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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

Best answer
44 votes
44 votes

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.

edited by
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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
Answer:

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