I came across this probability problem in a placement exam a few days ago , and i was not able to solve it. so, i am trying to solve it since then but i couldn't come up with a solution so please provide the solution and have fun solving it.
Problem Statement : We have an infinite chessboard with a bishop and knight on the same row. A massive meteor storm erupted, landing a meteor in each square of the chessboard with probability $p$ randomly and independently. The bishop and knight were not damaged, although their mobility may have been hampered by the meteors. For what value of $p$ is the expected number of valid squares to which the bishop may advance(in a single move) equal to the expected number of squares to which the knight can travel (in a single move)? Note that both the knight and bishop can't move to a square stuck with a meteor and the bishop can't move past a cell stuck with a meteor. For example if the bishop is on the cell (3,3) and the meteor strikes (5,5), then it can't move to (5,5),(6,6),(7,7) or any other cell beyond it after one move from (3,3)
Let $X,Y$ be the number of legal moves of the bishop and knight respectively. Consider the $4$ diagonal rays of valid moves of the bishop. For $i=1,2,3,4$ and $j\in\mathbb{Z}_+$, let $A_{ij}$ be $1$ if the $j$th square of the $i$th diagonal is accessible to the bishop, and $0$ otherwise. Then by symmetry we have, for all $i=1,2,3,4$ and all $j\in\mathbb{Z}_+$,
$$A_{ij} =^dA_{1j},$$
where the "$=^d$" indicates "equal in distribution". Now, $A_{1j}\sim$Bernoulli$((1-p)^j)$, so
$$E(X) = 4\sum_{j=1}^\infty E(A_{1j}) = 4\sum_{j=1}^\infty (1-p)^j = 4(1-p)/p.$$
On the other hand, $E(Y) = 8(1-p)$, so we find $p=1$ or $p=1/2$ as solutions.