Question

In a game two players A and B take turns in throwing a pair of fair dice starting with player A and the total of scores on the two dice, in each throw is noted. A win the game if he throws a total of 6 before B throws a total of 7 and B wins the game if he throws a total of 7 before A throws a total of six. The game stops as soon as either of the players wins. What is the probability of A winning the game?
 A. \[\dfrac{5}{{31}}\]
B. \[\dfrac{{31}}{{61}}\]
C. \[\dfrac{{30}}{{61}}\]
D. \[\dfrac{5}{6}\]

Answer 1

Hint: First find the sample spaces of throwing two dice, getting sum 7, and getting sum 6. Then find the probability of getting sum 6 and probability of getting sum 7. Then obtain the complement of probability of getting sum 6 and complement of probability of getting sum 7. After that apply the probability sum formula to find the probability of A wins.

Formula used:
The formula of sum of infinite geometric series \[a,ar,a{r^2},...\] is \[\dfrac{a}{{1 - r}}, - 1 < r < 1\] , where a is the first term and r is the common ratio.
The formula to obtain the complement of any probability K is \[P(\overline K ) = 1 - P(K)\].

Complete step by step solution:
The sample space is,
 \[\begin{array}{c}n(S) = {6^2}\\ = 36\end{array}\]
Sample space of getting sum 7 = {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}
Therefore, \[n(7) = 6\]
Sample space of getting sum 6 = {(1,5), (2,4), (3,3), (4,2), (5,1)}
Therefore, \[n(6) = 5\]
Now, the probability of getting sum 6 is \[P(A) = \dfrac{5}{{36}}\] .
And the probability of getting sum 7 is \[P(B) = \dfrac{6}{{36}}\] .
Here,
\[\begin{array}{c}P(\overline A ) = 1 - \dfrac{5}{{36}}\\ = \dfrac{{36 - 5}}{{36}}\\ = \dfrac{{31}}{{36}}\end{array}\]
And
\[\begin{array}{c}P(\overline B ) = 1 - \dfrac{6}{{36}}\\ = \dfrac{{36 - 6}}{{36}}\\ = \dfrac{{30}}{{36}}\end{array}\]
Probability of A wins is,
\[P({A_{wins}}) = P(A) + P(\overline A )P(\overline B )P(A) + P(\overline A )P(\overline B )P(\overline A )P(\overline B )P(A) + ....\]
\[ = \dfrac{5}{{36}} + \left( {\dfrac{{31}}{{36}}} \right)\left( {\dfrac{{30}}{{36}}} \right)\left( {\dfrac{5}{{36}}} \right) + ....\infty \]
=\[\dfrac{5}{{36}} + \dfrac{5}{{36}}.\dfrac{{155}}{{216}} + \dfrac{5}{{36}}.{\left( {\dfrac{{155}}{{216}}} \right)^2} + ....\infty \]
Here, \[\dfrac{5}{{36}} + \dfrac{5}{{36}}.\dfrac{{155}}{{216}} + \dfrac{5}{{36}}.{\left( {\dfrac{{155}}{{216}}} \right)^2} + ....\infty \] is an infinite geometric series with first term \[\dfrac{5}{{36}}\] and the common ratio \[\dfrac{{155}}{{216}}\] . Let us apply the infinite sum formula to obtain the required sum.
\[\dfrac{{\dfrac{5}{{36}}}}{{1 - \dfrac{{155}}{{216}}}}\]
\[ = \dfrac{{\dfrac{5}{{36}}}}{{\dfrac{{61}}{{216}}}}\]
\[ = \dfrac{5}{{36}} \times \dfrac{{216}}{{61}}\]
\[ = \dfrac{{30}}{{61}}\]
Hence the Required value is \[\dfrac{{30}}{{61}}\] .

Note: Students sometimes thought that the probability of A is the answer and wrote the answer as \[\dfrac{5}{{36}}\] but this is not correct. Here we have to calculate the probability of B and the complement probabilities of A and B to calculate the probability of A winning and compute the correct answer \[\dfrac{30}{{61}}\].