Question

Suppose the probability for A to win a game against B is 0.4. If A has an option playing either a “best of 3 games” or a “best of 5 games” match against B. Which option should A choose so that the probability of his winning the match is higher? (No game ends in draw)
A. Best of 5 games
B. Best of 3 games
C. Either best of 3 games or best of 5 games
D. None of the above

Answer 1

Here we use the concept of combinations to find out the probabilities of winning Best of 3 games and winning best of 5 games. We assume the probability of success of A as given in the question and using the formula for probability of failure we find probability of failure of A. Best of 3 games can have 2 or 3 wins in set of 3 games and best of 5 games can have 3, 4 or 5 wins in a set of 5 games.

Formula used: Combination formula is given by $^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}$, where n is total number of items w are choosing from and r is number of items we are selecting.
Probability of an event is given by dividing the number of favorable outcomes by total number of outcomes.

Complete step-by-step answer:
We have two players A and B
Probability of A winning a game against B is 0.4.
Probability that player A will lose the game is given by $1 -$probability of winning game by player A.
We denote winning the game by player A by $A'$and losing by player A by $A''$.
$P(A') = 0.4$
Then,$P(A'') = 1 - 0.4 = 0.6$
We can write probabilities in fraction form as
$P(A') = \dfrac{4}{{10}}$and $P(A'') = \dfrac{6}{{10}}$
For best of three games:
We have total number of games, n as 3
Number of games to be won to have best of three games is 2 and 3
Therefore, we use the formula of combination to find the probability of best of three.
Use the formula $^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}$
Use the factorial formula $n! = n(n - 1)!$ and substitute $P(A') = \dfrac{4}{{10}},P(A'') = \dfrac{6}{{10}}$
$\Rightarrow {P_1} = \dfrac{{3 \times 2!}}{{(1)!2!}}{\left( {\dfrac{4}{{10}}} \right)^2}\left( {\dfrac{6}{{10}}} \right) + \dfrac{{3!}}{{(0)!3!}}{\left( {\dfrac{4}{{10}}} \right)^3}$
Cancel the same terms from numerator and denominator, and put $0! = 1,1! = 1$
$\Rightarrow {P_1} = 3{\left( {\dfrac{4}{{10}}} \right)^2}\left( {\dfrac{6}{{10}}} \right) + {\left( {\dfrac{4}{{10}}} \right)^3}$
Open the bracket.
$\Rightarrow {P_1} = \dfrac{{3 \times 4 \times 4 \times 6}}{{10 \times 10 \times 10}} + \dfrac{{4 \times 4 \times 4}}{{10 \times 10 \times 10}}$
Multiply values in numerator and denominator.
$\Rightarrow {P_1} = \dfrac{{288}}{{1000}} + \dfrac{{64}}{{1000}}$
Take LCM
$\Rightarrow {P_1} = \dfrac{{352}}{{1000}}$
Write the fraction in decimal form
$\Rightarrow {P_1} = 0.352$
For best of five games:
We have total number of games, n as 5
Number of games to be won to have best of three games is 3, 4 and 5
Therefore, we use the formula of combination to find the probability of best of three.
Use the formula $^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}$
We know any value having power zero is equal to one.
Use the factorial formula $n! = n(n - 1)!$ and substitute $P(A') = \dfrac{4}{{10}},P(A'') = \dfrac{6}{{10}}$
$\Rightarrow {P_2} = \dfrac{{5 \times 4 \times 3!}}{{(2)!3!}}{\left( {\dfrac{4}{{10}}} \right)^3}{\left( {\dfrac{6}{{10}}} \right)^2} + \dfrac{{5 \times 4!}}{{(1)!4!}}{\left( {\dfrac{4}{{10}}} \right)^4}\left( {\dfrac{6}{{10}}} \right) + \dfrac{{5!}}{{(0)!5!}}{\left( {\dfrac{4}{{10}}} \right)^5}$
Cancel the same terms from numerator and denominator, and put $0! = 1,1! = 1$
$\Rightarrow {P_2} = 10{\left( {\dfrac{4}{{10}}} \right)^3}{\left( {\dfrac{6}{{10}}} \right)^2} + 5{\left( {\dfrac{4}{{10}}} \right)^4}\left( {\dfrac{6}{{10}}} \right) + {\left( {\dfrac{4}{{10}}} \right)^5}$
Open the bracket.
$\Rightarrow {P_2} = \dfrac{{10 \times 4 \times 4 \times 4 \times 6 \times 6}}{{10 \times 10 \times 10 \times 10 \times 10}} + \dfrac{{5 \times 4 \times 4 \times 4 \times 4 \times 6}}{{10 \times 10 \times 10 \times 10 \times 10}} + \dfrac{{4 \times 4 \times 4 \times 4 \times 4}}{{10 \times 10 \times 10 \times 10 \times 10}}$
Multiply values in numerator and denominator.
$\Rightarrow {P_2} = \dfrac{{23040}}{{100000}} + \dfrac{{7680}}{{100000}} + \dfrac{{1024}}{{100000}}$
Take LCM
$\Rightarrow {P_2} = \dfrac{{23040 + 7680 + 1024}}{{100000}}$
$\Rightarrow {P_2} = \dfrac{{31744}}{{100000}}$
Write the fraction in decimal form
$\Rightarrow {P_2} = 0.31744$
Now on comparing the values of ${P_1}$and ${P_2}$
$\Rightarrow 0.352 > 0.31744$
$\Rightarrow {P_1} > {P_2}$
So, the probability of A winning is higher in the best of three games.

So, the correct answer is “Option B”.

Note: Students might get confused in the concept of best of three and best of 5 games. Best of refers to more wins than losses, so in 3 match games a best will be if the player won more than he lost i.e. won 2 or 3. Similarly in a 5 match game a best will be if the player won more than he lost i.e. won 3, 4 or 5. Students are advised to keep the values of probabilities in fraction form as it is easier to add this way rather than decimal numbers.
Also, we use combinations here because the games can be switched like if a player lost the first game then he won the last two games, he lost the last game then he won the first two games etc.