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.)
I give the contestant about a 48% chance of taking the prize. Here's why:
For each of the four digits:
The contestant, lets give him a name, Erik, rolls the die.
If Erik rolls a "1" or "6" (1/3 chances):
He will guess with 6/6 accuracy
"higher" for "1" (he progresses if the digit is 1,2,3,4,5 or 6)
"lower" for "6" (he progresses on 6,5,4,3,2 or 1)
The chance that Erik rolls a "1" or "6" and guesses correctly is 1/3 * 6/6 = 6/18
If Erik rolls a "2" or "5" (1/3 chances):
He will guess with 5/6 accuracy
"higher" for "2" (he progresses on 2,3,4,5 or 6)
"lower" for "5" (he progresses on 5,4,3,2 or 1)
The chance that Erik rolls a "2" or "5" and guesses correctly is 1/3 * 5/6 = 5/18
If Erik rolls a "3" or "4" (1/3 chances):
He will guess with 4/6 accuracy
"higher" for "3" (he progresses on 3,4,5 or 6)
"lower" for "4" (he progresses on 4,3,2 or 1)
The chance that Erik rolls a "3" or "4" and guesses correctly is 1/3 * 4/6 = 4/18
So the chance of Erik progressing after each guess is 6/18 + 5/18 + 4/18 = 15/18 = 5/6
Erik has to do this four times to take the prize: (5/6)^4 = 625/1296 (about a 48% chance)

Posted by Bender
on 20051104 08:05:57 