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.
What is the average number of times it would it take to roll a fair 6-sided die and get all numbers on the die?  The order in which the numbers appear does not matter.  I had this questions explained to me by a professor (not math professor), but it was not clear in the explanation.  We were given the answer $(1-(\frac56)^n)^6 = .5$ or $n = 12.152$  Can someone please explain this to me, possibly with a link to a general topic?
What is the best way to store fresh herbs like parsley or cilantro?  I find I have about the same luck with them whether they are left in a plastic bag or I put them in a cup of water. I know that the usual ordering of $\mathbb R$ is not a well-ordering but is there an uncountable $S\subset \mathbb R$ such that S is well-ordered by $&lt;_\mathbb R$?   Intuitively I'd say there is no such set but intuitively I'd also say there is no well-ordered uncountable set at all, which is obviously wrong. I still struggle to grasp the idea of an uncountable, well-ordered set. Report this Ad Feature Why not to extend the set of natural numbers to make it closed under division by zero? We add negative numbers and zero to natural sequence to make it closed under subtraction, the same thing happens with division (rational numbers) and root of -1 (complex numbers).  Why this trick isn't performed with division by zero? Is there is a way to exclude nested subdirectory specified by a path to that dir from grep recursive search? Or I have to use ack or find for this?  Here's what I want. Say I have this dir structure:  dir/nesteddir/match.txt otherdir/nesteddir/match.txt   How do I exclude first nesteddir only? Is it even possible using grep?  Most obvious answer $ grep -r --exclude-dir="dir/nesteddir" "stringToFind" . doesn't seem to work, nor do any of my experiments with various --exclude-dir paths using globbing, leading ./, etc.   Reading man page and googling didn't help, the only mention of this usecase I found is this question:   It was concluded that this is bug introduced in grep 2.12, but I'm using latest version (2.16) and grep's bugtracker () doesn't mention anything related to --exclude-dir, so I guess either it was fixed or it's an intended behaviour.   Would be really nice to get some clarification. I've been struggling with this all day today. I imagine it's not very hard, but my algebra skills are terrible. So, how can I show that if $m&gt;n$ and $a$ is a positive integer, then $$a^{2^n}+1 \mid a^{2^m}-1.$$ I just can't get a coherent picture of what I'm supposed to do. If this too (see yesterday's question) boils down to the Unique Factorization Theorem, I would be very grateful for any intuition on how to think about this! Integral of the square of the normal distribution. Why might one also use a blank constructor? I was reading some Java recently and came across something (an idiom?) new to me: in the program, classes with multiple constructors would also always include a blank constructor. For example:   public class Genotype {   private boolean bits[];   private int rating;   private int length;   private Random random;    public Genotype() {              //  &lt;= THIS is the bandit, this one right here     random = new Random();   }    /* creates a Random genetoype */   public Genotype(int length, Random r) {     random = r;     this.length = length;     bits = new boolean[length];      for(int i=0;i&lt;length;i++) {         bits[i] =random.nextBoolean();     }   }    /* copy constructor */   public Genotype(Genotype g,Random r) {     random = r;     bits = new boolean[g.length];     rating = g.rating;     length = g.length;      for(int i=0;i&lt;length;i++) {         bits[i] = g.bits[i];     }    } }   The first constructor doesn't seem to be a "real" constructor, it seems as though in every case one of the other constructors will be used. So why is that constructor defined at all? There exists $C\neq0$ with $CA=BC$ iff $A$ and $B$ have a common eigenvalue Question:  Suppose $V$ and $W$ are finite dimensional vector spaces over $\mathbb{C}$. $A$ is a linear transformation on $V$,  $B$ is a linear transformation on $W$. Then there exists a non-zero linear map $C:V\to W$ s.t. $CA=BC$ iff $A$ and $B$ have a same eigenvalue.  ===========This is incorrect==========  Clearly, if $CA=BC$, suppose $Ax=\lambda x$, then $B(Cx)=CAx=C(\lambda x)=\lambda (Cx)$, so $A$ and $B$ have same eigenvalue $\lambda$. On the other hand, if $A$ and $B$ have a same eigenvalue $\lambda$, suppose $Ax=\lambda x, By=\lambda y$. Define $C:V\to W$ s.t. $Cx=y$, then $BCx=By=\lambda y=C\lambda x=CAx$. But I don't kow how to make $BC=CA$ for all $x\in V$.   ====================================== Google Sheets QUERY won't display cell text if other cells have numbers I'm not sure what is happening in my Google Sheet query and could really use some help. I have  that demonstrates and explains the issue.   Basically, I'm using a query to pull data from a Google Spreadsheet that is populated by a Google Form. If the form submits only numbers (100), there is no problem. However, if someone submits a string ($100 or 100 dollars), the query will not display the string. It will only display form submissions that are 100% numbers.   Also, if the 1st submission is a string (not 100% numbers) the query will work.   Here is my query  =query(data!B2:CL,"select B, C, D, E, F, G, H, I, J, CI, BZ, CA, BA  where J = 'Industry Certification'",false)   Any ideas on why this is happening?  