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-1

What is the probability that you try to check your email during the first $x$ minutes of the $(x+y)$-minute cycle? If you do that, then your waiting time is $0$. What is the probability that you try to check your email when more than $w$ minutes, but fewer than $y$ minutes, remain of the $(x+y)$-minute cycle, where $0<w<y$? In that case, you need to ...


2

Python's random.sample method solves this problem, and chooses between two different ways depending on how large the sample is compared to the full deck. If the sample is relatively small, it will generate a bunch of random indexes into the array, and throw out duplicates (using the set type), until it has the right number of distinct indexes. If the ...


0

Because you're making a gaming system, I guess you might be comfortable with some simulations that require little programming. Here is a program with an approximate answer to your question about 3 rolls of a six-sided die with and without a bonus point. (The third decimal place may be off by 1 or 2.) The probability is about 0.546 that three rolls with a ...


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There are online utilities that can do the calculations for you. This one for example looks good to me: you can combine dice rolls (subtract 1d6 from 1d8 for example) to get a probability curve for all the possible results.


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For your first question, six-sided die vs. eight-sided, make a $6$ by $8$ table, with values $1, 2, 3, 4, 5, 6$ in one direction and $1, 2, 3, 4, 5, 6, 7, 8$ in the other direction. The resulting $48$ small squares in the table determine $48$ possible outcomes of the two dice. You can then see for how many squares does six-sided beat eight-sided; how ...


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For one group you have $E(X_1)=E(X_2)=E(X_2)=\ldots =E(X_r)=n-n(1-p)^n+1$. Now it is known that $E(X)=\underbrace{E(X_1)+E(X_2)+E(X_3)+\ldots E(X_r)}_{\text{r times}}$ This is the expected value for all r groups. $E(X)=r\cdot \bigg(n- n(1-p)^n+1 \bigg)$ Now you can use the relation $r=\frac{m}{n}$ $E(X)=\frac{m}{n}\cdot \bigg( n-n(1-p)^n+1 \bigg)$ ad ...


1

I will answer the first two parts. a) If a switch doesn't work with probability $p$ then it does work with probability $1-p$. $P(X=1)$ is the probability that all the switches are working which is equal to $(1-p)^n$. The only other value $k$ can take is $n+1$ so $P(X=n+1)$ is the same as $P(X=1)^c$. By the complement law of probability $P(X=1)^c = 1 - ...


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Chaos from Dynamical Systems can model randomness in a very close way. Consider the humble Logistic Recurrence... $$X_{n+1}=a \cdot X_n \cdot (1-X_n)$$ It looks innocent, but using the methods of dynamical systems and bifurcation theory, we find that after a series of period doublings, instances were the number of solutions doubles, the system becomes ...


2

Virtually all computer "random" number generators generate "pseudo-random numbers", meaning that in many senses the numbers generated do not follow any easily predictable pattern but the sequence is deterministic. They usually take a "seed" as input to start generating the sequence at an arbitrary point. For more non-deterministic behavior you can use the ...


1

There are two parts to your question. For the second part, assuming X(t) and Y(t') are independent, requires the cross spectral density to be zero. The cross spectral density is the Fourier transform of the cross correlation function. The cross correlation is the ensemble average of the time-shifted product of X(t) and Y(t'), and if these are independent ...


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The idea is to regard each choice of uniformly distributed random number $U_i \sim \operatorname{Uniform}(0,1)$ where $i = 1, 2, \ldots, N$, as associated with a Bernoulli trial $B_i \sim \operatorname{Bernoulli}(p)$, where $$B_i = \begin{cases}1, & 0 \le U_i < t \\ 0, & t \le U_i \le 1. \end{cases}$$ That is to say, $B_i = 1$ if and only if the ...


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Every choice of a number results in a 'success' if the number is under the threshold and a failure if not. Assuming that the numbers are independently chosen you deal with binomial distribution having parameters $N$ and $t$. If $X$ denotes the number of successes then you are looking for $P(X\geq n)$


1

All you need is a normal number. They should be easy to find, because almost all reals are normal. Unfortunately, it is hard to prove any given number is normal unless it is constructed to be so and then it probably won't look "random". The Champernowne constant $0.123456789101112131415...$ fits this. It is known to be normal in base $10$, so if you break ...


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OK, your problem with 1st point is your wrong interpretation of that. As I see, it means if you take a fixed number of outcomes the probability of every outcome is equal, i.e. if you generate 100 true-random numbers from 0-9 there are 10 0's, 10 1's and 10 of each (the values are uniformly distributed over a defined interval or set). This probability never ...


