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If there is a constant probability $p$ of winning a round of a game such that each game is independent of every other, what is the probability $P$ of losing $n$ consecutive rounds out of a total of $k$?

In the simple case where $k$=$n$, $P=(1-p)^n$. However, I was not able to work out a formula in any case where $k>n,k\in \mathbb{N}$.

I've tried several approaches, including:

  1. As events are independent, the streak is equally likely to occur at any point. There are $k-n$ points where the streak could start, and the probability of a streak occurring at any point is $(1-p)^n$, so $P=(k-n)(1-p)^n$. However, I believe this double counts streaks of length $n+1$ as two $n$ length streaks, and in some cases this may be greater than $1$ if $p$ is quite low and $k-n$ is high.
  2. A binomial distribution can model the probability that there are $x$ losses, if $x\geq n$. We can sum over all possible values of $x$ and multiply the probability of $x$ total losses by the probability of $n$ consecutive losses given $x$ total losses using permutations and combinations. However, I could not work out how to correctly calculate the probability of $n$ or more consecutive losses given $x$ total losses out of $k$ events.

Can anyone help me solve this deceptively simple problem? I am trying to perform risk analysis, where $n$ is the number of successive investments I can afford to lose out of a total of $k$. Currently I am estimating these values using a primitive computer simulation, but it would be much better if I could calculate (or at least approximate) the true probability in a reasonable amount of time.

Update: this answer is a computationally viable approach to solving an almost identical problem relying on Markov chains which I've tested and it generalises wonderfully.

Oliver
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  • I think I have an approach that will work for this, but it's a bit late so I need to work on it a bit more... What are the odds of hitting that streak starting with the first round? Second Round? Third Round? Do any of these change with the total number of rounds? – Kitter Catter Jun 16 '19 at 05:18
  • Maybe worth a read: https://math.stackexchange.com/questions/234062/probability-of-tossing-a-fair-coin-with-at-least-k-consecutive-heads?rq=1, https://math.stackexchange.com/questions/3124684/probability-of-getting-a-sequence-of-at-least-k-heads-in-n-coin-flips, https://stats.stackexchange.com/questions/21825/probability-of-a-run-of-k-successes-in-a-sequence-of-n-bernoulli-trials, https://math.stackexchange.com/questions/491836/probability-of-finding-at-least-k-consecutive-heads-in-n-coin-tosses – Minus One-Twelfth Jun 16 '19 at 07:15
  • Should it be exactly one sequence of lost rounds of length $n$? – user Jun 16 '19 at 08:06
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    @MinusOne-Twelfth thanks, I'll take a look. – Oliver Jun 16 '19 at 08:22
  • @user any number of sequences of length $\geq n$ qualifies. – Oliver Jun 16 '19 at 08:22
  • @YaniorWeg whilst that question is almost the same, the answers presented do not generalise to the case where $p \neq 0.5$, and they are computationally intractable. However, this answer, linked to by Minus One-Twelfth on stats.SE solves my problem. If you can't mark it as a duplicate of that question though, I'll close this question. – Oliver Jun 16 '19 at 11:51

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