Probability question of ants movement $2$ ants are $10$cm apart on a straight line. At each step, each ant can move 1cm to the left with a $0.5$ probability or the to the right with the $0.5$ probability. 

What is the probability that they collide after $7$ steps?

I understand for them to collide both ants can either move:


*

*$5$ steps towards each other or

*$6$ steps for $ant1$ towards $ant2$ and $4$ steps for $ant2$ towards $ant1$
or

*$4$ steps for $ant1$ towards $ant2$ and $6$ steps for $ant2$ towards $ant1$
or

*$7$ steps for $ant1$ towards $ant2$ and $3$ steps for $ant2$ towards $ant1$
or

*$3$ steps for $ant1$ towards $ant2$ and $7$ steps for $ant2$ towards $ant1$
Is this the right approach or is there something I'm missing? Maybe an easier way?
I am also confused as to how '$7$ steps' affect the problem?
Thank you in advance
 A: At each step, with probability $1/2$ the distance of the two ants remains unchanged.  With probability $1/4$ it increases by $2$ cm, and with probability $1/4$ it decreases by $2$ cm.
My reading of the problem is that we want the probability that the ants collide after exactly $7$ steps.  This is different from the probability of colliding in at most $7$ steps.  The basic ingredients of the solution are the same, but they are mixed in different proportions.
The two ants collide the first time the number of steps in which they approach exceeds the number of steps in which they move away from each other by $5$.
Moreover, the last step must decrease the distance.  This means that we are interested in ways to decrease the distance to $2$ cm in six step.  There are two ways to go about this:


*

*We may decrease the distance $4$ times and keep it the same twice.  This can be done in $\binom{6}{4} = \binom{6}{2}$ ways.

*We may increase the distance once within the first five steps, and decrease it the other five times.  This can be done in $\binom{5}{1}$ ways.


Since the events we have described are mutually exclusive, we just add the probabilities:
$$ \left[\binom{6}{2}\Big(\frac{1}{4}\Big)^4 \Big(\frac{1}{2}\Big)^2 + \binom{5}{1} \Big(\frac{1}{4}\Big)^5\Big(\frac{1}{4}\Big)\right]\frac{1}{4} \enspace, $$
where the $1/4$ outside the square brackets accounts for the seventh step.  This formula simplifies to $\frac{65}{4^7}$, which is approximately $0.004$.

If we also want to consider the probability of colliding after $5$ or $6$ steps, we include two more events:


*

*The ants move toward each other for $5$ consecutive steps.  There is just one way of doing this.

*The ants keep their distance in one of the first five steps, but otherwise move toward each other.  This can be done in $\binom{5}{1}$ ways.


The probability of the union of these disjoint events is:
$$ \Big(\frac{1}{4}\Big)^5 + \binom{5}{1}\Big(\frac{1}{4}\Big)^5\frac{1}{2} = \frac{56}{4^7} \enspace. $$
All in all, we get $\frac{121}{4^7}$, which is approximately $0.0074$.
