If I pick an arbitrary interval like (0,1), the probability of picking any given number (like 0.5) = $\lim_{x \to \infty}\frac{1}{x}=0$  I at first thought that this was a paradox, and questioning whether '$0$' really meant impossible. I realised that it is impossible to even pick a number between (0,1). A computer cannot do it as it would require every number in that range to chose from (infinite).  Is this really a paradox? Either interpretation of 0 being impossible or infinitesimal leads to it not being one.
35679
I read a comment under :     There are plenty of events that can occur that have zero probability.   This reminds me that I have seen similar saying before elsewhere, and have never been able to make sense out of it. So I was wondering    if zero probability and impossibility mean the same? if an event with zero probability doesn't mean that the event is impossible to occur, how probability theory represents/describes impossibility?   Thanks and regards!
120519
I realize the title of my question will probably cause the raising of some eyebrows, so let me explain. Not sure whether to file this under "math" or "philosophy". This also might be able to be explained away due to some possible misconceptions I have about irrational/transcendental numbers, so please correct me where I'm wrong, since I'm quite certain I am...but this is nagging me, so here it is:  Ok, so imagine a non-repeating decimal with an infinite number of digits in it. The numbers are random, but for the sake of argument, let's say the number starts with "0.2368210236..." and continues from there.  So I look at the first decimal digit - 2 - and think to myself "in the rest of this decimal, there are an infinite number of times that '2' shows up."  So I look at the first TWO decimal digits - 23 - and again, think to myself "due to the randomness and infinity of this decimal, there is statistically a 100% certainty that the numbers "23" show up, in that order, somewhere in the rest of this decimal.  Continuing this line of thought, I reach the conclusion that if I choose ANY number of digits (variable: x) that start off this infinite decimal number, then a) there is statistically a 100% certainty that particular sequence shows up somewhere else in the decimal, and b) that it shows up an infinite number of times.  If such is the case, then what happens as x approaches infinity? Would that not mean that, given enough length in the decimal, there would eventually come a point where the decimal starts repeating itself, ad infinitum? Would that not mean that what was originally thought of as an irrational number is actually rational? Would this line of thinking not apply to ALL irrational numbers?  I guess what I'm asking is this: could it be that there ARE no irrational numbers, just rational numbers whose scope and vastness outstrip our ability to see their rationality?  Alright, mathphiles; have at me!
52121
I need to generate a random integer between 1 and n (where n is a positive whole number) to use for a unit test. I don't need something overly complicated to ensure true randomness - just an old-fashioned random number.  How would I do that?
86065
Two numbers are chosen randomly and independently from the unit interval $(0,1)$.  What is the probability that they differ by more than $\frac{1}{2}$? Hint: Think of the $2$ numbers chosen as coordinates in the Cartesian plane. Draw a square with vertices at $(0,0)$,$(1,0)$,$(1,1)$, and $(0,1)$. Allowable choices for the $2$ points lie below the line $y=x-\frac{1}{2}$ or the above $y=x+\frac{1}{2}$. Find the combined area of these regions.  I need help on this. I know what $y=x-\frac{1}{2}$ and $y=x+\frac{1}{2}$ looks like but I am not understanding the problem as a whole
1939
Loosely, a number is describable if it can be unambiguously defined by a finite string over a finite alphabet. Numbers such as $\frac{1}{3}$, $\sum_{n=0}^\infty \frac{1}{n!}$, and "the ratio of circumference to diameter of a circle" are all describable numbers. One can  show that the set of all such numbers is countable.  Let $U$ be the set of indescribable real numbers. I "claim" $U$ is empty, and so every real number is describable. Suppose to the contrary. Place a well-order on $U$ (this can be done assuming the Axiom of Choice). The least element $u$ of $U$ admits the description "the least element of $U$ according to the specified well-order", contradicting the indescribability of $u$.  Obviously, something has gone awry here. I suspect it is some combination of my loose definition of describability and my self-referential "description" of $u$. My (slightly open-ended) question is "What's wrong here?".  P.S. There is a  addressing something rather like my question. As I understand it, the author's response is "the given description isn't really a description". If so, I'd enjoy some elaboration.
