I've read that the answer to such a problem is the inverse probability. So here getting a one has probably 1/6, so the number of tries you would be expected to run in order to get a 1 is 6.  I'm not sure I understand this.
On average, how many times must I roll a dice until I get a $6$?  I got this question from a book called Fifty Challenging Problems in Probability.   The answer is $6$, and I understand the solution the book has given me. However, I want to know why the following logic does not work: The chance that we do not get a $6$ is $5/6$. In order to find the number of dice rolls needed, I want the probability of there being a $6$ in $n$ rolls being $1/2$ in order to find the average. So I solve the equation $(5/6)^n=1/2$, which gives me $n=3.8$-ish. That number makes sense to me intuitively, where the number $6$ does not make sense intuitively. I feel like on average, I would need to roll about $3$-$4$ times to get a $6$. Sometimes, I will have to roll less than $3$-$4$ times, and sometimes I will have to roll more than $3$-$4$ times.  Please note that I am not asking how to solve this question, but what is wrong with my logic above.  Thank you!
Suppose I look at a parallel plate capacitor in its rest frame and calculate the electrostatic energy, $E$.  Next, I look at the same capacitor in a primed frame boosted in the direction perpendicular to the plane of the plates. In this frame, the $E$-field is the same strength, there is no magnetic field, and the volume over which the $E$-field extends is less by a factor $1/\gamma$. This suggests $E' = \frac{1}{\gamma} E$, but relativity states that energy transforms as $E' = \gamma E$.   Where is the missing energy? I just watched a TV show and the guy says:     I ain't trying to fight with nobody with a baby.   So here, I can't understand the "nobody" since it make no sense to me, but it sounds like anybody, and if it's "I ain't trying to fight with anybody with a baby." that makes sense to me. How come everyone easily takes or say the name of "Voldemort" without any fear in Scorpius's world? Show that the operator norm is submultiplicative We had in our lecture on numerical analysis the following: Let $\mathrm{Lin}(X,Y)$ be the set of all linear maps $X\rightarrow Y$. Let $A\in\mathrm{Lin}(\mathbb R^l,\mathbb R^n)$ and $B\in\mathrm{Lin}(\mathbb R^n,\mathbb R^m)$ and $\|C\|_{op}:=\max_{\|x\|\leq1}\|C(x)\|$. Then our lecturer followed $\|A\circ B\|_{op}\leq\|A\|_{op}\cdot\|B\|_{op}$. So he didn't prove it and so I've tried it by my own. My attempt: $$\begin{aligned} \|A\circ B\|_{\mathrm{op}} &amp;= \max_{\|x\| &lt; 1}\|(A \circ B)(x)\| \\ &amp; \leq \max_{\|x\| &lt; 1} \|A\|_{\mathrm{op}} \|B(x)\| \\ &amp;= \|A\|_{\mathrm{op}} \max_{\|x\|\leq1}\|B(x)\|\\ &amp;= \|A\|_{\mathrm{op}}\|B\|_{\mathrm{op}} \end{aligned}$$ But this seems too easy. I am really interested in a nice proof so anybody could help? Thanks a lot! Disallow closing of questions with a certain number of question / answer upvotes With reference to this currently open question   At the time of writing, it has 3 close votes yet has 7 question upvotes and the single answerer (11 upvotes) will have been awarded the enlightened badge (silver).  Therefore in the opinion of many (i) the question adds value and (ii) the answer adds value.  As a staring point for discussion, I suggest that 5 net question upvotes (and perhaps 5 answer net upvotes) should be enough to keep a question permanently open. Let's haggle over the numbers though. Thoughts? What can I do if a repository/PPA does not have a Release file? I have recently analysed my website with pagespeed addon on firebug. It suggested me to set expiration on CSS, JS and image files.  I am wondering how do i do this? Intuition behind the definition of a measurable set Are proofs by contradiction really logical? Let's say that I prove statement $A$ by showing that the negation of $A$ leads to a contradiction.   My question is this: How does one go from "so there's a contradiction if we don't have $A$" to concluding that "we have $A$"?  That, to me, seems the exact opposite of logical. It sounds like we say "so, I'll have a really big problem if this thing isn't true, so out of convenience, I am just going to act like it's true". wrong numbering for subcaption in scrbook When using the scrbook document class in combination with the subcaption  package, the numbering of subfigures seems broken. Consider:  \documentclass{scrbook} \usepackage{xcolor} \usepackage{caption} \usepackage{subcaption} \captionsetup{%   format=hang,%   font=small,%   labelfont={sf,bf},%   aboveskip=0.2\normalbaselineskip,%   position=top } \pagestyle{empty}  \begin{document} \begin{figure}   \caption{Black}   \label{fig1}   \centering \rule{2cm}{2cm} \end{figure}  \begin{figure}   \caption{More figures}   \label{fig2}   \centering   \begin{subfigure}{.3\textwidth}     \centering     \caption{Blue}     \label{fig2a}     \color{red}\rule{2cm}{2cm}   \end{subfigure}   \begin{subfigure}{.3\textwidth}     \centering     \caption{Green}     \label{fig2b}     \color{green}\rule{2cm}{2cm}   \end{subfigure}   \begin{subfigure}{.3\textwidth}     \centering     \caption{Blue}     \label{fig2c}     \color{blue}\rule{2cm}{2cm}   \end{subfigure} \end{figure}  Figure \ref{fig1} shows a black box. Figure \ref{fig2} shows colored boxes, where Figure \ref{fig2a} shows a red box, Figure \ref{fig2b} shows a green box, and Figure \ref{fig2c} shows a blue box. \end{document}   This results in the following display:    Clearly, the figure numbering in the text references is off by one for the subfigures. Instead, if I change the document class to article, it looks fine:     Why does this happen, and how can I prevent it? $AB-BA$ is a nilpotent matrix if it commutes with $A$ I saw this in a  and am putting it here for posterity.  