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I play $n$ games of chess against my friend. I have a probability $p$ of winning, my friend a probability of $1-p$ of winning a single game. Assume games are independent.

What is the probability that one of us wins $x$ successive frames in the $n$ games, where $ x \leq n$?

I thought about using the geometric distribution of some sorts as this seems natural, given we're on about successive trials, but I've not got anywhere with it.

the man
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  • Do you mean exactly $x$ successive games, or at least $x$ successive games? – Adam Rubinson Sep 23 '21 at 12:21
  • @AdamRubinson at least – the man Sep 23 '21 at 12:39
  • For a set $S$ with a finite number of elements, let $|S|$ denote the number of elements in $S$. For $k \in {1,2,\cdots,(n+1-x)}$, let $A_k$ denote the set of outcomes that includes you winning $x$ consecutive games, starting from game $k$. Let $B_k$ denote the set of outcomes that includes your friend winning $x$ consecutive games, starting from game $k$. Then, you will need to calculate $|A_1 \cup \cdots \cup A_{(n+1-x)} \cup B_1 \cup \cdots \cup B_{(n+1-x)}|$. I suggest Inclusion-Exclusion. – user2661923 Sep 23 '21 at 12:40
  • "What is the probability that one of us wins..." : my previous comment made two assumptions: [1] You were focusing on (for example) your having at least one streak of consecutive wins, where the streak was $\geq x$. [2] You want the combined probability that either you or your friend has such a streak. – user2661923 Sep 23 '21 at 12:44
  • For large values of $n$, the math can get ugly here. I suggest, that you practice with sample values like $n = 15$ and $x = 4$. Also, if you can use a programming language, like Java, C, or Python, then with your sample problem, I suggest that you sanity-check your work via a computer program that counts all of the favorable sequences. For example, your program won't have any problem checking $2^{(15)}$ various sequences of your wins combined with your losses, to see how many of these sequences satisfy the constraints. – user2661923 Sep 23 '21 at 12:50
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    By the way, per the problem's constraints, the probability that a specific chess game is drawn is $(0)$. – user2661923 Sep 23 '21 at 12:51
  • Yes but for example all the games could be "Armageddon" - whereby the game is either a win for White or Black (i.e. no draws). So this is not necessarily unrealistic. – Adam Rubinson Sep 23 '21 at 13:40
  • I don't see how else to do this other than using a long Inclusion-Exclusion calculation as in user2661923's first comment. If $n=50, x= 13$ for example, then the calculation will be overwhelming, so at that point I would just run a few thousand simulations with a Python program and see what percentage of outcomes satisfy the condition. – Adam Rubinson Sep 23 '21 at 13:47
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    This has been asked ton of times here. https://math.stackexchange.com/questions/59738/probability-for-the-length-of-the-longest-run-in-n-bernoulli-trials – leonbloy Sep 23 '21 at 15:13
  • @leonbloy Could you please reverse your decision? This is a distinct problem; the question you linked is for long runs of heads only. This question is for long runs of heads or tails. – Mike Earnest Sep 23 '21 at 15:16

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