Hi,

this is exactly what i just wanted to post :-)

I think a is inverted with b or viceversa :-))
If a considers the "exact" streak length then the average cant be 1.01 for probability 99% at each trial and 101 for "1 or more"

a and b inverted makes sense. It takes 101.01 trials to get a success followed by a failure. You ignore the series length of 1 for that.

Using BruceZ post #14 example of 10 in a row coin flips, the original OP Q,
the calculator shows 2046 and 2048, matching BruceZ answers for a run of 10 coin flips.

a and b look like the series length is not used as BruceZ mentioned only showing the average number of trials needed.
The light just went on for me

added:
the average number of trials for a streak does not use "n" only p and k to arrive at "n". I just ran a few sims, and the calculator numbers appear to be correct.
I wonder if the webpage author will make the results easier to understand.

Yes Guido, youre right. And if you look closer at the formulas you wrote, youll see that there is no n there. Actually the inverse (reciprocal) of the average is the probability.....

Bruce, shouldnt be a streak of exactly 1 succes of length 1 noted as L-W-L ?

Average in the sense it was written earlier of course, or average number of trials.

OK, here is what these actually mean. He is using formulas that I described in this thread for average waiting time, and they are not reversed. The difference between A and B is that A considers a sucessful streak to end with a failure, while B starts a new streak immediately on the next trial after k successes. That is why A is higher than B. With A, WWWL counts as 1 streak of length 1 or more, while B counts this as 3 streaks of 1. So A has a longer average waiting time. For low probabilities or long streaks, this doesn't matter as much, but it makes a huge difference for p=0.99 and k=1. B will be just over 1, but A will be over 100 because we typically get 100 W's which would all count as 1 streak, so you have to wait all that time before you can get another one, hence longer average.

A formula: 1/p^k * 1/(1-p)

B formula: 1/p^k * (1-p^k)/(1-p)

where k is the streak length, and p is the probability of success.

The first term is the average number of trials, while the second multiplied term is the average length of a trial from the geometric series. He uses the word "trial" differently from me in his labels. For him a trial is a W or an L. I consider a trial to be a string for which you either get k successes or you don't. The difference between A and B is where a sucessful trial ends. Neither of these is actually the average number for exactly WL or LWL as you were thinking.

EDIT: By "average waiting time", I am refering to the average over multiple streaks, and this is what he is computing for both A and B. B also corresponds to the time to the first streak because we would stop as soon as we got k successes.

What is not good? In light of my last post, the formulas look OK, and I don't see that anything changed.

Sorry, I have deleted my message now as i didnt read Your post.

Still, i think B should have a longer average waiting time if B defines the "exact" k length. Just think at events with 99% probability of success each time. Clearly these will come more often as a string of successes then a single one.

Or did i understand something wrong?

Yes, wrong. I just added some more explanation to that post which should make it clearer.

Thank You for the explanation, i understand now. The definition of terms is very important.

It is actually correct as it is labeled. They are both the average number of trials per streak where

average trials = (number of streaks) / (total trials)

where a trial is a W or L. It's just that A defines a streak as k or more, while B defines a streak as exactly k, just as it says.

I agree that *something* needs to be clarified to prevent the interpretation that both of us made initially. "Exactly k" does NOT refer to getting k successes followed by a failure and preceded by no success.

"Exactly k" is misleading. How about labeling both "average number of trials per success run in a very long series of trials", then define a success run for A as "k or more consecutive successes, and for B as "k consecutive successes". Examples of each would be nice.

I guess I started this confusion by writing about both methods. Personally, I think that A is a silly way to compute the average that people probably don't care about. That is, k successes followed by another k successes should count as 2 streaks, not 1. Counting it as 1 makes sense for probability, not for average waiting time. I described both methods mainly for completeness.

Can you tell me how to average an infinite series?

Insightful graphs about streaks in coin tossing have been produced by Frank Martin in his note "What are the odds of having such a terrible streak at the casino?"

http://wizardofodds.com/askthewizard/images/streaks.pdf

Nice material about streaks in coin tossing can also be found in the book Understanding Probability, Chance rules in everyday life.

http://www.bbc.co.uk/news/uk-scotlan..****and-13680995

I don't know how they estimate that.

They must have used the actual frequency of hole-in-one, from some gathered history either on that course or others. So:
sqrt(67million) = 8185 to 1 for a single one. Unfortunately the skill factor makes their number just a guess, as it isn't random or uniform across players.

Pretty sick bump i guess, but I was working on something related to this, and it really is amazing that this rather simplified reduction works. Was explaining this type of thing, relating to fib numbers as well, to a few kids, and it really makes me wonder if Fib was on a permanent acid trip.

IMO necro-bumps should be allowed if they are quoting BruceZ.