Why are the Binomial and Poisson probabilities different for frequency of a successful dice roll?

Question

Why do I get different results using a Binomial vs Poisson process for calculating the frequency of a successful roll of a 6 sided die? Should I get the same answer calculating the same scenario with both ways?

1. In scenario #1, a die is rolled 6 times per 1 minute. Rolling a "4" is considered a success. I want to calculate the probability of getting exactly one success (rolling a "4") in 1 minute. Using the Binomial formula, since there's six rolls in one minute, I calculated the probability of exactly one success in six rolls as...

$$\binom{6}{1}\left ( \frac{1}{6} \right )^{1}\left (\frac{5}{6} \right )^{5}=.402$$

But using Poisson to find the probability of one success, while rolling a die six times in one minute... $$\ \ \ P(roll\ a\ "4")\ =\frac{1}{6}\ \ \rightarrow\ \ np=6\times\frac{1}{6}=1\ \ \ \ \rightarrow\ \ \ \ \lambda=1\ \ \ \ \rightarrow\ \ \ \ \frac{e^{-1}{1}^1}{1!}=.368$$

2. In scenario #2, the frequency is increased to 60 rolls per 1 minute period. I want to calculate to probability of getting 10 successes in a 1 minute period. So with 60 descrete rolls, getting exactly 10 successes...

$$\binom{60}{10}\left ( \frac{1}{6} \right )^{10}\left (\frac{5}{6} \right )^{50}=.137$$

But using Poisson to find the probability of 10 successes, while rolling a die 60 times in one minute... $$\ \ \ P(roll\ a\ "4")\ =\frac{1}{6}\ \ \rightarrow\ \ np=60\times\frac{1}{6}=10\ \ \ \ \rightarrow\ \ \ \ \lambda=10\ \ \ \ \rightarrow\ \ \ \ \frac{e^{-10}{10}^{10}}{10!}=.125$$

So in scenario #2,the probabilities are in closer agreement. As I understand it, the Binomial distribution converges to Poisson as n gets very large and p gets very small. So is that what is happening here, or should both methods always agree, and I'm just calculating incorrectly here?

Answer 1

Poisson Distribution as a Limit of Binomial Distribution

The Poisson and binomial distribution connection is related to the limits

$$ \lim_{n\to\infty}\binom{n}{k}\cdot\left(\frac{\lambda}{n}\right)^{k} = \frac{\lambda^{k}}{k!} $$

and

$$ \lim_{n\to\infty}\left(1-\frac{\lambda}{n}\right)^{n-k}=e^{-\lambda} $$

Combining the two, we have

$$ \lim_{n\to\infty}\binom{n}{k}\cdot\left(\frac{\lambda}{n}\right)^{k}\left(1-\frac{\lambda}{n}\right)^{n-k} = \frac{\lambda^{k}e^{-\lambda}}{k!} $$

Notice that the LHS is the probability mass function of a binomial distribution while the RHS is the probability mass function of a Poisson distribution.

Remarks About The Scenarios

As we have seen from above, the binomial distribution converges to the Poisson distribution when $n\to\infty$, in other words, when the number of experiments becomes so large (while the expected value $\lambda$ remains finite). That's why the two are closer in the second scenario. Try to compute for larger and larger number of rolls to explore further.