# Fair coin probability of Heads. I have a fair coin which I toss many times. What is the probability that it takes 12 rolls until I get two heads in a row? I thought this was just standard geometric. But it doesn’t seem to work? Can someone please explain how to do this?

I have a fair coin which I toss many times. What is the probability that it takes 12 rolls until I get two heads in a row?
I thought this was just standard geometric. But it doesn’t seem to work? Can someone please explain how to do this?
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emerhelienapj
Step 1
So, the first ten flips have no consecutive heads, which means that we can look at the different flips sequences by filling in the 10 flips with either T or HT. It's well known that there are ${F}_{10}=89$ (F is the fibonacci sequence) ways for this to happen.
Step 2
Answer: $89\cdot {2}^{-12}$

vangstosiis
Step 1
Call a sequence admissible if it doesn't contain two consecutive heads. Then we must have an admissible sequence of length 9, followed by the sequence THH. Let ${a}_{n}$ be the number of admissible sequence of length n, ${t}_{n}$ be the number of such sequences that end in T, and let ${h}_{n}$ be the number of such sequences ending in H.
Clearly, ${a}_{n}={t}_{n}+{h}_{n}.$
Step 2
Also, ${h}_{n}={t}_{n-1},$ since the last H is an admissible sequence ending in H must be preceded by an admissible sequence ending in T. However, there is no such restriction on an admissible sequence ending in T, so ${t}_{n}={t}_{n-1}+{h}_{n-1}={t}_{n-1}+{t}_{n-2},$and we see the Fibonacci numbers.
Note that ${a}_{9}={t}_{9}+{h}_{9}={t}_{9}+{t}_{8}={t}_{10}=89.$