Partial Dice

A and B are playing a game with a fair dice.

In this, they both throw the dice one by one and the person who first gets a 6 wins.

A starts rolling it first.

Let P P denote the probability that B will win the game.

If P P can be expressed as p q \dfrac pq , where p p and q q are coprime positive integers, submit your answer as p + q p+q .


The answer is 16.

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2 solutions

Jesse Li
Dec 1, 2018

Let's call the probability of A winning x.

In that case, A has an x chance of winning and B has a 5 x 6 \frac{5x}{6} of winning, because A has a 5 6 \frac{5}{6} chance of not getting a 6 on the first turn, and if that happens, B now has an x chance of winning.

Since one player has to win, x + 5 x 6 = 1 x+\frac{5x}{6}=1 .

Solving the equation, we get x = 6 11 x=\frac{6}{11} .

Since we're solving for the probability that B wins, we need to multiply by 5 6 \frac{5}{6} , to get P = 5 11 P=\frac{5}{11} .

5 + 11 = 16 5+11=\boxed{16}

Mr. India
Dec 1, 2018

B can win on second/fourth/sixth.....(even turn).

The probability of A losing in first turn is 5 6 \frac{5}{6}

B wins in second turn with 1 6 \frac{1}{6} probability.

So, total probability of winning in second turn is 5 6 \frac{5}{6} × 1 6 \frac{1}{6}

A losing in third is again 5 6 \frac{5}{6}

B winning fourth is 1 6 \frac{1}{6}

Now total probability is 5 6 \frac{5}{6} × 5 6 \frac{5}{6} × 5 6 \frac{5}{6} × 1 6 \frac{1}{6} since B loses in second then only fourth comes

Similarly, probability increases with a factor of 25 36 \frac{25}{36}

So, total probability will be the sum of all these probabilities till infinity which is, a 1 r \frac{a}{1-r} a is first term , r is common difference

5 / 36 1 25 / 36 \frac{5/36}{1-25/36} = 5 11 \frac{5}{11} .

The answer is 5 + 11 = 16 5 + 11 =\boxed{16} .

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