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From a remote sensing point of view, we usually refer to a bounded but unpredictable process as stochastic. If the process were unbounded and unpredictable I would tend to use random, but this case doesn't occur very much in my world! :)


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There is a really nice algorithm of Myrvold and Rusky to rank/unrank permutations in linear time. You can download the (very short) paper here: http://webhome.cs.uvic.ca/~ruskey/Publications/RankPerm/RankPerm.html Essentially, the algorithm is derived from the method of generating a random permutation. Start with the identity, then randomly swap the first ...


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Using Uniform Random Numbers to Simulate Various Distributions Usually, by 'random number' or 'pseudorandom number', we mean an observation $U$ from a uniform distribution on the interval (0, 1). Other distributions are simulated by transforming $U$ to match some other distribution. A histogram of many observations $U$ would have several bars with bases in ...


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This algorithm generates whole numbers in the range $[0,m-1]$ You can change that to any range you want by linear scaling. If you want $[a,b]$, take the random $x$ that you get and return $a+\frac x{m-1}(b-a)$


1

There are many ways in which “random” floating-point numbers could be distributed with mean $77$ and standard deviation $5$. @MyUserIsThis gave an example where the numbers are chosen from a normal distribution. Here’s how you could choose them from a uniform distribution. The continuous uniform distribution of real numbers on the interval $[77-a,77+a]$ has ...


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As Java-Code: public static void main(String[] args) { Random random = new Random(); List<Float> lst = new LinkedList<>(); for (int i = 0; i < 5000; i++) { float X = random.nextFloat() * (float) Math.PI * 2; float Y = random.nextFloat(); float r = (float) (5 * Math.sqrt(-2 * Math.log(Y))); float x = ...


1

This is done by a method (or group of methods) called random sampling. For example, let's say we have a normal distribution $(\mu,\sigma)$, the distribution density is: $$f(x)=\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$$ In this case we have an infinite domain. we can do two simple things, the first one would be to cut it. For example ...


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I found the solution: If $f \in C_c(E)$, then $\forall t \in \mathbb{R}: t * f \in C_c$. For $f_1, ... f_n \in C_c(E), t \in \mathbb{R^n}$ set $f = (f_1, ..., f_n)$ $$\phi_X(t \cdot f) = \mathbb{E}[\exp(i t \cdot I_f(X))] = \int_E \exp(it \cdot x) d\mathbb{P}_{I_f \circ X}$$ From the uniqueness of the characteristic function $\mathbb{P}_{I_f \circ X}$ is ...


0

What we are looking for is $$\sum_{k=1}^x P(k,x)\cdot\frac{k}{i}$$ where $P(k,x)$ is the probability that exactly $k$ numbers have come up in $x$ moves so far. For example, if two numbers (say, three and five) have come up in five rolls of a die, the results will look like: ...


0

Argue something like this: Suppose $\phi_X(f) = \phi_{X'}(f)$ for every $f$. Pick sets $A_1,\ldots,A_n$ in $B_b$ and constants $t_1,\ldots,t_n$ and put $f=\sum t_i \chi_{A_i}$, where $\chi_{A_i}$ is the indicator function of set $A_i$ (taking value $1$ on $A_i$ and $0$ elsewhere). By assumption, $$\mathbb{E}[\exp(i\int f\, dX)] = \mathbb{E}[\exp(i\int f\, ...


2

There are many such algorithms, with different efficiencies. There is a very simple algorithm which I will demonstrate with two examples. First, consider the case where you want to use a six sided dice to generate a random number between 1 and 100. Roll the dice 3 times. There are 216 (6 x 6 x 6) possible results. If you roll 1,1,1 the random number is 1. ...


1

All problems of this structure are solvable. You can find the maximum number of X cabinets by checking each constraint. In your example, there is room for $12$ of them, so that is the maximum. Similarly, there is a maximum of $9$ Y cabinets. Now you can just loop over $x$ from $0$ to $12$ and $y$ from $0$ to $9$,check whether you can fit $x$ type X ...


3

As the people are interchangeable, your sample space is just the $20 \choose 6$ ways to choose the occupied rear seats. To have everyone see the movie, the occupied ones can be chosen in $\binom{14}{6}$ ways. To have one person not see the movie, you choose his seat in $6 \choose 1$ ways and the rest of the seats in $14 \choose 5$ ways. Finally, the ...



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