TITLE: Estimate total song ('coupon') number by number of repeats
QUESTION [6 upvotes]: If shuffle-playing playlist ×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 <- 50 # k number of songs on the disk indexed 1:k
n <- 100 # n number of random song selections
m <- 20 # m number of repeat experiments
colnum <- 10
mat <- matrix(data=NA,nrow=m,ncol=colnum)
df <- as.data.frame(mat)
for(i in 1:m){
  played <- 1+floor(k*runif(n)) # actual song indices (1:k) selected 
  freq <- sapply(1:k,function(x){sum(played==x)})
  # = number of times song with index x is being played
  histo <- sapply(1:colnum,function(x){sum(freq==x)});
  for(j in 1:colnum){
   df[i,j] <- 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 <- 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 <- df.lambda$lambda[ii]
      ssq <- 0
      for(i in 1:20){
        for(j in 1:10){
          ssq <- ssq + (df[i,j] - n*dpois(j,l))^2
        }
      }
      print(ssq)
      df.lambda$lambda[ii] <- l
  df.lambda$ssq[ii] <- ssq
    }
    df.lambda
    lambda.est <- df.lambda$lambda[which.min(df.lambda$ssq)]
lambda.est # 2.4
points(x <- 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)?!

REPLY [2 votes]: Using the notation of leonbloy, also let $N_1=10$ be the number of songs that were played once. A simple Good-Turing estimator for $M$ is then given by $$ \hat M = {{P} \over {1-{N_1 \over N} }}= {{40} \over {1-{10 \over 100}}}=44.4,$$ which agrees nicely with the maximum likelihood approach.