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

Home > Probability
Easy as 1 2 3 (Posted on 2003-10-15) Difficulty: 3 of 5
In a game show, there is a game in which you have to order the value of three prizes in order of least expensive to most expensive. You have to get all three right in order to win.

The only problem is you have your spouse do all the shopping, so you only know that the first prize is between 500 and 2000, the second prize is between 1000 and 2500, and the third prize is between 1500 and 3000.

Which order should you put them in so that you have the highest probability of winning, and what is the probability that you will win using this arrangement?

See The Solution Submitted by Gamer    
Rating: 3.8333 (6 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re: The Price Is Right | Comment 5 of 9 |
(In reply to The Price Is Right by Dan)

The problem is that not all the ways shown are equally likely. For example, the line

1500-2000 1500-1999 2000-2499 123 or 213

should be in total as likely as say

1500-2000 1000-1499 2500-3000 213

and as

1500-2000 1500-1999 1500-1999 any order

so in the case of the splits between two orders, each is entitled to only half a "way", and in the case of "any order" each of those is entitled to only 1/6 of a way. Alternatively multiply each of the individual lines as counting as 6 ways, each possible order in double lines as 3 ways and each of the 6 sharers in "any order" as 1 way.
  Posted by Charlie on 2003-10-16 08:52:08

Please log in:
Remember me:
Sign up! | Forgot password

Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (4)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information