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combinations R2 <- 100 # Times to determine an index to remove dets <- matrix(NA, R1, R2) for (i in 1:R1){ # Determine distribution parameters # n <- sample(seq(1, 10, 1), 1, replace = F) # Number of dice …

Let A be the expected number of rolls of a fair d6 until you roll two consecutive sixes, conditional on every roll up to the consecutive sixes being even. Let B be the expected number of rolls of a fa …

() { x = rgeom(6,1/6) # the number of extra rolls is geometric distributed k = 20 # simulation for 20 strategies strategy = rep(0,k) # vector tonkeep the score ### loop to simulate dice …

in range(len(w)): ui = random.uniform(0, 1) ki = ui ** (1.0/w[j]) if ki > maxU: maxU = ki imax = j return imax, x[imax] #Test: Let us emulate two dice …

I read an example where someone considered a biased-towards-even 6-faces dice and say the sample space is $\{1, 2, 3, 4, 5, 6\}$ while the support is only $\{2, 4, 6\}$. … But if the dice is assumed to be biased and only produces even outcomes, are the odd values still worth to be included in the sample space, defined as the set of possible outcomes ? …

Since the throw of this dice in itself does not tell us anything about $\mu$, it is ancillary. …

Example: if I keep collecting the sums of three dice, I will get a symmetric distribution. … Example: if I keep collecting the products of three dice, I will get a distribution which is very skewed to the right. …

The question refers to the minimum number of die rolls and/or coin fills. If Peter fails on the first two rolls, he must roll at least 2 more times for a minimum of 4 rolls. Such as: 1 - 2 - 1 - 1 If …

This question is from 40 Puzzles and Problems in Probability and Mathematical Statistics by Wolfgang Schwarz: Peter and Paula play a simple game of dice, as follows. …

With each die roll, the number of dice increases by 1 with a probability $p=1/3$ and decreases by 1 with a probability $q=2/3$. …

You continue rolling all your dice one by one until you have no die left. For clarity: Say you have 2 dice now, you roll one of them and get 4. … If you were to get a 6, you will get an extra die such that you now have 3 dice. The game ends when you are left with no die. All the dice have uniform probabilities of getting 1-6. …

The whole point is that I doubt you have homogenuous data (aka identically distributed), so any formal statistical test is basically pointless; you will need to slice and dice it until you can be reasonably …

Though not exactly the same situation, it reminded me of the sum of two dice, which produces such a curve. …

Suppose I play a game with dice with a complicated win condition, as follows: You start rolling a d12 If you roll a 12, you replace the d12 by a d6 when you roll a 6, you win (let's call this path 1 … How would one calculate the expected number of dice rolls of any kind to get to the win condition? …

outcomes from 6-sided dice: [ 1 2 3 4 5 6 outcomes from 8-sided dice: 1 2 3 4 5 6 7 8 outcomes from 12-sided dice: 1 2 3 4 5 6 7 8 9 10 11 12 ] This works. … You can fill in the probabilities of each of the 26 samples in this space, outcomes from 6-sided dice: [ a b c d e f outcomes from 8-sided dice: g h i j k l m n outcomes from 12-sided dice: o p q …