113008
Four dice are thrown, what's the probability that:  a) None of them fall higher than three?  b) None of them fall higher than four?  c) That four is the highest number thrown?  So for a, I wanna think about the denominator first. There are 6 possible outcomes for each dice, and we have four dice. So we're technically picking r out of n objects, aka picking 4 possible values out of 6. So the denominator/sample space should be $6^4$ right?  Now this is where I'm getting tripped up. Order technically shouldn't count, because we only care about the quadruplets that don't have a value higher than 3. I might be wrong, maybe I am, but could someone explain a bit more? My professor stated that typically in the sample space/denominator, we want order to count.   So I'm trying to think of the numerator now, so we want to find the probability that no values appear higher than 3, so some events that can occur are: $(1, 1, 2, 3), (1, 2, 2, 3), (1, 1, 1, 1)$ etc. Again, I can't see why order should count here, because if we only are concerned about what appears rather than how they appear, then we can say $(1, 1, 2, 3)$ is equal to $(1, 2, 1, 3)$? So if order does not count, and we have replacement, then does this lead the case where we use $$\binom{n-1+r}{r}$$to find out the probability?   I'm pretty sure if I can do a), I could prob do b), but if someone could maybe lead me in the right direction for c) I would appreciate that too!
26412
I have been in a deep confusion for about a month over the topic of limits! According to our book, the limit at $a$ is the value being approached by a function $f(x)$ as $x$ approaches $a$.  I have a doubt that in real number line we can never ever reach the closest value to $a$ because always a more closer value will exist.  Now when talking about our methods for calculating limit, what method comes in our mind when we have to calculate where the value approaches? Let's say we have to find what value $f(x)$ approaches when $x$ moves from $[0,a)$.  So we calculate the value of $f(x)$ at $x=0$, say $4$, and then at $x\to a$.   But the problem is that we don't know the value $x\to a$, so we say the value is $a-h$ where $h\to 0$ and calculate the value of $f(a-h)$, say $5-h$.  Now here is where my doubt starts! In the final step we put the value of $h=0$ say in that it is an infinitesimal quantity.  My doubt is that $h$ was tending to $0$ means that it was never equal to zero  maybe it is infinitesimally small, not a stationary value, not an imaginable value, but we know for sure that it is not  equal to $0$. Maybe it is the point closest to zero, but it is not equal to zero and when we use the result $5-h=5$ we are actually making an error which is tending to zero. Maybe the error is very small, but still, there is some error in that we cannot calculate it but we can see that there is this infinitesimal error present.  That means we don't get the exact limiting value or last value of $f(x)$.  $x$ belongs to $[0,a)$ but a value approximate to infinitesimal? Isn't it right! We get an approximated value?
