Question

A and B alternately cut a pack of cards which is shuffled after each cut. The game is started by A and continuous until one of the players cuts a club. The probability that B wins the game is
(a) $ \dfrac{3}{7} $
(b) $ \dfrac{4}{7} $
(c) $ \dfrac{5}{7} $
(d) $ \dfrac{6}{7} $

Answer 1

Hint: First, before proceeding for this, we must suppose the name of the terms that if A wins the game then probability is WA, if B wins the game probability is WB, if A loses the game then probability is LA and if B loses the game probability is LB. Then, we get the probability of winning A in first, third, fifth and so on rounds which gives us an infinite series. Then, for solving the infinite series we have the formula where a is the first term and r is the common ratio as $ {{S}_{\infty }}=\dfrac{a}{1-r} $ . Then after getting the probability of A wins subtract it from 1 to get probability of B wins which is the required result.

Complete step-by-step answer:
In this question, we are supposed to find the probability that B wins the game from the continuous cutting of the cards until anyone gets the club.
So, before proceeding for this, we must suppose the name of the terms that if A wins the game then probability is WA, if B wins the game probability is WB, if A loses the game then probability is LA and if b loses the game probability is LB.
Then, let us start with the assumption that A wins the game in the first round by getting a club.
So, the probability of getting a club is given by:
 $ \begin{align}
  & P\left( WA \right)=\dfrac{13}{52} \\
 & \Rightarrow P\left( WA \right)=\dfrac{1}{4} \\
\end{align} $
Now, as they are playing alternately so to get the condition of winning by A again, we need to wait for the third round as second round was played by B.
Moreover, one more condition is there that both A and B in the first and second round loses.
So, we get the probability of losing of A or B by subtracting the winning probability as:
 $ \begin{align}
  & P\left( LA \right)=P\left( LB \right)=1-\dfrac{1}{4} \\
 & \Rightarrow P\left( LA \right)=P\left( LB \right)=\dfrac{3}{4} \\
\end{align} $
So, probability of A winning in the third round is given as:
\[\begin{align}
  & P\left( A \right)=P\left( LA \right)\times P\left( LB \right)\times P\left( WA \right) \\
 & \Rightarrow P\left( A \right)=\dfrac{3}{4}\times \dfrac{3}{4}\times \dfrac{1}{4} \\
\end{align}\]
Now, for getting the winning condition of A in fifth round, we get:
\[P\left( A \right)=\dfrac{3}{4}\times \dfrac{3}{4}\times \dfrac{3}{4}\times \dfrac{3}{4}\times \dfrac{1}{4}\]
So, to get the total probability of winning of A, we add all the conditions as:
 $ P\left( A \right)=\dfrac{1}{4}+\dfrac{3}{4}\times \dfrac{3}{4}\times \dfrac{1}{4}+\dfrac{3}{4}\times \dfrac{3}{4}\times \dfrac{3}{4}\times \dfrac{3}{4}\times \dfrac{1}{4}+..... $
Now, for solving the infinite series we have the formula where a is the first term and r is the common ratio as:
 $ {{S}_{\infty }}=\dfrac{a}{1-r} $
Now, by using it for the above series, we get:
 $ \begin{align}
  & P\left( A \right)=\dfrac{\dfrac{1}{4}}{1-{{\left( \dfrac{3}{4} \right)}^{2}}} \\
 & \Rightarrow P\left( A \right)=\dfrac{\dfrac{1}{4}}{1-\left( \dfrac{9}{16} \right)} \\
 & \Rightarrow P\left( A \right)=\dfrac{\dfrac{1}{4}}{\dfrac{7}{16}} \\
 & \Rightarrow P\left( A \right)=\dfrac{1}{4}\times \dfrac{16}{7} \\
 & \Rightarrow P\left( A \right)=\dfrac{4}{7} \\
\end{align} $
Now, we get the probability of winning of A as $ \dfrac{4}{7} $ .
So, the probability of winning of B is given by:
 $ \begin{align}
  & P\left( B \right)=1-P\left( A \right) \\
 & \Rightarrow P\left( B \right)=1-\dfrac{4}{7} \\
 & \Rightarrow P\left( B \right)=\dfrac{3}{7} \\
\end{align} $
So, we get the probability of B wins as $ \dfrac{3}{7} $ .
So, the correct answer is “Option A”.

Note: Now, to solve these types of questions we need to know some of the basic summation of the series as it is important to solve these types of problems. So, let us we have a series with is tending towards infinity and we need to find the summation of that series with first term as a and r being the ratio of the consecutive terms as $ r=\dfrac{{{a}_{2}}}{{{a}_{1}}} $ where $ {{a}_{2}} $ is second term of the series and $ {{a}_{1}} $ is the first term of the series. So the summation of such a series is $ {{S}_{\infty }}=\dfrac{a}{1-r} $ .