Question-80

Carlsen and Anand play a chess match in which the first player to win a game wins the match. Carlsen wins with probability \(0.4\), Anand with probability \(0.3\), the game is a draw with probability \(0.3\). The match has at most ten games. If all ten games are draws, then the match is a draw.

Part-1

What is the probability that Carlsen wins the match? Enter your answer correct to two decimal places.

NoteAnswer

\(0.57\)

NoteSolution

The required probability is:

\[ 0.4 \times [1 + 0.3 + 0.3^{2} + \cdots + 0.3^{9}] \approx 0.57 \]

Part-2

If the duration of the match is \(T\), find its PMF.

• \(\displaystyle p( t) =\begin{cases} 0.3^{t-1} \times 0.7, & 1\leqslant t\leqslant 9\\ 0.3^{9} , & t=10\\ 0, & \text{otherwise} \end{cases}\)

• \(\displaystyle p( t) =\begin{cases} 0.4^{t-1} \times 0.7, & 1\leqslant t\leqslant 9\\ 0.4^{9} , & t=10\\ 0, & \text{otherwise} \end{cases}\)

• \(\displaystyle p( t) =\begin{cases} 0.3^{t-1} \times 0.4, & 1\leqslant t\leqslant 9\\ 0.3^{9} , & t=10\\ 0, & \text{otherwise} \end{cases}\)

NoteAnswer

• \(\displaystyle p( t) =\begin{cases} 0.3^{t-1} \times 0.7, & 1\leqslant t\leqslant 9\\ 0.3^{9} , & t=10\\ 0, & \text{otherwise} \end{cases}\)

• \(\displaystyle p( t) =\begin{cases} 0.4^{t-1} \times 0.7, & 1\leqslant t\leqslant 9\\ 0.4^{9} , & t=10\\ 0, & \text{otherwise} \end{cases}\)

• \(\displaystyle p( t) =\begin{cases} 0.3^{t-1} \times 0.4, & 1\leqslant t\leqslant 9\\ 0.3^{9} , & t=10\\ 0, & \text{otherwise} \end{cases}\)

NoteSolution

For \(1 \leqslant t \leqslant 9\), if \(t\) is the duration, then there are \(t - 1\) draws. The probability of this is \(0.3^{t - 1}\) since all of them are independent. The \(t^{th}\) game has a decisive result, which could be a victory for either player. This happens with probability \(0.3 + 0.4 = 0.7\). Hence, \(P(T = t) = 0.3^{t - 1} \times 0.7\). For \(T = 10\), there nine draws and the tenth could have any result. No matter what the result of the last game, the duration will end up being \(10\). Hence, this branch has a probability \(0.3^{9}\).