There is a long line of people waiting outside a theatre to buy tickets. The theatre owner comes out and announces that the first person to have a birthday same as someone standing anywhere before him in the line gets a free ticket. Where will you stand to maximize your chance?
121891
At a movie theater, the whimsical manager announces that a free ticket will be given to the first person in line whose birthday is the same as someone in line who has already bought a ticket.  You have the option of choosing any position in the line. Assuming that you don't know anyone else's birthday, and that birthdays are uniformly distributed throughout the year (365 days year), what position in line gives you the best chance of getting free ticket?
106242
Bonus question: If I am a proxy for multiple people, what's the collective noun for that group?
43810
You're standing in front of a ticket vending machine and it says "The ticket is printing". Is that correct or should it be "The ticket is being printed"?  EDIT:  If both are correct, which one should be preferred or is more common?
56862
In Monopoly, when you roll doubles a third time, you go to Jail.  But do you get to interact with the space first?  More specifically, do you get to do any of the following first?   Pass go  Draw a Chance or Community Chest card Pay Luxury Tax or Income Tax  Pay Rent
90215
If there are $12$ people in the room what the chances are that no two people will have the same birthday month?   I tried to think of it in this way:  The total number of options is $12^{12}$  The option that no one will have the same birthday month is:  $12 * 11 * ... 1$ which means $12!$  So I think the answer is:  $$\frac{12!}{12^{12}}=0.0000537$$  What do you think?   Thank you!
61169
After half a weekend spent on conquering Candy Crush Saga levels, I have finally arrived at level 36. Unfortunately, there seems to be a "barrier" of sorts to start playing this level, with two options:   Ask your Facebook friends for help Buy a ticket for $0.99   While option #2 is extremely cheap, I'm curious what option #1 entails. I am personally extremely annoyed when I get enormous amounts of requests from Facebook applications and I don't want to annoy my friends this way. While I know selecting option #1 sends a request to my Facebook friends, what happens after that?
9435
I heard an interesting question recently:     What is the minimum number of people required to make it more likely than not that all 365 possible birthdays are covered?   Monte Carlo simulation suggests 2287 ($\pm 1$, I think).  More generally, with $p$ people, what is the probability that for each of the 365 days of the year, there is at least one person in the group with that birthday?  (Yes, ignoring the leap-day.)
104163
It's something that I'm not sure if it was put in the books or was left outside the book for people to guess but... In any moment of the books, during the travel from the train between King Cross and Hogwarts, did they ask or checked the ticket? I remember the scene on first year/book that Hagrid almost forgot to give Harry the ticket, even when he did not tell about the 9 3/4, but I'm not sure if it was mentioned if they really need the ticket.
20556
My professor assigned a completely random question on our problem set. Basically, it goes:  Sophie is 30 years old and majored in philosophy. As a student, she participated in anti-nuclear demonstrations and was deeply concerned with social justice. Which is more probable?  1. Sophie is a bank teller. 2. Sophie is a bank teller and is active in the feminist movement.  I'm guessing the point the prof was trying to get across is that assuming you have two different events $A$ and $B$,  $A\cap B$ would always have a smaller probability than either. Is this correct?
108037
The birthday paradox says, if you have 23 people in the same room, the odds are greater than half that two people will share a birthday. I tried to calculate it using 1 - (number of ways 23 people can have birthdays that are ALL DIFFERENT)/(number of ways 23 people can have birthdays).  Here's how 23 people can have birthdays that are all different: it is just 365 choose 23.  To calculate all the different ways 23 people can have birthdays is more complicated. In this case, order doesn't matter, but it is okay for multiple people to have the same birthdays. The formula for this is (n + k - 1) choose k, so in this case: (387 choose 23).  However, 1 - (365 choose 23) / (387 choose 23) = 75%, not 50%. What am I doing wrong?
21919
I have been looking at the birthday problem (http://en.wikipedia.org/wiki/Birthday_problem) and I am trying to figure out what the probability of 3 people sharing a birthday in a room of 30 people is. (Instead of 2).  I thought I understood the problem but I guess not since I have no idea how to do it with 3.
106725
$m$ tickets are drawn out of $n$ tickets which are numbered from $1$ to $n$. If $X$ denote the sum of the numbers on the tickets drawn. Find $V(X)$.   $X = X_1+X_2+\cdots+X_m$ , if $X_i$ can be treated as the $i$th number drawn. Otherwise, $X_i$ can be treated as the indicator variable of the number $i=1,2,...,n$.  In either way, I am able to get expectation since dependence of variables does not matter. However, while calculating Variance, dependence does matter. While calculating $E(X_iX_j)$ the second draw is supposed to be dependent on the first draw since there is a constraint of the sum $X$.  Please answer.
67373
This is the gift that my dad sent me on my birthday.   Vs   This is the gift that my dad sent to me on my birthday.  Is to required?   Also   Presenting you guys, the trailer of the film.   Vs   Presenting to you guys, the trailer of the film.    Is to required?
28043
There is a party with 20 people, and everyone writes their name down   on a piece of paper and puts it into a bag. The bag is mixed up, and   each person draws one piece of paper. If you draw the name of someone   else, you are considered to be in his "group". What is the expected   number of groups after everyone draws?   So basically if we have a loop where each person draws someone else's name, and the last person draws the first person in that list's name, we have a group. Not quite sure how to approach this problem.  Thanks for any help.
37526
A cute riddle (but maybe not so easy!) from Gardner:  At a gathering of mathemagicians, the Grand Master and his 8 disciples are seated at a round table. The Grand Master will judge each of his displices on their newest trick. After he has seen all the performances, the Grand Master hands to each of his disciples a card with their score on it (the score is some integer number of points greater than $0$). In return, The Grand Master then performs the trick himself and allows his disciples to judge his own perforamce. The disciples agree on a score and give the Grand Master a card with his score on it. It turns out that each mathemagician at the table has a different score.  The Grand Master then remarks: "I can think of a number that divides the product of my own score and the score of anyone seated at this table, other than the two people sitting beside me.". Each of the disciples look at their own scorecard and then look around the table, and each discplie remarks: "I can also think of such a number!". All at once, everyone seated at the table announces the number they have in mind. Incredibly, they all say same the same number! "Now that is some trick!" the Grand Master laughs.   What is the smallest possible number the mathemagicians could have announced?
20333
It’s my turn. It is a two handed game of Canasta. I have three cards left in my hand. There is only one card left in the Stock pile. Can I pick up the one card and play my hand?
39722
I have never personally had to reconfirm the ticket, but have read that in some case you can lose a place on a flight, even if you paid for it!  Where does the idea of ticket reconfirmation comes from? How to find out that you are required to reconfirm the ticket? Does it only applies to agencies of also to bookings made via website? Can an agent reconfirm your ticket already at the time of purchase?