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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 $\frac{C(16,8)}{2^{16}}$

enter image description here

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?

Mathematics
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    Formatting note: if your exponent requires more than one character you need to isolate it with curly brackets as in "2^{16}". This is true even if it renders as a single character, as in "2^{\pi}" – lulu Nov 22 '17 at 16:40
  • \binom{16}{8} renders $$\binom{16}{8}$$ as does {16\choose8} – gen-ℤ ready to perish Nov 22 '17 at 16:42
  • Also, the rules are not clear to me. What do you mean by "opposite corners"? If, say, one ant starts in the upper right most corner, is it blocked from moving at all? – lulu Nov 22 '17 at 16:42
  • "Can we say that overall 8 rightward and 8 upward steps must be taken in all by two ants?" No. What if ant one makes 2 rightward moves and 6 upward moves, and ant two make 8 upward moves? – XRBtoTheMOON Nov 22 '17 at 16:44
  • @btcgrl Your example would not lead in their meeting – Mathematics Nov 22 '17 at 16:46
  • @lulu each ant will move rightward or upward with respect to itself. Imagine ant at right most corner facing other side of grid. – Mathematics Nov 22 '17 at 16:47
  • No idea what that means. If the ant can be "facing" in one direction or another, what are the rules governing that? If an ant starts at the upper right most corner, what options has it got? – lulu Nov 22 '17 at 16:48
  • Imo question should be 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 do not move in backward direction) – Ananya Nov 22 '17 at 16:52
  • @Mathematics, good point about not meeting. Then yes, it looks like there has to be 8 rightward and 8 upward total, which is the answer they gave. – XRBtoTheMOON Nov 22 '17 at 16:55
  • @lulu Added an image for clarification to show permissible direction. – Mathematics Nov 22 '17 at 16:58
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    Ok. So say that ant $i$ moves $(v_i,h_i)$ . Here $v$ stands for vertical, $h$ stands for horizontal. Then the upper ant winds up at height $8-v_1$ and the lower winds up at height $v_2$. For these to be equal we must have $v_1+v_2=8$ and similarly for $h$. – lulu Nov 22 '17 at 17:16

1 Answers1

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  1. 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 $\frac{1}{2}$. $\mathbb{P}(X_1 = i) = \binom{8}{i}\frac{1}{2^8}$.
  2. 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.
  3. 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.
Abhiram Natarajan
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