You have n balls, and n boxes. There is a pairing of each ball to a box (and vice versa). If you were to randomly place balls in boxes, what is the probability that none of the balls would go in "their" box?  I've solved this numerically for:  n | probability --------------- 1 | 0 2 | 1 / 2 3 | 1 / 3   I believe an analytic solution is possible, but I've yet to be able to formulate it.
4694
I am trying to figure out how many permutations exist in a set where none of the numbers equal their own position in the set; for example, $3,1,5,2,4$ is an acceptable permutation where $3,1,2,4,5$ is not because 5 is in position 5.  I know that the number of total permutations is $n!$. Is there a formula for how many are acceptable given the case that no position holds its own number?
32480
Say, we have 4 red balls, 4 blue balls, 4 green balls, 4 yellow balls. How many ways are there to form a sequence of 10 balls such that every color of ball occurs at least twice?  My thought: r = red b = blue g = green y = yellow * = undetermined  Every arrangments should in this form: rrbbggyy**  Case 1: two * are same  $\frac{\binom{4}{2}\binom{4}{2}\binom{4}{2}\binom{4}{2}\binom{4}{1}10!}{4!2!2!2!}$  Case 2: two * are different  $\frac{\binom{4}{2}\binom{4}{2}\binom{4}{2}\binom{4}{2}\binom{4}{2}10!}{3!3!2!2!}$  Result = Case 1 + Case 2  Any suggestions would be appreciated!
714
What is wrong with my method of find the probability of a 2 pair?  For the first card we have 52 choices, one this is picked( call it A) we have 3 choices(different suite than A but this same value) for the second card, ( lets call this card B). Now we have AB, to we have to throw away 2 more cards so we don't have 3 of a kind or a four of a kind. Therefore one our third card we have 48 choices and on our fourth card we have 3 choices again similar to above.  On our last card we only have 44 choices left. So the probablity of getting a 2 pair should be (52*3*48*3*44)/(52 choose 5).  But this isn't correct.     Note: I edited this to "2 pair"; the original version was wrong.
9235
This is a neat little problem that I was discussing today with my lab group out at lunch. Not particularly difficult but interesting implications nonetheless  Imagine there are a 100 people in line to board a plane that seats 100. The first person in line realizes he lost his boarding pass so when he boards he decides to take a random seat instead. Every person that boards the plane after him will either take their "proper" seat, or if that seat is taken, a random seat instead.  Question: What is the probability that the last person that boards will end up in his/her proper seat.  Moreover, and this is the part I'm still pondering about. Can you think of a physical system that would follow this combinatorial statistics? Maybe a spin wave function in a crystal etc...
87575
I have two boxes, one with ten balls, eight white and two black and the other with ten balls, four white and six black. Without seeing I choose a box and choose three balls from this.  What is the probability that the fourth ball I'll choose is black if the other three are not all white?    My question is: Can I find the probability that the fourth ball is black if the other three balls are all white and then count $ 1- $ this probability?   I solved my question. For the exercise, that's my trying:   $E$ = the fourth ball is black  $A$ = the three balls are ALL WHITE  Then $P(E) - P(E|A)$ will give me what I need.  For $P(E)$:  $C$ = I choose first box and $D$ = I choose second box  $P(E) = P(E|C) \cdot P(C) + P(E|D) \cdot P(D)$  I will find $P(E|C), P(E|D)$ from the right groups of four balls  So, if $a$ is a white ball and $b$ is a black ball then:  For the first box: $aaab, baab, abab, aabb$ I sum the possibilities for all these groups and I get $P(E|C)$.  For the second box:  $aaab, baab, abab. aabb, bbab, babb, abbb, bbbb$ I sum the possibilities for all these groups and I get $P(E|D)$.
91418
Occupancy problem with balls and boxes. Suppose there are $N$ balls and $M$ boxes. The balls are thrown to the boxes at random. What is the probability of $k$ boxes contain exactly $1$ ball? where $k=1,2,...\min(N,M)$
77554
I have a problem, I'm learning probability at the moment (I'm a programmer) and starting I have this:  (Source: Minka.) My neighbor has two children. Assuming that the gender of a child is like a coin flip, it is most likely, a priori, that my neighbor has one boy and one girl, with probability 1/2. The other possibilities—two boys or two girls—have probabilities 1/4 and 1/4.  a. Suppose I ask him whether he has any boys, and he says yes. What is the probability that one child is a girl?  b. Suppose instead that I happen to see one of his children run by, and it is a boy. What is the probability that the other child is a girl?    Now my reasoning is: BB = 1/4 = 0.25 BG = 1/4 = 0.25 GB = 1/4 = 0.25 GG = 1/4 = 0.25  So for a., the probability of G I get it just by summing p(B,G) + p(G,B) = 0.5 And for b. p(G|B) = p(G,B)/p(B) = 0.5/0.5 = 1 that is wrong but I'm not getting why.
