Three red balls and two blue balls are placed in a bag. All the balls are indistinguishable except for their colours. Mr Red and Mr Blue take turns to draw a ball from the bag at random with replacement. The first player to draw the ball whose colour matches his name wins the game and the game stops immediately. If Mr Red draws first, find the probability that

1. Determine the required sequence of outcomes for Mr Red to win on his third draw

For Mr Red to win on his third draw, the first four draws must not result in a win. The sequence of draws must be:
1st draw (Mr Red): Fails to win (draws Blue)
2nd draw (Mr Blue): Fails to win (draws Red)
3rd draw (Mr Red): Fails to win (draws Blue)
4th draw (Mr Blue): Fails to win (draws Red)
5th draw (Mr Red): Wins (draws Red)

2. Calculate the probability of this exact sequence occurring

Since the balls are drawn with replacement, the probability of drawing a red ball is always $\frac{3}{5}$ and a blue ball is $\frac{2}{5}$.
$\text{Probability} = \text{P(Blue)} \times \text{P(Red)} \times \text{P(Blue)} \times \text{P(Red)} \times \text{P(Red)}$
$\text{Probability} = \left(\frac{2}{5}\right) \times \left(\frac{3}{5}\right) \times \left(\frac{2}{5}\right) \times \left(\frac{3}{5}\right) \times \left(\frac{3}{5}\right)$
$\text{Probability} = \frac{108}{3125}$

1. Identify the outcomes required for Mr Blue to win on his $n^{\text{th}}$ draw.

For Mr Blue to win on his $n^{\text{th}}$ draw, the game must go on for $n-1$ full rounds without a winner, followed by Mr Red failing on his $n^{\text{th}}$ draw, and then Mr Blue winning.
In each of the first $n-1$ rounds:

• Mr Red must draw a Blue ball: $P(\text{Red draws B}) = \frac{2}{5}$
• Mr Blue must draw a Red ball: $P(\text{Blue draws R}) = \frac{3}{5}$
  On the $n^{\text{th}}$ round:
• Mr Red must draw a Blue ball: $P(\text{Red draws B}) = \frac{2}{5}$
• Mr Blue must draw a Blue ball to win: $P(\text{Blue draws B}) = \frac{2}{5}$

2. Calculate the probability of this specific sequence of events.

$P(\text{Mr Blue wins at his } n^{\text{th}} \text{ draw}) = \left( \frac{2}{5} \times \frac{3}{5} \right)^{n-1} \times \left( \frac{2}{5} \times \frac{2}{5} \right)$
$= \left(\frac{6}{25}\right)^{n-1} \times \frac{4}{25}$
$= \frac{4}{25}\left(\frac{6}{25}\right)^{n-1}$

1. Find the probability of each player winning exactly at their own 3rd draw.

From earlier parts, the probability of Mr Red failing to win on his turn is $\frac{2}{5}$ (drawing a blue ball), and Mr Blue failing is $\frac{3}{5}$ (drawing a red ball).

$\text{P}(\text{Mr Red wins at his 3rd draw}) = \left(\frac{2}{5} \times \frac{3}{5}\right)^2 \times \frac{3}{5} = \frac{108}{3125}$

$\text{P}(\text{Mr Blue wins at his 3rd draw}) = \left(\frac{2}{5} \times \frac{3}{5}\right)^2 \times \frac{2}{5} \times \frac{2}{5} = \frac{144}{15625}$

2. Use the conditional probability formula to find the required probability.

$\text{Required probability} = \frac{\text{P}(\text{Mr Red wins at his 3rd draw})}{\text{P}(\text{Mr Red wins at 3rd draw}) + \text{P}(\text{Mr Blue wins at 3rd draw})}$

$= \frac{\frac{108}{3125}}{\frac{108}{3125} + \frac{144}{15625}}$

$= \frac{\frac{540}{15625}}{\frac{540}{15625} + \frac{144}{15625}}$

$= \frac{540}{684}$

$= \frac{15}{19}$

1. Express the total probability of Mr Blue winning as the sum of an infinite geometric series.

The probability that Mr Blue wins is the sum of the probabilities that he wins on his $n^{\text{th}}$ draw for all $n \ge 1$.
$P(\text{Mr Blue wins}) = \sum_{n=1}^\infty \left(\frac{6}{25}\right)^{n-1} \left(\frac{4}{25}\right)$
This is an infinite geometric series with first term $a = \frac{4}{25}$ and common ratio $r = \frac{6}{25}$.

2. Calculate the sum to infinity to find the total probability.

$P(\text{Mr Blue wins}) = \frac{a}{1 - r} = \frac{\frac{4}{25}}{1 - \frac{6}{25}}$
$= \frac{\frac{4}{25}}{\frac{19}{25}} = \frac{4}{19}$