In a game show a contestant selects one of three doors:   Behind one of the doors there is a prize and behind other two there is no prize. Each door has equal probability of having a prize. After contestant selects a door, the game show host who knows what's behind each of the doors, deliberately opens one of the remaining doors and reveals there is no prize behind it. The host then asks the contestant whether he want to $\text{SWITCH}$ his choice to other unopened door or $\text{STICK}$ to his original choice.  Which of the following options is/are correct ?: $(1)$ Probability of winning the prize is same whether contestant $\text{STICK}$ or $\text{SWITCH}$ $(2)$ Probability of winning the prize when contestant $\text{STICK}$, is Image not present $(3)$ Probability of winning the prize when contestant $\text{SWITCH}$, is Image not present $(4)$ It is advantageous for contestant to $\text{SWITCH}$ than to $\text{STICK}$ The solution to this question is $(2)$, $(3)$, $(4)$ But according to me the solution should be only $(1)$ because the probability should be same for both the cases, either he $\text{SWITCHES}$ or $\text{STICKS}$ to his choice. Can anyone explain.
I'm confused about why we should change door in the Monty Hall Problem, when thinking from a different perspective  gives me equal probability.   Think about this first: if we have two doors, and one car behind one of them, then we have a 50/50 chance of choosing the right door.   Back to Monty Hall: after we pick a door, one door is opened and shows a goat, and the other door remains closed. Let's call the door we picked A and the  other closed door B. Now since 1 door has already been opened, our knowledge has changed such that the car can only be behind A or B. Therefore, the problem is equivalent to: given two closed doors (A and B) and one car, which door should be chosen (we know it's a 50/50 thing)? Then, not switching door = choosing A, and switching door = choosing B. Therefore, it seems that switching should be equally likely, instead of more likely.   Another way to think: no matter which door we choose from the three, we know BEFOREHAND that we can definitely open a door with a goat in the remaining two. Therefore, showing an open door with a goat reveals nothing new about which door has the car.   What's wrong with this thinking process? (Note that I know the argument why switching gives advantage, and I know experiments have been done to prove that. My question is why the above thinking, which seems legit, is actually wrong.) Thanks.
Prove that if $0\le p_n&lt; 1$ and $S=\sum p_n&lt; \infty$ then  $\prod (1-p_n)&gt;0$. Hint given: First show that if $S&lt;1$, then $\prod (1-p_n)\ge 1-S$.   Attempt: I was able to show the hint by using recursion setting $A_n=\prod_{i=1}^{n}(1-p_i)$ re expression the $\prod (1-p_n)$ as $1-S+\sum_{n_1, n_2=1,n_1&lt;n_2 }^{\infty} p_{n_1}p_{n_2}-..... $ and observed that every subsequent term is less than the previous term hence  $\prod (1-p_n)\ge 1-S$ which is $&gt;0$ if $S&lt;1$ but am unsure about how to extend it to $S\ge1$. Make links posted by users open in a new window Most of the time when I follow a user's link, I just want to see their site, look it over, then return to answer their question (not leave SO completely and not come back). Could we make the links users post open in a new window please?  EDIT: I originally wanted this implemented because it's what some other sites did, but I now definitely prefer SO's way. If I want to open a new tab, I middle-click/control-click (or whatever); if not, I just click. I'm in control of what happens. Would never want to go back now. Should we use plural or singular for a fraction of a mile? I have seen people say both 0.25 mile and 0.25 miles. Should we use plural or singular for a fraction of a mile? I have reviewed this:   but it does not answer my question. The suggested fixes in this question is to set up FB, IG, and other social media pages. We have had these for months. We have 576 Instagram followers, 293 followers on Facebook, 17,425 views on Vimeo, Youtube, Pinterest and LinkedIn presence. Here is the response in a YouTube search:  Let's be clear. If you search for my company name and product name YT changes the spelling to something that doesn't exist and suggests a video with no matching result terms (using either spelling) disparaging OTHER luxury handbags, while insinuating it is referring to my handbags. When a user searches for our business, , Google and Youtube changes the spelling to Aslan Vaughn and displays these results:  For the first few months it just seemed like it was taking some time to percolate, but over 7 months we have had 17,500 unique visitors and nearly 90 backlinks.  The only results for Aslan Vaugn is the birth announcement of the second child of a musician in New Jersey. It is a Facebook post and has 14 likes. Google is favoring a spelling that has only two results with 14+ views over dozens of links and tens of thousands of views. How is this possible? An FB post with 14 likes is that much more popular than a business with 17,500 unique visitors (despite the Google obfuscation)? Is my company on some list? On other search engines my company covers the first few pages in results. Through Google, you cannot find our company no matter how many pages in you go. Unless you search for the term in quotes, . In which case we cover the first few pages here as well: Multi-partition multi OS bootable USB I want to create a recovery flash drive. I have my 8 gb flash drive[Kingston data traveller] with me. I also have the required .iso files for a>Ubuntu 13.10 b>Boot-repair disk c>AVG Rescue CD  I Wish to create 4 partitions(3 for each bootable .iso and one for data storage. Also, I want a way in which I can select which OS to boot each time I boot from USB.   So, is there a way to install GRUB on USB as well as the three other live distros? And how to create such a setup? How do I add a Javascript library to a field formatter? I'm trying to include a Javascript gallery () on my new Drupal 8 site and wrote a field formatter for this purpose.   Now I need to add the jQuery lightGallery to pages using this field formatter and call the library when the page is ready. I am confused about how I am supposed to do this, though.   I defined the library in the module containing my field formatter, but I'm not sure how I attach it only on sites that contain the field.   