Feedback sought on finding a probability based on a permutation / binomial D

doc

New Member
#1
Hi everyone,

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.)

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))


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!!!
 

doc

New Member
#3
Hi Dason,

Thanks for your quick reply! I'm basically interested in the probabilities for all streaks, i.e. 2, 3, 4, 5, 6, 7, 8 in a row, depending on being in distribution 1 or 2. When I count for the "number of 3A's in a row possibilities / all permutations" for now I also count those possible sequences in the permutations that contain 4, 5, 6, 7, 8As since they all also contain 3As within them.

So.. I guess that when I'm referring to the probability of 3 in a row, it would be 3 or more in a row, yes.
 
#4
I would try to solve this problem using a Markov chain. You have four states:
3 A in a row already drawn (absorbing state)
2 A drawn in the previous two draws
1 A drawn in the previous draw
0 A drawn in the previous draw
Then you write the transition probabilities. You start from a vector where the last state has probability one and you multiply it by the transition matrix as many times as you have draws. The vector you obtain gives you the probability of being in the first state after n draws