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Here are some preliminary computations. Assume the reference distribution is $(p(n))$. For every finite subset $I$ of $\mathbb N$, introduce the finite number $r(I)\ge1$ such that $$ \frac1{r(I)}=1-\sum_{k\in I}p(k). $$ Obviously, $P(X_1=n)=p(n)$ for every $n$. Likewise, $P(X_2=n)=E(p(n)r(X_1);X_1\ne n)$ hence $$ P(X_2=n)=p(n)(\alpha-p(n)r(n)),\qquad \alpha=\sum\limits_kp(k)r(k). $$ This shows that $X_1$ and $X_2$ are not equidistributed (if they were, $a_1-p(n)r(n)$ \alpha-p(n)r(n)$ would not depend on $n$, hence $p(n)$ would not either, but this is impossible since $(p(n))$ is a measure with finite mass on an infinite set).

One can also compute the joint distribution of $(X_1,X_2)$ as $$ P(X_1=n,X_2=k)=p(n)r(n)p(k)[k\ne n], $$ and this allows to expand $$ P(X_3=n)=E(p(n)r(X_1,X_2);X_1\ne n,X_2\ne n), $$ as the double sum $$ P(X_3=n)=p(n)\sum_{k\ne n}\sum_{i\ne n}[k\ne i]r(k,i)p(k)r(k)p(i), $$ but no simpler or really illuminating expression seems to emerge.
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Here are some preliminary computations. Write $(p(n))$ for Assume the reference distribution . Introduce also, for is $(p(n))$. For every finite subset $I$ of $\mathbb N$, introduce the finite number $r(I)\ge1$ such that $$ \frac1{r(I)}=1-\sum_{k\in I}p(k). $$ Obviously, $P(X_1=n)=p(n)$. P(X_1=n)=p(n)$ for every $n$. Likewise, $p_2(n)=E(p(n)r(X_1);X_1\ne P(X_2=n)=E(p(n)r(X_1);X_1\ne n)$ hence $$ P(X_2=n)=p(n)(a_1-p(n)r(n)),\qquad a_1=\sum\limits_kp(k)r(k). P(X_2=n)=p(n)(\alpha-p(n)r(n)),\qquad \alpha=\sum\limits_kp(k)r(k). $$ This shows that $X_1$ and $X_2$ are not equidistributed (if they were, $a_1-p(n)r(n)$ would not depend on $n$, hence $p(n)$ would not either, but this is impossible since $(p(n))$ is a measure with finite mass on an infinite set).

One can also compute the joint distribution of $(X_1,X_2)$ as $$ P(X_1=n,X_2=k)=p(n)r(n)p(k)[k\ne n], $$ and this allows to expand $$ P(X_3=n)=E(p(n)r(X_1,X_2);X_1\ne n,X_2\ne n), $$ as the double sum $$ P(X_3=n)=p(n)\sum_{k\ne n}\sum_{i\ne n}[k\ne i]r(k,i)p(k)r(k)p(i), $$ but no simpler or really illuminating expression seems to emerge.
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Here are some preliminary computations. Write $(p(n))$ for the reference distribution. Introduce also, for every finite subset $I$ of $\mathbb N$, the finite number $r(I)\ge1$ such that $$ \frac1{r(I)}=1-\sum_{k\in I}p(k). $$ Obviously, $P(X_1=n)=p(n)$. Likewise, $p_2(n)=E(p(n)r(X_1);X_1\ne n)$ hence $$ P(X_2=n)=p(n)(a_1-p(n)r(n)),\qquad a_1=\sum\limits_kp(k)r(k). $$ This shows that $X_1$ and $X_2$ are not equidistributed (if they were, $a_1-p(n)r(n)$ would not depend on $n$, hence $p(n)$ would not either, but this is impossible since $(p(n))$ is a measure with finite mass on an infinite set).

One can also compute the joint distribution of $(X_1,X_2)$ as $$ P(X_1=n,X_2=k)=p(n)r(n)p(k)[k\ne n], $$ and this allows to expand $$ P(X_3=n)=E(p(n)r(X_1,X_2);X_1\ne n,X_2\ne n), $$ as the double sum $$ P(X_3=n)=p(n)\sum_{k\ne n}\sum_{i\ne n}[k\ne i]r(k,i)p(k)r(k)p(i), $$ but no simpler or really illuminating expression seems to emerge.