Suppose we have a coin tossing game. What is the probability that out of 5 coin tosses, you will get 3 tails consecutively?
For this problem, I was thinking of using the Binomial distribution PMF since it seems to describe the number of success out of n trials. The only problem is that I don't think it takes into account the order of the tosses ie 3 consecutive tails. So I was thinking maybe instead of
Px(k)= nCk * (p^k) * (1-p)^(n-k)
I would replace n choose k with n!.
Idk if this would work, or if I am headed in the right direction. Any thoughts on my approach?
edit: I forgot to put that in the problem, we are finding the probability of getting either 3 tails or 5 tosses, whichever comes first, thus my suggestion of the binomial distribution PMF.
Best Answer
The binomial distribution PMF isn't really describing the same thing - it looks for the possibility of getting a specific exact number of tails. If that's really what you're looking for (the probability that 3 are tails and the rest are heads), the replacement for nCk is n-k+1, or 3, because there are 3 places where there could be 3 consecutive tails (the beginning, the middle 3, or the last 3). Following this should get you 3/32.
On the other hand, if you're looking for the probability of there just being 3 consecutive tails (so that 5 tails in a row would also count), you're going to have a lot of difficulties with overcounting. The easiest method in this case is really just to count the possibilities (there are 8 - ttttt,tttth,tttht,ttthh,thttt,htttt,httth,hhttt). Divide this by the total number of possibilities (there are 32 ways to flip a coin 5 times) to get the overall probability of 1/4.
If you really want to do the second part in a slightly fancier way, you have to avoid counting sequences like ttttt multiple times. The way to do this is to look for an h followed by 3 t's - if you find httt, there can be no previous sequence of 3 t's, and you avoid overcounting. Therefore, you are looking for ttt??,httt?, and ?httt, which have probabilities of 1/8,1/16,and 1/16 respectively, summing to 1/4.
Note that this method won't work for significantly longer sequences, as it's possible to have 3 consecutive t's, then httt. More advanced methods are required.
TL;DR depending on what you mean, the answer is probably 1/4, but the method you described isn't particularly useful