I have a dataset of 1296 unique codes which can be numbered 1 through 1296. If numbers are selected at random, one at a time, with replacement. On average, how many iterations will it take to select a number that has already been selected?  Experimentally, (looping through the list of 1296 codes and creating a subset of selected codes using Python) it averages out at 45.875 times (this number includes the duplicate) but I would like to verify it with a calculation so any help would be appreciated.  This question has some similarities but I am unable to perform a calculation based on the answer:
62763
I was wondering if someone could critique my argument here. The problem is to find the probability where exactly 2 people in a room full of 23 people share the same birthday.  My argument is that there are 23 choose 2 ways times $\displaystyle \frac{1}{365^{2}}$ for 2 people to share the same birthday. But, we also have to consider the case involving 21 people who don't share the same birthday. This is just 365 permute 21 times $\displaystyle \frac{1}{365^{21}}$. To summarize:  $$\binom{23}{2} \frac{1}{365^2} \frac{1}{365^{21}} P\binom{365}{21}$$
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How many 4 digit numbers are there which contain not more than two different digits?  As usual, I will highlight my attempt:   I can have 9 numbers using only 1 digit, e.g. $1111, 2222, \dots 9999$ = 9 digits Using two digits out of 9 (excluding 0) (selected in $\,^{8}C_{2}$ ways), I can arrange them in $4!$ ways. Furthermore, I can use one digit out of the selected 2 once or thrice (e.g. $2111$ or $2221$)  or simply both digits 2 times (e.g. $2288$, $6677$). Hence I get: $$\,^{8}C_{2} \times \left(\frac{4!}{2!\times2!} + \frac{4!}{3! \times 1!}\times 2\right) = 392$$ Finally, I consider numbers with zeroes. With zero selected as a digit, I have 9 options for other digits. And using the same logic as in 2) I can have either 1 zero ($8880$), 2 zeroes (e.g. $8800$) or 3 zeroes in a number ($7000$). This gives me $$9\times \left( \frac{4!}{2!\times2!} + \frac{4!}{3!\times1!} + 1\right) = 99$$   the above 3 steps yield a total number of $9+392+99=500$  However, the answer i have with me is $576$.
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If each suit is represented then we will have two cards from one suit and one card each from the remaining suits. So I am counting the ways this can happen like so -  ${52 \choose 1}$ ways of selecting the first card.  ${12 \choose 1}$ ways of selecting a card from the same suit as the first.  ${13 \choose 1}$ ways of selecting a card for each of the 3 remaining suits.  So I have -  $${52 \choose 1}{12 \choose 1}{13 \choose 1}{13 \choose 1}{13 \choose 1}$$  Then I have the sample space as $52 P 5$ as we are counting with regard to order on the numerator so we need to do the same for the denominator.  So I have $$\frac{{52 \choose 1}{12 \choose 1}{13 \choose 1}{13 \choose 1}{13 \choose 1}}{52P5}$$  Does that look correct?
59325
The answer: 6, 670, 903, 752, 021, 072, 936, 960, according to this site:  I have tried to get this number using direct methods but basically I have found the question too hard. There are too many possibilities, but I am likely missing some strategic ways to solve the problem. Any help would be greatly appreciated, thanks.
