一个袋子中装有一个绿色圆盘、$r$ 个红色圆盘和 $2r$ 个蓝色圆盘，其中 $r>1$。

1. Find the probabilities of drawing a green disc and not drawing a green disc.

The total number of discs is $1 + r + 2r = 3r + 1$.

Let $G$ be the event of drawing a green disc, and $G'$ be the event of not drawing a green disc.
$\mathrm{P}(G) = \dfrac{1}{3r+1}$
$\mathrm{P}(G') = \dfrac{3r}{3r+1}$

2. Express the probability of Sim winning as an infinite series.

Sim wins if she draws a green disc on her 1st turn, or her 2nd turn (after Tim fails), or her 3rd turn, and so on. This represents an infinite sequence of mutually exclusive events:
$\mathrm{P}(\text{Sim wins}) = \mathrm{P}(G) + \mathrm{P}(G'G'G) + \mathrm{P}(G'G'G'G'G) + \dots$

Since the disc is replaced each time, the draws are independent:
$\mathrm{P}(\text{Sim wins}) = \left(\dfrac{1}{3r+1}\right) + \left(\dfrac{3r}{3r+1}\right)^2 \left(\dfrac{1}{3r+1}\right) + \left(\dfrac{3r}{3r+1}\right)^4 \left(\dfrac{1}{3r+1}\right) + \dots$

3. Evaluate the infinite geometric series to show the required probability.

This is an infinite geometric series with first term $a = \dfrac{1}{3r+1}$ and common ratio $R = \left(\dfrac{3r}{3r+1}\right)^2$.
Since $r > 1$, $0 < R < 1$, the sum to infinity converges.

$\begin{aligned} S_\infty &= \dfrac{a}{1 - R} \\ &= \frac{\dfrac{1}{3r+1}}{1 - \left(\dfrac{3r}{3r+1}\right)^2} \\ &= \frac{\dfrac{1}{3r+1}}{\dfrac{(3r+1)^2 - 9r^2}{(3r+1)^2}} \\ &= \frac{\dfrac{1}{3r+1}}{\dfrac{9r^2 + 6r + 1 - 9r^2}{(3r+1)^2}} \\ &= \frac{\dfrac{1}{3r+1}}{\dfrac{6r + 1}{(3r+1)^2}} \\ &= \frac{1}{3r+1} \times \frac{(3r+1)^2}{6r+1} \\ &= \frac{3r+1}{6r+1} \text{ (shown)} \end{aligned}$

1. Determine the possible values of $T$ by evaluating the absolute differences in points between all possible pairs of drawn discs.

The points are Red (R) = $5$, Blue (B) = $2$, Green (G) = $0$.
Since $2$ discs are drawn together, this is without replacement (drawing two Green discs is impossible as there is only $1$).
The possible pairs drawn and their corresponding score difference $T$:

• R and R: $T = |5 - 5| = 0$
• B and B: $T = |2 - 2| = 0$
• B and G (or G and B): $T = |2 - 0| = 2$
• R and B (or B and R): $T = |5 - 2| = 3$
• R and G (or G and R): $T = |5 - 0| = 5$

The possible values of $t$ are $0, 2, 3$, and $5$.

2. Calculate $\mathrm{P}(T=0)$ by summing the probabilities of drawing two Red discs and drawing two Blue discs.

Total discs initially $= 1 + r + 2r = 3r + 1$.
$\begin{aligned} \mathrm{P}(T=0) &= \mathrm{P}(\text{R, R}) + \mathrm{P}(\text{B, B}) \\ &= \left(\frac{r}{3r+1} \times \frac{r-1}{3r}\right) + \left(\frac{2r}{3r+1} \times \frac{2r-1}{3r}\right) \\ &= \frac{r^2-r}{3r(3r+1)} + \frac{4r^2-2r}{3r(3r+1)} \\ &= \frac{5r^2-3r}{3r(3r+1)} \\ &= \frac{r(5r-3)}{3r(3r+1)} \\ &= \frac{5r-3}{3(3r+1)} \end{aligned}$

3. Calculate $\mathrm{P}(T=2)$ from the probability of drawing one Blue and one Green disc.

$\begin{aligned} \mathrm{P}(T=2) &= \mathrm{P}(\text{B, G}) + \mathrm{P}(\text{G, B}) \\ &= \left(\frac{2r}{3r+1} \times \frac{1}{3r}\right) + \left(\frac{1}{3r+1} \times \frac{2r}{3r}\right) \\ &= \frac{2r}{3r(3r+1)} + \frac{2r}{3r(3r+1)} \\ &= \frac{4r}{3r(3r+1)} \\ &= \frac{4}{3(3r+1)} \end{aligned}$

4. Calculate $\mathrm{P}(T=3)$ from the probability of drawing one Red and one Blue disc.

$\begin{aligned} \mathrm{P}(T=3) &= \mathrm{P}(\text{R, B}) + \mathrm{P}(\text{B, R}) \\ &= \left(\frac{r}{3r+1} \times \frac{2r}{3r}\right) + \left(\frac{2r}{3r+1} \times \frac{r}{3r}\right) \\ &= \frac{2r^2}{3r(3r+1)} + \frac{2r^2}{3r(3r+1)} \\ &= \frac{4r^2}{3r(3r+1)} \\ &= \frac{4r}{3(3r+1)} \end{aligned}$

