Expected gaps between events

Now I understand why you were looking at Poisson, because its profile is similar to the distribution that you seek. Excel is all that you need for a decent graph.

For gap sizes (n) from zero to any sufficiently large number to be illustrative, perhaps 500K, do the following:

1. Plot [ (99/100)^(n) ] [ (1/100) ]
This graph will be a gently sloped monotonic curve from (1/100) down towards the x axis asymptotically. It will remind you of Poisson.

2. Plot the incremental summation of the above series of terms. This graph will be a gently sloped monotonic curve from (1/100) up towards 1.0 asymptotically. Observe the value of (n) at which the curve exceeds 0.5, which means that likelihood has been achieved.

Please forgive yourself for any predilection for bell curve distributions. All of us have been conditioned to expect them everywhere.

Quote: BasilTheBorder

What I am interested in is the profile of gaps between successful events. Disregarding the lead up to the initial event, I think Mental's explanation fits my scenario. I have a slot game in development with a random event occurring 1 in 100 games. Logging the gaps between events gives a similar shaped curve to using Mentals approach. For some reason I was expecting a bell shaped curve centered around the 100 game gap. I need to brush up on my statistical knowledge! So in answer to your question, 6, or possibly 3. The profile of gaps between successful events.
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I confess that I had the thought of the most probable gap being near 1/p for a second. The most probable gap has to be at 0 for the same reason that the most probable regular interval to see the first radioactive decay event is the first period. There are fewer nuclei after the first period ends.

An easier way to think about it is that after every hit, the probability of a gap of length zero,1,2,... is the same as the question pwcrabb answered. The function you seek is pwcrabb'c function divided by p. If p=1, then P(0) =1 and and P(x!=1)=0.

The answer only works for an infinite series. The answer is only approximate for a finite series if you add any initial partial gap to any ending partial gap. You can check this by counting the number strings of (wrapped) consecutive zeros in a sequence of binary digits 0-(2^n). If n=3, 000 001 010 011 100 101 110 111. p=0.5,
P(3) = 1/(8*3)
P(2) = 3/(8*3)
P(1) = 3(8*3)
P(0) = 6/(8*3)

In this case the most likely time it will happen is the next roll/spin/toss. To show this easier here are two graphs of tossing coins (waiting for a Head) and rolling a single die. For win on the first toss ("Heads") you need "H". To win on the second toss you need "TH", this series continues "TTH" "TTTH" .... "TT...TTH" ... Thus you can see that you need an increasing number of Tails before the Heads. This is why the probability keeps going down. With other similar events you also need a series of misses before the hit.

In your example there's a 10% chance of getting the feature retriggered within 11 spins, etc.

Chance	Spins
10%	11
20%	23
30%	36
40%	51
50%	69
60%	92
70%	120
80%	161
90%	230
95%	299
98%	390
99%	459
99.5%	528
99.8%	619
99.9%	688
99.99%	917

Thank You CharliePatrick for providing two illustrative graphs. Please notice that the second graph is useful for both one number on one die and the total of Seven on two dice. In both dice cases, P(success) is (1/6).

Thank You also for your report of results for the original problem for which P(success) is (1/100). Please notice especially that the cumulative probability reacbes 50% when (n) is 69, which seems correct at first glance. This would be the Median of the distribution.

Another way to get the median of 69 is ln(2) * 100, rounded to nearest integer

Quote: Ace2

Another way to get the median of 69 is ln(2) * 100, rounded to nearest integer
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Here is an approximation I get:

Let p be the probability of a particular trial being successful, and c the probability of being successful in n or fewer trials

For p = 1/100 and c = 1/2, n = 71.49

Are you sure the lower bound of the integral shouldn’t be zero? That would yield n=69.47, much closer to the correct figure

I realize success can’t actually happen in less than one trial, but the integral is for continuous time, not individual trials…which isn’t the actual case either. It’s an approximation

Incidentally, I’ve found that integrals of Poisson processes give the correct answer only when integrating over all time (zero to infinity)

Integrals not required. Merely logarithms. Define the accumulation function including the exponent (n) and set the accumulated value of f(n) at 0.50. At what value of (n) is the statement true? Decimals to any degree of precision.

Monte Carlo simulations are useful for corroborating precise analytical answers. I have read of several gritty practitioners who swear by simulations as close enough for their non-engineering work.