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$A$ and $B$ throw a dice alternatively till one of them gets a $6$ and wins the game. Find their respective probabilities of winning if $A$ starts first.
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Hint: Try to find out probability in all those possible cases in which $A$ wins and then take the union of all these cases by simply adding these probabilities. To add the probabilities, use a formula for the sum of infinite G.P.
In the question, it is given that a person wins if he gets $6$ on the dice. A dice has a total
of six sides of which one of the sides has $6$ on it. So, probability of getting $6$ is,
$P\left( 6 \right)=\dfrac{1}{6}$
Let us consider $X$ as an event which denotes getting $6$ on the dice and $Y$ as an event for not
getting $6$ on the dice.
$\Rightarrow P\left( X \right)=\dfrac{1}{6}$
And $P\left( Y \right)=\dfrac{5}{6}$
To find the probability that $A$ wins, let us make all the possible cases.
Case 1: $A$ wins at first throw of dice
$\begin{align}
& P\left( A-wins \right)=P\left( X \right) \\
& \Rightarrow P\left( A-wins \right)=\dfrac{1}{6} \\
\end{align}$
Since $A$ and $B$ throw the dice alternative, so now, $B$ will throw dice. Now for the third throw,
$A$ will throw the dice.
Case 2: $A$ wins at the third throw of dice
$\begin{align}
& P\left( A\text{ }wins \right)=P\left( no\text{ }six\text{ }by\text{ }A \right).P\left( no\text{ }six\text{
}by\text{ B} \right).P\left( six\text{ }by\text{ }A \right) \\
& \Rightarrow P\left( A\text{ }wins \right)=\dfrac{5}{6}.\dfrac{5}{6}.\dfrac{1}{6} \\
& \Rightarrow P\left( A\text{ }wins \right)={{\left( \dfrac{5}{6} \right)}^{2}}.\dfrac{1}{6} \\
\end{align}$
Now $B$ will throw on the fourth throw. Then again, $A$ will throw on the fifth throw of the dice.
Case 3: $A$ wins at the fifth throw
$\begin{align}
& P\left( A\text{ }wins \right)=P\left( no\text{ }six\text{ }by\text{ }A \right).P\left( no\text{ }six\text{
}by\text{ B} \right).P\left( no\text{ }six\text{ }by\text{ }A \right).P\left( no\text{ }six\text{ }by\text{ B}
\right).P\left( six\text{ }by\text{ }A \right) \\
& \Rightarrow P\left( A\text{ }wins
\right)=\dfrac{5}{6}.\dfrac{5}{6}.\dfrac{5}{6}.\dfrac{5}{6}.\dfrac{1}{6} \\
& \Rightarrow P\left( A\text{ }wins \right)={{\left( \dfrac{5}{6} \right)}^{4}}.\dfrac{1}{6} \\
\end{align}$
Similarly, we can find the probability of $A$ winning on the odd number of throws. The probability
of $A$ winning at ${{n}^{th}}$ thrown is equal to ${{\left( \dfrac{5}{6} \right)}^{n-1}}.\dfrac{1}{6}$ .
Since all the above listed cases are possible, the probability of $A$ winning is given by the union of
all these cases i.e. we have to add all these cases. Hence, the probability that $A$ wins is,
\[\begin{align}
& P\left( A\text{ }wins \right)=\dfrac{1}{6}+{{\left( \dfrac{5}{6} \right)}^{2}}\dfrac{1}{6}+{{\left(
\dfrac{5}{6} \right)}^{4}}\dfrac{1}{6}+{{\left( \dfrac{5}{6} \right)}^{6}}\dfrac{1}{6}+......... \\
& \Rightarrow P\left( A\text{ }wins \right)=\dfrac{1}{6}\left( 1+{{\left( \dfrac{5}{6}
\right)}^{2}}+{{\left( \dfrac{5}{6} \right)}^{4}}+{{\left( \dfrac{5}{6} \right)}^{6}}+.........
\right)............\left( 1 \right) \\
\end{align}\]
The above term is an infinite G.P. Since the common ratio of the above G.P. if \[{{\left( \dfrac{5}{6}
\right)}^{2}}\] which is less than $1$, we can find it’s sum.
