problem: coin toss game

My thanks to anyone who can help with the following problem. It's an
analog to a behavioral assessment (of basic discrimination ability)
with a "pass" criterion equivalent to the "win" criterion described
below. We have used the assessment for many years, but to my knowledge
no one has articulated the precise probability of passing by chance.

# The Game #
Toss a fair coin repeatedly, recording the result on each toss. The
game can end in two ways:
1. You land eight heads consecutively, thereby winning.
2. You land eight tails cumulatively, thereby losing.

The shortest possible game is 8 tosses: TTTT TTTT or HHHH HHHH.
The longest possible game is 64 tosses:

HHHH HHHT HHHH HHHT
HHHH HHHT HHHH HHHT
HHHH HHHT HHHH HHHT
HHHH HHHT HHHH HHH(T/H)

What the probability (relative frequency) of winning?

 

What's the probability of 8Hs in a row?
What's the probability of a T without 8Hs in a row?
What's the probability of 8Ts without 8Hs in a row?
What's the probability of 8Hs in a row before 8Ts?

I make the last one 1-(1-0.5^8)^8 = 0.030826

 

The game can be viewed as a sequence of tossing sequences each of
which has one of these 9 forms

T
HT
HHT
HHHT
HHHHT
HHHHHT
HHHHHHT
HHHHHHHT
HHHHHHHH p = 1/256


You lose if you toss eight times in a row one of sequeces #1..#8, This
probability is (255/256)^8. Therefore the probability of winning is
1-(255/256)^8

 

Sorry I'm being thick how come the longest game is 64 tosses? Tou could
have a sequence such as HHTHHTTHHTTTHHTH etc going on for a long time.

 

According to the rules the game is over after the 8th tail or after
the 8th head in a row.

 

That's excellent, thanks you guys.

(And for your explanation, Horst!)

 

Look at the wording of rule 1 vs. rule 2. It's "consecutive" heads, but
"cumulative" (total) tails. So the longest run is HHHHHHHT repeated 7 more
times.