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

Home > Probability
Game Show (Posted on 2024-07-25) Difficulty: 3 of 5
In a game show, there is a certain game in which there are four hidden digits. There are no numbers greater than six among them, and no zeros.

You roll a die and then guess if the first digit is higher or lower than what you rolled. (If the die you rolled is equal to the first digit, you win no matter what you said.) You then roll and guess for each of the other three digits.

If you use the best strategy each time when saying "higher" or "lower", what is the chance you will get all four right and win? (Keep in mind you have no idea what the 4 digit number is.)

No Solution Yet Submitted by K Sengupta    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution solution Comment 1 of 1
Strategy:  if the roll is 1, 2, or 3, say "higher", otherwise say "lower".
If the roll is a 1 or a 6, you are correct 100% of the time.
If the roll is 2 or 5, you are correct 5/6 of the time.
If the roll is 3 or 4, you are correct 2/3 of the time.

On average, for each roll, you are correct 5/6 of the time.

p(win) = (5/6)^4 = 625/1296 ~ 0.482

  Posted by Larry on 2024-07-25 11:10:43
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 (0)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information