I know we can try and count the number of ways that they can add to 20, however this is much easier when there is only 2 or 3 dice. I believe there is a way to do this using combinations but I am not sure. I thought it was a stars and bars, so I wrote out  20 like -> 1 + 1 + 1 + 1 |+ 1 + 1 + 1 + 1| + 1 + 1 + 1 + 1 + 1 + 1 |+ 1 + 1 + 1 |+ 1 + 1 + 1  Which makes a 20 choose 4, however, this is over counting as each block of 1's can't be greater than 6 or less than 1.   Thanks for your help
122782
I am trying to identify the general case algorithm for counting the different ways dice can add to a given number.  For instance, there are  with two 6-dice.  I've spent quite a bit of time working on this (for a while my friend and I were using Figurate numbers, as the early items in the series match up) but at this point I'm tired and stumped, and would love some assistance.  So far we've got something to this effect (apologies for the feeble attempt at mathematical notation - I usually reside on StackOverflow):  count(x):    x = min(x,n*m-x+n)    if x = n       1    else       some sort of (recursive?) operation   The first line simplifies the problem to just the lower numbers - where the count is increasing.  Then, if we're looking for the count of the minimum possible (which is also now the max because of the previous line) there is only one way to do that so it is 1, no matter the n or m.
22632
Suppose I have 3 dice. Each has some mechanism that can prevent other dice being in the same number as itself when they are rolled together. Now I roll the 3 dice at the same time. Then what is the probability of having a 2 appear?  Is the answer 1/6 + 1/6 + 1/6 = 1/2 since the event are mutually exclusive and each has a probability of 1/6? But 1/2 seems to be too high?
18569
I have come across numerous questions where I am given the picture such as the above one been asked "how many rectangles are there?". I have even come across some slightly different images that instead of rectangles you are supposed to find the number of triangles. Well, I was thinking whether there is any formula or strategy that is used to solve these problems without having to manually count every shape.  Help would be appreciated.  Thank you :)
106666
A 5x5 board has $25$ cells. The numbers $\{1,2,3,4,5\}$ are written on every row,every column and the two main diagonals without any repetition. If the sum of the numbers of the diagonal below the leading diagonal be represented as $A$. Show that $A\ne 20$. Also find the maximum possible value of $A$.    I could prove that $A\ne 20$ by showing that each of the cells of the diagonal below the leading diagonal cannot be $20$ as if such arrangement exists, then the board cannot have $5$ five times, which actually should be by the given condition of the question.    For the second part, I can imagine the maxiumum value to be $17$. For which I proved that $5$ can occur only $2$ times in the four cells and probably $4$ cannot occur twice. But I got no formal proof. Please help!
115174
How many combinations are possible in the game tic-tac-toe (Noughts and crosses)?  So for example a game which looked like: (with positions 1-9)  A1   --   B1  A2   --   B2  A3   --   --   [1][3][4][6][7] would be one combination
5024
Problem: Find the number of twelve-card hands containing four cards in each of three suits.   My solution:   pick a suit : 4 ways. choose 4 cards from the 13 in that suit: ${^{13}}{C}_{4}$ ways. pick another suit: 3 ways. choose 4 cards from the 13 in the new suit: ${^{13}}{C}_{4}$ ways. pick the final suit: 2 ways. choose 4 cards from the 13 in the final suit: ${^{13}}{C}_{4}$ ways.   Hence, total number of ways is $4\cdot{^{13}}{C}_{4}\cdot3\cdot{^{13}}{C}_{4}\cdot2\cdot{^{13}}{C}_{4} = {^{4}}{P}_{3}\cdot\left({^{13}}{C}_{4}\right)^3 $ total hands.   This is the incorrect answer. Why?
