Is it possible for example to sell a property you own to another player for 1$? I was playing with some friends and one of them had to pay me 1200$ and since he didnt have enough he sold all his houses reached 1200 gave me the money and then sold all his stuff for 1$ to another player.
I play with some serious board gamers, but we have neglected Monopoly for a number of years, so our knowledge is mostly based on high school &amp; house rules.  We played last week and enjoyed it, but there was a piece of contention. There are times where we decide that we don't like one of our friends for some reason and don't want them to win (i.e. we landed on their property a bunch, or they refused to make a deal, etc.).  So when player 1 has some property and player 2 lands on it and cannot pay the rent their assets (houses/hotels get liquidated), their property gets transferred to the other player. (Please correct me if I'm wrong.) In this case, though, player 3 "traded" the amount of money owing for a piece of property (way overpaid). In this case they did it so that player 1 would not get the properties. Player 2 had no chance of winning the game (all properties mortgaged, no cash to speak of and many of the other properties with 3+ houses/hotels on them).  To summarize: Is it OK for a third player to purchase/trade property or cash to/from a player who is about to lose?
Ideal contained in a finite union of prime ideals Let $I \subset R$ be an ideal and $P_i$  $(i=\{1,...,n\})$ prime ideals with $I\subseteq\bigcup_{i=1}^nP_i$. Prove that then $I$ is contained in one $P_i$.   I don't know how to show this because I don't have any approach.  So I am looking for something to start with or something like a sequence of tips (Since I know this is a large proof)  Thanks in advance! Fastest way to try all passwords Suppose you have a computer with a password of length $k$ in an alphabet of $n$ letters.  You can write an arbitrarly long word and the computer will try all the subwords of $k$ consecutive letters. What is the smallest word that contains all combinations of $k$ letters as subword? (i.e. the fastest way to hack the computer :) )  The smallest word that contains $n^k$ subwords of size $k$ has length $k-1+n^k$ and based on some easy cases, we would like to prove that it is in fact possible to find a word of such length that contains all possible passwords. The problem can be translated into a problem in graph theory, by taking as vertices all words of length $k$.  We tried $k=2$, where you can prove the conjecture by induction. For $n=2$ and small $k$ it also works. Feature request: Add pronouns to usercards? Installing Guest Additions causing problems What is the difference between '/' and '//' when used for division? How to set spawnpoint? What gifts would be welcome for a Swedish host family? Why would I vote for an increase in the number of authorized shares? Restore console session after remote desktop disconnects? I have a random sample of Bernoulli random variables $X_1 ... X_N$, where $X_i$ are i.i.d. r.v. and $P(X_i = 1) = p$, and $p$ is an unknown parameter.  Obviously, one can find an estimate for $p$: $\hat{p}:=(X_1+\dots+X_N)/N$.  My question is how can I build a confidence interval for $p$? How to Regex search/replace only first occurrence in a string in .NET? It seems the .NET Regex.Replace method automatically replaces all matching occurrences.  I could provide a MatchEvaluator delegate that returns the matched string after the first replacement, rendering no change, but that sounds very inefficient to me.  What is the most efficient way to stop after the first replacement? Why isn't there a centre of charge? When determining the gravitational attraction between 2 solid bodies, we can simplify computations by taking their masses to be concentrated at their respective centres of mass. However, had they been electrically charged bodies and we needed to compute electrostatic attraction, there is no "centre of charge" notion available to us. ( We would evaluate a 2-dimensional integral, or equivalently, apply Gauss's law. )  Is there any simple reason for the absence of such a centre of charge? Frequently, I've got the answer: "charge just doesn't work that way". Updating .class file in jar Why is the electric field zero inside a hollow conducting sphere? I  want to download free app in my iPod touch device and i have the following doubts.  Kindly resolve me doubts.   when i try to download it asks my credit card security number. since i am trying to download free app , why do i need to provide my credit card security code? Will i be debited for downloading the free app from app store?   I cant able to find any information regarding this on apple.  Please provide me some references about this from apple and please resolve my doubts.  Thanks a lot. The names of the derived functors $\operatorname{Tor}$ and $\operatorname{Ext}$ seem quite cryptic to me. Does anyone know what these abbreviations stand for? I would be glad if someone could tell me where these names come from. Favorite questions in the Android app Proving this can be done as follows: consider a finite group G and elements $g_i \in G$ for some integer $i$. Now consider $\langle g_i \rangle = \{g_i^n: n\geq 0\}$, a generator. It can be proved that $\langle g_i \rangle \leq G$ and that the order of $g_i$ is equal to the order of $\langle g_i \rangle$, so $|\langle g_i \rangle| \leq |G|$.  We can now use Lagrange's theorem which states that if $H \leq G$ then $|H|$ divides $|G|$ and we're done. But...  Is there a simpler way to prove this fact? I'm using Android 4.4.2 (on ). I've a Windows PC at home which has highspeed broadband internet connection. I don't have wireless router or access point. I just want to use PC's internet connection on my phone. I've tried multiple methods but none seems to work.  Method 1: Using native Android 4.4.2 option  In my phone when I connect my phone, in Settings &gt; Wireless &amp; networks &gt; USB Internet (sharing Windows PC internet via USB cable) checkbox, which I checked then it asked me to choose my Windows version (which is Windows 8.1 but since it didn't show that option) I chose Windows 8.  It gives me further instructions as showing in the following picture:    Here is the problem, there is no sharing tab (see the following screen shot)   Method 2: As described in   I learned on this site that it is called Reverse tethering. I read and followed the procedure mentioned in    I followed all the steps described in that answer, until executing the command.   Install USB drivers from Android SDK [DONE] Connect USB cable and activate USB Tethering (You should see on a new network interface.) [DONE] Bridge the 2 network interfaces [DONE] On your computer execute adb shell netcfg usb0 dhcp [PROBLEM]   When I execute adb shell netcfg usb0 dhcp it says: error: device not found. I tried the same thing with USB Debugging in Developer settings, ON &amp; OFF. The error remains the same.  Method 3: Rooting the device and using an app  I read here and there that there is another method, in which we can directly use an app but it requires rooting the device. My phone is a new phone and warranty becomes void if I root it.   EDIT: I've tried the method suggested in answer, at the 4th step of method 2, I'm facing this problem: How do I get rid of eggs?