Lets flip a fair coin. If it's head, I get $1 and if its tail I loose $1. Let "k" be the number of trials when I choose to stop. Let S_k be the total sum (amount) till the "kth" trial (inclusive).
Assume given that the probability S_k = "n" (some positive integer) before S_k is "-m" (some negative integer) for the first time = m/(n+m).
Question: What is the probability that Sum, S_k shall hit every integer infinite number of times.
Hint: use borel-cantenelli lemma.
Here is my attempt-
Pr(S_k = n before S_k = -n) = 1/2.
Pr(S_k+1 = 0 before S_k+1 = -n = 1).
These are independent events. Hence, Pr(S_k = n and S_k+1 =0) = 1/2.
This would mean Pr(S_k hits number "n" infinite times) = 1/2*1*1/2*1...
However, using Borel Cantenelli Lemma, I can say that 1/2+1/2 +.. = 1+1+.. = infinity. Hence, Pr(Sum hits "n" infinite times) = 1.
Since, we chose "n" arbitrary, the Pr that sum hits every integer infinitely often is = 1*1*.. = 1.