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.)
Contradicting the solution, as the problem is written there are two ways to win:
1) By rolling the number of the first digit
2) By correctly guessing each digit as higher or lower (thus,
a roll equal to the digit, other than on the first roll, could
only be a losing roll).
There is a 1/6 chance that you would roll the number of the first digit. If not having won on the first roll, there is a 4/6 chance you may continue rolling, with the same odds for each subsequent roll. That is, there is a (4/6)4 = 16/81 chance of winning by guessing each digit as higher or lower. Therefore, there is a 1/6 + 16/81 = 59/162 chance of winning the game.
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Posted by Dej Mar
on 2006-08-28 15:06:10 |