Also, I'm aware that I can limit the form to numbers only. I'm trying to avoid that and learn 'why' this is happening. Is there a nice solution to get a "presenter mode" for Latex presentations? I really like the way Keynote and MS Office (and also OpenOffice &amp; similar) support a presentation mode that displays the current slide, the next (and maybe previous) slide, the elapsed and current time, and also any notes attached to the slides on one screen and the presentation itself on the beamer/second screen.  It enables me to give good, professional presentations without having to know everything by heart (I suck at that).  I found a , but I am looking for alternatives, especially OS-independent would be great. Not having to create 3 different PDFs is a boon.  Any hints? It's the first time I'm using Tikz on Latex. I have to draw this block diagram (this is a low resolution snip from a paper)   In Tikz I have done this:   That is very similar but I need some adjustments, like the u_i arrow and the node sum after the H(s) block.  Is there anybody that can help me with this?  my code:      \tikzstyle{block} = [draw, fill=white, rectangle,  minimum height=2em, minimum width=3em] \tikzstyle{sum} = [draw, fill=white, circle, radius=1mm, node distance=1.5cm] \tikzstyle{input} = [coordinate] \tikzstyle{output} = [coordinate] \tikzstyle{pinstyle} = [pin edge={to-,thin,black}]  \begin{tikzpicture}[auto, node distance=2cm,&gt;=latex'] \node [input, name=input] {}; \node [input, above of=input](input1) {}; \node [sum, right of=input] (sum) {}; \node [block, right of=sum] (controller) {$K(s)$}; \node [sum, right of=controller] (sum1) {}; \node [block, right of=sum1] (filterinv) {$H^{-1}(s)$}; \node [block, right of=filterinv] (system) {$G(s)$}; \node [output, right of=system] (output) {}; \node [output, above of=output] (output1) {}; \node [block, below of=filterinv] (filter) {$H(s)$}; \node [block, above of=controller] (delay) {$D(s)$}; \node [sum, below of=sum1] (sum2) {};  \draw [-&gt;] (filterinv) -- node[name=u] {$u_{i}$} (system); \draw [draw,-&gt;] (input) -- node {$s_{i-1}$} (sum); \draw [-&gt;] (sum) -- node {$e_{i}$} (controller); \draw [-&gt;] (system) -- node [name=q] {$q_{i}$}(output); \draw [-&gt;] (q) |- (filter); \draw [-&gt;] (filter) -- node {} (sum2); \draw [-&gt;] (controller) -- node {} (sum1); \draw [-&gt;] (sum1) -- node[name=xi] {$\xi_{i}$} (filterinv); \node [input, below of=sum1](input2) {}; \draw [draw,-&gt;] (input2) -- node {$r_{i}$} (sum2); \draw [-&gt;] (sum2) -| node[pos=0.99] {$-$}      node [near end] {$\tilde{s}_i$} (sum); \draw [-&gt;] (delay) -| node[] {}      node [near end] {} (sum1); \draw [draw,-&gt;] (input1) -- node[pos=0.15] {$u_{i-1}$} (delay); \draw [-&gt;] (u) |- node[pos=0.99] {$u_i$}      node [near end] {} (output1);  \end{tikzpicture} I am non-EU citizen working in Germany. Can I travel to UK without a visa? I am from Non-EU country (from Nepal to be precise) and recently moved to Germany for work. I am applying for BlueCard. I would like to travel to UK later this year, can I go there without visa as I am resident in another EU country?  I can't find this information online. Any links/references will be much appreciated. Change style of labels of Tikz nodes of a given style Invert Color Ramp in QGIS How to determine the amount of experience needed for leveling up Related questions under this topic might include:   How should I base how much experience points it takes to get to the next level? What factors are involved in this? How do I keep the player not entertained long enough till he levels up? How can I vary the way the player gets experience points? I edited a question, on Stack overflow and when submitting preview, got red bar telling me my account isn't allowed to make suggested edits.  On the Sandbox android question it works fine, but that is editing an answer to my question.   The message I receive is:     Account is not allowed to suggest edits   Are the rules for suggested edits different for Android app? Based on the table Encounter XP Multipliers from the Basic Rules' DM section, Page 57, when you're building encounters in 5E, you include an XP modifier depending on the number of creatures your PCs will be fighting: 1.5x for a pair, 2x for 3-6 and so on. However, the PCs only get the base XP as a reward.     This doesn't change the actual xp reward ... just your calculations of how difficult the encounter is.   Why is this? After acknowledging that battles vs multiple opponents are more difficult, why not reward them as such? Proving the so-called "Well Ordering Principle" Is there anything wrong with the following proof?  Theorem. Every non-empty subset $B \subset\mathbb{N}$ has a least member.  Proof. Assume not. Then, of necessity, we'd have to have $B=\varnothing$, for if $B$ contained even one $b\in\mathbb{N}$, then that $b$ would satisfy $b=min(B). $ Contradiction, end proof.  I get the sense there is something wrong here, but I can't seem to define exactly what. Also, is there a context where one can simply take this principle as an axiom, and not have to prove it? After all, it is extremely intuitive. Thanks. How to line up the pieces at the coast near the shipwreck? In The Witness, I was walking along the coast, and I came upon this:    Now, since I've learned      that mazes can be embedded in the environment,   I see the puzzle here, but I just can't get the pieces to line up well enough.  Standing on the couch and walking all around it doesn't seem to get me a       functional maze.   Where do I need to stand to activate the puzzle successfully?