Question

Two persons A and B take turns in throwing a pair of dice. The first person to throw 9 from both dice will be awarded the prize. If A throws first and the probability that B wins the game is $\dfrac{k}{{17}}$, then find k.

Answer 1

Hint: Write the sample space and the favourable outcomes. Then calculate the probability of getting a 9. Consider all the cases where the second person can win, i.e., the second person can win in their first try, second try, third try, and so on. Add the probabilities of all such cases.

Formula Used: Sum to infinity of a geometric progression is $\dfrac{a}{{1 - r}}$ where $a,r$ are the first term and the constant ratio respectively.

Complete step by step Solution:
A player will win the prize if the numbers on the pair of dice that they throw add up to 9. Let us see how many such cases are there.
Sample space - \[\begin{array}{*{20}{c}}
  {\left( {1,1} \right),}&{\left( {1,2} \right),} \\
  {\left( {2,1} \right),}&{\left( {2,2} \right),}
\end{array}\,\,\,\,\,\,\begin{array}{*{20}{c}}
  {\left( {1,3} \right),}&{\left( {1,4} \right),} \\
  {\left( {2,3} \right),}&{\left( {2,4} \right),}
\end{array}\,\,\,\,\,\,\begin{array}{*{20}{c}}
  {\left( {1,5} \right),}&{\left( {1,6} \right),} \\
  {\left( {2,5} \right),}&{\left( {2,6} \right),}
\end{array} \\
  \begin{array}{*{20}{c}}
  {\left( {3,1} \right),}&{\left( {3,2} \right),} \\
  {\left( {4,1} \right),}&{\left( {4,2} \right),}
\end{array}\,\,\,\,\,\,\begin{array}{*{20}{c}}
  {\left( {3,3} \right),}&{\left( {3,4} \right),} \\
  {\left( {4,3} \right),}&{\left( {4,4} \right),}
\end{array}\,\,\,\,\,\,\begin{array}{*{20}{c}}
  {\left( {3,5} \right),}&{\left( {3,6} \right),} \\
  {\left( {4,5} \right),}&{\left( {4,6} \right),}
\end{array} \\
  \begin{array}{*{20}{c}}
  {\left( {5,1} \right),}&{\left( {5,2} \right),} \\
  {\left( {6,1} \right),}&{\left( {6,2} \right),}
\end{array}\,\,\,\,\,\,\begin{array}{*{20}{c}}
  {\left( {5,3} \right),}&{\left( {5,4} \right),} \\
  {\left( {6,3} \right),}&{\left( {6,4} \right),}
\end{array}\,\,\,\,\,\,\begin{array}{*{20}{c}}
  {\left( {6,5} \right),}&{\left( {5,6} \right),} \\
  {\left( {6,6} \right),}&{\left( {6,6} \right)}
\end{array} \]

Favourable outcomes - $\left\{ {\left( {3,6} \right),\left( {4,5} \right),\left( {5,4} \right),\left( {6,3} \right)} \right\}$
Total number of outcomes = 36
Number of favourable outcomes = 4
The probability of an event occurring equals the number of favourable outcomes divided by the total number of outcomes. Let the event of getting a 9 be event A and let the event of not getting a 9 be event B. Therefore,
$P(A) = \dfrac{4}{{36}} = \dfrac{1}{9}$
Since the probability of getting a 9 and probability of not getting a 9 add up to 1, $P(B) = 1 - P(A)$.
Therefore, $P(B) = \dfrac{8}{9}$
We know that A throws the dice first and that A and B throw the dice alternately.
Let’s consider the following cases where the second person wins:
a) First person doesn’t throw 9 on their first try and the second person throws 9 on their first try. Let us call this event ${C_1}$. $P({C_1}) = P(B)P(A) = \dfrac{8}{{81}}$
b) First person doesn’t throw 9 in their first try, the second person doesn’t throw 9 in their first try, the first person doesn’t throw 9 in their first try and the second person throws 9 in their second try. Let us call this event ${C_2}$. $P({C_2}) = P(B)P(B)P(B)P(A) = \dfrac{8}{{81}}.\dfrac{{64}}{{81}}$
The probability of the second person winning is the sum of all such probabilities as in the above two cases
$P(winning) = P({C_1}) + P({C_2}) + P({C_3}) + .......$
$P(winning) = \dfrac{8}{{81}} + \dfrac{8}{{81}}.\dfrac{{{8^2}}}{{{9^2}}} + \dfrac{8}{{81}}.\dfrac{{{8^4}}}{{{9^4}}} + ....$
$P(winning) = \dfrac{8}{{81}}\left( {\dfrac{{{8^0}}}{{{9^0}}} + \dfrac{{{8^2}}}{{{9^2}}} + \dfrac{{{8^4}}}{{{9^4}}} + ....} \right)$
Simplifying further,
\[P(winning) = \dfrac{8}{{81}}\left( {\dfrac{1}{{1 - \dfrac{{{8^2}}}{{{9^2}}}}}} \right)\]
\[P(winning) = \dfrac{8}{{81}}\left( {\dfrac{{81}}{{17}}} \right)\]
\[P(winning) = \dfrac{8}{{17}}\]
Therefore, the value of k is 8.

Note: Probability of second person winning = 1 – Probability of first-person winning
$P\left( {{\text{First person win}}} \right) = \dfrac{1}{9} + \dfrac{8}{9}.\dfrac{8}{9}.\dfrac{1}{9} + \dfrac{8}{9}.\dfrac{8}{9}.\dfrac{8}{9}.\dfrac{8}{9}.\dfrac{1}{9} + ...$
If we calculate it, we get
$P\left( {{\text{First person win}}} \right) = \dfrac{9}{{17}}$
Therefore, the probability that the second person wins will be $\dfrac{8}{{17}}$.