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You have to play three games with opponents $A$ and $B$ in a specified sequence. You win the series if you win two consecutive games. $A$ is a stronger player than $B$. Which sequence maximizes your chance of winning the series?

  1. $AAB$
  2. $ABA$
  3. $BAB$
  4. $BAA$
  5. All are the same.
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3 Answers

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41 votes
Best answer

Let the three games in a series be called $G_1, G_2$ and $G_3$ respectively, and the probability of winning the game $x$ be denoted as $P(x)$

You can win the series if and only if: You win $G_2$ and you win atleast one of $\{G_1, G_3\}$.

$\large\boxed{P\left ( \substack{\text{winning}\\\text{the series}\\ \\{G_{1}G_{2}G_{3}}} \right ) = P(G_2) \times \Bigl ( P(G_1) + P(G_3) - P(G_1) \times P(G_3) \Bigr )\;}$


Let the probability of winning against player $A$ be $a$ and the probability of winning against player $B$ be $b$.

Then, $a < b.(A \text{ is a stronger player than} B,\text{ so probability of winning against } A$

$ \text{ is smaller compared to}\;B)$

Let $P(xyz)$ be the probability of winning the series in which the games played are against $x, y$ and $z$ in order.

  1. $P(AAB) = a(a+b-ab) = a^2 + ab - a^2b$
  2. $P(ABA) = b(a+a-aa) = 2ab - a^2b$
  3. $P(BAB) = a(b+b-b^2) = 2ab - ab^2$
  4. $P(BAA) = a(b+a-ba) = a^2 + ab - a^2b$

We can see that not all probabilities are equal, so option E is not correct.

We can also see that options A and D result in the same value, so they are not correct either.

Comparing option B and option C.

Since $a < b$ and $a,b\geq 0$, we have that $2ab - a^{2}b > 2ab - ab^{2}$

Hence, option B is the correct answer.

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@Pragy Agarwal 

Beautiful !!

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reshown by

Beautiful explanation... @Pragy Agarwal Sir

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 Sir why Option A and D cant be the answer? Why twi different sequences can have same probability

2. How to solve if the question would have asked for Finding the sequence with highest probability amongst all possible sequence in which atleast one A must be there

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4 votes
4 votes

 

Sloppier method but works :D May take 2 minutes during a test.

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0 votes
One change to the question..

If the options given include BBA then P(BBA) = $b^{_{2}} + ab - ab^{_{2}} > all other options$

substitute some values like A = 0.1, B = 0.2 you will get the answer
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