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If shuffle-playing playlist &times;100 resulted in [10 13 10  3  2  2] different songs being repeated [1 2 3 4 5 6] times, what is the estimate for the total number of songs? (assuming shuffle play was completely random)  Update: (R code)  k &lt;- 50 # k number of songs on the disk indexed 1:k n &lt;- 100 # n number of random song selections m &lt;- 20 # m number of repeat experiments colnum &lt;- 10 mat &lt;- matrix(data=NA,nrow=m,ncol=colnum) df &lt;- as.data.frame(mat) for(i in 1:m){   played &lt;- 1+floor(k*runif(n)) # actual song indices (1:k) selected    freq &lt;- sapply(1:k,function(x){sum(played==x)})   # = number of times song with index x is being played   histo &lt;- sapply(1:colnum,function(x){sum(freq==x)});   for(j in 1:colnum){    df[i,j] &lt;- histo[j]   } } df   Resulting in: e.g. 20 distributions (V1=number of single plays, V2=number of double plays, etc):     V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 1  15 13 11  1  2  2  0  0  0   0 2  15 12 10  1  4  0  1  0  0   0 3  12 14  7  6  3  0  0  0  0   0 4  17 16  6  4  2  0  1  0  0   0 5  17 10 12  5  0  0  1  0  0   0 6  13 15 11  6  0  0  0  0  0   0 7  10 14  9  3  2  1  1  0  0   0 8  12 17  5  6  3  0  0  0  0   0 9   9 19  8  3  1  2  0  0  0   0 10 13  9 11  6  1  0  1  0  0   0 11 16  9 12  5  2  0  0  0  0   0 12 15  9 11  6  2  0  0  0  0   0 13 19  9  7  4  4  1  0  0  0   0 14 17 11  4  7  3  1  0  0  0   0 15 11 20  8  1  3  1  0  0  0   0 16 14 12 10  5  0  2  0  0  0   0 17  9 12  8  7  3  0  0  0  0   0 18 10 15  9  4  2  0  1  0  0   0 19 14 11 12  7  0  0  0  0  0   0 20 16 14 11  3  1  1  0  0  0   0   Now I need to get from here to the Poisson modelling--my R is a bit rusty (?lmer)...--Any help would be appreciated...  Attempted Poisson modelling: disappointing fit?!  plot(1:colnum,df[1,1:colnum],ylim=c(0,30),    type="l",xlab="repeats",ylab="count") for(i in 1:m){   clr &lt;- rainbow(m)[i]   lines(1:colnum,df[i,1:colnum],type="l",col=clr)   points(1:colnum,df[i,1:colnum],col=clr) }  df.lambda=data.frame(lambda=seq(1,5,0.1),ssq=c(NA));df.lambda for(ii in 1:dim(df.lambda)[1]){   l &lt;- df.lambda$lambda[ii]       ssq &lt;- 0       for(i in 1:20){         for(j in 1:10){           ssq &lt;- ssq + (df[i,j] - n*dpois(j,l))^2         }       }       print(ssq)       df.lambda$lambda[ii] &lt;- l   df.lambda$ssq[ii] &lt;- ssq     }     df.lambda     lambda.est &lt;- df.lambda$lambda[which.min(df.lambda$ssq)] lambda.est # 2.4 points(x &lt;- 1:10, n*dpois(1:10,lambda.est),type="l",lwd=2)  100*dpois(1:10,3) n/lambda.est   the estimated lambda stays around 2.3, with an n estimate of around 43; the fitted curve seems very discrepant, and seems to worsen with rising n !?  Doesn't this have to do with the fact, that our repeats are different from the 'classical' Poisson distributions: it's not just ONE event that repeats itself x number of times, but the sum of repeats of different items (songs)?!
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I have the following problem: A class has 20 students, 16 females, 4 males. Find all possibilities of choosing 6 students so that at least one is male.  I did the following, pick one male from the 4, you have 4 possibilities, then you are left with 19 students, pick 5, so I get 4*${19}\choose{5}$ possibilities.  However, one can notice that if you want no one to be male, then the possibilities are ${16}\choose{6}$, so, one could argue that the solution is instead ${20}\choose{6}$ - ${16}\choose{6}$.  So, which is solution is correct, and why?  $$4* {{19}\choose{5}}$$  or $${{20}\choose{6}} - {{16}\choose{6}}$$
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I am just looking to understand how mtry works in random forests.  Please correct me if I am wrong.  When you specify mtry (say 10), it takes 10 random variables from your data set and examines them for one tree.  So the next tree would take 10 more random variables, examine them, so on and so forth until it runs through the ntrees that you specify and then returns the average estimates for the best/most important variables?
6251
Choose a random number between $0$ and $1$ and record its value. Do this again and add the second number to the first number. Keep doing this until the sum of the numbers exceeds $1$. What's the expected value of the number of random numbers needed to accomplish this?
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?
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For example:  &gt;&gt;&gt; x = [1, 1, 2, 'a', 'a', 3] &gt;&gt;&gt; unique(x) [1, 2, 'a', 3]   Assume list elements are hashable.  Clarification: The result should keep the first duplicate in the list. For example, [1, 2, 3, 2, 3, 1] becomes [1, 2, 3].
