Infinite number of students Consider countably infinite number of students. They all sit on an infinite line.
They all try to solve a problem. The probability of the event "student $N$ solves the problem" is $0.5$ $\forall N$. Every student can cheat and write off the correct solution from left or right neighbour. The probability of successful cheating is $0.25$.
Question: what is the probability of a fixed student to gain a correct solution to the problem?
My answer is $\frac{2}{3}$ but something tells me this is wrong.
 A: Assuming that a student can only try to copy from one neighbour: $p$ is the probability of a single student finding the solution, and $q$ is the probability of a student finding the solution given that one of his neighbours does not (at least not before he does). Equations for $p$ and $q$ are
$$p=0.5+(1-0.5)\cdot \Big(1-(1-q)^2\Big)\cdot0.25,$$
and
$$q=0.5+(1-0.5)\cdot q\cdot0.25.$$
This should be interpreted as follows: The student finds the solution himself with probability $0.5$. If he fails that (with probability $1-0.5$), he will look at his neighbours. Since he has at this point not solved the problem, his neighbours will have solved it with probability $q$. The probability that a given neighbour fails is $1-q$, so the probability that both neighbours fail is $(1-q)^2$. Thus the probability that at least one of our students neigbours finds the solution is $1-(1-q)^2$. If a neigbour finds the answer, then our student can successfully copy this answer with probability $0.25$.
The calculation for $q$ is similar, but here there is only one neigbour that can find the answer (if our student is looking at the neighbour on the right, then the neighbour won't find the answer on his left).
These equations gives $q=\frac47$ and $p=\frac{59}{98}\approx0.602$.

If a student can try to copy from one neighbour if he fails to copy from the other, we get a different equation for $p$:
$$p=0.5+(1-0.5)\cdot\Bigg(2\cdot q\cdot(1-q)\cdot0.25+q^2\cdot\bigg(1-(1-0.25)^2\bigg)\Bigg)$$
The change can be explained as follows: If the student fails by himself there are two possibilities where the student can try to cheat: either exactly one of his neighbours succeed, or both do. The probability of exacly one neighbour succeeding is $2\cdot q\cdot(1-q)$ (left neighbour succeeding and right neighbour failing plus vice versa). In that case he can try to copy once and succeeds with probability $0.25$. The probability of both neighbours succeding is $q^2$, and in this case the student can manage to cheat with probability $1-(1-0.25)^2$.
The equation for $q$ doesn't change, so in this case we get $p=\frac{62}{98}\approx0.633$.
