# Math Help - Find the probability that player 2 wins

1. ## 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.

Thanks

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

$P(\text{\#2 wins}) \:=\:\frac{42}{127} \:\approx\:33.1\%$ . . . 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.

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