SOLUTION: Jin and Gary are playing a game with a 3 fair dice , each with 9 faces and with numbers 1-9 written on the faces . If a person gets the number 7 or above on all 3 dice ,he wins . A

Question 1138547: Jin and Gary are playing a game with a 3 fair dice , each with 9 faces and with numbers 1-9 written on the faces . If a person gets the number 7 or above on all 3 dice ,he wins . Assuming Jin goes first , what is the probability that Jin will win ?

The answer is 27/53 . Please help to solve this . Thanks
Answer by greenestamps(13206) (Show Source): You can put this solution on YOUR website!

3 of the 9 numbers on each die are 7 or above, so the probability of each die showing a number 7 or above is 3/9, or 1/3. Then the probability that all three dice show 7 or above is (1/3)^3 = 1/27.

So each person on each play has a probability of 1/27 of winning and a probability of 26/27 of not winning.

The first way Jin can win is if he gets the winning result on his first turn. The probability of that happening is 1/27.

The second way Jin can win is if he doesn't win on his first play and neither does Gary, and then Jin wins on his second play. The probability of that happening is (26/27)(26/27)(1/27).

Similarly, the probability that Jin wins on his third play is (26/27)(26/27)(26/27)(26/27)(1/27).

Theoretically, the game could go on forever; then the probability that Jin wins is the sum of an infinite geometric series:

1/27)+(1/27)(26/27)^2+(1/27)(26/27)^4+...

The sum of an infinite geometric series is

where a is the first term and r is the common ratio:

------------------------------------------------------------------

We want to find the overall probability that Jin is the first to roll the dice and get 7 or more on all three.

So Gary isn't the one who first rolls a winning set of dice.

Let P(J) represent the probability that Jin wins on his first roll; let P(JGJ) represent the probability that Jin wins on his second throw; and similarly P(JGJGJ) represents the probability that Jin wins on his third throw; and so on.

The probability of each non-winning play is 26/27; the probability of the winning play (which ends the game) is 1/27. Then

P(J) = (1/27)
P(JGJ) = (26/27)(26/27)(1/27) = (1/27)*(26/27)^2
P(JGJGJ) = (26/27)(26/27)(26/27)(26/27)(1/27) = (1/27)*(26/27)^4
... etc....

Then the overall probability is the sum of those individual probabilities:

(1/27)+(1/27)(26/27)^2+(1/27)(26/27)^4+...
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