For an infinite G.P. $a,ar,a{{r}^{2}},a{{r}^{3}}.......$ the sum is given by,’
${{S}_{\infty }}=\dfrac{a}{1-r}...........\left( 2 \right)$
Using formula $\left( 2 \right)$ to find the sum of infinite GP in equation $\left( 1 \right)$ with
$a=1,r={{\left( \dfrac{5}{6} \right)}^{2}}$, we get,
\[\begin{align}
& P\left( A\text{ }wins \right)=\dfrac{1}{6}\left( \dfrac{1}{1-{{\left( \dfrac{5}{6} \right)}^{2}}} \right)
\\
& \Rightarrow P\left( A\text{ }wins \right)=\dfrac{1}{6}\left( \dfrac{1}{1-\dfrac{25}{36}} \right) \\
& \Rightarrow P\left( A\text{ }wins \right)=\dfrac{1}{6}\left( \dfrac{1}{\dfrac{11}{36}} \right) \\
& \Rightarrow P\left( A\text{ }wins \right)=\dfrac{1}{6}\left( \dfrac{36}{11} \right) \\
& \Rightarrow P\left( A\text{ }wins \right)=\dfrac{6}{11}...........\left( 3 \right) \\
\end{align}\]
In probability, we have a property,
$P\left( event \right)+P\left( complement\text{ }of\text{ }the\text{ }event \right)=1.......\left( 4
\right)$
Since complement of $A$ wins is $B$ wins, from $\left( 4 \right)$, we have,
$\begin{align}
& P\left( A\text{ }wins \right)+P\left( \text{B }wins \right)=1 \\
& \Rightarrow P\left( \text{B }wins \right)=1-P\left( A\text{ }wins \right) \\
\end{align}$
From $\left( 3 \right)$ substituting \[P\left( A\text{ }wins \right)=\dfrac{6}{11}\] in the above
equation, we get,
$\begin{align}
& P\left( \text{B }wins \right)=1-\dfrac{6}{11} \\
& \Rightarrow P\left( \text{B }wins \right)=\dfrac{5}{11} \\
\end{align}$
Note: One must know that the formula for the sum of infinite G.P. is applicable only for those G.P. in which the common ratio is less than $1$. If the common ratio is greater than $1$, we cannot use that formula to calculate sum.
In the question, it is given that a person wins if he gets $6$ on the dice. A dice has a total
of six sides of which one of the sides has $6$ on it. So, probability of getting $6$ is,
$P\left( 6 \right)=\dfrac{1}{6}$
Let us consider $X$ as an event which denotes getting $6$ on the dice and $Y$ as an event for not
getting $6$ on the dice.
$\Rightarrow P\left( X \right)=\dfrac{1}{6}$
And $P\left( Y \right)=\dfrac{5}{6}$
To find the probability that $A$ wins, let us make all the possible cases.
Case 1: $A$ wins at first throw of dice
$\begin{align}
& P\left( A-wins \right)=P\left( X \right) \\
& \Rightarrow P\left( A-wins \right)=\dfrac{1}{6} \\
\end{align}$
Since $A$ and $B$ throw the dice alternative, so now, $B$ will throw dice. Now for the third throw,
$A$ will throw the dice.
Case 2: $A$ wins at the third throw of dice
$\begin{align}
& P\left( A\text{ }wins \right)=P\left( no\text{ }six\text{ }by\text{ }A \right).P\left( no\text{ }six\text{
}by\text{ B} \right).P\left( six\text{ }by\text{ }A \right) \\
& \Rightarrow P\left( A\text{ }wins \right)=\dfrac{5}{6}.\dfrac{5}{6}.\dfrac{1}{6} \\
& \Rightarrow P\left( A\text{ }wins \right)={{\left( \dfrac{5}{6} \right)}^{2}}.\dfrac{1}{6} \\
\end{align}$
Now $B$ will throw on the fourth throw. Then again, $A$ will throw on the fifth throw of the dice.
