Each Chocolate Frog comes with one collectable illustrated wizard card (very cool and not dorky at all, honest). There are equal odds of each card being in a pack (i.e., they have all been produced and distributed evenly). How many packs must we buy in order to have an 80% chance of having obtained all 12 cards? How about 90%? Thanks.
I'm trying to solve the well known Coupon Collector's Problem by explicitly finding the probability distribution (so far all the methods I read involve using some sort of trick). However, I'm not having much luck getting anywhere as combinatorics is not something I'm particularly good at.   The Coupon Collector's Problem is stated as:  There are $m$ different kinds of coupons to be collected from boxes. Assuming each type of coupon is equally likely to be found per box, what's the expected amount of boxes one has to buy to collect all types of coupons?  What I'm attempting:  Let $N$ be the random variable associated with the number of boxes one has to buy to find all coupons. Then $P(N=n)=\frac{|A_n|}{|\Omega _n|}$, where $A_n$ is the set of all outcomes such that all types of coupons are observed in $n$ buys, and $\Omega _n$ is the set of all the possible outcomes in $n$ buys. I think $|\Omega _n| = m^n$, but I'm not even sure about that anymore, as all my attempts so far led to garbage probabilities that either diverged or didn't sum up to 1.
How does Cantor's diagonal argument work? I'm having trouble understanding Cantor's diagonal argument.  Specifically, I do not understand how it proves that something is "uncountable".  My understanding of the argument is that it takes the following form (modified slightly from the  article, assuming base 2, where the numbers must be from the set $ \lbrace 0,1 \rbrace $):  \begin{align} s_1 &amp;= (\mathbf{0},1,0,\dots)\\ s_2 &amp;= (1,\mathbf{1},0,\dots)\\ s_3 &amp;= (0,0,\mathbf{1},\dots)\\ \vdots &amp;= (s_n \text{ continues}) \end{align}  In this case, the diagonal number is the bold diagonal numbers $(0, 1, 1)$, which when "flipped" is $(1,0,0)$, neither of which is $s_1$, $s_2$, or $s_3$.  My question, or misunderstanding, is: When there exists the possibility that more $s_n$ exist, as is the case in the example above, how does this "prove" anything?  For example:  \begin{align} s_0 &amp;= (1,0,0,\mathbf{0},\dots)\ \ \textrm{ (...the wikipedia flipped diagonal)}\\ s_1 &amp;= (\mathbf{0},1,0,\dots)\\ s_2 &amp;= (1,\mathbf{1},0,\dots)\\ s_3 &amp;= (0,0,\mathbf{1},\dots)\\ s_4 &amp;= (0,1,1,\mathbf{1},\dots)\\ s_4 &amp;= (1,0,0,\mathbf{1},\dots)\ \ \textrm{ (...alternate, flipped } s_4\textrm{)}\\ s_5 &amp;= (1,0,0,0,\dots)\\ s_6 &amp;= (1,0,0,1,\dots)\\ \vdots &amp;= (s_n \text{ continues}) \end{align}  In other words, as long as there is a $\dots \text{ continues}$ at the end, the very next number could be the "impossible diagonal number", with the caveat that it's not strictly identical to the "impossible diagonal number" as the wikipedia article defines it:     For each $m$ and $n$ let $s_{n,m}$ be the $m^{th}$ element of the $n^{th}$ sequence on the list; so for each $n$,      $$s_n = (s_{n,1}, s_{n,2}, s_{n,3}, s_{n,4}, \dots).$$      ...snip...      Otherwise, it would be possible by the above process to construct a sequence $s_0$ which would both be in $T$ (because it is a sequence of 0s and 1s which is by the definition of $T$ in $T$) and at the same time not in $T$ (because we can deliberately construct it not to be in the list). $T$, containing all such sequences, must contain $s_0$, which is just such a sequence. But since $s_0$ does not appear anywhere on the list, $T$ cannot contain $s_0$.      Therefore $T$ cannot be placed in one-to-one correspondence with the natural numbers. In other words, it is uncountable.   I'm not sure this definition is correct, because if we assume that $m = (1, \dots)$, then this definition says that "$s_n$ is equal to itself"&mdadsh;there is no "diagonalization" in this particular description of the argument, nor does it incorporate the "flipping" part of the argument, never mind the fact that we have very clearly constructed just such an impossible $T$ list above.  An attempt to correct the "diagonalization" and "flipping" problem:  $$s_n = (\lnot s_{m,m}, \lnot s_{m,m}, \dots) \quad \text{where $m$ is the element index and} \quad\begin{equation}\lnot s_{m,m} = \begin{cases}0 &amp; \mathrm{if\ } s_{m,m} = 1\\1 &amp; \mathrm{if\ } s_{m,m} = 0\end{cases}\end{equation}$$  This definition doesn't quite work either, as we immediately run in to problems with just $s_1 = (0),$ which is impossible because by definition $s_1$ must be $ = (1)$ if $s_1 = (0)$, which would also be impossible because...   Or more generally, with the revised definition there is a contradiction whenever $n = m$, which would seem to invalidate the revised formulation of the argument / proof.  Nothing about this argument / proof makes any sense to me, nor why it only applies to real numbers and makes them "uncountable". As near as I can tell it would seem to apply equal well to natural numbers, which are "countable".  