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I solved this trivially. We can extend both ranges by employing $y_i = \xi_i / x$ in case of the range from $[0,x]$ and $y_i = \frac{\xi_i - x}{1 - x}$.


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That's a start... but you're forgetting the possibility of vanity plates, plates with 7 characters or 5 characters or even fewer, and obscene plates. In the United States, I've seen many license plates with any number of characters between 1-7. The DMV also won't approve obscene vanity plates, so you have to subtract an arbitrary guess of the number of ...


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Many linear congruential PRNG can be used. Take $a$ with $\gcd(a,n)=1,\;$ i.e. $a^{-1} \bmod n$ exists and define $$f(x) = 1 + ((ax+b) \bmod n),\;$$ where $b$ is any integer. Then $f$ maps $[1,n]$ to $[1,n]$ and is invertible: from $y=f(x)$ you compute $$x=a^{-1}(y-1-b)\bmod n,$$ and if $x=0$ you should set $x=n,\,$ this comes from your unusal domain ...


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The program given computes Monte Carlo integration of $$\int_0^1 x^2dx$$ We can also compute this exactly, as $\frac{x^3}{3}|_0^1=\frac{1}{3}$. More generally, replacing $2$ by $n$, the integral will equal $\frac{1}{n+1}$ (provided $n\neq-1$).


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See Liggett (1991), Stochastic Interacting Systems, part I, section 3, p.71. I'll provide some insight below in how to think about it. This should be helpful in understanding the difference in extinction time on the extreme ends of super- and sub-criticality. Of course, near criticality, the analysis becomes much harder. Subcritical: In the subcritical ...


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As pointed out above by @gnometorule, if you generate a number using an algorithm, it is not going to be random. Instead, you need to look at pseudorandom numbers. There are interesting ways to understand pseudorandom generators in the context of computational complexity. Very loosely speaking, if ahead of time you define your resources for determining ...


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If you want to generate real random number then you must use current time as a variable and do some operation with them. But still they have a algorithm. A random number is free from algorithm


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Finally I found another way of obtaining this kind of sequences! Reviewing at OEIS, some kind of partition problems provide sequences whose gaps show also this kind of pseudorandom behavior I was looking for. For instance: The strictly increasing elements of the multiplicative partition function: number of ways of factoring n with all factors greater ...


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How about numbers that are the product of two distinct primes. 6, 10, 14, 15, 21, 22, ...


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Consider the sequence $a_n=\lfloor{r\cdot{b^n}}\rfloor$, where: $r$ is any irrational number $>1$ $b$ is any natural number $>1$ For example, for $a_n=\lfloor\pi\cdot10^n\rfloor$ we get: $a_0=3$ $a_1=31$ $a_2=314$ $a_3=3141$ $a_4=31415$ $a_5=314159$ $a_6=3141592$ $a_7=31415926$ $a_8=314159265$ $a_9=3141592653$ $\dots$ For generally smaller ...


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What you wrote down isn't the probability for the $K$-th smallest element (you wrote "largest" but it seems you meant "smallest") to be less than $x$; it's the probability that $K$ particular elements are less than $x$ and the other $N-K$ are greater. But the $K$-th smallest element is less than $x$ even if all $N$ elements are less. You can tell that your ...


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Your idea is on the right path but you are missing something. $F(i)$ in your notation is the probability that some $U_{i_1},\ldots,U_{i_K}$ values are smaller than $x$, however you can pick indexes $i_1,\ldots,i_K$ in $\binom{n}{K}$ ways therefore the probability that $U_{(K)}$ ($K^{\text{th}}$ order statistic = $K^{\text{th}}$ smallest element) is smaller ...


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A random process is unpredictable such as the movement of the tip of a feather In wind. If we assume that the movement of a roller coaster is deterministic, then a stochastic process would be the movement of the tip of a feather attached to a moving roller coaster. That is to say, stochastic processes have components that are both deterministic AND random; ...


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I'm assuming you mean $A$ is selecting from the numbers $1, 2, \dots, 29, 30$. We're going to have to assume $A$ chooses uniformly at random (that is, with probability $\frac{1}{30}$ for each number). Then write $X_a$ for the number of guesses $B$ makes, given that $A$ picked number $a$. We have $X_a = a$, because if $A$ picked $1$ then $B$ guesses once ...


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Here is a symmetry argument for you. Call an ordering of $\{0,1,2,\ldots,31\}$ a strategy. Two sample strategies are $0,1,2,\ldots, 31$ and $0,31,1,30,\ldots,15,16$. Each strategy leads to a payoff, depending on what number A picked. Now, for each strategy, there is an "opposite" strategy, which consists of the same numbers in opposite order. The ...


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The grading system that is proposed here is unfair in the sense that it is random. Suppose a win chance of $p$ and a lose chance of $q$ against the random bot, where $p+q\le1$. Then the probability of getting a high score in ten games becomes better when $p$ gets higher and/or $q$ gets lower. However, suppose you have a very high $p$ and a very low $q$. ...


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You need to define what you mean by the grading system being wrong. With short runs like this it is reasonably likely that the results will not match the expected result over a long run of matches. It sounds like the grading system had an announced algorithm for scoring and you have not presented anything that would suggest the algorithm was not ...



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