I cannot prove that $P(win) \approx {1 \over \sqrt{n}}$, but here is a proof that:
Claim: $P(win) \le {2 \over \sqrt{n}}$
which of course implies $P(win) \to 0$, answering the OP question.
Proof: For convenience, let $G =$ the OP's original game. Consider some large, fixed $n$. Define:
We will separately bound $P(A), P(B)$ both $\le {1 \over \sqrt{n}}$. The main claim then follows because $P(win) = P(A)+P(B)$.
Lemma: $P(A) \le {1\over \sqrt{n}}$: Imagine a modified game $G'$ where the die is always rolled for all $n$ rounds, and the game result is the first winning or losing roll. This modified game $G'$ is clearly equivalent to the OP's truncated version $G$.
Let random variable $X=$ no. of winning rolls among the initial $\sqrt{n}$ rounds of $G'$. To win on or before $\sqrt{n}$ rounds, it is necessary (though not sufficient) that $X\ge 1$, i.e. $P(A) \le P(X\ge 1)$. Meanwhile, for any non-negative integer r.v.s, we have:
$$P(X\ge 1) = \sum_{k=1}^\infty P(X=k) \le \sum_{k=1}^\infty kP(X=k) = E[X]$$
In this case, by linearity, $E[X] = {1\over n} \sqrt{n} = {1 \over \sqrt{n}}.$ Combining, we have:
$$P(A) \le P(X\ge 1) \le E[X] = {1 \over \sqrt{n}} \;\;\; \square$$
Lemma: $P(B) \le {1 \over \sqrt{n}}$: First, define:
- event $E= $ game $G$ is inconclusive (neither won nor lost) after $\sqrt{n}$ rounds
Note that event $B$ requires event $E$, i.e. $P(B) = P(B \cap E) = P(E) P(B \mid E)$.
Now imagine two different modified games. In $G_1$, the game starts with $1$ to $\sqrt{n}$ as the losing numbers and adds one more losing number every round. By construction, $P(\text{win }G_1) = P(B \mid E)$.
In $G_2$, the game starts with $1$ to $\sqrt{n}$ as the losing numbers and the set of losing numbers doesn't change for the rest of the game. Clearly, $G_2$ is easier to win than $G_1$ (e.g. via a sample-point by sample-point dominance argument).
The modified game $G_2$ is not limited to $n$ rounds, but it does terminate with probability $1$. Therefore, the ratio:
$${P(\text{win }G_2) \over P(\text{lose }G_2)} = {P(\text{win }G_2 \mid G_2 \text{ terminates}) \over P(\text{lose }G_2 \mid G_2 \text{ terminates})} = {\text{no. of winning rolls} \over \text{no. of losing rolls}} = {1 \over \sqrt{n}}$$
The easiest proof of the above is to consider $G_2$ as a $3$-state Markov chain with $2$ absorbing states, for win and loss. Alternately one can consider there to be $1+\sqrt{n}$ terminating states, by symmetry all equally likely, and then color exactly $1$ of them as "winning" and the others as "losing".
Combining everything, we have:
$$P(B) = P(E) P(B \mid E) \le P(B \mid E) = P(\text{win }G_1) \le P(\text{win }G_2) = {1 \over 1 + \sqrt{n}} < {1 \over \sqrt{n}} \;\;\; \square$$
Best Answer
Clearly $p_A + p_B = 1$ (somebody wins). And if $A$ misses on the first throw, then $B$ is in the position that $A$ was originally in; so $p_B = (5/6)p_A$. Putting these together, you find that $p_A=6/11$ and $p_B=5/11$.