The question can apparently be clarified as follows: Let $(X_1,Y_1),\dots,(X_n,Y_n)$ be iid pairs of real-valued random variables with a common continuous joint cumulative distribution function (cdf) $F$, so that $F(x,y)=P(X_i<x,Y_i<y)$ for all $i$ and all real $x,y$. What is then the probability \begin{equation} p:=P(\exists i\ \forall j\ne i\ X_i>X_j, Y_i>Y_j)? \end{equation}

To answer this question, note first that the random pairs $(X_1,Y_1),\dots,(X_n,Y_n)$ are exchangeable and also, because of the continuity of $F$, $P(X_i=X_j)=0=P(Y_i=Y_j)$ for all distinct $i$ and $j$. So,
\begin{equation} p=np_1,\quad p_1:=P(\forall j\ne 1\ X_1>X_j, Y_1>Y_j). \end{equation} Next, \begin{equation} p_1=Eg(X_1,Y_1),\quad g(X_1,Y_1):=P(\forall j\ne 1\ X_1>X_j, Y_1>Y_j|X_1,Y_1). \end{equation} Further, for all real $x,y$ \begin{equation} g(x,y)=P(\forall j\ne 1\ x>X_j, y>Y_j)=F(x,y)^{n-1}. \end{equation} Thus, \begin{equation} p_1=Eg(X_1,Y_1)=\int_{\mathbb R^2}F(x,y)^{n-1}\,dF(x,y) \end{equation} and \begin{equation} p=n\int_{\mathbb R^2}F(x,y)^{n-1}\,dF(x,y). \end{equation}

In particular, if $X_1$ and $Y_1$ are independent, so that $F(x,y)=G(x)H(y)$ for all real $x,y$ (where $G$ and $H$ are the cdf's of $X_1$ and $Y_1$), then \begin{multline} p=n\int_{\mathbb R^2}G(x)^{n-1}H(y)^{n-1}\,dG(x)\,dH(y) \\ =n\int_{\mathbb R}G(x)^{n-1}\,dG(x)\; \int_{\mathbb R}H(y)^{n-1}\,dH(y) \\ =n\int_0^1 u^{n-1}\,du\;\int_0^1v^{n-1}\,dv =\frac1n, \end{multline} which also follows immediately because the random pairs $(X_1,Y_1),\dots,(X_n,Y_n)$ are exchangeable.

The question can apparently be clarified as follows:

Let $(X_1,Y_1),\dots,(X_n,Y_n)$ be iid pairs of real-valued random variables with a common continuous joint cumulative distribution function (cdf) $F$, so that $F(x,y)=P(X_i<x,Y_i<y)$ for all $i$ and all real $x,y$. What is then the probability \begin{equation} p:=P(\exists i\ \forall j\ne i\ X_i>X_j, Y_i>Y_j)? \end{equation}

To answer this question, note first that the random pairs $(X_1,Y_1),\dots,(X_n,Y_n)$ are exchangeable and also, because of the continuity of $F$, $P(X_i=X_j)=0=P(Y_i=Y_j)$ for all distinct $i$ and $j$. So,
\begin{equation} p=np_1,\quad p_1:=P(\forall j\ne 1\ X_1>X_j, Y_1>Y_j). \end{equation} Next, \begin{equation} p_1=Eg(X_1,Y_1),\quad g(X_1,Y_1):=P(\forall j\ne 1\ X_1>X_j, Y_1>Y_j|X_1,Y_1). \end{equation} Further, for all real $x,y$ \begin{equation} g(x,y)=P(\forall j\ne 1\ x>X_j, y>Y_j)=F(x,y)^{n-1}. \end{equation} Thus, \begin{equation} p_1=Eg(X_1,Y_1)=\int_{\mathbb R^2}F(x,y)^{n-1}\,dF(x,y) \end{equation} and \begin{equation} p=n\int_{\mathbb R^2}F(x,y)^{n-1}\,dF(x,y). \end{equation}

In particular, if $X_1$ and $Y_1$ are independent, so that $F(x,y)=G(x)H(y)$ for all real $x,y$ (where $G$ and $H$ are the cdf's of $X_1$ and $Y_1$), then \begin{multline} p=n\int_{\mathbb R^2}G(x)^{n-1}H(y)^{n-1}\,dG(x)\,dH(y) \\ =n\int_{\mathbb R}G(x)^{n-1}\,dG(x)\; \int_{\mathbb R}H(y)^{n-1}\,dH(y) \\ =n\int_0^1 u^{n-1}\,du\;\int_0^1v^{n-1}\,dv =\frac1n, \end{multline} which also follows immediately because the random pairs $(X_1,Y_1),\dots,(X_n,Y_n)$ are exchangeable.

