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As I'm not too familiar with mathematics, I hope someone can provide a fairly straight answer to my question. For the purpose of the question I'll use the following example:

  • A random natural number between 1 and 10 is picked x times.
  • If the picked number is e.g. 1, then what are the chances that the same number is picked again after exactly 5 more picks?
  • If the number 1 is picked again after more than 5 picks, then what is the probability the same number is picked again after exactly 5 more picks?
  • etc for the next occurrence.

So far I've only found out that: 0.1^10 = 0.35, which makes it 65% less likely to not be 1 after 10 picks. This doesn't look at the recurrence rate at all so it's only part of the answer.

I'd like to put the answers in a spreadsheet so formulas are also welcome. Basically I'm trying to find out what the percentual likelyness is of number 1 being repeated by another 1 after 5 picks, everytime this number occurs.

Example sequence of randomly picked numbers between 1 and 10: 2,5,8,10,2,1,8,3,7,2,9,9,3,1,4,8,1,5,3,8,0,1,4,10,2,8,1,2,8,9,4,3

As you can see, the 3rd sequence of 5 numbers that starts with a 1 also ends with a 1. I'd like to know the percentual probability of that happening.

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    Not following. If all you are asking is "what is the probability that I will see the next $1$ in exactly five trials" the answer is $\left( \frac 9{10}\right)^4\times \frac 1{10}\approx .0651$. But maybe you are asking something else? – 2017-02-28
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    Assuming trials are independent, the five intervening trials are irrelevant unless you forbid them to be matches. The probability of any one 'match' is $1/10$ and the probability of $n$ of them is $(1/10)^n.$ However, I agree with @lulu that it's not clear what you're asking. – 2017-03-01

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