74131
I was doing some software engineering and wanted to have a thread do something in the background to basically just waste CPU time for a certain test.  While I could have done something really boring like for(i &lt; 10000000) { j = 2 * i }, I ended up having the program start with $1$, and then for a million steps choose a random real number $r$ in the interval $[0,R]$ (uniformly distributed) and multiply the result by $r$ at each step.   When $R = 2$, it converged to $0$. When $R = 3$, it exploded to infinity.    So of course, the question anyone with a modicum of curiosity would ask: for what $R$ do we have the transition. And then, I tried the first number between $2$ and $3$ that we would all think of, Euler's number $e$, and sure enough, this conjecture was right. Would love to see a proof of this.    Now when I should be working, I'm instead wondering about the behavior of this script.   Ironically, rather than wasting my CPUs time, I'm wasting my own time. But it's a beautiful phenomenon. I don't regret it. $\ddot\smile$
97568
I'm working on the following to get the upper limit of my integration, as I know I need to have some idea where the point might occur but in real it can be any value between 0 and infinity. May I know is there any improvement can be done so that it can work within 0 and infinity?   findx = function(f, interval, target) {   optimize(function(x) {     abs(integrate(f, 0, x)$value-target)   }, interval)$minimum }  pdf = function(x, a=2, b=0.5, lam=-1.2, del=1.5){   s = besselK(2*sqrt(a*b), lam/del, expon.scaled = FALSE)   pdf = del/2 * (a/b)^(lam/(2*del)) / s * x^(lam-1)*exp(-a*x^del-b*x^(-del))   return(pdf) }   findx(pdf, interval=c(0,Inf), target=runif(1, min = 0, max = 1))   The following is the error I get when I have my interval=c(0, Inf)  &gt; findx(pdf, interval=c(0,Inf), target=runif(1, min = 0, max = 1)) Error in optimize(function(x) { : invalid 'xmax' value   EDIT:  Actually, I'm referring to technique in the link of  but it mentioned that "optimize needs an interval in which to search for a solution so you'll want to have some idea of where the point might occur" but my x is from 0 to infinity in which I wouldn't know where will it occur. While I need to find the minimum upper limit for my integration.
29799
Possible Duplicate:          This is the Bertrand's Box Paradox I read about on Wikipedia:     Assume there is three boxes:       a box containing two gold coins,       a box with two silver coins       and a box with one of each.       After choosing a box at random and withdrawing one coin at random,       if that happens to be a gold coin,       the probability is actually 66% instead of 50%.      And the problem is equivalent to asking the question       "What is the probability that I will pick a box with two coins of the same color?".   No matter how hard I try, I just couldn't comprehend this..   How is the possibility of picking a gold coin the same as the probability of picking a box with two coins of the same color?  Does this imply there is a 66% chance of picking a gold coin and a 66% chance of picking a sliver coin?   If so, can we just say there is 50% chance of picking either one of them since both stand a 66% chance....?! and suddenly everything makes no sense..  [UPDATES] It is actually the probability of the remaining coin to be gold is 66% but not the probability of obtaining the gold coin is 66%.. I've misread it....   And everything makes sense now :D !
110955
Forgive me if this has been asked before, but I searched and could not find an answer.  I am trying to show that $\lim\limits_{x \rightarrow 0} \frac{1}{x}$ does not exist. If the limit did exist, and was equal to $l$, then for every $\varepsilon &gt; 0$ there would exist some $\delta &gt; 0$ such that for all $x$,  $$0 &lt; |x| &lt; \delta \implies \left| \frac{1}{x} - l \right| &lt; \varepsilon$$  If the limit doesn't exist, then for every number $l$, there exists some $\varepsilon &gt; 0$ such that for all $\delta &gt; 0$,  $$0 &lt; |x| &lt; \delta \implies \left| \frac{1}{x} - l \right| \geq \varepsilon$$  So, in my efforts to find some unattainable $\varepsilon$, I have found that  \begin{align} \varepsilon &amp;\leq \left| \frac{1}{x} - l \right| \\\ &amp;= \left| \frac{1}{x} - \frac{lx}{x} \right| \\\ &amp;= \left| \frac{1-lx}{x} \right| \\\ &amp;= \frac{|1-lx|}{|x|} \\\ &amp;\implies \varepsilon |x| \leq |1 - lx| = |1 + (-lx)| \leq |1| + |-lx| \\\ &amp;\implies 1 \geq \varepsilon |x| - |l| |x| = (\varepsilon - |l|) |x| \\\ &amp;\implies |x| \leq \frac{1}{\varepsilon - |l|} \end{align}  So I have found that if $|x| &lt; \frac{1}{\varepsilon - |l|}$, then $\left| \frac{1}{x} - l \right| \geq \varepsilon$, for any $\varepsilon$. But, I need this to be true for any choice of $\delta$, for some specific $\varepsilon$.  Intuitively, I can think "as $\varepsilon$ gets closer to $|l|$, then the quotient becomes larger, meaning that $|x| &lt; \frac{1}{\varepsilon - |l|}$ is satisfied for more choices of $\delta$", but I don't know how to formalize this. Any ideas for a next step? Am I completely off base with my reasoning?