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Recently I needed to do weighted random selection of elements from a list, both with and without replacement. While there are well known and good algorithms for unweighted selection, and some for weighted selection without replacement (such as modifications of the resevoir algorithm), I couldn't find any good algorithms for weighted selection with replacement. I also wanted to avoid the resevoir method, as I was selecting a significant fraction of the list, which is small enough to hold in memory.  Does anyone have any suggestions on the best approach in this situation? I have my own solutions, but I'm hoping to find something more efficient, simpler, or both.
74210
Here is the problem I'm trying to solve:  In order to send spam, a spammer generates fake nicknames, by picking random girl names (and appending a random number to it). I suppose it randomly and fairly picks names from a fixed list, and I get to observe the outcome. Over time, names eventually start repeating.  From the distribution of the repetition count of every distinct name, is there a way to estimate the size of the fixed list ?  Intuitively, if there was no repetition among the names I have observed, I would have no information about the size of the list, but the minimum bound given by the actual distinct names observed.  On the other hand, if the distribution of the repeat counts is far away from zero, I can be reasonably confident that the sample size is much larger than the word list, and that I probably have observed the full list already.  The problem lies in the middle zone. So far the distribution of repeat counts looks like this:  $$ \begin{matrix}       n &amp; k \\    114 &amp; 1  \\     66 &amp; 2 \\     30 &amp; 3 \\      4 &amp; 4 \\      2 &amp; 5 \\      1 &amp; 6 \end{matrix} $$  Here $n$ is the count of distinct names appearing $k$ times in the sample.  Is there a simple estimator for the size of the list?
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Duplicate:    I want an pseudo random number generator that can generate numbers with no repeats in a random order. For example: random(10) might return 5, 9, 1, 4, 2, 8, 3, 7, 6, 10 Is there a better way to do it other than making the range of numbers and shuffling them about, or checking the generated list for repeats?  Edit: Also I want it to be efficient in generating big numbers without the entire range.  Edit: I see everyone suggesting shuffle algorithms. But if I want to generate large random number (1024 byte+) then that method would take alot more memory than if I just used a regular RNG and inserted into a Set until it was a specified length, right? Is there no better mathematical algorithm for this.
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Let $X = \{1, 2, 3, . . . , 25\}$. If a student selects a function randomly from the set of all functions from $X$ onto $X$, then what is the probability that the selected function maps prime numbers to prime numbers ? No idea how to proceed, please guide.
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Let's say I have numbers each taken in a set $A$ of $n$ consecutive naturals, I ask myself : how can I found what are all the unique multisets, which could be created with $k$ elements of this set $A$?  For example I've got $A=[1,2,\dots,499]$. If I wanted to create unique multiset of 3 elements, I would search all the multisets $\{a,b,c\}$ such as $a\leq b\leq c$ and then have all the unique multisets. As such for each elements $a$ there is $499-(a-1)=500-a$ possibilities for $b$ and $500-b$ possibilities for $c$. Unfortunately I'm stumped here and can't find the number of possible combination, as I didn't do any maths for years. I know that I should have some kind of product, but I don't know how to find the product anymore.  So first, I would like to know if I am right in my original assumption, or if I am looking in the wrong direction. Second I would like to approach that as if I were doing an homework, as I would like to understand the logic behind it: What would be the formula to give the numbers of multisets of $k$ elements from a set $A$ of $n$ consecutive naturals?  P.S. I'm totally new to math.SE, as such, I'm not sure I tagged the post appropriately.
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Say there is a contest to collect a full set of $M$ coupons.  Each box of product has $N$ distinct coupons of the $M$ possible coupons in it, selected uniformly without replacement from the $M$ possible coupons, the same $N$ in number for all boxes.  I'm interested in the probability that after opening $K$ boxes of product I have at least one complete set of the $M$ coupons.  My attempt has been to take the probability of a coupon not being in a box, use that to determine the probability of that coupon not being in any of the $K$ boxes, and then... I get flummoxed, and don't think this is the correct approach.  I searched the site for coupon collector problems, there are a couple that deal with similar situations (multiple in a box), but with the box having coupons selected with replacement, so duplication in a box is possible.  Right now I'm using a CAS to get probability of sums of hypergeometric samples having all bins covered to check my ideas, but that's utterly impractical for other than tiny cases.   Is this not as trivial as I think it is, or is there a direct way to calculate this?
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A number is removed from the set of integers from $1$ to $n$. Now, the average of remaining numbers turns out to be $40.75$. Which integer was removed?  By some brute force, I got $61$. I want to know if there's any analytic approach?