Probability of non-event is therefore (99/100)

P(gap size zero) is therefore [ (99/100)^0 ] (1/100)

P(gap size one) is therefore [ (99/100)^1 ] (1/100)

P(gap size nineteen) is therefore [ (99/100)^19 ] (1/100)

Sum of all probabilities of gaps from zero to infinity is 1

Obtained by infinite series summation

(Edited 7:23 pm PST 03 Feb)

I was understanding the OP's question differently.Quote:pwcrabbProbability of event is (1/100)

Probability of non-event is therefore (99/100)

P(gap size one) is therefore [ (99/100)^1 ] (1/100)

P(gap size nineteen) is therefore [ (99/100)^19 ] (1/100)

Sum of all probabilities of gaps from zero to infinity is 1

Obtained by infinite series summation

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You say the P(1) = [ (99/100)^1 ] (1/100). This is the probability of one miss before a hit when you are starting at a hit.

In an infinite sequence, the hits come only once in 100 games. The probability of seeing a sequence of back to back hits starting at any particular game is 1 in 10,000. A sequence with two hits separated by exactly one miss appears with a probability 0.000099, etc.

Perhaps the OP can clarify which number he is seeking.

"how do I determine the chance of the gap between events happening?" leads me to believe it was my interpretation.Quote:BasilTheBorderIf there is an event occurring, say 1 in 100 times for ease of use, how do I determine the chance of the gap between events happening? So basically a "games since last event" counter. How many times would you expect it to occur on successive games? How many with exactly a 1 game gap? How many with exactly a 2 game gap ? How many with a 1000 game gap? etc., I'm thinking an infinite number of samples, rather than a small set. Is there a statistical formula to use? Any ideas very much welcomed.

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By pure coincidence there's a similar discussion using craps dice asking how often would it take to throw a "2" (1-1) and then mentions other totals including a "Natural Winnerl" (7 or 11). The enclosed picture (I hope ChumpChange doesn't mind me copying it here) shows how the frequencies decrease. In this case there are 8 ways out of 36 to throw a "Natural Winner". I've since replied by comparing this to an example of a long road where some of the buildings are pubs and the rest private houses, asking how far is it to the next pub. https://wizardofvegas.com/forum/gambling/craps/37930-roll-frequencies-after-69-6k-rolls-on-wincraps/#post880610

Quote:ChumpChange

OP specifically asked about "gap between events" but maybe OP meant gaps before the first event. These two are certainly not the same question.Quote:charliepatrickIf you start out afresh and ask what are the chances of an event with pr=1/100 happening on the 1st, 2nd etc. attempt; then the answer is 1st = 1/100, 2nd = 99/100 * 1/100 etc. This gives a distribution of "gaps" which just gradually decreases as the gap gets larger.

Basil, are you analyzing for:

1. The probability of first success on exactly a specific future trial?

2. The cumulative probability of first success on or before a specific future trial?

3. The expected number of gaps, or consecutive failures, of a certain size within a large universe of experiments?

4. The expected size of the next gap?

5. The mean size of all gaps?

6. Something else?