A gambler rolls two unbiased dice and stands to loss of $ \$2$ if he fails to throw a six, to win $ \$4$ if he throws one six and to win $ \$10$ if he throws two sixes. Is the game fair?  I can find probability for all three cases but to decide whether game is fair or not? Could someone help me with this?
Suppose $X_n$ is the fortune of a gambler after $n$ th game. Then the game is called fair (Breiman 1968) if   $$E[X_{n+1} \mid X_1, \dots, X_n] = X_n \forall n$$  My question is why a fair game is not defined as the following   $$E[X_{n+1}] = E[X_n] \forall n$$ i.e. $$E[X_{n+1}- X_n]=0$$. This should be the proper definition as a fair game is where avg. gain is zero. Nothing conditioning should be there.
Is there a single word for a "friend of a friend"? I am looking for a word that I could use instead of "friend of a friend". On certain occasions, it happens that you may want to talk about a friend of a friend with someone else.   In such cases how would you refer to this friend? Is there a general formula for finding all subgroups of dihedral groups? 32-bit vs. 64-bit systems What are the differences between 32-bit and 64-bit systems?  If you have used both of them, what kind of sharp differences have you experienced?   Would it be a problem to use 32-bit programs on 64-bit systems in some cases? Proof of linear independence of $e^{at}$ Given $\left\{ a_{i}\right\} _{i=0}^{n}\subset\mathbb{R}$ which are distinct, show that $\left\{ e^{a_{i}t}\right\} \subset C^{0}\left(\mathbb{R},\mathbb{R}\right)$,  form a linearly independent set of functions.   Any tips on how to go about this proof. I tried a working from the definition of an exponential and combining sums but that didn't seem to get me anywhere. I saw a tip on the internet that said write it in the form  $\mu_{1}e^{a_{1}t}+\dots+\mu_{n}e^{a_{n}t}=0$ to try to show $\mu_{1}=\dots=\mu_{n}=0$ considering each term of the left hand side must be positive, but I can't get my head around that because while I understand $e^{x}&gt;0\forall x\in\mathbb{R}$ I cannot see why $\mu_{i}$ must be positive in any case. I have thought about differentiating but that doesn't seem to help. The question did originally ask for a "rigourous" proof but I'll take any hints right now and the provided the solution of 'is obvious' is most unhelpful to me.  Any input would be fantastic. Thank you. Infinite distinct factorizations into irreducibles for an element Consider the factorization into irreducibles of $6$ in $\mathbb{Z}[\sqrt{-5}]$. We have $6=2 \times 3$ and  $6=(1+\sqrt{-5}) \times (1-\sqrt{-5})$, i.e. $2$ distinct factorizations. And,  $$6^2=3 \times 3\times2\times2$$ $$=(1+\sqrt{-5}) \times (1-\sqrt{-5}) \times (1+\sqrt{-5}) \times (1-\sqrt{-5})$$ $$=(1+\sqrt{-5}) \times (1-\sqrt{-5})\times3\times2.$$ More generally, $6^n$ will have $n+1$ distinct factorizations into irreducibles in $\mathbb{Z}[\sqrt{-5}]$ by a simple combinatorial argument. But, can we construct a ring in which there exists an element that has an infinite number of distinct factorizations into irreducibles? To make life harder, can we construct an extension of $\mathbb{Z}$ in which this happens? I have been thinking about this for a while and have managed to find no foothold.. Any help is appreciated. I'm completely new and lost in Ubuntu 12.04, I'm having trouble navigating to my Sublime folder which I know is in my /opt folder. Yet I've tried to no avail to navigate into it via Terminal window. The directory is colored in blue, no idea what this means...but I am able to access the folder using the GUI explorer...?  k@k-Ubuntu:~$ cd /opt k@k-Ubuntu:/opt$ ls Sublime Text 2 k@k-Ubuntu:/opt$ cd Sublime Text 2 bash: cd: Sublime: No such file or directory k@k-Ubuntu:/opt$ cd /Sublime Text 2 bash: cd: /Sublime: No such file or directory php Replacing multiple spaces with a single space Convert a string to an enum in C# What's the best way to convert a string to an enumeration value in C#? I have an HTML select tag containing the values of an enumeration. When the page is posted, I want to pick up the value (which will be in the form of a string) and convert it to the corresponding enumeration value. In an ideal world, I could do something like this: StatusEnum MyStatus = StatusEnum.Parse(&quot;Active&quot;);  but that isn't a valid code. Why are more powerful tools using higher voltage? AFAIK each of the world leading cordless power tools manufacturers produces several "product lines" of cordless power tools with different voltages. For example, Bosch currently produces tools with Li-Ion batteries with 10,7V, 14,4V, 18V and 36V output and the higher the voltage the more powerful a tool is.  Now those tools are powered by batteries that are assembled from cells with lower voltage (something like 3,7 volts for Li-Ion cells I guess) and cells are connected in sequence until the target voltage is reached.  They could instead connect cells in parallel. They would have the same voltage, but higher current and that would again yield higher power.  Why do they choose higher voltage over same voltage and higher current to get higher power in electric tools? Which Ubuntu to use alongside Windows 7? Serving wp-includes (front-end) javascript from a different domain? The WordPress Codex points to an extremely simple way to change the URL of the wp-content directory, which is, adding this in wp-config.php:  define( 'WP_CONTENT_URL', 'http://example.com/wp-content');   And that does the job of serving most of the JavaScript and CSS files on my website from the different domain, except two.  http://mywebsite.com/wp-includes/js/comment-reply.js http://mywebsite.com/wp-includes/js/quicktags.js   The reason is very clear — these JavaScript files are served from within the wp-includes directory, which is not affected by the aforementioned rule.  