Why the hint is required depends on the way you derived your result, not just on the result you post.
However, here's an alternative solution.
Denote
$$ p_{k,n}:=P(\text{Player #$k$ wins an $n$ round tournament}).$$
As a preliminary step, we should eliminate the cases $p=\frac12$ as then everything is symmetric, i.e. the probability that player #2 (or anybody else) wins, is exactly $\frac1n$. Similarly, if $p=1$, then #1 wins the tournament a.s., hence $p_{2,n}=0$. And if $p=0$, then the max player will win a.s., i.e. then $p_{2,n}=0$ except that $p_{2,1}=1$.
After these remarks, we don't get into trouble if we wish to divide by $p$, or $1-p$ or $1-2p$ below.
Next, let us compute the probability that player $1$ wins the tournament. To do so, he must survive the first round and then win a tournament of $2^{n-1}$ players among which he (after renumbering the players to fill the gaps without changing relative order) is still number one. Then we have
$$ p_{1,n}=\begin{cases}1&\text{if }n=0,\\
p\cdot p_{1,n-1}&\text{if }n>0,\end{cases}$$
In other words
$$p_{1,n}=p^n.$$
Now what about player $2$? To win, he must survive the first round; that is he
- either plays player #1 (with probaility $\frac1{2^n-1}$) and wins with probaility $1-p$, after which he is automatically the #1 of the remaining players
- or plays somebody else (with $1-\frac1{2^n-1}$) and wins with probability $p$. Player #1 also wins with porbability $p$, and then player #2 is still #2 of the remaining players
- or as before, but player #1 loses his match, so our #2 becomes #1 of the survivors.
That is, for $n>0$ (otherwise, a player #2 doesn't even participate) we have
$$\begin{align}p_{2,n}&=\frac1{2^n-1}\cdot(1-p)\cdot p_{1,n-1}+\left(1-\frac1{2^n-1}\right)\cdot\left( p^2\cdot p_{2,n-1}+ p\cdot (1-p)\cdot p_{1,n-1}\right)
\\&=\frac{p^{n-1}(1-p)}{2^{n}-1} +\frac{(2^{n}-2)(1-p)p^n}{2^{n}-1}+\frac{(2^{n}-2)p^2}{2^{n}-1}\cdot p_{2,n-1}.
\end{align}$$
(Note that this immediately gives $p_{2,1}=1-p$).
If we substitue $p_{2,n}=\frac{p^n}{2^n-1}(a_n+c)$, we obtain
$$ a_n +c= \frac{1-p}{p}+(2^n-2)(1-p)+2p(a_{n-1}+c),$$
hence by chosing $c=\frac{1-p}{p(1-2p)}$
$$ a_n = (2^n-2)(1-p)+2pa_{n-1}.$$
Note that $p_{2,1}=1-p$ implies $a_1=\frac{2(1-p)}{2p-1}$, so we can extend this to $a_0=\frac{1-p}{(2p-1)p}$.
Substitute again, $a_n=2^np^nb_n$, to find
$$ b_n = \frac{(2^n-2)(1-p)}{2^np^n}+b_{n-1}= (1-2^{1-n})\frac{(1-p)}{p^n}+b_{n-1}.$$
Hence
$$\begin{align} b_n &= b_0+\sum_{k=1}^n(1-2^{1-k})\frac{(1-p)}{p^k}\\&=b_0 + (1-p)\sum_{k=1}^np^{-k}-2(1-p)\sum_{k=1}^n(2p)^{-k}\\&
= b_0 + (p^{-n}-1) -2(1-p) \frac{(2p)^{-n}-1}{1-2p}\\
&= \frac1{p^{n}}-\frac{2(1-p)}{(1-2p)(2p)^n} +\left(b_0-1+\frac{2(1-p)}{1-2p}\right)
\\&=\frac1{p^{n}}-\frac{2(1-p)}{(1-2p)(2p)^n}-\frac1p\end{align}$$
Therefore,
$$ a_n = 2^n-\frac{2(1-p)}{1-2p}-2^np^{n-1}$$
and ultimately,
$$ p_{2,n} = \left(2^n-2^np^{n-1}+\frac{1-p}{p}\right)\frac{p^n}{2^n-1}.$$
Your strategy looks right. Let us check your RJR calculation by using a different method, which I hope is accessible.
Let random variable $X_1$ be equal to $1$ if you win the first game, and $0$ if you lose the first game. Define $X_2$ and $X_3$ analogously. Then the number $Y$ of games won is given by $Y=X_1+X_2+X_3$. By the linearity of expectation we have $E(Y)=E(X_1)+E(X_2)+E(X_3)$.
We have $E(X_1)=E(X_3)=0.1$ and $E(X_2)=0.4$, so $E(Y)=0.6$.
Best Answer
The only thing we need to remember is that there are $n-1$ matches, each one between a different pair of opponents, and the possible sequences of events are completely symmetrical: for any possible history of the tournament (who played whom in each pairing of each round), any permutation of the player's roles in that history is equally likely to have occurred. In particular, no two players are more likely to meet than any other two players.
Give the players identifying numbers from $1$ to $n$ inclusive. For $1 \leq j < k \leq n,$ let $X_{j,k}$ be $1$ if player number $j$ ever plays player $k$ during the tournament, $0$ otherwise. Then $X_{j,k}$ is a Bernoulli random variable.
Since each $X_{j,k}$ is a Bernoulli variable, $E(X_{j,k}) = P(X_{j,k} = 1)$. Since no two players are more likely to meet than any other two players, no probability $P(X_{j,k} = 1)$ is greater than any other, so no expected value is greater than any other, and therefore $$ E(X_{1,2}) = E(X_{1,3}) = E(X_{2,3}) = \cdots = E(X_{n-1,n}). $$
There are $C_2^n$ such variables $X_{j,k}$. Their sum is
$$ Y = X_{1,2} + X_{1,3} + X_{2,3} + \cdots + X_{n-1,n} = n - 1 $$
because exactly $n - 1$ pairs must play, $X_{j,k} = 1$ for each of those pairs, and $X_{j,k} = 0$ for every other pair.
By linearity of expectation, \begin{multline} E(X_{1,2}) + E(X_{1,3}) + E(X_{2,3}) + \cdots + E(X_{n-1,n}) \\ = E(X_{1,2} + X_{1,3} + X_{2,3} + \cdots + X_{n-1,n}) = n - 1 \end{multline} since the expectation of a constant is just the constant.
But since the expected values of $E(X_{j,k})$ for each pair of players are all equal, $$ E(X_{1,2}) + E(X_{1,3}) + E(X_{2,3}) + \cdots + E(X_{n-1,n}) = C_2^n E(X_{A,B}) $$ where $A$ and $B$ are the numbers assigned to the two players in the question.
That is, $$ C_2^n E(X_{A,B}) = n - 1, $$ therefore $$ P(X_{A,B} = 1) = E(X_{A,B}) = \frac{n - 1}{C_2^n} $$
And of course $P(X_{A,B} = 1)$ is the probability that $A$ and $B$ meet.