Lost in a "one directional" concert
At step $k$, draw a random number $q$ uniformly between $1$, $2$, and $3$.
  
  
  if $q = 1$, walk $2^{k-1}$, wait $2^{k+1}$, walk $2^{k-1}$
  if $q = 2$, wait $2^{k-1}$, walk $2^{k-1}$, wait $2^{k}$, walk $2^{k-1}$, wait $2^{k-1}$
  if $q = 3$, wait $2^k$, walk $2^k$, wait $2^k$
  


At each step $k$, both friends will walk $2^k$ steps. If $k &lt; \log_2(x) + 1$, they won't meet during that step, however if $k &gt;= \log_2(x) + 1$, they will meet if and only if they don't draw the same number. The probability that this doesn't happen is only $1/3$ at each step.

Hence the expected walking distance is (bounded above by):

$$2\left(\sum_{k=0}^{\lceil\log_2(x)\rceil}2^k+3^{\lceil\log_2(x)\rceil}\sum_{k=\lceil\log_2(x)\rceil+1}^{\infty}\left(\frac{2}{3}\right)^k\right)$$

Which is finite, and equal, if my napkin maths are to be trusted, to $2^{\lceil\log_2(x)\rceil+3}-2 \le 16x$.

By the way, if $d$ is the random variable representing the distance walked, we still have that $\forall D&gt;0, \mathrm{P}(d&gt;D) &gt; 0$, i.e. the distance is unbounded and can end up being arbitrarily high. Luckily this probability vanishes fast enough to ensure that the infinite sum $\sum_{D=0}^{\infty}\mathrm{P}(d=D)\cdot D = \mathbb{E}[d]$ converges. Having a finite upper bound for $d$ is a much stronger property and I reckon it's not possible to find a solution satisfying it.