Odd Escape
 

You are being held prisoner in a small room. Only you and a gaurd are inside the room. There are three identical doors that lead out of the room. The gaurd offers a choice: you may exit through one of the three doors. Two lead to your execution, one leads to escape. You have no knowledge of the doors and so you choose randomly, door #1. You indicate your choice to the gaurd and he says, "I cannot tell you how to escape, but I can tell you that door #2 will NOT lead to escape." Assuming that the gaurd is impartial and is telling the truth, what should you do?

-Stick with your original choice, door #1.
-Change to door #3.
-It doesn't matter, it's 50/50.

This is not an easy one, but it is also not a trick question by any means. Please justify (mathematically, if possible) any answer.

Re: Odd Escape
 

i say go with 3

Re: Odd Escape
 

change to door #3.

when you chose door number 1 initially, you were chosing from out of three doors, and therefore had a 33% chance of getting the right one. with one of the wrong doors now eliminated, there are only 2 remaining. you are, however, still unaware as to wether or not your door leads to escape. if you stay with your door, you stick with the 33% chance you had originally. however, if you switch to door #3, you will have picked from 2, and your chances are now 50%.

i know it doesn't at first make sense, but I think that its right.

Re: Odd Escape
 

ah - ha

its the old Monty Hall problem from my statistics book.

Monty Hall had a game show "Let's Make a Deal" which dealt with the contestant choosing from 1 of 3 doors and then he'd reduce the number to 2 doors and you could choose again... or something like that. i don't really remember.

Re: Odd Escape
 

its easy ask the guard wich doors do not lead to escape

Re: Odd Escape
 

u picked 1 your onyl option left is 3 so how would picking three improve your chance? you still picked one of the 2 proper choices by choseing over again you dont improve your odds

EDIT: yes jake that just proves you are the head wombat cuz ur the smarter one

Re: Odd Escape
 

Solace, you've got it spot on. Your 33% never changes, but the other door "takes on" the odds from the one that is eliminated, becoming 67%. It's a hard one to grasp, but you got it.

Another way to look at it: Imagine 1,000 doors, one that is correct and 999 that are wrong. You pick one randomly. The gaurd eliminates 998 others. Which do you think is more likely the right one? The one you chose randomly or the one the gaurd left?

I've seen this riddle argued on another forum for several pages, but there is no denying the answer....I was very surprised someone got it so quickly. Goes to show you that this forum has some amazingly intelligent people.

How about this one:

Option A) Play the same set of lottery numbers every week for 52 weeks.
Option B) Play 52 different sets of lottery numbers in one week.

Of course your odds of winning are pretty slim either way, but is one better than the other?

Re: Odd Escape
 

does that mean i get a cookie? I like cookies.

Re: Odd Escape
 

Score one cookie for Solace.

Re: Odd Escape
 

hmm.... i so do not agree with that answer.... its more phycological.


Anyway... both options are the same.
There is an equal chance that a set of numbers will turn up and there is a chance that the same set of numbers turn up twice in a row. Some people just believe that it will be more likely to turn up if the same numbers are used every time.
Did you say 52 numbers? is the 52 numbers + a bonus ball or what?
cos here in england there is choice of49, and 6 numbers are drawn plus a bonus ball.
So really ,,, it makes no difference.

Re: Odd Escape
 

This question causes more heated debate than any other popular probability-based problem I know.

The answer is simple if you start out with the assumption that your first pick has one chance in three of being correct. From that, it should be obvious that your first pick has two chances in three of being incorrect, right? So if you switch, you have a 2/3 chance of switching away from a losing door.

After the other losing door is revealed and removed from your options, you end up having a 2/3 chance of switching to a winning door. It's good to switch doors. 1/3 of the time you'll pick the winning door to begin with and switch to a loser, but 2/3 of the time you'll start with a losing door and switch to the winner.

Re: Odd Escape
 

If I were the one asking the questions then I would throw in the free door to get the person to reconsider their first choice. By the way you are chosing then I would always have the upper hand. Therefore I always stay with my first choice as it is most likely right.

Remember that if you have a 67% chance of chosing the right door then you also have a 67% chance of chosing the wrong one. Optimist of pesimist your choice. :rolleyes:

Re: Odd Escape
 

Regarding:
>How about this one:
>
>Option A) Play the same set of lottery numbers every week for 52 weeks.
>Option B) Play 52 different sets of lottery numbers in one week.
>
Here's a counter example:
Supose that there were seventy million possible sets of numbers to be played. Consider the following two options:
Option A') Play the same set of lottery numbers every week for seventy million weeks.
Option B') Play all seventy million different sets of lottery numbers in one week.
In option A' it is possible that your set of lottery numbers never comes up but in option B' you are certain to hit. Option B is better.

Re: Odd Escape
 

Score the second cookie for Frank Toussaint. ;)

It's so true:

http://www.gtplanet.net/forum/showth...threadid=34178

Re: Odd Escape
 

yea and after playing seventy millinon diffrent combos in one week you will need to win the lotto after spending 70 million dollars on tickets.