There are 400 people in a room. I pick two people at random. What is the probability that they have the same birthday?   I know that there must be two people in the room who share the same birthday through pigeonhole principle.   But if I pick two people at random I am not sure how to calculate the probability.
122990
Suppose a room contains $n$ people. What is the probability that at least two people share the same birthday?   Let $A$ be the probability that at least two people have the same birthday. I know that the way to solve this question is actually to find the complement of A and solve $1 - P(A^c)$. However, I'm confused on why $A^c$ is the probability that no one shares the same birthday (everyone has different birthdays), and not the probability that at most two people share the same birthday. Isn't the opposite or complement of "at least two people share the same birthday" equal to "at most two people share the same birthday?"
16382
Four cards are face down on a table. You are told that two are red and two are black, and you need to guess which two are red and which two are black. You do this by pointing to the two cards you’re guessing are red (and then implicitly you’re guessing that the other two are black). Assume that all configurations are equally likely, and that you do not have psychic powers. Find the probability that exactly j of your guesses are correct, for j = 0, 1, 2, 3, 4. Hint: Some probabilities are 0. My professor worked out this example in class and I know the answers for j = 1 and j = 3 are zero, j = 0 is 1/6, j = 2 is 2/3, and j = 4 is 1/6 but I do not understand the process or concept behind the question. I do not understand where the numbers are coming from and why only the even js have a probability but not the odd js. The j guesses and pointing is confusing me. Can someone please help explain the question, I would appreciate it.
82881
Before reading the Wikipedia article, my idea to calculate the probability was as follows:  $1-\left(\frac{364}{365}\right)^{_nC_{2}}$  Basically, I thought to compare all combinations of pairs ($_nC_{2}$) and require them to have different birthday ($\frac{364}{365}$). It gave a very similar result to the wiki method, but is slightly different and I can't tell how  my way of thinking is wrong.  Can anyone tell me what I'm missing?
105174
I am trying to calculate the odds of in a group of 14 people, the percentage of there being 2 different birthday matches (2 people having one Date of Birth, then another 2 people having a different Date of Birth).   I know the odds are about 20% of 2 people out of 14 having the same birthday.   So what would the odds be of there being 2 different birthday matches out of a group of 14?
75151
The problem is quite simple to formulate. If you have a large group of people (n > 365), and their birthdays are uniformly distributed over the year (365 days), what's the probability that every day of the year is someone's birthday?  I am thinking that the problem should be equivalent to finding the number of ways to place n unlabeled balls into k labeled boxes, such that all boxes are non-empty, but C((n-k)+k-1, (n-k))/C(n+k-1, n) (C(n,k) being the binomial coefficient) does not yield the correct answer.
85843
A worker carries two bags. Each of the bags initially contains N nails. Whenever the worker needs a nail, he takes it from a bag picked at random. At some point the worker picks an empty bag. Find the probability that the other bag contains exactly m nails.  My reasoning:  The desired probability: $C_2^1 \cdot P(|A| = 0\ and\ |B| = m)$, where A, B - sets of nails in each bag. Then we can "divide" this probability (I suppose that I deal with independent events):  $P(|A| = 0\ and\ |B| = m) = P(|A| = 0) \cdot P(|B| = m);$  $P(|A| = 0) = (\frac{1}{2})^N$, where $|A| = |B| = N$ (initial condition).  $P(|B| = m) = (\frac{1}{2})^{N - m}$  Result: $C_2^1 \cdot P(|A| = 0\ and\ |B| = m)$ = $C_2^1 \cdot (\frac{1}{2})^{2N - m}$  Am I wrong or it is correct?
43369
During a period of $n$ days, two persons drink beers, one each. There are $n$ different beers $B_1, B_2, B_3,\ldots, B_n$, where $n\geqslant 1$ is an integer. t Person1 drinks the beers in order, i.e., on the $i$-th day, she drinks beer $B_i$. Person2 takes a uniformly random permutation $a_1, a_2, \ldots, a_n$ of the set $\{1, 2,\ldots, n\}$ and drinks beer $B_{a_i}$ on the $i$-th day. Let $X$ be the random variable whose value is the number of days during which person 1 and person 2 drink the same beer. Determine the expected value $\mathbb E(X)$ of $X$.  I know that I should let $X$ be the number of days during which both persons drink the same beer. And  $$ X_i = \left\{\begin{array}{rc} 1,&amp;\text{if they drink the same beer on the }i\text{-th day} \\ 0,&amp;\text{if they drink different beers on the }i\text{-th day}\end{array}\right. $$  Then  $$\begin{align} \mathbb E[X_i] &amp;= 1\cdot P(X=1) = P(\text{Person 1 and Person 2 drink beer }i) \\&amp;= P(\text{Person1 drinks beer }i)P(\text{Person 2 drinks beer }i) \\ &amp;= 1\cdot P(\text{Person 2 drinks beer }i) = \frac{1}{(n-i)!} \end{align}$$   So, is my approach right? If it's will the next step be: $\mathbb E(X) = n\cdot\frac{1}{(n-i)!}$?
88068
What is the chance that at least two people were born on the same day   of the week if there are 3 people in the room?   I'm wondering if my solution is accurate, as my answer was different than the solution I found:  Probability that there are at least 2 people in the room born on the same day = 1 - (No one was born on the same day) - (Exactly one person was born on the same day)  There are (3 choose 2) different pairs of couples. Each couple has the same birthday as another couple with the chances of 1/7 and different with chances 6/7. Thus:  $$1 – (6/7)^3 – 3(1/7)(6/7)^2 = 0.0553$$  Thanks for any help!
