Three people, A,B,C play a game. A rolls the die.
Then, in order of "B,C,A,B,C,A..." they each roll the die. They keep going until someone wins. To win, you have to get the same number as the previous number rolled on the die. ( A can't win with his first roll because there was no roll before to compare it too.)
What is the probability that each person will win?
(In reply to
solution by SilverKnight)
I got essentially the same answer as SilverKnight, except that (surprise, surprise !!) his solution is more intelligent than mine.....
They roll A,B,C,A,B,C,A,B,C,........ It is mathematically certain beyond a negligible doubt that there will be a winner in the first 16 rolls of the dice, so only those rolls need be considered.
Roll 1: A has 0 chance to win
Roll 2: B has 1/6 chance to win
Roll 3: C has (5/6)*(1/6) chance
Roll 4: A has [(5/6)^2]*(1/6) chance (^ means "raised to the power of")
Roll 5: B has [(5/6)^3]*(1/6) chance
....etc.....
Roll 16: A has [(5/6)^14]*(1/6) chance
A's total chance in the first 16 rolls is 26%
B's is 38%
C's is 36%
Edited on November 11, 2003, 1:33 am

Posted by Dan
on 20031110 18:51:53 