[f97tosc]

Hmmm... maybe your approach is right but I think the answer is wrong... we can bound the answer as follows.

Define
P1 = P(frog never passes 0)
P2 = P(frog never passes 1)
P3 = P(frog passes first 1, then 0)

Since these are mutually exclusive, we have P1+P2+P3<=1.

Also by symmetry, P1=P2.

Furthermore, P3>P1, because these two together form the event that the frog hasn't stopped at 0 when it first reaches 1, and if this happens, the frog is more likely (>50%) to eventually go back to 0 as in P3, than to never go back to 0, as in P1.

Then 1 >= P1+P2+P3 > 3 P1

P1<1/3

Hence, at the very most 1/3 of integers can be skipped.