Question

Suppose the probability for $A$ to win a game against $B$ is $0.4,A$ has two options of playing matches against $B$, One 'the best of $3$ games' and the other 'the best of $5$ games'. Which of the options, $A$ should choose?(No game ends in a draw).

• A. best of $3$ games
• B. best of $5$ games
• C. both options are equally probable
• D. can't say

Answer 1

Answer: (A)

best of $3$ games

Case$I$ when $A$ plays $3$ games against $B$. In this case, we have $n=3,p=0.4$ and $q=0.6$

Let $X$ denote the number of wins. Then,
$PP(X=r){=}^3{C}_r{\left(0.4\right)}^r{\left(0.6\right)}^{3-r};r=0,1,2,3$

$P_1=$Probability of winning he best of three games
$=P(X\ge 2)=P(X=2)+P(X=3)$
$={}^3{C}_2{\left(0.4\right)}^2{\left(0.6\right)}^1{+}^3{C}_3{\left(0.4\right)}^3{\left(0.6\right)}^0$
$=0.288+0.064=0.352$Case$II$when $A$ plays $5$ games against $B$. In this case, we have $n=5,p=0.4$ and $q=0.6$

Let $X$ denote the number of wins in $5$ games. Then,
$PP(X=r)={}^5{C}_r{\left(0.4\right)}^r{\left(0.6\right)}^{5-r};r=0,1,2,3,...,5$

$P_{12}=$Probability of winning he best of five games
$=P(X\ge 3)=P(X=3)+P(X=4)+P(X=5)$
$={}^5{C}_3{\left(0.4\right)}^3{\left(0.6\right)}^2{+}^5{C}_4{\left(0.4\right)}^4\left(0.6\right){+}^5{C}_5{\left(0.4\right)}^5{\left(0.6\right)}^0$
$=0.2304+0.0768+0.1024=0.31744$
Clearly, $P_1\ge P_2$.