Probability of a consecutive event over (x) trials?

Question

How do I calculate the chance/probability of an event with a 35% chance of success occurring 5 times in a row over the course of (x) number of trials?

So over 100 trials how many consecutive 5 successes are there? 1,000 trials? 1 million trials?

Answer 1

Hi Jason M.,

This one is a little tricky. Break that 100-trials universe down into segments of trials 1-5, 2-6, 3-7.... 96-100. These are all the possible consecutive runs of 5 E (call E the desired result, with the 0.35 probability/trial) events happening (note that there are 96 of them, not 95 or 100!). Now for just one of these, the probability of getting EEEEE is just (0.35)^5 ~ 0.005252, right? But, you can't just multiply that by the number of segments, b/c eventually that cumulative value would exceed probability 1!

So, knowing the P(EEEEE) ~ 0.005252, therefore the P(NOT EEEEE) ~= 1 - 0.005252 ~ 0.994748 for any segment. The P of getting 96 of those results in a row is (0.994748)^96, or ~ 0.6032 . That's the P of not having gotten ANY (EEEEE) string in the first 100 trials, so the P(at least one EEEEE string) = 1 - 0.6032 or ~ 0.3968 . You might wonder if these events are really independent, since they overlap -- but they are; you are only asking "did I get an EEEEE, or not?", not "were there only DDDDD" where D = anything but E. So you don't care what you got on any particular individual trial, only that you didn't have an EEEEE over a particular 5-trial set.

Similar calculations enable your 1000 trials and so on. Remember to pay attention to the exact number of possible runs, it's not 1000 and so on.

-- Cheers, -- Mr. d.

Answer 2

If the trials are independent, you just use the counting principal and the probability of that event. So first we need P(5 in a row) which is (.35)^5.

No in N, trials, we need to figure out how many different possiblities there are for the 5 positions. We could have the 1st 5, or maybes trials 2-6.... If we have 10 trials, its 1-5,2-6, 3-7, 4-8, 5-9, 6-10. there were 6, so the last one will be n-4,n-3,n-2,n-1,n.

So you just multiple (n-4)(.35^5_ for however many trials