? We have no choice at that stage, and assuming that A was wrong, we *must* win. Therefore, by changing, we have a 2/3 chance of being right. The only way we could lose is if A was right in the first place, which is a 1/3 chance.

no it's Wrong !!!!

 

The chances of it being right are 1/3 up until the point where B is eliminated. Then A and C become 1/2.

:D

Originally posted by Gio

Dave's right. Here's another way of answering the question:

no he's not and now you are wrong as well. :p

 

 

And as for those links you put up, you can't do this on a program because it can change the option you picked and you'll never know. When we meet up myself and James will explain this to you all.

 

Then and only then will you understand LMAO !!!!

OK, last try! :D

 

If I choose A and I don't change, I'll only win if I guessed right in the first place. So 1/3 chance.

 

If I decide to change, I'll win *unless* my original guess was right. So a 2/3 chance.

My head is bursting. Looked on the net and it's obviously a pretty well known puzzle. Although it's dependant on the host knowing what's behind each door,it seems that the explanation Dave gave in his last post is correct.

I admit to not being able to get my head round it, has anyone got a mathematical proof(forumla) for it. Card school and the ABC sequences cancelled.

LOL!!

 

Geeks... :p

Bollox :D

Originally posted by Dave Marley

OK, last try! :D

 

If I choose A and I don't change, I'll only win if I guessed right in the first place. So 1/3 chance.

 

If I decide to change, I'll win *unless* my original guess was right. So a 2/3 chance.

 

I'm sorry, I cannot follow the logic in that! LMFAO. I think its how you work it out in your own head that gives you your answer.

 

"If I decide to change, I'll win *unless* my original guess was right. So a 2/3 chance" ........so say he takes one away and you haven't guessed yet.........he forgot to ask you!.........if you could just tell me which one is 1/3 likely to be right and which one is 2/3???? ...............you can't.............so why does it make any difference if you guess before one gets taken away??????

This is driving me nuts............I'm off. LOL

:D

Originally posted by Robtor

LOL!!

 

Geeks... :p

You're not at school any more, young man. It's perfectly OK to have an intellectual debate! :D

Originally posted by james300

........so say he takes one away and you haven't guessed yet.........he forgot to ask you!.........if you could just tell me which one is 1/3 likely to be right and which one is 2/3???? ...............you can't.............so why does it make any difference if you guess before one gets taken away??????

If he takes one away before you guess, then it's a simple 50/50, as it's just a choice of 2.

 

If you guess first, and *then* he takes one away, he can't take away the one you originally chose, so that alters things.

 

Can't you 2 just try it out yourselves, and see if there's a difference between always changing your mind, and always sticking with your original choice? It really does work. Honestly! :p

 

:D

Originally posted by Dave Marley

If he takes one away before you guess, then it's a simple 50/50, as it's just a choice of 2.

But this is exactly what you are doing. He is taking one away and then saying do you want to guess again !!!. Therefore you got a 50/50 chance.

 

And it's at this point that you are asked "does the contestant have a better, worse or equal chance of winning if he/she chooses the other box"

 

I wish I could change the font size of my smilie :D LOL:D

Originally posted by smw1

But this is exactly what you are doing. He is taking one away and then saying do you want to guess again !!!. Therefore you got a 50/50 chance.

 

And it's at this point that you are asked "does the contestant have a better, worse or equal chance of winning if he/she chooses the other box"

The point is that you chose *before* he took one away. That makes the 2 situations totally different and it isn't a simple 50/50 chance.

 

Take the example we had before. Choose 15 letters - 5 As, 5 Bs and 5 Cs in whichever order you like. I then choose A every single time, so you'll be removing B or C each time. If I don't change my mind, I'll get 5/15 right. If I do change my mind, I'll get 10/15 right! Simple! :D

1. A B C

 

2. A B C

 

3. A B C

 

4. A B C

 

5. A B C

 

6. A B C

 

7. A B C

 

8. A B C

 

9. A B C

 

10. A B C

 

11. A B C

 

12. A B C

 

13. A B C

 

14. A B C

 

15. A B C

A C C C A C A B B B A A B B C

 

The red letter has the has the prize, using Dave's technique

 

1. Lose

2.Win

3.win

4. Win

5. lose

6. Win

7. lose

8. Win

9. Win

10. Win

11. lose

12. Lose

13. Win

14. Win

15. Win

 

Try it, pick a letter, remove one 'empty' letter change your original choice for the one left (ie pick one of two a 50/50 chance)

and youll find it's not 50/50.

You need to pick the same letter every time but that's only to eliminate the randomness, the principle applies.