56137
There is a probabilistic method to solve it. But I am not familiar with probability. I am trying to compute it by analytic method, such as using L Hospital's rule or Stolz formula, but they are not working.
90539
Apparently, the solution to the  is that a uniform probability distribution over the positive real numbers doesn't exist. Can anyone explain why this is the case and what probability distributions can exist over the positive real numbers (it seems that this would be quite limited, given that such a simple distribution is impossible)?
87732
If I have an "experiment", the results of which can be classified clearly into "outcomes" (like rolling a die), then I can make a concrete and verifiable empirical claim that "if you repeat this experiment many times, the frequency of this outcome will tend towards this number". This is what I mean when I say that the probability of something is such and such a number.  But then what is the meaning of the phrase "there is a 25% chance of rain tomorrow"? If I repeat the experiment of it being tomorrow many many times... what? It just doesn't make any sense, the definition of probability given above just doesn't apply.  To take an even more concrete example, let's say someone presents you a box containing one million black balls and one white. You get to take out one ball (and only one). If it's white, they give you a thousand dollars. Otherwise, you give them a thousand dollars. They're very stubborn and won't let you refuse to play unless you can give a satisfying philosophical explanation as to why you don't want to. What could you say?  Is there a name for this "paradox"? Is there a solution to it?
1129
I need to generate random numbers within a specified interval, [max;min].  Also, the random numbers should be uniformly distributed over the interval, not located to a particular point.  Currenly I am generating as:  for(int i=0; i&lt;6; i++) {     DWORD random = rand()%(max-min+1) + min; }   From my tests, random numbers are generated around one point only.  Example min = 3604607; max = 7654607;   Random numbers generated:  3631594 3609293 3630000 3628441 3636376 3621404   From answers below: OK, RAND_MAX is 32767. I am on C++ Windows platform. Is there any other method to generate random numbers with a uniform distribution?
101854
It is a common saying that there are more real numbers than integers, since you can not uniquely map integers to real numbers - even after using up all integers, there will still be at least one real number left that you haven't mapped to any integer.  So I applied the same argument to the two intervals in question. One obvious way to map number from $[0,1]$ to $[0,2]$ is $\times 2$ - $0$ is still $0$, $0.25$ becomes $0.5$, and $1$ becomes $2$. Good enough.  However, if we change the second interval to an open interval, namely $(0,2)$, then the number $0$ and $1$ in the original mapping is now the "remaining" number, while all numbers in $(0,2)$ have already been "used". Does that mean that $[0,1]$ have more real numbers than $(0,2)$? (even though $(0,2)$ seems to be bigger than $[0,1]$&hellip;)  Does my argument even make sense? I did some Googling but didn't found anything useful.
93645
So, I came across the following paradox:  At $1$ minute before noon, put in balls $1 \sim 10$ and take out ball number $1$. At $1/2$ minute before noon, put in balls $11\sim20$ and take out ball number $2$ and so on. How many balls are there at noon?  None.  At $1$ minute before noon, put in balls $1 \sim 10$ and randomly take out a ball. At $1/2$ minute before noon, put in balls $11\sim20$ and randomly take out another ball and so on. How many balls are there at noon?  None.  Okay, so I understand the first paradox because one can describe the exact moment each ball was taken out. But, you can't apply a similar argument to the second paradox because we randomly take out a ball.   I feel as if it's like infinitely summing $\frac{1}{n}$ and eventually there would be too many balls.  Can someone explain to me mathematically why this is the case?