Randomly place 100 balls, labeled from 1 to 100, in the 100 boxes, also labeled. What is the chance that at least one box contains the ball with matching label?
Wouldn't this be a binomial distribution since we have 1 trial of randomly placing one ball in each box and each box has a probability of (1/100) of getting a ball of the same label? If so, then the probability is:
P(x = 2) = (100 choose 1) (1/100)^1 (1-(1/100)^99 = .0037
X ~ Bin(1, 1/100) since there are 100 boxes and each boxes has a probability of 1/100 of getting the ball with the same number.
BUT, when I approximate it using the Poisson variable(lamda = 1), I get : P(i>= 1) = 1 -P(i=0) = 1 - e^-1 = .63
What am I missing here? Shouldn't the approximations be close?
This reduces to finding the probability of a given permutation having at least one fixed point (i.e it not being a derangement). This post details how to compute the probability of $k$ agreements in labels. The TL;DR of it is:
$$ p = 1 - \sum_{n=0}^{100}\frac{(-1)^n}{n!} $$
which indeed is extremely well-approximated by $1-e^{-1}$.