Case 3: $A$ wins at the fifth throw
$\begin{align}
& P\left( A\text{ }wins \right)=P\left( no\text{ }six\text{ }by\text{ }A \right).P\left( no\text{ }six\text{
}by\text{ B} \right).P\left( no\text{ }six\text{ }by\text{ }A \right).P\left( no\text{ }six\text{ }by\text{ B}
\right).P\left( six\text{ }by\text{ }A \right) \\
& \Rightarrow P\left( A\text{ }wins
\right)=\dfrac{5}{6}.\dfrac{5}{6}.\dfrac{5}{6}.\dfrac{5}{6}.\dfrac{1}{6} \\
& \Rightarrow P\left( A\text{ }wins \right)={{\left( \dfrac{5}{6} \right)}^{4}}.\dfrac{1}{6} \\
\end{align}$
Similarly, we can find the probability of $A$ winning on the odd number of throws. The probability
of $A$ winning at ${{n}^{th}}$ thrown is equal to ${{\left( \dfrac{5}{6} \right)}^{n-1}}.\dfrac{1}{6}$ .
Since all the above listed cases are possible, the probability of $A$ winning is given by the union of
all these cases i.e. we have to add all these cases. Hence, the probability that $A$ wins is,
\[\begin{align}
& P\left( A\text{ }wins \right)=\dfrac{1}{6}+{{\left( \dfrac{5}{6} \right)}^{2}}\dfrac{1}{6}+{{\left(
\dfrac{5}{6} \right)}^{4}}\dfrac{1}{6}+{{\left( \dfrac{5}{6} \right)}^{6}}\dfrac{1}{6}+......... \\
& \Rightarrow P\left( A\text{ }wins \right)=\dfrac{1}{6}\left( 1+{{\left( \dfrac{5}{6}
\right)}^{2}}+{{\left( \dfrac{5}{6} \right)}^{4}}+{{\left( \dfrac{5}{6} \right)}^{6}}+.........
\right)............\left( 1 \right) \\
\end{align}\]
The above term is an infinite G.P. Since the common ratio of the above G.P. if \[{{\left( \dfrac{5}{6}
\right)}^{2}}\] which is less than $1$, we can find it’s sum.
For an infinite G.P. $a,ar,a{{r}^{2}},a{{r}^{3}}.......$ the sum is given by,’
${{S}_{\infty }}=\dfrac{a}{1-r}...........\left( 2 \right)$
Using formula $\left( 2 \right)$ to find the sum of infinite GP in equation $\left( 1 \right)$ with
$a=1,r={{\left( \dfrac{5}{6} \right)}^{2}}$, we get,
\[\begin{align}
& P\left( A\text{ }wins \right)=\dfrac{1}{6}\left( \dfrac{1}{1-{{\left( \dfrac{5}{6} \right)}^{2}}} \right)
\\
& \Rightarrow P\left( A\text{ }wins \right)=\dfrac{1}{6}\left( \dfrac{1}{1-\dfrac{25}{36}} \right) \\
& \Rightarrow P\left( A\text{ }wins \right)=\dfrac{1}{6}\left( \dfrac{1}{\dfrac{11}{36}} \right) \\
& \Rightarrow P\left( A\text{ }wins \right)=\dfrac{1}{6}\left( \dfrac{36}{11} \right) \\
& \Rightarrow P\left( A\text{ }wins \right)=\dfrac{6}{11}...........\left( 3 \right) \\
\end{align}\]
In probability, we have a property,
$P\left( event \right)+P\left( complement\text{ }of\text{ }the\text{ }event \right)=1.......\left( 4
\right)$
Since complement of $A$ wins is $B$ wins, from $\left( 4 \right)$, we have,
$\begin{align}
& P\left( A\text{ }wins \right)+P\left( \text{B }wins \right)=1 \\
& \Rightarrow P\left( \text{B }wins \right)=1-P\left( A\text{ }wins \right) \\
\end{align}$
From $\left( 3 \right)$ substituting \[P\left( A\text{ }wins \right)=\dfrac{6}{11}\] in the above
equation, we get,
$\begin{align}
& P\left( \text{B }wins \right)=1-\dfrac{6}{11} \\
& \Rightarrow P\left( \text{B }wins \right)=\dfrac{5}{11} \\
\end{align}$
Note: One must know that the formula for the sum of infinite G.P. is applicable only for those G.P. in which the common ratio is less than $1$. If the common ratio is greater than $1$, we cannot use that formula to calculate sum.
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