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The Dice Game (Posted on 2005-11-04) Difficulty: 2 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.)

See The Solution Submitted by Gamer    
Rating: 3.6667 (3 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
sean correct - another approach | Comment 11 of 12 |
A roll of 6 or 1 on presents 6/6 probability; 5 or 2, 5/6; 4 or 3, 4/6.
30/36 = .833333333.
1 roll as above
to 4th pwr./4 rolls = .48225
This is the same probability {30/36} as that of rolling any of the numbers between 4 and 10 in conventional craps (i.e. all except the rolls of 2/12/3/11.
  Posted by Charles Archibals on 2006-08-31 02:34:15
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