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I know that this form of the Geometric distribution gives the probability that for a success probability $p$, the kth trial out of k trials is the first success.

$Pr(X=k) = (1-p)^{k-1}p$

My question is, conditioned on the first success happening on the kth trial, what would be the probability of a subsequent success occurring after the mth trial out of m trials. To give some numbers, let's say that after 9 failures, we have a success on the 10th trial. Now, given the success on the 10th trial, what would be the probability of say having 4 subsequent failures before a second success on the 5th trial? In a sense, I'm asking about the probability of getting an interval of 4 failures between two successes.

TSP
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  • Do you mean after the $k$th trial, e.g. on the $k+4$th trial? Or do you mean on the $k$th trial? If so, look into the gambler's fallacy. – Arya McCarthy Mar 29 '21 at 20:12
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    What do you mean by "a success"? Do you mean "the next success"? Or just any success? If the latter, it's a trick question. – Stephan Kolassa Mar 29 '21 at 20:19
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    @StephanKolassa Or, if not a trick, the question is an invitation to prove the memoryless property of the geometric distribution. – Sycorax Mar 29 '21 at 20:20
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    @Sycorax True -- but the situation is even simpler than that, because it comes down to independence of outcomes. After observing a success on trial $k,$ reset the count on the next trial from $k+1$ to $1$ and proceed. In terms of the new count the question becomes "what is the chance the next $m-k$ trials are all failures?" – whuber Mar 29 '21 at 20:50
  • Possible duplicate of https://stats.stackexchange.com/q/377083/6633 – Dilip Sarwate Mar 29 '21 at 20:50
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    @whuber I think the title and the body are mismatched -- the title says we've observed a success on the $k$th trial, which suggests independence, but the body says "given the first success happens after the $k$th trial," which is, I think, a question about memorylessness. – Sycorax Mar 29 '21 at 20:57
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    @Sycorax Good point -- there do appear to be multiple reasonable interpretations of the problem. – whuber Mar 29 '21 at 21:35
  • Thanks everyone for the comments. I modified the question. I'm essentially asking about the probability of observing a certain number of failures between two successes. – TSP Mar 30 '21 at 19:54
  • @AryaMcCarthy I mean on the kth trial – TSP Mar 30 '21 at 19:55
  • @whuber I think the way you phrased it is what I was looking for – TSP Mar 30 '21 at 19:55
  • @whuber Would you say that the answer to this involves the memoryless property of the geometric distribution? – TSP Apr 02 '21 at 02:18
  • I would say instead it involves the concept of independence: the chances of none of the outcomes after trial $k$ depend on the outcomes on or before trial $k.$ – whuber Apr 02 '21 at 13:30

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