BasilTheBorder Joined: Feb 3, 2023
• Posts: 3
Thanks for this post from: February 3rd, 2023 at 6:44:20 AM permalink
If 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.
pwcrabb Joined: May 15, 2010
• Posts: 174
February 3rd, 2023 at 8:56:50 AM permalink
Probability of event is (1/100)
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)
Last edited by: pwcrabb on Feb 3, 2023
"I suppose I was mad. Every great genius is mad upon the subject in which he is greatest. The unsuccessful madman is disgraced and called a lunatic." Fitz-James O'Brien, The Diamond Lens (1858)
BasilTheBorder Joined: Feb 3, 2023
• Posts: 3
February 3rd, 2023 at 9:15:52 AM permalink
Thanks. I was heading in that direction but had been diverted by Poisson theory which was a bit of a red herring!
pwcrabb Joined: May 15, 2010
• Posts: 174
February 3rd, 2023 at 9:31:12 AM permalink
Yes, Poisson is very useful in other circumstances but not here.
"I suppose I was mad. Every great genius is mad upon the subject in which he is greatest. The unsuccessful madman is disgraced and called a lunatic." Fitz-James O'Brien, The Diamond Lens (1858)
Mental Joined: Dec 10, 2018
• Posts: 476
February 3rd, 2023 at 3:02:46 PM permalink
Quote: pwcrabb

Probability 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

I was understanding the OP's question differently.

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.
Quote: BasilTheBorder

If 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.

"how do I determine the chance of the gap between events happening?" leads me to believe it was my interpretation.
charliepatrick Joined: Jun 17, 2011
• Posts: 2793
February 3rd, 2023 at 4:29:24 PM permalink
If 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.

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 link to original post

ChumpChange Joined: Jun 15, 2018
• Posts: 3907
February 3rd, 2023 at 4:57:15 PM permalink
I started my thread after reading the OP's post of this thread. It wasn't quite coincidence, but I didn't want to clog up the OP's thread. The replies here are beyond my typical math abilities.
Mental Joined: Dec 10, 2018
• Posts: 476
February 3rd, 2023 at 6:26:15 PM permalink
Quote: charliepatrick

If 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.

OP specifically asked about "gap between events" but maybe OP meant gaps before the first event. These two are certainly not the same question.
pwcrabb Joined: May 15, 2010
• Posts: 174
February 3rd, 2023 at 7:07:58 PM permalink
CharliePatrick and I do not disagree. His pub scenario can be modeled similarly to Basil's scenario, except with differing probabilities for P(success) and P(failure). Mental requests from Basil a clarification and a specific problem, and I agree. Precise problem specification is essential.

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?
Last edited by: pwcrabb on Feb 3, 2023
"I suppose I was mad. Every great genius is mad upon the subject in which he is greatest. The unsuccessful madman is disgraced and called a lunatic." Fitz-James O'Brien, The Diamond Lens (1858)
BasilTheBorder Joined: Feb 3, 2023