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 The Dice Game (Posted on 2003-11-10)
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?

 See The Solution Submitted by Gamer Rating: 4.2500 (4 votes)

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 solution - not so nice as that of SK | Comment 5 of 6 |
If we inspect the probabilities of winning on 1st move, we get A=0, B=1/6, C=5/6^2. In 2nd move we get A=5^2/6^3, B=5^3/6^4 and C=5^4/6^5. Final probabilities for A, B, C will be sum of partial probabilities in 1st, 2nd...infinite roll for A, B and C respectively. These probabilities are infinite geometrical sequences with quocient of 5^3/6^3 but with different first sequence members (1/6 for B, 5/6^2 for C and 5^2/6^3 for A). Then from formula for sum of infinite geometrical sequence sum= a/(1-q), where a is first member and q is quocient we get A=25/91, B=36/91, C=30/91.
 Posted by Saso on 2003-11-11 09:45:20

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