# probability that 2 heads do not occur consecutively when a fair coin is tossed 10 times? A sequence of n > 2 coin tosses has no consecutive heads if, and only if:

1. It begins with a tail, and is followed by n−1 tosses with no consecutive heads; or
2. It begins with a head, then a tail, and is followed by n−2 tosses with no consecutive heads. cindybrown wrote: 9 . A fair coin is tossed 10 times. What is the probability that two heads do not occur consecutively?
cindybrown wrote: 9 . A fair coin is tossed 10 times. What is the probability that two heads do not occur consecutively? 300+ Views | 300+ Likes
45.6 K+ views | 91.5 K+ people like this Alternatively, we could argue that the 1st coin has probability 1/2 to come up heads or tails, the 2nd coin has probability 1/2 to come up heads or tails, and so on for the 3rd and 4th coins, so that the probability for any one particular sequence of heads and tails is just (1/2)x(1/2) x (1/2)x(1/2)=(1/16).
Since the probability to flip a head is the same as the probability to flip a tail, the probability of outcome (i) must be equal to the probability of outcome (ii). So both must be equal to 1/2.

So Pr(3) = 1- Pr(1) – Pr(2) = 1- (1/2)^N – (1/2)^N
= 1 – power(1/2,N-1)
I consider that someone would only continue flipping if he/she has not flipped two heads in a row. That is, is not in the done state.
That is, for any toss level N, it wrong to assume that number of nodes is equal to 2^N. It will be strictly smaller as of level N = 3.
Anyway, after 3-4 levels drawn, one should see a pattern, that is, the number of tails, or initial state nodes (is), at a level N is equal to the number of is’s I had at level N-1 + the number of is’s I had at level N-2. We can thus write: