Expected number of coin toss to get two consecutive tails with maximum of six trials

Question

We toss a fair coin for a maximum of six times. During this process, if we obtain two consecutive “tails”, we stop tossing the coin. What is the expected number of coin tosses?
I know that if we can have an infinite many trials the expected number of tosses would be:
TT = 1/4
HTT = 1/8
HHTT, THTT = 1/8
HHHTT, HTHTT, THHTT = 3/32
HHHHTT, HTHHTT, HHTHTT, THHHTT, THTHTT = 5/64

So the expected number coin tosses is:
2*(1/4)+3*(1/8)+4*(1/8)+5*(3/32)+6*(5/64) = 2.3125
Is this way correct?

Answer 1

Your (second editted) way is correct, but do note: Because it is the maximum allowed, you must stop at 6 tosses whether the last two were consecutive tails or not.

• 2 tosses (TT) = $1/4$
• 3 tosses (HTT) = $1/8$
• 4 tosses (HHTT, THTT) = $2/16$
• 5 tosses (HHHTT, HTHTT, THHTT) = $3/32$
• 6 tosses (anything else) = $1-(3+4+4+8)/32 = 13/32$