This question is an extension of the already fairly well discussed problem Does the drunk man fall off the cliff?

A few people raised this question of why the probability should not be 1, i.e. why does the drunk man not always fall off given he has infinite steps to take.

As quoted by a user in his answer.

If allowed to randomly step indefinitely means he keeps stepping until he falls off the cliff.

Some arguments against this were:

  1. "There are many infinite sequences of steps which never cross the cliff."
  2. "So If you could somehow collect an infinitely large set of computers and run a simulation, I maintain that a nonzero fraction of them would run without halting forever!"

To this I would ask: what about the case when q < 1 / 2. In that case the agreed upon answer is 1. So aren't there still infinite sequences of steps which never cross the cliff?

  • $\begingroup$ @DanielSank you seem to have clarity on this. Could you explain? $\endgroup$ Commented Feb 27, 2018 at 8:03
  • $\begingroup$ There are lots of possible sequences which never cross the cliff, but the cumulative probability of all the sequences is zero. The possible sequences of steps are the same whatever the probability of left and right - the probabilities govern which sequences are most likely to be taken. $\endgroup$ Commented Feb 27, 2018 at 8:19
  • $\begingroup$ So if there are 8 such sequences, but infinitely many, where he does fall - you can't claim the probability is non-zero. $\endgroup$
    – dEmigOd
    Commented Feb 27, 2018 at 8:21
  • $\begingroup$ When $p$ and $q$ are not equal (i.e. not both $\frac12$), different sequences have different likelihoods $\endgroup$
    – Henry
    Commented Feb 27, 2018 at 8:31
  • $\begingroup$ 2. If you had uncountably many computers (one for each possible path), then infinitely many of them would run without halting forever, but the "size" (measure) of this infinite set of non-halting computers with respect to the total amount of computers will be zero, like the measure of $\mathbb{N}$ inside $\mathbb{R}$. (Here we would be in the case $p=1/2$, because each path corresponds to a computer which is like all the others). $\endgroup$
    – 57Jimmy
    Commented Feb 27, 2018 at 8:31

1 Answer 1


Of course there are still infinitely many sequences never crossing the cliff, but they make up a set of measure zero in the set of all sequences.

  • $\begingroup$ What about this argument when q > 1/2. in this case the motion will have a clear drift eastward and after many steps we will see a normal distribution with a peak at some point to the right of 0 and when time increases indefinitely the proportion of men to the left of 0 will tend to 0 and hence the probability of dying tends to 0? $\endgroup$ Commented Feb 28, 2018 at 4:53
  • $\begingroup$ No: In the above thought experiment there is just a red line, hence nobody will fall off the cliff. In the real experiment the fraction $p$ of the men will fall off the cliff on their first step. $\endgroup$ Commented Feb 28, 2018 at 9:00

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