The question can apparently be clarified as follows: Let $(X_1,Y_1),\dots,(X_n,Y_n)$ be iid pairs of real-valued random variables with a common continuous joint cumulative distribution function $F$, so that $F(x,y)=P(X_i<x,Y_i<y)$ for all $i$ and all real $x,y$. What is then the probability \begin{equation} p:=P(\exists i\ \forall j\ne i\ X_i>X_j, Y_i>Y_j)? \end{equation}

To answer this question, note first that the random pairs $(X_1,Y_1),\dots,(X_n,Y_n)$ are exchangeable and also, because of the continuity of $F$, $P(X_i=X_j)=0=P(Y_i=Y_j)$ for all distinct $i$ and $j$. So,
\begin{equation} p=np_1,\quad p_1:=P(\forall j\ne 1\ X_1>X_j, Y_1>Y_j). \end{equation} Next, \begin{equation} p_1=Eg(X_1,Y_1),\quad g(X_1,Y_1):=P(\forall j\ne 1\ X_1>X_j, Y_1>Y_j|X_1,Y_1). \end{equation} Further, for all real $x,y$ \begin{equation} g(x,y)=P(\forall j\ne 1\ x>X_j, y>Y_j)=F(x,y)^{n-1}. \end{equation} Thus, \begin{equation} p_1=Eg(X_1,Y_1)=\int_{\mathbb R^2}F(x,y)^{n-1}\,dF(x,y) \end{equation} and \begin{equation} p=n\int_{\mathbb R^2}F(x,y)^{n-1}\,dF(x,y). \end{equation}

The question can apparently be clarified as follows: Let $(X_1,Y_1),\dots,(X_n,Y_n)$ be iid pairs of real-valued random variables with a common continuous joint cumulative distribution function (cdf) $F$, so that $F(x,y)=P(X_i<x,Y_i<y)$ for all $i$ and all real $x,y$. What is then the probability \begin{equation} p:=P(\exists i\ \forall j\ne i\ X_i>X_j, Y_i>Y_j)? \end{equation}

To answer this question, note first that the random pairs $(X_1,Y_1),\dots,(X_n,Y_n)$ are exchangeable and also, because of the continuity of $F$, $P(X_i=X_j)=0=P(Y_i=Y_j)$ for all distinct $i$ and $j$. So,
\begin{equation} p=np_1,\quad p_1:=P(\forall j\ne 1\ X_1>X_j, Y_1>Y_j). \end{equation} Next, \begin{equation} p_1=Eg(X_1,Y_1),\quad g(X_1,Y_1):=P(\forall j\ne 1\ X_1>X_j, Y_1>Y_j|X_1,Y_1). \end{equation} Further, for all real $x,y$ \begin{equation} g(x,y)=P(\forall j\ne 1\ x>X_j, y>Y_j)=F(x,y)^{n-1}. \end{equation} Thus, \begin{equation} p_1=Eg(X_1,Y_1)=\int_{\mathbb R^2}F(x,y)^{n-1}\,dF(x,y) \end{equation} and \begin{equation} p=n\int_{\mathbb R^2}F(x,y)^{n-1}\,dF(x,y). \end{equation}

In particular, if $X_1$ and $Y_1$ are independent, so that $F(x,y)=G(x)H(y)$ for all real $x,y$ (where $G$ and $H$ are the cdf's of $X_1$ and $Y_1$), then \begin{multline} p=n\int_{\mathbb R^2}G(x)^{n-1}H(y)^{n-1}\,dG(x)\,dH(y) \\ =n\int_{\mathbb R}G(x)^{n-1}\,dG(x)\; \int_{\mathbb R}H(y)^{n-1}\,dH(y) \\ =n\int_0^1 u^{n-1}\,du\;\int_0^1v^{n-1}\,dv =\frac1n, \end{multline} which also follows immediately because the random pairs $(X_1,Y_1),\dots,(X_n,Y_n)$ are exchangeable.