*I've just signed up and I'm posting this here (and not in the other probability section) since it's not a course or homework problem. Nonetheless it may appear as simple and straightforward to someone versed in probability stuff... It's for a research project I'm assisting and it's been some time since I've dealt with probabilities.*

I have had some thoughts on it but I'd like to confirm them before actually putting them to use.

**I'm interested in finding probabilities based on the following parameters:**

- 2 symmetric binomial distributions (e.g. (1) 70% A's / 30% B's vs. (2) 30% A's + 70% B's)

- A sequence of 8 (with replacement) is drawn from either distribution 1 or 2.

Now, I'd like to find the probability of getting 3A's in a row, both for distribution (1) and (2).(E.g. an example sequence for 3A's in a row is BAAABABA.)

- 2 symmetric binomial distributions (e.g. (1) 70% A's / 30% B's vs. (2) 30% A's + 70% B's)

- A sequence of 8 (with replacement) is drawn from either distribution 1 or 2.

Now, I'd like to find the probability of getting 3A's in a row, both for distribution (1) and (2).

In words,

**I'd like to find the P(3As in a row in a sequence of 8|Distribution 1) and P(3As in a row in a sequence of 8|Distribution 2)**. Once I have that I can proceed to calculate probabilities for different possible streaks such as 4A's, 5A's etc...

My thoughts on this revolved around some sort of combination of a permutation + x and the use of the binomial function (n, successes, p) but it never quite took everything I needed into account...

Finally, I figured 2 to the power of 8 (=256) gives me all possible sequence permutations. Adding all those permutations that include 3A's and dividing that number by the total (256) gives me the probability of getting 3As in a row. However, that would only be true for a 50/50 distribution so I'm lacking some weights...

In the end I came up with this:

**For distribution 1:**

(.7 to the power of 3) * ((number of 3A in a row possibilities)/(256))

For distribution 2:

(.3 to the power of 3) * ((number of 3A in a row possibilitites)/(256))

(.7 to the power of 3) * ((number of 3A in a row possibilities)/(256))

For distribution 2:

(.3 to the power of 3) * ((number of 3A in a row possibilitites)/(256))

I believe that not taking the order into account the probability of getting 3A's in a row is simply .7 (or .3, depending on the distribution) to the power of 3 and I've tried to adjust that probability by the second term, taking into account the length of 8 sequence and the ratio of getting 3As in all possible permutations.

How wrong am I? Could this actually be correct or is there something missing / wrong in my little calculation?

Many thanks for your help!!!