I'm not a mathematician just an ordinary person, I would like to know, what is the mathematical chance for this to happen: A computer randomly chose 1 number out of 1024, how many times you need to try to find the number? After each unsuccessful try the computer will randomly chose another number out of 1024 which you have to guess and so on until you find the right number. Thanks
122047
Is there anything that can be said about how many attempts it will take to correctly guess a random number out of 1000 numbers? If the number wouldn't change the probability would just increase every guess, giving in the end a probability of 1.  If the number changes every guess tough, you just have a 1/1000th chance everytime. Can it be calculated how many attempts it would probably take, on average?  (I'm sorry, i'm not very skilled in probability maths)
28618
A coin is tossed $n$ times. What is the probability of getting odd number of heads?   I started this chapter sometimes ago and faced in front of a tough problem. At first I started considering cases. Case-I : The probability of getting 1 head.Case-II : The probability of getting 3 head and so on. But there are many cases. So how can I solved this . Please help me. Thank you!
14953
Suppose the winning combination consists of $7$ digits, each digit randomly ranging from $0$ to $9$. So the probability of $1111111$, $3141592$ and $8174249$ are the same. But $1111111$ seems (to me) far less likely to be the lucky number than $8174249$. Is my intuition simply wrong or is it correct in some sense?
40571
I pick a random subset $S$ of $\{1,\ldots,N\}$, and you have to guess what it is. After each guess $G$, I tell you the number of elements in $G \cap S$. How many guesses do you need?
21919
I have been looking at the birthday problem (http://en.wikipedia.org/wiki/Birthday_problem) and I am trying to figure out what the probability of 3 people sharing a birthday in a room of 30 people is. (Instead of 2).  I thought I understood the problem but I guess not since I have no idea how to do it with 3.
82881
Before reading the Wikipedia article, my idea to calculate the probability was as follows:  $1-\left(\frac{364}{365}\right)^{_nC_{2}}$  Basically, I thought to compare all combinations of pairs ($_nC_{2}$) and require them to have different birthday ($\frac{364}{365}$). It gave a very similar result to the wiki method, but is slightly different and I can't tell how  my way of thinking is wrong.  Can anyone tell me what I'm missing?
21465
When I say totally random, I mean absolutely random, not pseudorandom. If I want to say "totally random" numbers such as 1,26,17,4,1 and 27, although I see them to be totally random, they aren't. These are numbers that I think are influenced by my childhood, ideology and everything that I've seen. So, what do you think? Is there anything completely random?
714
What is wrong with my method of find the probability of a 2 pair?  For the first card we have 52 choices, one this is picked( call it A) we have 3 choices(different suite than A but this same value) for the second card, ( lets call this card B). Now we have AB, to we have to throw away 2 more cards so we don't have 3 of a kind or a four of a kind. Therefore one our third card we have 48 choices and on our fourth card we have 3 choices again similar to above.  On our last card we only have 44 choices left. So the probablity of getting a 2 pair should be (52*3*48*3*44)/(52 choose 5).  But this isn't correct.     Note: I edited this to "2 pair"; the original version was wrong.
12282
In a family with two children, what are the chances, if one of the children is a girl, that both children are girls?  I just dipped into a book, The Drunkard's Walk - How Randomness Rules Our Lives, by Leonard Mlodinow, Vintage Books, 2008. On p.107 Mlodinow says the chances are 1 in 3.   It seems obvious to me that the chances are 1 in 2. Am I correct? Is this not exactly analogous to having a bowl with an infinite number of marbles, half black and half red? Without looking I draw out a black marble. The probability of the second marble I draw being black is 1/2.
69401
Suppose we uniformly and randomly select a permutation from the 20! Permutations of 1,2,3,...,20. What is the probability that 2 appears at an earlier position than any other even number in the selected permutation?  My approach :   2 in first position ==> 1*19! (1 is the position of 2)  2 in second position ==> 10 *1 *18!  2 in third position ==> 10*9 *1 *17!  . . .  2 in 11th position ==> 10*9*8*7*6*5*4*3*2*1*1*9!  But this was asked as a multiple choice question in GATE 2007 and the options were  (A) 1/2   (B) 1/10  (C) 9!/20!  (D) None of these  I am not able to reduce this analysis to a given option.
30265
I have been trying to solve , but in vain. Please help.     You are given two boxes with a number inside each box. The two numbers   are different but you have no idea what they are. You pick one box to   open; read the number inside; and then guess if the number in the   other box is larger or smaller. You win if you guess correctly, and   lose otherwise. Is there anyway that you can win the game with more   than 50% chances no matter what the two numbers are?
119731
I am mainly confused about my reasoning/solution  Problem: I have a bag of numbers ranging from $1,2... n$. I randomly choose a number and then I throw out all numbers less than that. Then I keep choosing numbers until I've found the largest one. What's the expected number of things I have chosen?  My Argument: I expect on average to have chosen a number of $n/2$ on each pick. Hence, everytime I choose a number, I expect to discard half of them. Hence, the $E(X) = \log_2(n)$  Note that I devised this problem as an equivalent one while trying to solve the "cluster of cars on one long road" question. If this is not equivalent, please let me know but I am confident that it is. As a result of this equivalency, I know my solution is incorrect. However, I cannot pinpoint where the issue arises.
50454
Several people are playing a game with random events and require a way to produce a random number. (Such as dice rolls or a lottery.)  Can this be done such that each player has the power to be reasonably sure that the random number was fairly selected, without having to trust anyone else?  This question stems from a discussion in comments to my answer on the . I came up with a protocol but someone else spotted a flaw. This site seems more appropriate. Please feel free to make use of my answer to that question.
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?
8069
I was watching the movie 21 yesterday, and in the first 15 minutes or so the main character is in a classroom, being asked a "trick" question (in the sense that the teacher believes that he'll get the wrong answer) which revolves around theoretical probability.  The question goes a little something like this (I'm paraphrasing, but the numbers are all exact):  You're on a game show, and you're given three doors. Behind one of the doors is a brand new car, behind the other two are donkeys. With each door you have a $1/3$ chance of winning. Which door would you pick?  The character picks A, as the odds are all equally in his favor.  The teacher then opens door C, revealing a donkey to be behind there, and asks him if he would like to change his choice. At this point he also explains that most people change their choices out of fear; paranoia; emotion and such.  The character does change his answer to B, but because (according to the movie), the odds are now in favor of door B with a $1/3$ chance of winning if door A is picked and $2/3$ if door B is picked.  What I don't understand is how removing the final door increases the odds of winning if door B is picked only. Surely the split should be 50/50 now, as removal of the final door tells you nothing about the first two?  I assume that I'm wrong; as I'd really like to think that they wouldn't make a movie that's so mathematically incorrect, but I just can't seem to understand why this is the case.  So, if anyone could tell me whether I'm right; or if not explain why, I would be extremely grateful.
36924
Is there a reasonable statistic test one can do to standard random number generators (say, one of those that come built in in Python libs) which shows they are not really random?  (By reasonable I mean to exclude silly things like «generate a looong list and see that it is periodic»)  Later: Notice I am looking for a test that fails on standard generators.
76514
folks,  i am new to this forum and not a math expert. so please bear with me if am asking silly questions.   The question is "probability of getting 50 heads from tossing a coin 100 times".  So the answer for this is, I guess, ${100 \choose 50} (2 ^{-100})$.  So all am trying to get is easier way to calculate ${100 \choose 50}$, or another approach to the parent problem only.  Thanks all, appreciate that.  raj