5. Calculate $\mathrm{P}(T=5)$ from the probability of drawing one Red and one Green disc.

$\begin{aligned} \mathrm{P}(T=5) &= \mathrm{P}(\text{R, G}) + \mathrm{P}(\text{G, R}) \\ &= \left(\frac{r}{3r+1} \times \frac{1}{3r}\right) + \left(\frac{1}{3r+1} \times \frac{r}{3r}\right) \\ &= \frac{r}{3r(3r+1)} + \frac{r}{3r(3r+1)} \\ &= \frac{2r}{3r(3r+1)} \\ &= \frac{2}{3(3r+1)} \end{aligned}$

1. Calculate the probabilities for the non-zero values of $T$ (derived from part (b)(i)).

The total number of ways to draw 2 discs from $3r+1$ discs is $\binom{3r+1}{2} = \frac{3r(3r+1)}{2}$.
We compute the probability $\mathrm{P}(T=t)$ for each possible non-zero absolute difference in scores:

• $T = |2-0| = 2$ (1 Blue, 1 Green): $\mathrm{P}(T=2) = \frac{2r \times 1}{\frac{3r(3r+1)}{2}} = \frac{4}{3(3r+1)}$
• $T = |5-2| = 3$ (1 Red, 1 Blue): $\mathrm{P}(T=3) = \frac{r \times 2r}{\frac{3r(3r+1)}{2}} = \frac{4r}{3(3r+1)}$
• $T = |5-0| = 5$ (1 Red, 1 Green): $\mathrm{P}(T=5) = \frac{r \times 1}{\frac{3r(3r+1)}{2}} = \frac{2}{3(3r+1)}$

(Cases where $T=0$ will contribute $0$ to the expectation, so they do not need to be evaluated for this part.)

2. Apply the expectation formula to show the required result.

$\mathrm{E}(T) = \sum t \mathrm{P}(T=t)$
$\mathrm{E}(T) = 2\left(\frac{4}{3(3r+1)}\right) + 3\left(\frac{4r}{3(3r+1)}\right) + 5\left(\frac{2}{3(3r+1)}\right)$
$\mathrm{E}(T) = \frac{8 + 12r + 10}{3(3r+1)}$
$\mathrm{E}(T) = \frac{12r + 18}{3(3r+1)}$
$\mathrm{E}(T) = \frac{6(2r+3)}{3(3r+1)}$
$\mathrm{E}(T) = \frac{2(2r+3)}{3r+1}$

1. Determine the value of $r$ by formulating and solving the probability equation from part (c).

Let $G$, $B$, and $R$ be the events of drawing a green, blue, and red disc respectively.
For May to end up with more money than Jun, the mutually exclusive winning combinations for (May, Jun) are $(G, B)$, $(G, R)$, and $(B, R)$.
Since their draws are independent with replacement:
$\mathrm{P}(\text{May} > \text{Jun}) = \mathrm{P}(G)\mathrm{P}(B) + \mathrm{P}(G)\mathrm{P}(R) + \mathrm{P}(B)\mathrm{P}(R)$
$\frac{27}{112} = \left(\frac{1}{3r+1}\right)\left(\frac{2r}{3r+1}\right) + \left(\frac{1}{3r+1}\right)\left(\frac{r}{3r+1}\right) + \left(\frac{2r}{3r+1}\right)\left(\frac{r}{3r+1}\right)$
$\frac{27}{112} = \frac{2r^2+3r}{(3r+1)^2}$
$112(2r^2 + 3r) = 27(9r^2 + 6r + 1)$
$224r^2 + 336r = 243r^2 + 162r + 27$
$19r^2 - 174r + 27 = 0$
$(r - 9)(19r - 3) = 0$
Since $r > 1$ and is the integer count of discs, $r = 9$.

2. Calculate the expected winnings of the game for a player.

Using $r = 9$, the total number of discs is $3(9) + 1 = 28$.
Let $W$ be the player's winnings in dollars.
$\mathrm{E}(W) = 0.15\,\mathrm{P}(G) + 0.10\,\mathrm{P}(B) - 0.25\,\mathrm{P}(R)$
$\mathrm{E}(W) = 0.15\left(\frac{1}{28}\right) + 0.10\left(\frac{18}{28}\right) - 0.25\left(\frac{9}{28}\right)$
$\mathrm{E}(W) = \frac{0.15 + 1.80 - 2.25}{28}$
$\mathrm{E}(W) = -\frac{0.30}{28} = -\frac{3}{280} \approx -0.0107$

3. Conclude whether the game is fair based on the expected winnings.

A game is considered fair if the expected winnings are exactly $\char36{}0$.
Since $\mathrm{E}(W) = -\frac{3}{280} \neq 0$, the game is not fair for a player.