# Estimate Number of Party Guests

You are throwing a party and want to know how many guests you can expect.

You are inviting $$N$$ persons, each person has a probability $$x_i$$, $$1 \leq i \leq N$$ of showing up and $$x_i$$ is between 0 and 1. The probabilities $$x_i$$ are given. I want to write a small script (preferably in Python), to calculate a $$1-\alpha$$ - confidence interval for the number of guests to be expected ($$\alpha = 0.1$$ or $$0.05$$).

Could somebody point me to some resources for this problem? I feel like this is a very simple, introductory problem, yet, I could not find anything. Alternatively, could somebody explain (conceptually) a solution?

• In my opinion, you are looking for a prediction interval, not a confidence interval Sep 6 '20 at 20:55
• – whuber
Sep 6 '20 at 21:38
• Why do you have the "mcmc" tag? How does your question relate specifically to Markov chain Monte Carlo? Sep 7 '20 at 0:22
• @Glen_b I gave it the mcmc tag, because I thought this kind of problem would be a prime candidate for bein solved by mcmc. Evidently, I was wrong, because the posted solutions work out nicely.
– r0f1
Sep 7 '20 at 12:53

My computations will be in R (not Python), I hope you can translate. In R, functions dbinom,pbinom, and qbinom designate a binomial PDF, CDF, or quantile (inverse CDF) function. Also, with appropriate parameters, rbinom simulates binomial and Bernoulli samples.

Begin by assuming $$n=20$$ guests are invited, of whom each has probability $$p = .7$$ of attending the party. Then the binomial probability distribution of the number who attend is $$X \sim \mathsf{Binom}(n, p).$$ It's distribution (to 5 decimal places) is tabled below. (Ignore row labels in brackets [ ].)

k = 0:20; pdf = round(dbinom(k, 20, .7), 5)
cbind(k, pdf)
k     pdf
[1,]  0 0.00000
[2,]  1 0.00000
[3,]  2 0.00000
[4,]  3 0.00000
[5,]  4 0.00001
[6,]  5 0.00004
[7,]  6 0.00022
[8,]  7 0.00102
[9,]  8 0.00386
[10,]  9 0.01201
[11,] 10 0.03082
[12,] 11 0.06537
[13,] 12 0.11440
[14,] 13 0.16426
[15,] 14 0.19164
[16,] 15 0.17886
[17,] 16 0.13042
[18,] 17 0.07160
[19,] 18 0.02785
[20,] 19 0.00684
[21,] 20 0.00080


In order to get a 95% confidence interval (CI), you need to 'trim' $$2.5\%$$ of the probability from each tail of this (discrete) distribution, to leave $$95\%$$ on both sides of $$X = 14,$$ where the highest probability lies.

Roughly, the boundaries of the CI will be 10 and 18.

qbinom(c(.025,.975), 20, .7)
[1] 10 18


It seems you have a choice between intervals $$[11,18]$$ or $$[10,17].$$

diff(pbinom(c(10,18), 20, .7))
[1] 0.9444008
diff(pbinom(c(9,17), 20, .7))
[1] 0.9473721


If the $$n$$ people invited may each have a different attendance probability $$p_i, i = 1, 2, \dots n,$$ then a simulation may be helpful (as suggested by tags on your Question). Suppose the 20-vector p is defined as below:

 p = seq(.5,1, len=20); p
[1] 0.5000000 0.5263158 0.5526316 0.5789474 0.6052632
[6] 0.6315789 0.6578947 0.6842105 0.7105263 0.7368421
[11] 0.7631579 0.7894737 0.8157895 0.8421053 0.8684211
[16] 0.8947368 0.9210526 0.9473684 0.9736842 1.0000000


Then the R function rbinom(20, 1, p) simulates individual attendees at one particular party and sum(rbinom(20, 1, p)) simulates the total number attending.

set.seed(906)
a = rbinom(20, 1, p);  a
[1] 1 0 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1
x = sum(a);  x
[1] 18


Replicating this kind of count for $$100\,000$$ gives us the simulated distribution of the random number $$X$$ attending such parties. (Because of the set.seed statement, we have already seen that $$X = 18$$ attend the first of these many parties. An approximate 95% CI for the number attending is $$[12, 18].$$

set.seed(906)
p = seq(.5,1, len=20)
x = replicate(10^5,  sum(rbinom(20, 1, p)))
summary(x)
Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
7.00   14.00   15.00   15.01   16.00   20.00
quantile(x, c(.025, .975))
2.5% 97.5%
11    18
mean(x >= 12 & x <= 18)
[1] 0.95119

hist(x, prob=T, br = (6:20)+.5, col="skyblue2",
main="Simulated Attendees")


Note: If you're only interested in getting a 95% (or 99%) CI for the number $$X$$ attending, you might get $$E(X)$$ and $$V(X)$$ by considering $$X$$ as a sum $$X = \sum_{i=i}^n B_i$$ of independent Bernoulli random variables each with its own $$p_i.$$ Then assume approximate normality (note the symmetry in the summary above). Finally, make a confidence interval based on the normal distribution.

• (+1) great effort Sep 6 '20 at 21:23
• Simulation is the tedious approach, because exact calculation is fast and easy. This is explored in various other threads. See, inter alia, stats.stackexchange.com/questions/41247 .
– whuber
Sep 6 '20 at 21:41

You'll generate $$N$$ Bernoulli RVs, and count how many guests are present. Then, you'll repeat this experiment, say, $$T$$ times, and record every result. This will be your distribution. Then, pick the percentiles $$\alpha/2$$ and $$1-\alpha/2$$ to come up with a confidence interval. Theoretically, you can compare to Binomial distribution's CI when probabilities are equal.

• Ok, so this is a simulation approach. I think that the percentiles give you a prediction interval, not a confidence interval though. Sep 6 '20 at 20:58
• Yes, I did it via only simulation because the tag was mcmc. Sep 6 '20 at 21:22