What am I missing? Filtering points in GeoServer layer A proof for the inequality $\sum_{i=0}^{t-2}{\frac{1}{t+3i}} \leq \frac{1}{2} $ for all $t \geq 2$ I'm struggling with proving the following inequality: $$\sum_{i=0}^{t-2}{\frac{1}{t+3i}} \leq \frac{1}{2}$$  for all $t \geq 2$. I think it is monotonic non-increasing in $t$, which would suffice.  Thanks in advance! I want to find a set of integer solutions of Diophantine equation: $ax + by = c$, and apparently $\gcd(a,b)|c$. Then by what formula can I use to find $x$ and $y$ ?   I tried to play around with it: $x = (c - by)/a$, hence $a|(c - by)$.     $a$, $c$ and $b$ are known. So to obtain integer solution for $a$, then $c - by = ak$, and I lost from here, because $y = (c - ak)/b$. I kept repeating this routine and could not find a way to get rid of it? Any hint?  Thanks, Chan Why do Lambertian Reflectors Privilege the Observer? How do I disable auto-maximizing of newly launched windows in GNOME? Each time I open any application, it zooms to fill-up the entire desktop and so I am always forced to use "Unmaximize" using Alt+Space menu. Is there any hint of of the first Wizarding War being an international event? Is there any hint that the MoM or anywhere else got support in combating voldemort the first time around? Or is everything completely UK-centric?  I can't remember anyone in the second war being specifically sent from europe to help, even though we see international wizards in book 4. But I don't think all the events of the first war are covered in the books.  So are other bodies like the MoM involved in the first war, or was it isolated in the UK? The Bash Manual says   { list; }        Placing a list of commands between curly braces causes the list to be executed in the current shell context. No subshell is   created. The semicolon (or newline) following list is required.      The braces are reserved words, so they must be separated from the list   by blanks or other shell metacharacters.       The parentheses are operators, and are recognized as separate tokens   by the shell even if they are not separated from the list by   whitespace.   If I remove the semicolon, like this:  $ { date } &gt;    Why does it expect stdin input? A metacharacter is a character that separates words.  The semicolon and whitespace are both "shell metacharacters". Why can't whitespace separate words date and }? Why do we need a semicolon instead of a whitespace for separating the words? When I was using ubuntu 14.04, it's easy to set title. Just press F2 or move mouse to top menu. But there is no menu on ubuntu 16.04 and press F2 can't change the title. What should I do? Can I run out of disk space by creating a very large number of empty files? It is well-known that empty text files have zero bytes:    However, each of them contains , which according to my research, is stored in , and do use space.   Given this, it seems logical to me that it is possible to fill a disk by purely creating empty text files. Is this correct? If so, how many empty text files would I need to fill in a disk of, say, 1GB?    To do some checks, I run df -i but this apparently shows the % of inodes being used(?) rather than how much they weigh.  Filesystem             Inodes  IUsed    IFree IUse% Mounted on udev                   947470    556   946914    1% /dev tmpfs                  952593    805   951788    1% /run /dev/sda2            28786688 667980 28118708    3% / tmpfs                  952593     25   952568    1% /dev/shm tmpfs                  952593      5   952588    1% /run/lock tmpfs                  952593     16   952577    1% /sys/fs/cgroup /dev/sda1                   0      0        0     - /boot/efi tmpfs                  952593     25   952568    1% /run/user/1000 /home/lucho/.Private 28786688 667980 28118708    3% /home/lucho Why can't I destroy or place blocks? I'm playing on a Survival world with cheats on, "easy" difficulty, and at night in-game. I didn't have a sword or anything to make a sword with, and I was starving, so I switched to Creative to give myself a stack of food and a bed to set my spawn point. When I switched back to Survival, I couldn't destroy or place blocks. I'm on single player, and my Internet is fine, and I can destroy blocks still if I switch to creative. I've tried resetting the button one, and reloading the map, and it still doesn't work. I've even tried deleting everything I gave myself in Creative. This happened before on a different world and I ended up deleting it.  This only happens on this world. How do I fix it?  If I know I'm in the correct game mode, what else can cause this?  Are there different causes in multiplayer? How could Luke hide from Vader during the duel on Death Star II? In Return of the Jedi, Vader throws his lightsaber at Luke causing the catwalk come crashing down. Luke uses this in his advantage and hides from him under the throne room platform. Unable to locate his son, Vader tried to lure Luke out of hiding.  