Problem: Let $A$ and $B$ by square matrices and set $C=AB-BA$. If $AC=CA$, prove $C$ is nilpotent. How do I effectively defeat the Wall of Flesh? I've been having trouble defeating this monster. Even with lots of buff potions, health potions, high tier armor (Molten), and ranged weapons with plenty of ammo (Minishark), I can't seem to kill this boss.  Worse, when I get it down to about 2000 hp, even if I'm fairly far away, it'll suck me in and continually rise towards the skybox, which doesn't allow me to do anything for some reason.  How do I defeat the Wall of Flesh? What gear should I be using and what kind of potions do I need? How to render a pixelated, occluded wireframe with freestyle? I'd like to combine these features:    The pixelated look of a wire material with anti-aliasing disabled. Proper occlusion of hidden lines.   A wireframe material produces the first feature, freestyle produces the second. Can a line set be created that produces neither double pixels nor gaps? Line thickness alone does not seem to be the answer.  There is the trick with two materials and z-offset, but I have trouble getting a really clean solution at the outline of the object. So a freestyle solution would be best.    The main problem is that freestyle (to the right) produces double pixels: Let $a_n$ be a sequence of positive reals, such that the partial sums $S_n = \sum_{i = 1} ^ n a_i$ diverge to $\infty$. For given $\epsilon &gt; 0$ do we have $$\sum_{n = 1} ^ \infty \frac{a_n}{S_n^{1 + \epsilon}} &lt; \infty?$$  For $\epsilon \ge 1$ we can resolve this quickly by noting $$\frac{a_n}{S_n ^ 2} \le \frac 1 {S_{n - 1}} - \frac 1 {S_n}$$ so for sufficiently large $n$ we can bound $\frac{a_n}{S_n^{1 + \epsilon}}$ by $\frac 1 {S_{n - 1}} - \frac 1 {S_n}$ as well. I'm wondering if this is true for arbitrary $\epsilon &gt; 0$. I know that the series in question diverges for $ \epsilon = 0$, so all that is missing is what happens in $(0, 1)$. Splitting all lines in layer at vertices using QGIS I want to automatically split all lines in a QGIS layer at their vertices (). I've done a Google search and found , which is quite close but the workflow is complicated. Because I'll have to train my co-workers (who are unfamiliar with GIS) to use this tool, I want something as simple as a plugin or something that doesn't require GRASS. Since it's been a long time from 2010 - the time of , I wonder if there is any better way to achieve the same result? I want to construct a complex power series with the radius of convergence $R=1$ that diverges in:  1) $k$ points on the circle $\{|z|=1\}$.  2) countable number of points on the circle  I have already crafted series that converges on the whole circle and diverges on the whole circle.  Can anybody help with the problem?  Update: I have googled a series $\sum{\frac{z^{kn}}{kn}}$ that is an answer for 1). Though I still need help for the countable case. What's the intuition behind difference of $\neg(S\land E)$ and $\neg S \land \neg E$? Let $S$ represent the English statement "Sales will go up."And let $E$ represent the English statement "Expenses will go up." Now my problem is that I have trouble telling the difference between these two statements: $\neg(S\land Q)$ and $\neg S \land \neg E$. When these two statements are translated into English they sound,to me, the same. However, when you look at their truth tables they are different.   Here is my best translation of these two statements:  1) $\neg(S\land E)$ represented in English, "Both sales and expenses will not go up."(Correct me if I translated it wrong)  2)$\neg S \land \neg E$ represented in English, "Expenses will not go up and sales will not go up."(Again, correct me if I'm wrong)  I still do not understand the difference between these two statements in the English representation. They both sound the same thing to me.  Once more, I have the same trouble I have with understanding with the or-connective( $\lor$ ). Again let E and S be the same statements as before.Tell me if I might have translated them wrong.  1) $\neg (S \lor E)$ represented in English is "Neither sales nor expenses will go up."  2)$\neg S \lor \neg E$ represented in English is "Sales will not go up, or Expenses will not go up."  Yet I still have trouble understanding the difference between the English translation.  In short, can you give me some alternative example that better explains the difference, something intuitive that gives me a clear picture of the difference of the meaning between the statements I showed with the connective-or and connective-and. Thank you. Last activity name still missing from migrated question According to this post  this should be fixed. But it happened again.  See Image below (link to )  There was also another migrated question () in the list where the name was missing, but it was edited on meta. I have heard rumors that you can compile documents online, and more specifically that Google has a free online compiler, but I have never been able to find any. Is there a way to compile documents online, so that I can write and compile documents even if I don't have a TeX distribution installed on my computer? What is the correct way to create a single-instance WPF application?