All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Probability
Pick a box! (Posted on 2002-03-28) Difficulty: 3 of 5
You are shown three boxes, and told that one of them contains a prize. You are then asked to pick one box, and if that box is the one with the prize, you will win it. After picking a box, you are shown that one of the other two boxes is empty, and offered a chance to change your selection.

Should you do this? Would changing your choice to the other remaining box affect your odds of winning? Why or why not?

See The Solution Submitted by levik    
Rating: 4.2857 (14 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Two Solutions | Comment 39 of 40 |

If the person opening one of the unpicked boxes is guaranteed to open an empty box, then we should switch. If the person randomly picks from the remaining boxes and it happens to be empty, then it doesnít matter what we do. If this seems confusing, no worries; you have a lot of good company in the math world.

The trick is to realize that there are two pieces of information being given; (1) how the person is determining which box to open and (2) the results of that opening. Letís attack this via dissecting our sample space into equiprobable events. Letís number the boxes, starting with the one we first chose. There are three, equally likely (1/3) ways the prize can be situated:

1-P, 2-E, 3-E (1/3) --- 1-E, 2-P, 3-E (1/3) --- 1-E, 2-E, 3-P (1/3)
where E=Empty and P=Prize.

Assuming (1) the person is picking a box at random, these events each split into 2 equally likely events:

1-P, 2-E-X, 3-E (1/6) --- 1-P, 2-E, 3-E-X (1/6)
1-E, 2-P-X, 3-E (1/6) --- 1-E, 2-P, 3-E-X (1/6)
1-E, 2-E-X, 3-P (1/6) --- 1-E, 2-E, 3-P-X (1/6)
where the X indicates the box the person opens.

Knowing that (2) the person did not reveal the prize eliminates two of these events:

1-P, 2-E-X, 3-E (1/4) --- 1-P, 2-E, 3-E-X (1/4)
1-E, 2-P, 3-E-X (1/4)
1-E, 2-E-X, 3-P (1/4)

We see that there is a 1/2 chance of getting the prize, which means switching is of no use.

Now assume (1) that you know the other person would not reveal the prize box. We still have three initial possibilities:

1-P, 2-E, 3-E (1/3) --- 1-E, 2-P, 3-E (1/3) --- 1-E, 2-E, 3-P (1/3)

But before he even reveals a box, we know there is only one possibility for each of the last two cases:

1-P, 2-E-X, 3-E (1/6) --- 1-P, 2-E, 3-E-X (1/6)
1-E, 2-P, 3-E-X (1/3)
1-E, 2-E-X, 3-P (1/3)

In this case, seeing the revealed box is empty (2) adds no new information. Thus, there is a 2/3 chance of getting the prize if you switch.

On the Monty Hall Show often a contestant was asked to pick one of three doors, a goat (or some other lesser prize) was revealed behind one of the remaining doors, and then the contestant was asked whether or not they wanted to switch choices. It is assumed that for television purposes, the producers would deliberately not open a door that hid the grand prize, so in this case the contestant should switch. And actually, since there is no disadvantage to switching in either case, contestants should switch regardless of their certainty about how the producers are selecting the doors.


  Posted by owl on 2005-08-17 14:36:31
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2020 by Animus Pactum Consulting. All rights reserved. Privacy Information