Let’s first consider the joint density function \[ p(x_1,\dots,x_n|\theta) = \begin{cases} \prod 2x_i/\theta^2 & \text{if } x_{(1)}, x_{(n)} \in(0,\theta) \\ 0 & \text{otherwise} \end{cases} \]
We can rewrite this to \[ p(x_1,\dots,x_n|\theta) = \mathbb{1}_{(0,\theta)}(x_{(0)}) \mathbb{1}_{(0,\theta)}(x_{(n)}) 2\theta^{-2n}\prod x_i = \\ \color{green}{ \mathbb{1}_{(-\infty,\theta)}(x_{(n)}) 2\theta^{-2n}} \color{blue}{ \mathbb{1}_{(0,\infty)}(x_{(0)}) \prod x_i} \] By the factorization theorem, \(x_{(n)}\) is a sufifcient statistic.
The probability of any one \(x_j\leq t\) is the distribution function \[ \mathbb{P}(x_j\leq t|\theta) = \int_{-\infty}^t p(x|\theta)dx = \int_0^t 2x/\theta^2 dx = \left[x^2/\theta^2\right]_0^t = t^2/\theta^2 \] if \(t\in(0,\theta)\).
So the probability that all \(x_j\) are at most \(t\) is by independence \[ \mathbb{P}(x_1\leq t, \dots, x_n\leq t|\theta) = \mathbb{P}(x_{(n)}\leq t) = \prod_j \int_0^t p(x|\theta)dx = t^{2n}/\theta^{2n} \]
The density is the derivative of this: \[ p(x_{(n)}|\theta) = \frac{d}{dt}\frac{t^{2n}}{\theta^{2n}} = 2nt^{2n-1}/\theta^{2n} \]
To prove completeness, suppose \(\mathbb{E}g(x_{(n)}) = c\) for all \(\theta\). Then \[ \int_0^\theta g(t)2nt^{2n-1}dt/\theta^{2n} = c \\ \int_0^\theta g(t)t^{2n-1}dt = c\theta^{2n}/2n \]
Take the \(\theta\)-derivative of both sides (because the integral is difficult): \[ g(\theta)\theta^{2n-1} = c\theta^{2n-1} \]
So \(g(\theta)=c\) for almost every \(\theta\). Hence \(g(t)=c\) for almost every \(t\).
A sequence of games that team \(X\) wins will terminate with \(X\) and have some sequence of exactly two wins for \(X\) and at most two wins for \(Y\) preceding it. The same holds for a \(Y\) win.
Such a sequence corresponds to the outcome \(2\) from a binomial distribution on \(n=2, 3, 4\). Its probability is given by \[ \mathbb{P}(k|n,p) = {n\choose k}p^k(1-p)^{n-k} \]
Summing the options up, we get \[ \mathbb{P}(x,y|p) = \begin{cases} \theta{x+y\choose2}\theta^2(1-\theta)^{x+y-2} &\text{if $x=3$ and $y<3$} \\ (1-\theta){x+y\choose2}\theta^{x+y-2}(1-\theta)^2 &\text{if $y=3$ and $x<3$} \\ 0 & \text{otherwise} \end{cases} \]
We can rewrite this to \[ \mathbb{P}(x,y|p) = {x+y\choose 2}\delta_{3,\max(x,y)}\left( \theta^3(1-\theta)^{x+y-2}\mathbb{1}(x>y) + (1-\theta)^3\theta^{x+y-2}(1-\mathbb{1}(x>y)) \right) \]
\[ \theta^3{x+y\choose 2}(1-\theta)^{x+y-2}\mathbb{1}_{3}(x)\mathbb{1}_{[0,2]}(y) + (1-\theta)^3{x+y\choose 2}\theta^{x+y-2}\mathbb{1}_{3}(y)\mathbb{1}_{[0,2]}(x) \]