Fair coin probability of HeadsI 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?

Question

emerhelienapj · Accepted Answer

Step 1So, 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&#039;s well known that there are       F          10        =  89 (F is the fibonacci sequence) ways for this to happen.Step 2Answer:   89  ⋅      2          −      12

vangstosiis · Accepted Answer

Step 1Call a sequence admissible if it doesn&#039;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 2Also,       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.

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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