112474
So earlier today I came across , and I am having some trouble understanding the solution. The question is as follows:     You throw a fair six-sided die until you get 6. What is the expected   number of throws (including the throw giving 6) conditioned on the event   that all throws gave even numbers?   Quoted from Jimmy Jin in "Elchanan Mossel’s dice problem"  In the paper it goes on to state why a common wrong answer is $3$. Then afterwards explains that this problem has the same answer to,   "What is the expected number of times you can roll only $2$’s or $4$’s until you roll any other number?"  I don't understand why this is the case. If the original problem is asking for specifically a $6$, shouldn't that limit many of the possible sequences?  I also attempted to solve the problem using another method, but got an answer different from both $3$ and the correct answer of $1.5$.  I saw that possible sequences could have been something like:  $$\{6\}$$ $$\{2,6\}, \{4,6\}$$ $$\{2,2,6\}, \{2,4,6\}, \{4,2,6\}, \{4,4,6\}$$ $$\vdots$$  To which I set up the following summation and solved using :  $$\text{Expected Value} =\sum_{n=1}^\infty n\left( {\frac{1}{6}} \right)^n 2^{n-1} = 0.375$$ Obviously this is different and probably incorrect, but I can't figure out where the error in the thought process is.  Any help on understanding this would be greatly appreciated.  A blog post discussing the problem can be found
98913
The probability of two dice summing to k is simple enough, make a diagram of the throws,  $$ \begin{array}{c}  &amp;|&amp;1&amp;2&amp;3&amp;4&amp;5&amp;6\\\hline 1&amp;|&amp;2&amp;3&amp;4&amp;5&amp;6&amp;7\\ 2&amp;|&amp;3&amp;4&amp;5&amp;6&amp;7&amp;8\\ 3&amp;|&amp;4&amp;5&amp;6&amp;7&amp;8&amp;9\\ 4&amp;|&amp;5&amp;6&amp;7&amp;8&amp;9&amp;\color{green}{10}\\ 5&amp;|&amp;6&amp;7&amp;8&amp;9&amp;\color{green}{10}&amp;11\\ 6&amp;|&amp;7&amp;8&amp;9&amp;\color{green}{10}&amp;11&amp;12\\ \end{array} $$  and the probability of a sum is just the length of the corresponding diagonal divided by $36$. For instance, $\mathrm{P}(\mathrm{sum} = 10) = \frac{3}{36} = \frac{1}{12}$. However, it's not as easy to draw a diagram for three dice or more, so how would a general formula look?
47902
How do I calculate the number of options for $k$ different balls in $n$ cells, with at least $1$ ball in each cell?   Lets say $6$ different balls in $3$ different cells, with at least $1$ ball in each cell.   like this:  $$x_1+x_2+x_3 = 6$$  I would like to see an explanation please and not just the answer.  Thanks!
29781
I have the following problem: A class has 20 students, 16 females, 4 males. Find all possibilities of choosing 6 students so that at least one is male.  I did the following, pick one male from the 4, you have 4 possibilities, then you are left with 19 students, pick 5, so I get 4*${19}\choose{5}$ possibilities.  However, one can notice that if you want no one to be male, then the possibilities are ${16}\choose{6}$, so, one could argue that the solution is instead ${20}\choose{6}$ - ${16}\choose{6}$.  So, which is solution is correct, and why?  $$4* {{19}\choose{5}}$$  or $${{20}\choose{6}} - {{16}\choose{6}}$$
59477
Over the holidays I played a dice game where each player had 3 to 7 d6 to roll each turn.  The game gave certain advantages to doubles and triples.  I wanted to know the odds of rolling doubles or triples given N dice to understand the importance of "upgrading" to more dice.  My question is really about how to reason about such problems:  It's easy to see that the odds of rolling doubles with 2 d6 are 1/6 (first die can be anything, and 1/6 of the time the second will match).  It's also easy to see that with 7 dice the odds are 1.0 (6 dice could be 1..6, but the 7th must match one of them).  The middle cases are fuzzier to me.  I ended up writing a program to enumerate all of the possibilities (even 6^7 is only 279,936 cases to evaluate) because that was the only way I felt I could verify any closed formula I came up with.  This also enabled me to distinguish cases of multiple doubles (or triples).  I would like to know how to derive closed forms and how to validate such solutions without brute-force simulation.
