2019 AIME I Problems/Problem 5 Regarding the AIME Problem:
Problem Link
Tried to solve backward - going from 4,4 and moving towards the origin and recording the number of ways to reach different points- then find the number of ways to each point of the coordinate axis and calculate the probability of hitting the origin. This approach seems to be clearly wrong as the answer in the  source is different. Can somebody point out the fault in my approach. 
My Trial Solution
 A: If you had used probabilities rather than counts, it would have worked.

For example, the probability of reaching position $(3,4)$ is ${\large{\frac{1}{3}}}$, and the same for position $(4,3)$.

But the probability of reaching position $(3,3)$ is 
$${\small{\frac{1}{3}}}+{\small{\frac{2}{9}}}={\small{\frac{5}{9}}}$$
which has nothing to do with the count you showed for that position. 

Counts are not applicable here since not all paths from $(4,4)$ to $(0,0)$ are equally likely.

To see that using probabilities rather than counts would have worked, let $P(a,b)$ be the probability of reaching position $(a,b)$ from the initial position $(4,4)$.

The goal is to find $P(0,0)$.

By symmetry, we have $P(b,a)=P(a,b)$, so it suffices to compute $P(a,b)$ for $a\le b$.

Starting from $(4,4)$ and working towards $(0,0)$ we get
\begin{align*}
P(4,4)&=1\\[6pt]
P(3,4)&=\frac{1}{3}P(4,4)=\frac{1}{3}\\[6pt]
P(3,3)&=\frac{1}{3}P(4,3)+\frac{1}{3}P(3,4)+\frac{1}{3}P(4,3)=\frac{5}{3^2}\\[6pt]
P(2,4)&=\frac{1}{3}P(3,4)=\frac{1}{3^2}\\[6pt]
P(2,3)&=\frac{1}{3}P(3,4)+\frac{1}{3}P(2,4)+\frac{1}{3}P(3,3)=\frac{1}{3}\\[6pt]
P(1,4)&=\frac{1}{3}P(2,4)=\frac{1}{3^3}\\[6pt]
P(2,2)&=\frac{1}{3}P(3,3)+\frac{1}{3}P(2,3)+\frac{1}{3}P(3,2)=\frac{11}{3^3}\\[6pt]
P(1,3)&=\frac{1}{3}P(2,4)+\frac{1}{3}P(1,4)+\frac{1}{3}P(2,3)=\frac{13}{3^4}\\[6pt]
P(1,2)&=\frac{1}{3}P(2,3)+\frac{1}{3}P(1,3)+\frac{1}{3}P(2,2)=\frac{73}{3^5}\\[6pt]
P(1,1)&=\frac{1}{3}P(2,2)+\frac{1}{3}P(1,2)+\frac{1}{3}P(2,1)=\frac{245}{3^6}\\[6pt]
P(0,0)&=\frac{1}{3}P(1,1)=\frac{245}{3^7}\\[6pt]\\[6pt]
\end{align*}