58816
I have this simple looking question: "In tennis tournament, there are $2n$ participants. So, there must be $n$ pairs for the first round. In how many ways can such a pairing be arranged??". Now, shouldn't the answer be $^{2n}C_2$? But the book says it is $1 \cdot 3 \cdot 5 \cdots (2n - 1)$. Where am I going wrong?
72003
You have 6 red balls, 6 blue, and 6 white. You randomly select a sample of 3 balls without replacement. The balls, save for the color, are indistinguishable from each other. What are the odds that your sample contains at least a ball from each color?  I thought about finding the chance that the sample will not contain at least one color and subtracting that from one:  $$P(\text{sample is missing at least one color})=P(\text{sample is missing one color})+P(\text{sample is missing two colors}) $$  Essentially the second probability statement on the right hand side is the probability that all the balls in the sample are the same color. The case where the sample does not contain three colors doesn't make sense, so I didn't include it. $$=\binom{3}{1} \left[\left(\frac{12}{18}\right) \cdot \left(\frac{11}{17}\right) \cdot\left(\frac{10}{16}\right)\right]+\binom{3}{2}\left[\left(\frac{6}{18}\right) \cdot \left(\frac{5}{17}\right) \cdot\left(\frac{4}{16}\right)\right] \approx 0.882353$$  When all of this is done, we simply subtract the probability from one to get $0.117647$.  What am I doing wrong?
53473
I have one white ball, one yellow ball, one red ball, one black ball. I put the four balls in a nontransparent box. I pick a ball from the box to see its color and put it back to the box.  Assuming picking is random, how many times on average do I need to pick in order to see all four colors?  If I'm lucky, I only need to pick four times. If I'm out of luck, I may get red, red, red, red, red, red,....  But what is it on average?
10845
Suppose there are $m$ red balls and $n$ blue balls in an urn. We randomly choose $p:m&lt;p&lt;n$ balls uniformly from the urn. What is the probability that exactly $q$ red balls are chosen?  Note:- Normally the answer would be $\frac{{m}\choose{q}}{{m+n}\choose{q}}$. However, since the number of balls that are chosen are provided, it is confusing me out. Any hints for the answer will also be appreciated.
115723
A Professor prepares n letters and puts them in envelopes and closes them. Then realizes he has to write addresses, now he randomly writes addresses in the closed envelopes. What is the probability that he sends exactly k letters with correct addresses?  Attempt: Fix the k correct letters. We can choose them in $ n\choose k$ ways. Total number of combination is n!. Could I say answer is $\frac {n\choose k} {n!}$.  I am just getting started with probability, please post answers considering this.
7970
I a programmer not a mathematician so please excuse my ignorance and please dumb it down for me.  My research indicates that this is a variation of the Coupon Collector problem but I really don't know how to get the answer.  Assuming $n$ equally likely desired selections (Probability for each=$1/m$) how many selections with replacement must be made to have an $X$ probability of getting at least one of each of the $n$ desired objects?
94802
I'm trying to find a way to calculate a problems such as this: if you have $n$ objects and $k$ indistinguishable boxes, how do you put in $n$ objects such that each box has no more than $C$, where $C \le k$.  For example, I have $6$ objects that need to be placed in $4$ boxes, where no box can have more than $4$ objects in it. I tried $\displaystyle\binom{n+k-1} {k-1}$. I got $84$ ways. But that is too big. When I do it by hand, I get $74$.   Would like actual explanation.
59448
The question is: n balls are distributed into n holes at random. What is the probability that exactly one hole remains empty. I came up with $$P\left(A\right)=\frac{\:\dbinom{n\:}{1}\:\dbinom{n\:-1}{1}\:\left(n-2\right)!}{n^n}$$ but was told I’m way way off. Yes, I do realize I’m awful at probability. Thanks for the help.
28875
Choose n points randomly from a circle, how to calculate the probability that all the points are in one semicircle? Any hint is appreciated.
31846
This is a modified version of the question  and asked based on the clarifications obtained .   If 5 squares are chosen at random from a chess board, what is the probability that they lie on a diagonal line (squares can overlap and all 204 possible squares shall be considered)  Note: Assume squares lie on a single diagonal line if their diagonals are parallel and overlapping. for example, squares such as the ones given in the image are in the same diagonal .