The documentation of the library uses the jQuery $(document).ready function to call the libary, how would that work inside a Drupal module? Moving here does not mean moving to a new city or something like that. I am talking about the people that just can't stay at one place. It is impossible for them to sit put at one place and relax. What is the secondary CPU Usage Graph line in the Windows 10 Task Manager I was looking at the task manager in Windows 10 and for the first time noticed a secondary dashed line below the solid? What does this line indicate? How to create a grain / noise effect in Adobe Illustrator I want to create a noise effect using Adobe Illustrator.  Inside the hexagonal shape there is some kind of texture effect, I don't know what the name of this effect is:    How can I create such an effect in Adobe Illustrator? or is it created using Photoshop? How to clone ArrayList and also clone its contents? how to host multiple pages from diferent domains all together? What is the "show desktop" keyboard shortcut? I just upgraded to 11.10 from 11.04. Previously the shortcut to show the desktop was Super +D (same as in Windows), But now it is not working. What is the new shortcut? Values of $\sum_{n=0}^\infty x^n$ and $\sum_{n=0}^N x^n$ Why does the following hold:  \begin{equation*} \displaystyle \sum\limits_{n=0}^{\infty} 0.7^n=\frac{1}{1-0.7} = 10/3\quad ? \end{equation*}  Can we generalize the above to      $\displaystyle \sum_{n=0}^{\infty} x^n = \frac{1}{1-x}$ ?   Are there some values of $x$ for which the above formula is invalid?  What about if we take only a finite number of terms? Is there a simpler formula?     $\displaystyle \sum_{n=0}^{N} x^n$   Is there a name for such a sequence?    This is being repurposed in an effort to cut down on duplicates, see here: Coping with  duplicate questions.  and here: . I installed or upgraded to Ubuntu 16.04, but I'm having some problems with my graphics card.   The display is blank The display flickers Items on the screen do not display correctly Certain programs complain of graphics issues The computer boots to low-graphics mode Your specific issue may not be here, but try the solutions described in the answers anyway.   This question is very broad, but each one of these issues likely has one solution.    If you are experiencing a login loop, check the last part of my answer.    If you experience one of the listed issues and have your own solution (mine didn't work, or there's a faster way), feel free to post another POLITE answer, describing your problem and how you fixed it.    Those attempting to install Ubuntu, please refer to the section titled: If you are unable to enter a TTY or are trying to install Ubuntu. Show that $A \setminus ( B \setminus C ) \equiv ( A \setminus B) \cup ( A \cap C )$ Show that  $A \setminus ( B \setminus C ) \equiv ( A \setminus B) \cup ( A \cap C )$  This is an exercise I was trying to do ( not homework ) and I got stuck as follows:  Working from $A \setminus ( B \setminus C )$:  $x \in A \wedge x \notin ( B \setminus C )$ $x \in A \wedge ((x \in B \wedge x \in C) \vee (x \notin B \wedge x \in C) \vee (x \notin B \wedge x \notin C) )$  At this point I'm not sure how to proceed.  Working from the other side $ ( A \setminus B) \cup ( A \cap C ) $:  $ x \in ( A \setminus B) \cup ( A \cap C ) $  $ x \in ( A \setminus B) \vee x \in ( A \cap C ) $  $ ( x \in A \wedge x \notin B ) \vee ( x \in A \wedge x \notin C ) $  $ x \in A \wedge (x \notin B \vee x \notin C ) $  $ x \in A \wedge x \notin ( B \cap C ) $  $ A \setminus ( B \cap C ) $  I could also use some formating tips like how display logical not and how to get the lines to display properly without blank lines inbetween. :-) How can I rotate about an arbitrary point in 3D (instead of the origin)? I have some models that I want to rotate using quaternions in the normal manner, except instead of rotation about the origin, I want it to be offset slightly. I know that you don't say, in 3d space, that you rotate about a point; you say you rotate about an axis. So I'm visualizing it as rotating about a vector whose tail is positioned not at the local origin.  All affine transformations in my rendering / physics engine are stored using SQT (scale, quaternion, translation; an idea borrowed from the book Game Engine Architecture.) So I build a matrix each frame from these components and pass it to the vertex shader. In this system, translation is applied, then scale, then rotation.  In one specific case, I need to translate an object in world space, scale it, and rotate it about a vertex not centered at the object's local origin.  Question: Given the constraints of my current system described above, how can I achieve a local rotation centered about a point other than the origin? Automatic upvote to anyone who can describe how to do this using only matrices as well :) unicode-math broken I've been using the idiom below for some time now. And it seems to be the most wide-spread, at least on the sites I've visited.  Is there a better/different way to read a file into a string in Java?  private String readFile(String file) throws IOException {     BufferedReader reader = new BufferedReader(new FileReader (file));     String         line = null;     StringBuilder  stringBuilder = new StringBuilder();     String         ls = System.getProperty("line.separator");      try {         while((line = reader.readLine()) != null) {             stringBuilder.append(line);             stringBuilder.append(ls);         }          return stringBuilder.toString();     } finally {         reader.close();     } } How do I convert two lists into a dictionary? Imagine that you have the following list. keys = ['name', 'age', 'food'] values = ['Monty', 42, 'spam']  What is the simplest way to produce the following dictionary? a_dict = {'name': 'Monty', 'age': 42, 'food': 'spam'} Lewis Carroll's Pillow Problem via Bayes' Theorem In Pillow Problems and a Tangled Tale Lewis Carroll presents the following problem:     A bag contains a counter, known to be either white or black. A white counter is put in, the bag is shaken, and a counter is drawn out, which proves to be white. What is now the chance of drawing a white counter?   One can easily reason that in the end, the bag must be in one of three states:   Black remains, new white taken New white remains, original white taken Original white remains, new white taken   In two of those states, a white counter remains to be drawn. Therefore, the chance is 2/3.  How can one arrive to this answer using Bayes' theorem?