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According to the red-sauce theorem, the possibility of you becoming green before seventy years of age is just as low as you becoming purple. I hope this is thoughtful enough.


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This just adds some information about things you've already mentioned -- I'm quite interested as well in any methods. I note: To the authors’ knowledge, the only known prime generation algorithms for which the statistical distance to the uniform distribution can be bounded are the one proposed by Maurer [19,20] on the one hand, and the trivial ...


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For any vector $\mathbf x$ let $r^k(\mathbf x)$ be the vector obtained by rotating the components by $k$ positions. Then $\sum\mathbf a\cdot\sum\mathbf b=\sum_{k=0}^{N-1}\langle \mathbf a,r^k(\mathbf b)\rangle$. Hence at least if the distribution for the various components are the same and are independent (but otherwise unspecified), the approximation ...


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There is a standard algorithm for this: Let $S$ (the selected entry) be the first entry. If there is a second entry, generate a random number $x$ between 0 and 1. If $x\lt \frac12 $ discard the previous selection and let $S$ be the second entry instead. If there is a third entry, generate a random number $x$ between 0 and 1. If $x\lt \frac13$ discard ...


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You have to relabel the elements: use a bijection $\phi$ from $\mathbb N$ to $\mathbb N^2$ and define $B_n:=A_{\phi(n)}$. The assumption gives that $\mathbb P(\limsup_n B_n)=0$ hence $\mathbb P(\mbox{number of } n \mbox{ such that } B_{n} \mbox { occurs} < \infty) =1$.


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A run in a binary string $s\in\{0,1\}^n$ is a maximal consecutive substring of ones in $s$. Denote by $E_0(n)$ the expected number of runs in a random string of length $n$ ending in $0$, by $E_1(n)$ the expected number of runs in a random string of length $n$ ending in $1$, and let $$E(n):={1\over2}\bigl(E_0(n)+E_1(n)\bigr)$$ be the number we are after. ...


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The probability of the program lasting precisely $n$ experiments is $$ (0.2)^{n-1}\cdot (0.8) $$ Break the problem into two regions: Find the expected number and cost of expensive experiments, and add the cost of inexpensive experiments. The expected cost of the expensive experiments, each of which requires $n-1$ previous failures to be needed, is $$ 10 ...


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If Pr(N) is the probability of reaching N trials before a successful outcome. Assuming you mean the Expected cost of any single trial. E(cost) = 10*( Pr(1) + Pr(2) + Pr(3) + Pr(4) + Pr(5)) + 5* sum(Pr(n)) N=6 to inf


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Whether 2n − 1 is prime isn’t very important if we look for multiplication by certain constant in F2n . Multiplicative groups of finite fields are always cyclic, and any cyclic group has elements of the full order, prime or composite. As follows from inspection the code, mathematics of this example is iteration of a linear operator on 33-dimensional vector ...


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Let $P_i=1$ if $i$-th symbol in bitstring starts a maximal run of ones and $P_i=0$ otherwise (for example, for $1000111110100111$ $P_i=1$ for $i=1,5,11,14$ and $P_i=0$ for any other $i$). Note that $X=\sum_{i=1}^nP_i$. Then use linearity of expectation: $E(X)=E\left(\sum_{i=1}^nP_i\right)=\sum_{i=1}^nE(P_i)$ Because $P_i$ is indicator random variable, ...


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Let $p$ be the probability that $s_i = 1$. Let $X_n$ be the number of maximal runs of ones in a sequence of $n$ bitstrings. Compare $X_n$ with $X_{n-1}$. Imagine the last bit is hidden. The only way that $X_n$ will increase is that if the $n-1$ bit is 0 and the $n$ bit turns out to be 1. So, given $X_{n-1}$, $X_n = X_{n-1} + 1$ with probability $p(1-p)$, ...


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I do not know if it is what was intended, but you could have your indicator registering if the $n-1^\text{th}$ bit is $0$ and the $n^\text{th}$ bit is $1$. You may need to take the unseen bit before the first to be $0$. Then use linearity of expectation.


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The primes are not randomly distributed. They are completely deterministic in the sense that the $n$th prime can be found via sieving. We speak loosely of the probability that a given number $n$ is prime $(p~(n\in P) \approx 1/\log n)$ based on the prime number theorem but this does not change matters and is largely a convenience. Some confusion is maybe ...


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Let $X_i$ be the time that it takes Adrian to go to school on the $i$-th day of the week, for $i=1,2,3,4,5$. Then, the total weekly journey time $X$ of Adrian can be expressed as $$X=X_1+X_2+X_3+X_4+X_5$$ Due to linearity of expectation $$E[X]=E[X_1]+E[X_2]+E[X_3]+E[X_4]+E[X_5]=20+\ldots+20=5\cdot20=100$$ Due to independence of journey times ...



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