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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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ERROR MY SOLUTION | Comment 6 of 12 |

I made an error in counting favourable outcomes, should of been 30 not 28, so my revised solution is as follows. For each roll of the dice while adhering to the optimum strategy there are 30 favourable outcomes out of a total outcomes of 36.

so plugging these values into the formula we have 30/36 or 5/6

now given that each roll is an indepentant event we use the multiplication law for independant events.

so for the overall probability we multiply the independant probabilities by each other and we get 5/6x5/6x5/6x5/6

this equals   625/1296 or 48.2% chance of a favourable outcome.

  Posted by Leigh Lillico on 2005-11-07 04:19:21
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