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Math Help - Find the probability that player 2 wins

  1. #1
    Yan
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    Find the probability that player 2 wins

    A jar contains marbles of 7 different colors. Each player draws a marble on his/her turn, and returns it to the jar. A player wins the game by drawing the same color marble on two consecutive turns (so no one can win on the ¯rst turn). If there are three players, what is the probability that player 2 wins? Give your answer as a fraction.

    Can you show all your work, please!!
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  2. #2
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    Hello, Yan!

    A jar contains marbles of 7 different colors.
    Each player draws a marble on his/her turn, and returns it to the jar.
    A player wins the game by drawing the same color marble on two consecutive turns.
    If there are three players, what is the probability that player 2 wins?
    Give your answer as a fraction.
    I had to crank out a few scenarios to discern a pattern.
    Call the players #1, #2, and #3.

    On the first draw, each player can draw any color.

    Suppose #2 wins in the second round.
    . . #1 must not get his first color: . \text{prob} = \tfrac{6}{7}
    . . Then #2 gets his color: . \text{prob} = \tfrac{1}{7}
    P(\text{\#2, 2nd round}) \:=\:(\tfrac{1}{7})(\tfrac{6}{7})

    Suppose #2 wins in the third round.
    . . None of them get their number in the second round: . \text{prob} = (\tfrac{6}{7})^3
    . . #1 does not get his second color in the 3rd round: . \text{prob} = \tfrac{6}{7}
    . . Then #2 gets his second color: . \text{prob} \ \tfrac{1}{7}
    P(\text{\#2, 3rd round)} \:=\:(\tfrac{1}{7})(\tfrac{6}{7})^4

    Suppose #2 wins in the fourth round.
    . . None of them get their first numbers in the second round: . \text{prob} = (\tfrac{6}{7})^3
    . . None of the get their second numbers in the third round: . \text{prob} = (\tfrac{6}{7})^3
    . . #1 does not get this 3rd color in the 4th round: . \text{prob} = \tfrac{6}{7}
    . . Then #2 gets his 3rd color: . \text{prob} = \tfrac{1}{7}
    P(\text{\#2, 4th round}) \:=\:(\tfrac{1}{7})(\tfrac{6}{7})^7

    See the pattern?


    P(\text{\#2 wins}) \;=\;\frac{1}{7}\left(\frac{6}{7}\right) + \frac{1}{7}\left(\frac{6}{7}\right)^4 + \frac{1}{7}\left(\frac{6}{7}\right)^7 + \frac{1}{7}\left(\frac{6}{7}\right)^{10} + \hdots

    . . . . . . . . = \;\frac{1}{7}\!\cdot\!\frac{6}{7}\underbrace{\left[1 + \left(\frac{6}{7}\right)^3 + \left(\frac{6}{7}\right)^6 + \left(\frac{6}{7}\right)^9 + \hdots \right]}_{\text{geometric series}}

    The geometric series has: first term a = 1, common ratio r \,=\, (\tfrac{6}{7})^3 \,=\,\tfrac{216}{343}

    . . Its sum is: . \frac{1}{1 - \frac{216}{343}} \:=\:\frac{343}{127}


    Therefore: . P(\text{\#2 wins}) \;=\;\frac{6}{49}\cdot\frac{343}{127} \;=\;\frac{42}{127}

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  3. #3
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    The answer made me curious.

    P(\text{\#2 wins}) \:=\:\frac{42}{127} \:\approx\:33.1\% . . . slight less than \tfrac{1}{3} of the time.

    If the game were "fair", each player would win 33\tfrac{1}{3}\% of the time.
    So one of the players has an advantage . . . I'd suspected it's Player #1.


    So I worked out the probabilities for each player.

    . . \begin{array}{|c||c|c|}<br />
\text{Player} & \text{Prob} & \text{Percent} \\ \hline \hline \\[-4mm]<br />
\#1 & \frac{49}{127} & 39\% \\ \\[-4mm] \hline \\[-4mm]<br />
\#2 & \frac{42}{127} & 33\% \\ \\[-4mm] \hline \\[-4mm]<br />
\#3 & \frac{36}{127} & 28\%\\ \hline<br />
 \end{array}

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