Suppose I roll a six-sided die 1000 times and write down the number of times each face comes up.  How do I test whether the die is fair?  Can I use a chi-squared test where the expected number of each face is 1000/6=167?  There also appears to be a , but that seems less likely to be baked into stats packages and software.  .
33729
How can I test the fairness of a twenty sided die (d20)?  Obviously I would be comparing the distribution of values against a uniform distribution.  I vaguely remember using a Chi-square test in college.  How can I apply this to see if a die is fair?
92075
An interview problem is like the following:     Given a coin you don’t know it’s fair or unfair. Throw it 6 times and    get 1 tail and 5 head. Determine whether it’s fair or not. What’s your   confidence value?   I came out the following solution:     $H_0:$ the coin is fair   $H_a:$ the coin is unfair      $X$: is the number of heads      Rejection region: $|X - 3| &gt; 2$, i.e., $X = 0,1,5,6$      Significance level alpha:      $\alpha = P(\text{reject }H_0 \mid H_0 \text{ is true})$   $=P(X=0,1,5,6 \mid H_0 \text{ is true})$   $= (\binom{6}{0}+\binom{6}{1}+\binom{6}{5}+\binom{6}{6})*(1/2)^6$    $= (1+6+6+1)*(0.5^6) = 0.21875$      because $\alpha &gt; 0.05$, we do not have enough evidence to reject $H_0$, and   we  accept $H_0$, so the coin is fair   Is the above test a good one? And I did not know how to calculate the confidence value?
58384
I wish to draw integers from 1 to some specific $N$ by rolling some number of fair six-sided dice (d6). A good answer will explain why its method produces uniform and independent integers.  As an illustrative example, it would be helpful to explain how a solution works for the case of $N=150$.  Furthermore, I wish for the procedure to be as efficient as possible: roll the least number of d6 on average for each number generated.  Conversions from  to decimal are permissible.    .
106836
If I roll the dice 50 times, how do I calculate the chance that I will roll 6 at least 5 times in a row? Why this problem is hard  With 5 tries this would be easy: take $(1/6)$ to fifth power. With 6 tries this is manageable; take the probability of rolling 6 the first five times, add the probability of rolling 6 the last five times, then subtract the overlap (all six results are 6). Given two overlapping sets of 5 rolls, the probability that one is all 6's is not independent from the probability that the other is all 6's. In principle this could be continued, but inclusion-exclusion gets out of hand. There has to be a better way; what is it?
7299
In the past I have used the Chi-squared test to check the statistical randomness of my generator. Is this a good test to use? Are there other tests?
72323
Ok, I know this is wrong, but I want someone to tell me why.   Let's take a normal heads tails example of a fair coin. The probability of getting head = 1/2. And I write this is because, either it will be heads, or not. Hence two cases that makes it 1/2.  Now, I know I can't do this to all the cases. For example, the probability of getting a 2 when I roll a dice. I know the answer is 1/6 but why can't I do, either the outcome will be 2 or not. And in that case, my probability is 1/2.
116273
In the UK there are 90 bingo balls. A bingo card consists of 9 columns and 3 rows. A row contains exactly five numbers and four blanks. A column consists of one, two or three numbers and never three blanks. The columns are occupied by the number chosen from 1-9 (only nine), 10-19 (ten), 20-29 ... 70-79 and importantly 80-90 (eleven). Numbers in columns are ordered from top to bottom.  How many possible unnumbered card layouts are there?  How many possible numbered cards are there?  I have started to approach this problem by first choosing the blanks for the first two rows and then attempting to work out a compensating row. This was hard.  Next I begun to understand that there were a number of equivalence classes as columns could be reordered for the purposes producing different layouts.  Next I looked into the most convoluted card which had five numbers followed by four blanks in the first row and four blanks followed five number in the second. This satisfied the "no blank column" requirement. Now the final row could be arranged however it wanted, though there was one confusing overlap to deal with.   I considered another convoluted card which had as many columns of three as possible. It turns out it's possible have a maximum of three columns of three which also gives rise to the maximum of six columns of one. It also turns out you can have a maximum of six columns of two. I think the starting point might be to determine the classes of card initially by column combinations and then permute them.  Having gotten all muddled by the layouts, I turned to the numbers. With a single number, it was one of 9, 10 or 11 depending on the column and that was easy. For the two number case, let's say with a column of ten, the first ($i$) was chosen from 9 and the next from $10-i$. It suddenly occurred to me that it was just combinations, phew. How to fit that with the whole business of column choice lost me.  It seems possible that for each possible "set" of columns, there will be a way of determining the distribution of possible first columns and adding in an appropriate factor for the number combinations.  Frankly, it's all getting bigger than my coffee table.   Any simplifications from higher maths I'm missing?
