I have a system where there are 5 players. At each of times $t$ and $t+1$, each of players 2 to 5 generates $0$ or $1$, where the probability to get $1$ is $p$.
If a player gets at least one $1$, then he participates to a drawing at each of times $t+2$ and $t+3$. In this case, at each of $t+2$ and $t+3$, the player draws randomly a seat among $R$ available seats; the $R$ seats at $t+2$ are different from the $R$ seats at $t+3$.
Player 1 always participates to the drawings, i.e. at each of times $t+2$ and $t+3$, he randomly draws a seat.
I am looking for the probability that, at each of times $t+2$ and $t+3$, (at least) a player chooses the same seat as player 1.
Attempt: Let $P_r$ be the probability we are looking for.
$P_r= \sum_{n=1}^4 P(\text{same choices } | n) P_n $, where:
(a) $P( \text{same choices }| n)$ is the probability that at least one the $n$ players chooses the same seat as player 1 at each of $t+2$ and $t+3$, given that $n$ players participate to the drawings, and (b) $P_n$ is the probability to have $n$ players participating to the drawings.
$P(\text{same choices } | n)= (1- (1- 1/R)^n)^2$.
$P_n= \frac{4!}{ n! (4-n)! } (1- (1-p)^2)^n ((1-p)^2)^{4-n} $.
You attempt is OK provided $t$ means the unique moment (that is we don’t repeat the procedure at moments $t+1$, $t+2$, and so on), the choices of sets have uniform distribution (that is for each player and each seat the probability that a player (if draws) chooses a set is $1/R$), and there are no dependencies between events which are not mentioned in the problem conditions.