So, is there a similar or simple method to have them served from example.com i.e. the domain of my choice? or is unregistering and enqueuing the scripts from a custom URL (done in functions.php) the only way? Keeping chunks loaded in minecraft I have built an iron farm somewhere far away from my actual spawn chunks in my single player world. In hindsight, this was a dumb idea, but I can't exactly change it anymore. I'd like to keep the thing running all the time, so I'm looking for a way to keep the chunks loaded.  I'm running minecraft 1.12.1, is it possible at the moment? I have looked at some chunk loader mods, but chicken chunk loader hasn't been updated yet, and the others simply didn't work.  I have tried moving the spawn chunks, but that hasn't done anything either.  Edit: I have worked this out with IronAnvil, and it ended up being a render setting in optifine which didn't load them correctly somehow. It has been working fine after I reset my optifine settings to default. Did she judge him "wrong" or "wrongly"? Which one is the correct use?  She judged him wrong. She judged him wrongly.  Or, are both correct, but have slightly different meanings? Let $G$ be a finite group and $N$ a normal subgroup of $G$ of index $4$. Prove that $G$ has a subgroup of index $2$ (hence normal) and if $G/H$ is not cyclic then $G$ is equal to the union of three proper subgroups.   I understand that $G/H$ is isomorphic to $\mathbb{Z}_4$ or $\mathbb{Z}_2\times \mathbb{Z}_2$. Let $K^*$ be a subgroup of $G/H$ of order $2$. Then $K^*\cong K/H$ for some $K$ such that $H$ is normal in $K$ and hence $[G:K]=[G/H:K/H]=4/2=2$.  If $G/H$ is not cyclic then $G/H\cong \mathbb{Z}_2\times \mathbb{Z}_2$. I don't know how to prove $G$ is the union of three subgroups. Could it be true that as $G/H$ is the union of three proper subgroups $K_1^*,K_2^*,K_3^*$ then the corresponding groups $K_1,K_2,K_3$ have union equal to $G$? Would anyone give me a hint to prove this?  Thank you. Pluralization rule for "five-year-old children", "20 pound note", "10 mile run" Is an expanding map on a compact metric space continuous? Do downvotes on Meta truly mean something different? Where $\phi(n)$ is the Euler phi function, how do you find all $n$ such that $\phi(n)|n$? For which Natural $n\ge2: \phi(n)=n/2$ My first issue was when I tried to apt-get update or apt-get upgrade. On upgrade I get the following error:  You might want to run 'apt-get -f install' to correct these. The following packages have unmet dependencies: linux-image-server : Depends: linux-image-3.2.0-27-generic but it is not installed E: Unmet dependencies. Try using -f.   I tried running apt-get install -f and this was the output(after saying yes on prompt)  (Reading database ... 186183 files and directories currently installed.) Unpacking linux-image-3.2.0-27-generic (from .../linux-image-3.2.0-27-generic_3.2.0-27.43_amd64.deb) ... Done. dpkg: error processing /var/cache/apt/archives/linux-image-3.2.0-27-generic_3.2.0-27.43_amd64.deb (--unpack):  failed in write on buffer copy for backend dpkg-deb during `./boot/System.map-3.2.0-27-generic': No space left on device  No apport report written because the error message indicates a disk full error                                                                           dpkg-deb:    error: subprocess paste was killed by signal (Broken pipe) Examining /etc/kernel/postrm.d . run-parts: executing /etc/kernel/postrm.d/initramfs-tools 3.2.0-27-generic   /boot/vmlinuz-3.2.0-27-generic run-parts: executing /etc/kernel/postrm.d/zz-update-grub 3.2.0-27-generic /boot/vmlinuz-3.2.0-27-generic Errors were encountered while processing: /var/cache/apt/archives/linux-image-3.2.0-27-generic_3.2.0-27.43_amd64.deb E: Sub-process /usr/bin/dpkg returned an error code (1)   I tried running apt-get autoremove and it gives me the same error as apt-get upgrade.  When I run df, I get this for /boot:  /dev/sda1                    233191     230297         0 100% /boot   So, I read elsewhere that I should try to purge old kernels. I checked for what kernels I had with:  $ dpkg -l linux-image-\* | grep ^ii ii  linux-image-2.6.38-13-server  2.6.38-13.52  Linux kernel image for version 2.6.38 on x86_64 ii  linux-image-3.0.0-13-server   3.0.0-13.22   Linux kernel image for version 3.0.0  on x86_64 ii  linux-image-3.0.0-14-server   3.0.0-14.23   Linux kernel image for version 3.0.0  on x86_64 ii  linux-image-3.0.0-15-server   3.0.0-15.26   Linux kernel image for version 3.0.0  on x86_64 ii  linux-image-3.0.0-16-server   3.0.0-16.29   Linux kernel image for version 3.0.0  on x86_64 ii  linux-image-3.0.0-17-server   3.0.0-17.30   Linux kernel image for version 3.0.0  on x86_64 ii  linux-image-3.2.0-24-generic  3.2.0-24.39   Linux kernel image for version 3.2.0  on 64 bit x86 SMP ii  linux-image-3.2.0-25-generic  3.2.0-25.40   Linux kernel image for version 3.2.0  on 64 bit x86 SMP ii  linux-image-3.2.0-26-generic  3.2.0-26.41   Linux kernel image for version 3.2.0  on 64 bit x86 SMP   When I try to remove the oldest with this:  $ sudo apt-get purge linux-image-2.6.38-13-server Reading package lists... Done Building dependency tree Reading state information... Done You might want to run 'apt-get -f install' to correct these: The following packages have unmet dependencies: linux-image-server : Depends: linux-image-3.2.0-27-generic but it is not going to be     installed E: Unmet dependencies. Try 'apt-get -f install' with no packages (or specify a solution).   How can I free up or extend boot without messing up my installation?