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Team $A$ is playing team B in a series of three games. Team $A$'s probability of winning any particular game is nonzero, independent of other games, and is $1.6$ times as large as its probability of winning the series. What is the probability of team $A$ winning the series?

My approach :

Let the probability of winning any game be $x$, then $$x = 1.6 \left( x^3 + 3x(1 -x) \right)$$ because the series can be won as WWW, WLW, LWW, WWL.

Please help.

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2 Answers 2

up vote 4 down vote accepted

Yes your approach is indeed correct. However, the equation you have obtained is incorrect. You will obtain $$x = 1.6(x^3 + 3\color{red}{x^2}(1-x)).$$ Since you are given that the probability of $A$ winning a game is non-zero, we can cancel off the $x$ to get $$5 = 8 (x^2 + 3x -3x^2) = -16x^2 + 24x \implies 16x^2 - 24x + 5 = 0$$ Now solve the above quadratic equation, with the constraint the $0 <x < 1$.

$$16x^2 - 24x + 5 = 0 \implies x = \dfrac{24 \pm \sqrt{24^2 - 4 \times 16 \times 5}}{2 \times 16} = \dfrac34 \pm \dfrac12 = \dfrac14, \dfrac54.$$ Since $x$ denotes, a non-zero probability, we get that the desired probability is $x = \dfrac14$.

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Yes sir Marvis, sorry, it was a a typo. I'm getting answer as 5/32. Thanks for helping me. You're awesome. –  Bazinga Jun 30 '12 at 15:58
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You equation isn’t quite right: the probability that $A$ wins a $WWL$ series is $x^2(1-x)$, not $x(1-x)$. Thus, the equation should be $x=1.6\big(x^3+3x^2(1-x)\big)$. Multiply out the righthand side to get $1.6(x^3+3x^2-3x^3)=1.6(3x^2-2x^3)=4.8x^2-3.2x^3$, and combine everything on one side of the equation: $3.2x^3-4.8x^2+x=0$. You can get rid of the decimals by multiplying through by $5$ to get $16x^3-24x^2+5x=0$. Now factor out the obvious common factor: $x(16x^2-24x+5)=0$. One solution is $x=0$, but you’ve been told that this isn’t the solution, so it must be the case that $16x^2-24x+5=0$. That’s just a quadratic, so it’s easy to solve.

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