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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)

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Solution Solution given assumptions. | Comment 1 of 12

Assumptions:  Each digit is equally likely to be 1, 2, 3, 4, 5, or 6.  The individual digits are independent.  You have no idea what the actual number is.

The optimum strategy would be to guess "higher" on rolling a 1, 2, or 3 and "lower" on a 4, 5, or 6.

If a digit is a 6 we will be correct if we roll a 1, 2, 3, or 6;
If a digit is a 5 we will be correct if we roll a 1, 2, 3, 5, or 6;
If a digit is a 3 or 4 we will always be correct;
If a digit is a 2 we will be correct if we roll a 1, 2, 4, 5, or 6;
If a digit is a 1 we will be correct if we roll a 1, 4, 5, or 6.

(4/6 + 5/6 + 6/6 + 6/6 + 5/6 + 4/6)/6 = 5/6 so this is our total chance of being correct on any given digit.

Since there are 4 independent digits the overall probability of being correct on all 4 is (5/6)^4 = 625/1296  = .48225


  Posted by Jer on 2005-11-04 07:47:24
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