Define the probabilities of success and failure for a single toss.

A tetrahedral die has 4 faces (1, 2, 3, 4). A success is throwing a 4.
Let $p$ be the probability of success (throwing a 4) and $q$ be the probability of failure (not throwing a 4).

$$p = P(\text{throwing a 4}) = \frac{1}{4}$$ $$q = P(\text{not throwing a 4}) = 1 - p = 1 - \frac{1}{4} = \frac{3}{4}$$

In the context of a geometric distribution, "tosses required before a game is lost" can be interpreted as the average number of failures before the first success (throwing a 4).
Let $X$ be the number of tosses until the first 4 appears. $X$ follows a Geometric distribution with parameter $p$. The expected value of $X$ is $E(X) = \frac{1}{p}$.
The average number of failures before the first success is $E(X) - 1 = \frac{1}{p} - 1 = \frac{1-p}{p} = \frac{q}{p}$.

$$E(\text{failures}) = \frac{3/4}{1/4} = 3$$

Player A throws first. A wins if they throw a 4 on their turn.
A can win on their 1st toss, or if A fails, B fails, then A wins on their 2nd turn (3rd toss overall), or if A fails, B fails, A fails, B fails, then A wins on their 3rd turn (5th toss overall), and so on.
The probabilities are:

• A wins on 1st toss: $p$
• A wins on 3rd toss (A fails, B fails, A succeeds): $q \cdot q \cdot p = q^2 p$
• A wins on 5th toss (A fails, B fails, A fails, B fails, A succeeds): $q \cdot q \cdot q \cdot q \cdot p = q^4 p$ This forms an infinite geometric series with first term $a = p$ and common ratio $r = q^2$. The sum of an infinite geometric series is $\frac{a}{1-r}$.

$$P(A \text{ wins}) = \frac{p}{1-q^2}$$

Substitute the values of $p = \frac{1}{4}$ and $q = \frac{3}{4}$:

$$P(A \text{ wins}) = \frac{1/4}{1-(3/4)^2} = \frac{1/4}{1-9/16}$$ $$P(A \text{ wins}) = \frac{1/4}{(16-9)/16} = \frac{1/4}{7/16}$$ $$P(A \text{ wins}) = \frac{1}{4} \times \frac{16}{7} = \frac{4}{7}$$

The final answers are:
i. The average number of tosses required before a game is lost: $\boxed{\text{3 tosses}}$
ii. The probability that A wins the game: $\boxed{\frac{4}{7}}$