111650
I'm struggling with the following problem:      I have $N$ balanced 6-sided dice. I roll the dice simultaneously, and remove any sixes that occur. I roll the remaining dice again, and remove any more sixes. I repeat the process until there are no dice remaining. What is the expected number of rolls this will take?   So far I have calculated that by the $n^{th}$ dice roll there will be $N\left(\frac56\right)^n$ dice remaining: If we start with $N$ dice, on the first roll we expect $\frac{N}{6}$ sixes. Therefore, on the second roll we expect to have $N-\frac{N}{6}$ dice and hence expect $\frac{5N}{36}$ sixes, meaning after the second roll there are $\frac{25}{36}N$ dice remaining.  Repeating this calculation gives the sequence $N, \frac{25}{36}N, \frac{125}{216}N...$ which is equal to $N\left(\frac56\right)^n$ where $n$ is the roll number. I'm struggling with the next part: My thinking is we must find the expected number of rolls $n$ such that $N\left(\frac56\right)^n\lt0.5$, and therefore $n&gt;\frac{\ln(\frac{0.5}{N})}{\ln(\frac56)}$. Taking $N$ to be $8$, the expected number of rolls is then about $15.21$. I used MATLAB to run the experiment $200,000$ times, and it gave me an average number of rolls of $15.4$. I seem to be close to the right answer, but I'm not sure what I've done wrong. What is the solution?
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?
16441
This is one of those popular pictures on sites like Facebook. I always see a huge variation of answers such as $8, 9, 16, 17, 24, 28, 30, 40, 41, 52,$ etc., yet I've never seen a definitive answer on any website.    I counted $16$ whole squares ($4$ going down on left side, $4$ going down on right side, $8$ small squares in the centre). With this method the answer could also be $17$ if you want to count the entire shape as one big square.  I can also see the answer being $24$ if you count the smaller squares separately, as if they were floating above the main grid instead of intersecting with it.  Is there a right answer, or is it possible that there is no 'right' answer and it just depends on your counting method?
42849
We roll a die until each side appears at least once, then we stop.     What is the probability of rolling exactly $n$ dice?   I guess the answer is  $$6-6\left(\dfrac{5}{6}\right)^n\;,$$  but this may be wrong.  In trying to solve this problem, I read , but I couldn't solve the problem.
110894
Suppose that I roll $3$ dice and write down the outcome as the coefficients $a,b,c$ in the polynomial $ax^2+bx+c$ respectively. What is the probability that this polynomial has no real roots?  So, I have to count the number of triples $(a,b,c)$ such that $b^2 &lt; 4ac$, where $a,b,c \in \{1,2,3,4,5,6\}$. I'm not sure how I can do that. Please give me a hint first. This problem is from a high school probability course, so I think it must have a very basic solution.
113008
Four dice are thrown, what's the probability that:  a) None of them fall higher than three?  b) None of them fall higher than four?  c) That four is the highest number thrown?  So for a, I wanna think about the denominator first. There are 6 possible outcomes for each dice, and we have four dice. So we're technically picking r out of n objects, aka picking 4 possible values out of 6. So the denominator/sample space should be $6^4$ right?  Now this is where I'm getting tripped up. Order technically shouldn't count, because we only care about the quadruplets that don't have a value higher than 3. I might be wrong, maybe I am, but could someone explain a bit more? My professor stated that typically in the sample space/denominator, we want order to count.   So I'm trying to think of the numerator now, so we want to find the probability that no values appear higher than 3, so some events that can occur are: $(1, 1, 2, 3), (1, 2, 2, 3), (1, 1, 1, 1)$ etc. Again, I can't see why order should count here, because if we only are concerned about what appears rather than how they appear, then we can say $(1, 1, 2, 3)$ is equal to $(1, 2, 1, 3)$? So if order does not count, and we have replacement, then does this lead the case where we use $$\binom{n-1+r}{r}$$to find out the probability?   I'm pretty sure if I can do a), I could prob do b), but if someone could maybe lead me in the right direction for c) I would appreciate that too!