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 smw1

Why would changing from box A to C increase his chances of winning ?

Haven't you been paying attention! :rolleyes:

 

:D

3 boxes, 1000 boxes, a fookng million boxes.....if it comes down to 2 then it is a 50/50 chance you have chosen the right box. So NO advantage of swapping.........unless its your wife.

Originally posted by james300

3 boxes, 1000 boxes, a fookng million boxes.....if it comes down to 2 then it is a 50/50 chance you have chosen the right box. So NO advantage of swapping.........unless its your wife.

J*sus wept, I give in! :mad:

 

I'm right though! :D

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

Originally posted by Dave Marley

 

I'm right though! :D

 

 

Nope..........not having that and haven't read anything to change my mind! :p

the only other option is box C so the odds of it winning are 2/3

Dave,ste

So using this theory you would be prepared to bet £2 to get £3 back everytime you get the correct box and therfore if this theory is correct you would break even, yes?

Originally posted by ste

congratulations dave (and pete in the end) - it is better to swap. its already been explained pretty well but in a nutshell:

 

three boxes, A B C

 

contestant picks box A, it has a 1/3 chance of winning

 

box B is then shown to be empty but this does not change te odds of the original bet

 

the choice then becomes

 

1/3 that the prize is in box A

or

2/3 that the prize is not in box A

 

the only other option is box C so the odds of it winning are 2/3

 

"THE CHOICE THEN BECOMES"...........the prize is in one of 2 boxes (yes?) you have chosen one of them (yes?) Tell me how 2 things in front of you isn't a 50/50 chance of being correct??????

Once one box is revealed empty the odds change. End of.

Originally posted by Peter Hood

Dave,ste

So using this theory you would be prepared to bet £2 to get £3 back everytime you get the correct box and therfore if this theory is correct you would break even, yes?

Statistically, yes, you would break even.

Originally posted by james300

....the prize is in one of 2 boxes (yes?)

 

nope, its in one of three boxes when you make your choice

 

Originally posted by james300

you have chosen one of them (yes?)

 

yep

 

Originally posted by james300

Tell me how 2 things in front of you isn't a 50/50 chance of being correct??????

 

if you were presented with two boxes in the first place then this would be perfectly true

 

Originally posted by james300

Once one box is revealed empty the odds change. End of.

 

sort of, this changes the odds on one box to 0/3. the first is still 1/3 so the other must be 2/3

Oh goody. I'll post a word doc with fifteen sequences of ABC's with one highlighted to indicate the right one. You bet £2 every round, I'll give you £3 everytime your right. You post which box you want, I'll remove an empty box, post which one I removed, you change or not, post your final answer and we'll see who's up after 15 rounds.

Originally posted by Peter Hood

Oh goody. I'll post a word doc with fifteen sequences of ABC's with one highlighted to indicate the right one. You bet £2 every round, I'll give you £3 everytime your right. You post which box you want, I'll remove an empty box, post which one I removed, you change or not, post your final answer and we'll see who's up after 15 rounds.

Assuming you pick A 5 times, B 5 times and C 5 times, then yes, we will come out even at the end. If not, there's always going to be some natural variability over only 15 rounds.

 

You choose them, I'll pick A every time and I'll change my choice once you remove one box.

 

Have fun! :D

Originally posted by Dave Marley

Assuming you pick A 5 times, B 5 times and C 5 times, then yes, we will come out even at the end. If not, there's always going to be some natural variability over only 15 rounds.

 

You choose them, I'll pick A every time and I'll change my choice once you remove one box.

 

Have fun! :D

 

using this method, dave would win 2 times out of 3 - guarunteed!

Originally posted by ste

 

sort of, this changes the odds on one box to 0/3. the first is still 1/3 so the other must be 2/3

 

The first one isn't still 1/3 cos there aren't 3 left in the equation, there's 2 which is 1/2, which is half, which is 50/50.

He he heh we're never going to agree LOL! its good though.

There are only 9 possibilities:

 

Pick A. A is right. If we don't change, we win. If we change, we lose.

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

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

 

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

Pick B. B is right. If we don't change, we win. If we change, we lose.

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

 

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

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

Pick C. C is right. 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?

:rolleyes: :D

Surely you must take one away in each group for the box that has been proved to be empty? Cos we know now that one of those is definaltey not right.

 

Leaving 6 possibilities........winning 3 out of 6, which is half, which is 50/50.

 

 

 

:rolleyes: :D

I'm not convinced by this at all! I'm pretty sure that there's no advantage.

 

WHATEVER door the contestant chooses, another door will ALWAYS be opened, whether or not the correct door has been chosen, right?

 

So, in the end it will ALWAYS come down to a choice of two doors. The contestant can't know whether the door (s)he has chosen has the prize or not, and it's completely random whether it has. The third door never actually comes into the probability equation - it's ALWAYS 50/50.

 

 

So it makes no difference in probability whether the contestant chooses the other box or not.

 

Like the coin toss argument. After 10 heads, the probability that the coin will hit heads on the next toss is 50/50 - as it always is, BUT the probability that it will hit heads AGAIN is I think 0.00000095 (where 1 is certainty and 0 is no chance).

 

In the door situation, this is NOT a cumulative probability, and the first door has no effect on the outcome at all.

 

What do you say to that?!

Russ.

Originally posted by james300

Surely you must take one away in each group for the box that has been proved to be empty? Cos we know now that one of those is definaltey not right.

 

Leaving 6 possibilities........winning 3 out of 6, which is half, which is 50/50.

 

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?

Ohhhh. This is getting very complicated. For sure I'm not right. Gonna have to think about this some more.

Originally posted by russtic

What do you say to that?!

I say that you're wrong! :D

 

Try the situation were you start with 1000 doors. It should be much more obvious in that case what you should do!

 

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?

 

 

 

No I can't, frustratingly! I think I'm gonna vanish in a puff of logic...

 

Russ.

Nope, further reflection assures me that I'm completely wrong. I take it all back. Dave is right!

 

...remind me never to play cards with you Dave!

 

:D

 

Russ.

There are 9 possibilites BEFORE the incorrect possibilty is taken away.........once you do take that away, in my mind that leaves 6 possibilities!

3 correct guesses out of 6 is half......which is.......... LOL!

 

Ok. I'm really trying to make my point in simple terms but I know you won't come around! How can the first one be 1 in 3 still when they are only 2 left??? that has now become 1 in 2...ie 1/2 !! ie 50/50.

 

This is making me laugh........and get a headache :D

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?

Originally posted by james300

There are 9 possibilites BEFORE the incorrect possibilty is taken away.........once you do take that away, in my mind that leaves 6 possibilities!

3 correct guesses out of 6 is half......which is.......... LOL!

If you just assume it's 50/50 and change half the time and don't change the other half, then yes, you'll win 50% of the time.

 

However, if you take into account the information you already have, i.e. that there were originally 9 possibilities, and you only had a 1/3 chance of being right with your first guess, you can double your chances of winning by changing to the remaining one of the ones you didn't choose.

 

Ok. I'm really trying to make my point in simple terms

Trust me, I know you're being simple! :p :D

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

 

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!