58836
Suppose, we roll a fair die $n$ times. The expected value of how many times each side of the die will appear is $\frac{n}{6}$. The question is, what is the probability that any side of the die will come up more than $k$ times, where $n&gt;k&gt;\frac{n}{6}$? Or you could put it this way: what is the probability that no side will come up more than $k$ times? Clearly these two probabilities add up to 1.
92945
If I roll an n-sided die m times, what is the probability of getting only one face, two faces, .... or n faces?  Let us assume that I have a 6-sided die and I roll it 10 times. I want to calculate different probabilities. The probability of getting only one face from all rolls. The probability of getting only 2 faces in all rolls and so on until calculating the probability of the appearance of the 6 faces (at least one time for each face) in all rolls.
89448
A fair die is thrown 3 times. Find the probability  that the largest outcome of the three throws is a 3.    I understand that the problem isn't as simple as $ \frac{3^3}{6^3}=1/8$, being the answer. That would mean getting any of the values from $1-3$ for all of the throws is favorable, which includes series of throws that do not even have $3$ as an outcome.  What's the solution to ensure that there is at least one throw with $3$ as the outcome, with the $2$ other throws being anywhere from $1-3$?
13065
I have a 5-sided dice, and I want to use it to simulate a 7-sided dice. Is there a way to do this with a finite number of rolls?
66589
This is a dice problem  1) I want to calculate the probability to have more than X throwing 3 dice of 6 faces. X = addition of the result of the 3 dice.  2) This is the first step but if you can also provide me a solution to calculate the probability to have more than X with Y dices of Z faces it would be really great.
87902
Original Question:  Paul, Dave and Sarah are rolling a fair six sided die. Paul will go first, always followed by Dave, who is always followed by Sarah, who is always followed by Paul, and so on... What is the probability that Sarah will be the first one to roll a six?  Here's my work, I found the following formula in s, the number of sequences Sarah did not roll a six, terms  (5/6 * 5/6 * 5/6)^s * (5/6 * 5/6 *1/6)  (5^3/6^3)^s * (5^2/6^3)  (5^3s/6^3s) * (5^2/6^3)  $\dfrac{5^{(3s+2)}}{6^{(3s+3)}}$ = $\dfrac{5^{(3s+2)}}{6^{(3)(s+1)}}$  $\dfrac{5^{(3s+2)}}{216^{(s+1)}}$  And then we would use the summation formula for  $\displaystyle \sum_{s=0}^{\infty} \frac{5^{3s+2}}{216^{s+1}}$.  Would this be an acceptable answer or do I need to find an exact value
59865
I'm trying to make a game where the damage you deal is rolled by a dice. It all works. However, it keeps on giving the same values in a row so it isn't entirely random  ini_open("Health.ini"); attack = ini_read_real( "Attacks", "Felix", 0); ini_close(); global.damage=0 repeat(attack){ global.damage=irandom(6) }   if (global.damage ==0){instance_create_depth(300, 400, 0, obj_numbers0)} if (global.damage ==1){instance_create_depth(400, 400, 0, obj_numbers1)} if (global.damage ==2){instance_create_depth(500, 400, 0, obj_numbers2)} if (global.damage ==3){instance_create_depth(600, 400, 0, obj_numbers3)} if (global.damage ==4){instance_create_depth(700, 400, 0, obj_numbers4)} if (global.damage ==5){instance_create_depth(800, 400, 0, obj_numbers5)} if (global.damage ==6){instance_create_depth(900, 400, 0, obj_numbers6)}   The attack value is 1 because it is only one dice that is used for damage for now. The software I am using is Game Maker Studio 2
108922
First player rolls a 6-sided die 100 times and the second player 101 times. What are the odds of the second player getting more odd numbers than the first one?  I tried solving this via listing the options which led to a nasty sum that I wasn't able to solve neither by hand nor by using wolfram alpha. Clearly there has to be a more clever solution using the properties of a binomial distribution, but so far I haven't come across anything useful.
2079
After realizing that my knowledge of how probabilities function has waned significantly, I'm wondering what the probability of rolling multiple 6's or higher (6-10) on four 10-sided dice, five 10-sided dice, etc. up to nine 10-sided dice.   As well, what the probability of rolling a 7 or higher, 8 or higher, 9 or higher on the same sets of dice as mentioned above.   Something like; %chance rolling 3 of 6-10 on 4d10, %chance rolling 3 of 7-10 on 4d10, %chance rolling 3 of 8-10 on 4d10, %chance rolling 3 of 9-10 on 4d10, etc.  %chance rolling 3/4/5 of 6-10 on 5d10... etc.  I've tried creating some functions on AnyDice, but I either can't get the function(s) to work, or I fear I may be misunderstanding my results.  How do I make this work?
7728
Repeatedly rolling a six sided die four times and summing the highest three results gives you a distribution with what mean and standard deviation?  I've only taken AP statistics, but I would like to learn how to do this.