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?

Livia Cardenas 2022-07-18 Answered
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?
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Answers (2)

emerhelienapj
Answered 2022-07-19 Author has 14 answers
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 2 12

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vangstosiis
Answered 2022-07-20 Author has 3 answers
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.

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