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Let's say I have an n-bit string, where I flip any bit with probability $1-p$. Call the uncorrupted string $\mathbf{b} = (b_1,...,b_n)$, and the corrupted one $\tilde{\mathbf{b}} = (\tilde{b}_1,..)$. Now define the "bit-flip" distance between $\mathbf{b}$ and $\tilde{\mathbf{b}}$ with: \begin{align*} X = || \mathbf{b} - \tilde{\mathbf{b}} ||_1. \end{align*}

I want the expected value of $X$ to be less than some value $d$ where $d = 0,1,2,..$. So we have: \begin{align*} E[X] = \sum_{x=1}^n x\binom{n}{x} p^{n-x} (1-p)^x = n(1-p). \end{align*} And note this expected value is correct, see Expected Value of a Binomial distribution?. So we solve for $p$ and have: \begin{align*} n(1-p) \leq d \implies p \geq 1 - \frac{d}{n}. \end{align*} So if we set constraint where $\mathbf{b}$ and $\tilde{\mathbf{b}}$ do not differ in more than 10% of the bits, then say $d=10$ and $n=100$. Then this means $p \geq 1 - 0.10 = 0.90$. This places a very high constraint on the value of $p$. Am I missing something here?

chibro2
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    Can you go more into why you think that's confusing? You want at least 90% of the bits to be correct, so the probability of not flipping should be at least 90%? Also I should note that in the title you say that the probability of flipping is $p$, but in your first sentence the probability of flipping is $1 - p$. – S.C. Nov 11 '24 at 13:22
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    This is obvious by the Law of Large Numbers: the probability is the number of successes over the number of attempts, $\frac{90}{100}$. What more do you want? –  Nov 11 '24 at 13:28
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    @S.C. fixed. Yves Daoust , it make sense intuitively, but I was somewhat disappointed by the triviality of the result. – chibro2 Nov 11 '24 at 13:31

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