The question states:
A biased coin with probability p, $0 < p < 1$, of success (heads) is tossed
until for the first time, the same result occurs three times in succession
(that is, three heads or three tails in succession).
Find the probability that the game will end at the seventh throw.
Here is my work: just looking to see if this is correct. So the sample space, assuming we end with the last 3H (by symmetry we can calculate the last 3T) includes $\{HTH-THHH, HHT-THHH,TTH-THHH,THH-THHH,THT-THHH \}$
The last 4 throws must be THHH to stop the game on the 7th toss. My original answer was $\frac{2 *5}{2^7}$.However after Lulu comment brought to my attention that I was not taking into account the probability p. So I edited my answer to the below result.
Final Answer: here I use the sample space previously defined to include the probability p. so is the probability of ending the game on the 7th throw = $3p^5(1-p)^2+2(1-p)^3p^4+3(1-p)^5p^2+2p^3(1-p)^4$
The result you give in the edited question is correct. Here's a graph of it.