55887
Given there are 365 days in a year and $n$ people, the probability that at least two people have the same birthday is given by  $$\frac{365^n - 365 \cdot 364 \cdots (365-n+1)}{365^n}.$$ In this calculation, we treat the $n$ birthdays we selected as a sequence where their order matters.   If we think the $n$ dates that randomly selected as a set where the order does not matter, I believe we would have the probability $$\frac{{365+n-1 \choose n} - {365\choose n}}{{365+n-1 \choose n}}.$$   For the total number of outcomes, we model by throwing $n$ balls uniformly randomly into $365$ boxes. Essentially, there are $n$ balls and $365-1$ walls (between the boxes). For example $$ \circ |\;\;\; |\;\;\;|\circ \circ |\;\;\;|\circ |\cdots |$$ would mean one person was born on day $1$, two people were born on day $4$, one person was born on day $6$, and so on. This is how I came up with the ${365+n-1\choose n}$. For the case that everyone has a distinct birthday, it would be ${365 \choose n}$.   What is wrong with this approach?
89588
Question:  Consider a group of n people with the following properties:  • no person is friends with everyone, • any pair of strangers share exactly one friend in common,  • no three people are mutually friends. Show that everyone has the same number of friends.  I want to solve this using Ramsey's theorem but I'm struggling to formulate it in a way that would make it straightforward.. Any help would be greatly appreciated.
52652
I don’t know how to handle this 3 players case, may anyone give a hand? Thanks!     When playing card game, there are 3 players playing 32 cards, each   receive 10 cards, 2 remind on the table. Overall are 4 aces and 4   kings in the game. How likely is P (E | D)? D = "The table remains 2   aces" E = "I (at least) get one king".   I guess I should have steps as follows.   P(E) = P(1King)+P(2K)+P(3K)+P(4K) P(D) = (4/32) * (4/32) = 1/256 P (E | D) = P(E) + P(D)   Equations may be used:  Counting combinations      The number of combinations of n objects taken r at a time is     nCr = n(n - 1)(n - 2) . . . (n - r + 1)/r! = n! / r!(n - r)!                 Number of favorable outcomes P(I) =     Total number of possible outcome
65411
Say 4 or more people are sitting around a table. Each has a sheet of paper. Devise an algorithm to pass these papers between these people that guarantees:  Each person passes and signs every piece of paper exactly once, no more, no less Every time the papers are passed, each person receives a paper from a person they have not received it from before The papers must all be passed the exact same number of times as there are people (i think this follows from 1, just making sure) People can only have one paper at a time; in other words, no person can be signing two papers at the same time.  Questions:  Is there a mathematical name for this problem? Or can anyone find internet pages where this problem has been discussed before? Is this possible for every number of players? If so, is there an algorithm to generate this? If not, is there an algorithm that can at least minimize the number of times people receive a paper from someone they have received it from before?  This has real world implications!!! I would like to use this theoretical algorithm in my game Drawphone, more info  about how it fits in. Thanks! :)
39087
I am struggling with this probability/combinatorics problem.  the problem states: "11 people are seated around a round table. At some point, everyone stands up and sits again at a randomly chosen seat. Show that the expected value of the number of pairs which swapped places (A sat at B's place and B sat at A's place) is 1/2."  I struggle with calculating the number of orderings in which pairs switch places. I know we have 10! possible permutations for 11 people sitting in a circle, and there are $11\choose2$ possible pairs that could switch places. But I'm not sure how to count for example the number of permutation in which only one couple switches places.
76514
folks,  i am new to this forum and not a math expert. so please bear with me if am asking silly questions.   The question is "probability of getting 50 heads from tossing a coin 100 times".  So the answer for this is, I guess, ${100 \choose 50} (2 ^{-100})$.  So all am trying to get is easier way to calculate ${100 \choose 50}$, or another approach to the parent problem only.  Thanks all, appreciate that.  raj
84038
Given $20$ people, what is the probability that, among the $12$ months   in the year, there are $4$ months containing exactly $2$ birthdays and   $4$ months containing exactly $3$ birthdays?   I am less concerned about knowing the answer than making sure I have thought about the problem correctly.   The general problem seems to be finding the distinct integer-valued vectors that satisfy $$x_1 + x_2 + ... x_{12} = 20\ \ \ \ \ \ \ \ \ x_i \ge 0, i = 1, ..., 12$$  It can be proven that there are $\binom{20+12-1}{12-1}$ such vectors. I am presuming this to be the sample space.  (Please correct me if I'm wrong)  $4\cdot 2 + 4\cdot 3 + 4 \cdot 0= 20$, so there are $3$ distinct groups of $4$ months. Therefore we have $\binom{12}{4\ 4\ 4}$ possible ways to get the desired distribution of months.   The probability I get is $$\frac{\binom{12}{4\ 4\ 4}}{\binom{20+12-1}{12-1}} $$. The answer in the back of the textbook is $1.0604 \times 10^{-3}$ so my answer is off. I just want to know where I went wrong.
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In my daughter's class of $23$, three students and the teacher all share the same birthday. Of course, there are $365$ days in the year, and the first case of the shared birthday is not counted in the probability. But is the likelihood of this just $$\frac{1}{365}\times 3 = 0.0082?$$ For my curiosity, leap years can be omitted.
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In the traditional Birthday Paradox the question is "what are the chances that two or more people in a group of $n$ people share a birthday".  I'm stuck on a problem which is an extension of this.  Instead of knowing the probability that two people share a birthday, I need to extend the question to know what is the probability that $x$ or more people share a birthday.  With $x=2$ you can do this by calculating the probability that no two people share a birthday and subtract that from $1$, but I don't think I can extend this logic to larger numbers of $x$.  To further complicate this I also need a solution which will work for very large numbers for $n$ (millions) and $x$ (thousands).