But how could Luke conceal himself from Vader in the first place? Did he use a Force power of some kind to blur Vader's sensors? I mean, the underneath of the platform was pretty much an open space with very little possibilities for hiding. The only places I see in the movie are the openings on the both sides of the floor (pointed by the yellow arrows):    The background of a close-up to Luke doesn't give much of a clue (although there is a round shape which could be the bottom of a window frame):    But as Luke reveals himself, he appears to be beneath of the staircase to the dais, right in the middle of it:    There's no way he could have hidden there all the time, because just a couple of seconds before, Vader walked towards and stared at the very same spot:    My first ideas were that Luke used the Force to make himself invisible, or to teleport him to the position. But after taking a look at one of the cross sections of the Emperor's throne room, I assume Luke could have slid almost straight under the floor of the lower level (yellow arrow) and then later  back to the staircase (although we don't see it in the movie, and I'm not sure if Luke knew the ability):    Does someone know better? Why do I see a saddle in this picture of a computer screen? I am not entirely sure this is an appropriate question for PSE, however since many of you have such diverse backgrounds, I'll give it a shot.  I have noticed that when one takes a picture of a computer, TV screen with a phone, you can "see" what looks like a saddle phase portrait.     Why does this happen? Does it have any relevance to differential equations or other areas of math? If a spell is cast on a moveable creature/object and lasts longer than a round, and the creature/object moves, does the spell move with it? Proving $M_n(I)\lhd M_n(R)$ and proofs about ideal rings. I'm trying to do a few proofs about rings: $(1)$ I want to show that for a ring $R$ and the ring of matrices $M_n(R)$, if $I\lhd R\implies M_n(I)\lhd M_n(R)$ and that all ideals of $M_n(R)$ are of the form $M_n(I)$. $(2)$ Prove that if $R$ is a ring with division then $M_n(R)$ is a simple ring.  The relation $I\lhd R$ implies that with $r\in R$ and $x\in I$ we have $rx\in I, xr\in I$. But how can I prove the statement for matrices?, here's what I would do: Consider $\psi \in M_n(R)$, if is given any matrix $\phi\in M_n(I)$ it must be $\psi \phi \in M_n(I),\phi \psi \in M_n(I)$, if we denote the element $(i,j)$ of each matrix by $\phi_{ij}$ and $\psi_{ij}$, then $(\psi \phi)_{ij}=\sum_{k=1}^n \psi_{ik}\phi_{kj}$ and $(\phi \psi)_{ij}=\sum_{k=1}^n \phi_{ik}\psi_{kj}$.  Since $I\lhd R$ is $\psi_{ik}\phi_{kj}\in I$ and $\phi_{ik}\psi_{kj}\in I$, which means that $I\lhd R\implies M_n(I)\lhd M_n(R)$.  To show that every ideal of $M_n(R)$ are of the form $M_n(I)$ with $I$ ideal: Suppose $I$ is not an ideal of $R$, then exists $r\in R$ such that $rx\notin I$ or $xr\notin I$, then with taking a matrix $\phi\in M_n(I)$ such that $x$ is an entry of $\phi$ and a matrix $\psi$ such that $r$ is an entry of $\psi$ we get that the matrices $\phi$ and $\psi$ have an entry $xr$ or $rx$, in either case this is not an element of $I$ then the matrix is not in $M_n(I)$. This concludes my proof (is it?) for $(1)$.  For $(2)$ I'd like to proceed by absurdum . Let's say that $M_n(R)$ is not a simple ring, then it has at least one ideal. Denote by $\varphi$ this ideal in $I$, then for any matrix $\psi$ we have $\varphi\psi \in M_n(I)$ and $\psi \varphi \in M_n(I)$.  Now, can $R$ be a ring with division?.I'm not sure how to prove that it can't. I believe I could use this theorem: "If $I\lhd R$ and $1\in I$ then $I=R$. The case $I\neq R$ implies $1\neq R$ and since $\varphi \psi$ and $\psi \varphi$ are in $M_n(I)$ then there are at entries in the product matrix which belong to $I$ (using $I \lhd R$) but are not $1$ (1\notin I); then there are elements in $I$ (and therefore in $R$ by $I\subset R$) that are not invertible, follows that $R$ is not a ring with division. But if $I=R$?, here I can't see a why $R$ can't be a ring with division. I was browsing through the apt folder to try my proxy settings and I was unable to find the apt.conf file in Ubuntu 11.04. Is there any alternative file for it?  Adding proxy settings to 99-synaptic in apt.conf.d does not help as it gets overwritten everytime I restart the system. How do you effectively study physics? How does one read a physics book instead or just staring at it for hours?   (Apologies in advance if the question is ill-posed or too subjective in its current form to meet the requirements of the FAQ; I'd certainly appreciate any suggestions for its modification if need be.) Beamer: Change only several bullets in the list I would like to change the bullets in the second itemize to arrows.  \begin{alertblock}{The Aim of this Research}      \begin{itemize}         \item Item 1         \item Item 2             \begin{itemize}                 \item Item 2-1                 \item Item 2-2             \end{itemize}     \end{itemize}  \end{alertblock}   Form above, I would like to make:  ・Item 1  ・Item 2    =&gt; Item 2-1  (change bullet to arrow)    =&gt; Item 2-2   I've tried inserting \renewcommand{\labelitemii}{$\rightarrow$} before and after the second \begin{itemize} and , but they don't work. You can reach this IP address  (if my computer is online). I want to assign a domain to it so I wonder how is that possible when I have no DNS? I do own my domain which is ".com". My IP is static.  Thanks If the $n^{th}$ partial sum of a series $\sum a_n$ is $S_n = \frac{n-1}{n+1}$, find $a_n$ and the sum.