Friday, March 30, 2007
Mathematical Diversions - 2
There are 3 boxes. Each of the three boxes contains 2 balls. In one there are two white balls, in another two black balls, and in the third one ball is black, the other is white. Boxes have been labeled to indicate their contents. However, whoever did the job got all labels wrong. The task is to straighten things out. You may select 1 box and blindly pick up a ball out of it and then have to put the correct label on the boxes.
Subscribe to:
Post Comments (Atom)
4 comments:
Pick the box with WB on it. Pick a ball.
If the ball is white, put WW label on WB, WB label on BB and BB label on WW.
if the ball is black, put BB label on WB, WB on WW and WW on BB.
A puzzle for you? you mite have heard of this one... but still, worth a try.
You are standing in front of 3 closed doors. You are told that one of them has a gift behind it for you. You are asked to choose a door. Once you have chosen, I open one of the other 2 doors that does not have a gift behind it.
After that, you are again given a chance to choose a door that might have the gift. Would you stay at the same door, or would you switch to the other closed door?
In mathematical terms, which door has the higher probability of having the gift behind it? Is it the door you initially chose or the door you did not choose? or is it the same?
you got it..
Reply to your puzzle: I would not change my choice of the door after you have opened one of the doors which does not have a gift behind it. Probability of having a gift behind a door is 1/3 and is independent of the door you open.
No... your answer is not correct. Though my explanation might sound counter-intuitive, mathematically, it is correct.
At the start, before you choose anything...
Prob. of gift being behind your chosen door = 1/3
prob of gift being behind the door opened to you = 0 (since the door opened never has the gift behind it)
Total probability is always 1...
so, prob. of gift being behind the 3rd door =
1 - (prob of your door) - (prob of door opened)
= 1 - 1/3 - 0
= 2/3
So, you should switch.
There are better explanations for this problem using Bayes theorem (conditional probability). I guess it's better I post links rather than typing them out :-)
http://montyhallproblem.com/
http://www.remote.org/frederik/projects/ziege/
http://en.wikipedia.org/wiki/Monty_Hall_problem
Post a Comment