Independent flips of a coin that lands on heads with probability p are made.
What is the probability that the pattern T, H, H, H occurs before the pattern H, H, H, H?
Hint: How can the pattern H, H, H, H occur first?
my approach is $\sum_{n=0}^\infty (1-p^4)^n(1-p)^3$ then i get $\frac {(1-p)p^3}{1-(1-p^4)}$
However, the suggested answer is
Am I wrong? What is the rationale behind the suggested answers?
Another answer I found online is: Our first observation is that for any 0 < p < 1 we will eventually see the pattern (H, H, H, H). Suppose that the first such pattern starts at the nth flip. If n > 1, then the n − 1th flip cannot be H since then the first (H, H, H, H) pattern would have started before the nth flip. Hence in this case the n − 1th flip is necessarily T, and starting with the n − 1th flip we see the pattern (T, H, H, H, H). In that sequence, (T, H, H, H) appears before (H, H, H, H). Summarizing, (H, H, H, H) can only appear before (T, H, H, H) if it starts immediately at the n = 1st flip, that is if all four first flips are heads. The probability of that is $p^4$.