Let Aj
be the event that the drunk reaches k
before reaching −j.
Then, Aj⊆Aj+1
since to reach −(j+1)
the drunk needs to pass −j.
Note that ∞⋃j=1Aj
is equivalent to the event that the drunk ever reaches k,
since the complement of this event, namely the event that the drunk
reaches −j
before reaching k
for all j
implies that the drunk never has the time to reach k.
By assumption, P(∞⋃j=1Aj)=limn→+∞P(An).P(An)
can be found as a result of a gambler’s ruin problem.