一个均匀的正方体骰子，其各个面上分别标有四个数字 1、2、3 和 4 之一。投掷该骰子一次，朝上的面上的数字为得分 $T$。已知 $T$ 的众数是 2，且 $\text{P}(T = 1) = \text{P}(T = 4) = \frac{1}{6}$。求 $T$ 的概率分布。

1. Deduce the probability distribution of $T$ from the main stem.

From the main stem, the fair cubical die has 6 faces, so probabilities are multiples of $\frac{1}{6}$.
Given $\text{P}(T=1) = \text{P}(T=4) = \frac{1}{6}$, there is 1 face labelled '1' and 1 face labelled '4'.
This leaves $6 - 1 - 1 = 4$ faces for '2' and '3'.
Since the die contains all four numbers and the mode is 2, '2' must appear on more faces than '3'.
Thus, there are 3 faces for '2' and 1 face for '3'.
Therefore:
$\text{P}(T=1) = \frac{1}{6}$
$\text{P}(T=2) = \frac{3}{6} = \frac{1}{2}$
$\text{P}(T=3) = \frac{1}{6}$
$\text{P}(T=4) = \frac{1}{6}$

2. Find the probability of throwing an odd number and the probability of throwing a '2'.

The player throws an odd number if $T=1$ or $T=3$.
$\text{P}(\text{odd}) = \text{P}(T=1) + \text{P}(T=3) = \frac{1}{6} + \frac{1}{6} = \frac{1}{3}$
The player continues to throw if $T=2$, which has probability $\text{P}(T=2) = \frac{1}{2}$.

3. Formulate the probability $\text{P}(Y=0)$ as an infinite series.

The event $Y=0$ occurs if the player loses the game. This happens if an odd number is thrown before a '4'.
This corresponds to throwing a sequence of $k-1$ consecutive '2's followed by an odd number, for any $k \ge 1$.
$\text{P}(\text{lose on } k\text{-th throw}) = \left[ \text{P}(T=2) \right]^{k-1} \times \text{P}(\text{odd}) = \left(\frac{1}{2}\right)^{k-1} \left(\frac{1}{3}\right)$
Summing these probabilities for all possible values of $k$:
$\text{P}(Y=0) = \sum_{k=1}^\infty \left(\frac{1}{2}\right)^{k-1} \left(\frac{1}{3}\right)$
$\text{P}(Y=0) = \frac{1}{3} + \left(\frac{1}{2}\right)\left(\frac{1}{3}\right) + \left(\frac{1}{2}\right)^2\left(\frac{1}{3}\right) + \dots$

4. Evaluate the sum to infinity to prove the given result.

This is an infinite geometric series with first term $a = \frac{1}{3}$ and common ratio $r = \frac{1}{2}$.
Using the sum to infinity formula $S_\infty = \frac{a}{1 - r}$:
$\text{P}(Y=0) = \frac{\frac{1}{3}}{1 - \frac{1}{2}}$
$\text{P}(Y=0) = \frac{\frac{1}{3}}{\frac{1}{2}} = \frac{2}{3} \text{ (shown)}$

1. Deduce the probability $\text{P}(T=2)$ from the initial context.

A cubical die has 6 faces. Given $\text{P}(T=1) = \text{P}(T=4) = \frac{1}{6}$, these outcomes account for 2 faces.
The remaining 4 faces are labelled with 2 and 3, meaning $\text{P}(T=2) + \text{P}(T=3) = 1 - \frac{2}{6} = \frac{4}{6}$.
Since 2 is the mode, it must appear on more faces than any other number. Hence, 2 is on 3 faces and 3 is on 1 face.
$\text{P}(T=2) = \frac{3}{6} = \frac{1}{2}$

2. Determine the general formula for $\text{P}(Y=y)$ when $y \ge 1$.

For the player to win and end the game on the $y$-th throw, they must roll a '2' for the first $y-1$ throws and a '4' on the $y$-th throw.
Since the throws are independent:
$\text{P}(Y = y) = [\text{P}(T=2)]^{y-1} \cdot \text{P}(T=4) = \left(\frac{1}{2}\right)^{y-1} \left(\frac{1}{6}\right) \quad \text{for } y \ge 1$

3. Express $\text{E}(Y^2)$ as an infinite series.

The definition of the expected value of $Y^2$ is:
$\text{E}(Y^2) = 0^2 \cdot \text{P}(Y=0) + \sum_{y=1}^{\infty} y^2 \text{P}(Y=y)$
Substituting the expression for $\text{P}(Y=y)$:
$\text{E}(Y^2) = 0 + \sum_{y=1}^{\infty} y^2 \left(\frac{1}{6}\right) \left(\frac{1}{2}\right)^{y-1} = \frac{1}{6} \sum_{y=1}^{\infty} y^2 \left(\frac{1}{2}\right)^{y-1}$

4. Evaluate $\text{E}(Y^2)$ using the given identity.

The given identity is $1^2 + 2^2 x + 3^2 x^2 + 4^2 x^3 + \dots = \frac{1+x}{(1-x)^3}$, which can be written as $\sum_{y=1}^{\infty} y^2 x^{y-1} = \frac{1+x}{(1-x)^3}$.
Substituting $x = \frac{1}{2}$:
$\sum_{y=1}^{\infty} y^2 \left(\frac{1}{2}\right)^{y-1} = \frac{1 + \frac{1}{2}}{\left(1 - \frac{1}{2}\right)^3} = \frac{\frac{3}{2}}{\left(\frac{1}{2}\right)^3} = \frac{\frac{3}{2}}{\frac{1}{8}} = 12$
Thus, $\text{E}(Y^2) = \frac{1}{6}(12) = 2$.

5. Compute $\text{Var}(Y)$ using the standard variance formula.

Using the variance formula $\text{Var}(Y) = \text{E}(Y^2) - [\text{E}(Y)]^2$ and the given $\text{E}(Y) = \frac{2}{3}$:
$\text{Var}(Y) = 2 - \left(\frac{2}{3}\right)^2 = 2 - \frac{4}{9} = \frac{14}{9}$