Probability that ants meet after $8$ steps Two ants are on the opposite corners of a grid of size $8 × 8$ if they move then what is the probability that they
will meet after each travelled eight steps (Assuming that they move rightward or upward).
Now for each step, both ants have 2 option, move upward or rightward. So for 8 steps, total cases=$2^8 \cdot 2^8=2^{16}$ but how to calculate favorable cases?
Given answer is $\frac{C(16,8)}{2^{16}}$

I hope the image provides some clarification
Can we say that overall $8$ rightward and $8$ upward steps must be taken in all by two ants?
 A: *

*Focus on the first ant (bottom left). Let the ant's position be (0, 0). Observe that if we know the final x-coordinate of the first ant as $X_1$, we know that the y-coordinate is $8 - X_1$. $X_1$ is a binomial random variable with parameters $8$ and $\frac{1}{2}$. $\mathbb{P}(X_1 = i) = \binom{8}{i}\frac{1}{2^8}$.

*Similarly the second ant. Let the x-coordinate be $X_2$. $X_2$ is $8 - X_2'$, where $X_2'$ is a binomial random variable with the same parameters as above. Due to symmetry of the binomial distribution, we have the same distribution for $X_2$ as for $X_2'$. This in turn means that $X_1$ and $X_2$ have the same distribution.

*Thus, for the ants to meet, we need $X_2 = X_1$, where both are binomial random variables with parameters $8$ and $\frac{1}{2}$. We obtain the final probability as $$\sum_{i=0}^8 \mathbb{P}[X_1 = i] \cdot \mathbb{P}[X_2 = i] = \sum_{i=0}^8 \left[\binom{8}{i}\frac{1}{2^8}\right] \cdot \left[\binom{8}{i}\frac{1}{2^8}\right] = \frac{1}{2^{16}}\sum_{i=0}^8 \binom{8}{i}^2 = \frac{\binom{16}{8}}{2^{16}}.$$ I used this result for the last equality.

