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As in the $2$-dimensional problem linked, the probability that a particular interval much shorter than $n^{\alpha-1}$ contains $n^\alpha$ points is very small, and we can use the union bound over a sm …

Will Sawin's idea seems good. Here is a slightly simpler way to get a similar pairwise independent distribution where the maximum load with under a binary choice is still $\Theta(\sqrt{n}).$ Let $\lb …

For any random variable $X$ taking values in $\mathbb{N}$, $E[X]= \sum_{n=0}^\infty P(X\gt n)$. In this case, the probability that the sum of the first $n$ numbers is less than $1$ is $1/n!$, the volu …

Here are some nonconstructive estimates on the number of tries needed to find the exact answer, not just to get $90\%$ or more correct. See this paper by Erdős and Rényi for similar analysis of a rela …

As Pat Devlin commented, you can compute the expected value. A similar calculation gives you the higher moments. $$E(Y^2) = \sum_{i,j} I_{X_i = 0} I_{X_j = 0} \\\ = \sum_i I_{X_i = 0} + 2 \sum_{i\lt …

It's not immediate whether rock can win at all, although I think it should be possible to work that out. For example, if $(r,p,s) = (3,4,1)$ then rock can't win, but at $(2,4,2)$ rock wins with probab …

One strategy is to start removing marbles whenever you have drawn a blue marble. Let $a(p,n)$ be the expected number of removals under this strategy before reaching a collection of $n$ red marbles. …

Ori Gurel-Gurervich's comment suggests a very simple way to use a martingale (an example of a Wald martingale) to evaluate the final question in which the probability of gaining a dollar is $p \ne \fr …

While the exact answer by Mark Meckes is nice, it's worth pointing out that if you condition on not repeating elements, the conditional distributions are equal by symmetry, and your condition $n \gt\g …

You want $\frac15 = \sum_t |P_1(count=t) - P_2(count=t)|$. where $P_1$ has a binomial distribution and $P_2$ is hypergeometric. The difference between these distributions is shown in this Mathemati …

The expected number of picks needed equals the sum of the probabilities that at least $t$ picks are needed, which means that $t-1$ subsets left at least one value uncovered. We can use inclusion-exclu …

Edit: I've filled in a few more details. The Hoeffding bound from expressing $Y-X$ as the sum of $n$ differences between Bernoulli random variables $B_q(i)-B_p(i)$ is $$Prob(Y-X \ge 0) = Prob(Y-X + …

The effect of adding a Bernoulli$(p)$ on the parity applies this matrix $$M=\left(\begin{array}{cc} 1-p & p \\\ p & 1-p \end{array}\right)$$ to $(prob(even), prob(odd))$. We can decompose the ini …

Earlier, I gave the following sketch: Evenly distributed blocks implies evenly distributed sequences: Estimate the upper and lower densities of a sequence of length $k$ from the frequencies of block …

Monte Carlo methods are appropriate for analyzing some systems involving chance, not incomplete information. Monte Carlo methods tell you nothing about how to model a poker strategy. For general game …