[Math] Rolling dice and Poisson distribution

dicepoisson distributionprobability

I was doing a dice related calculation, while encountering this interesting problem.

Assuming I'm rolling 2 dice, and an outcome of 1 or 2 count as a success ($\lambda = 1/3\cdot2$). A CDF of poisson distribution at $k = 2$ gives a result of $0.9697878915060072$, but we all know it should add up to $1$ since there could be no other outcomes other than $0,1$ and $2$.

How did this happen?

p.s. If it helps, I get the CDF calculated using the function scipy.stats.poisson.cdf(2, 2/3)

Best Answer

Let $X$ be the number of Successes (1 or 2) when rolling two fair dice independently. Then $X \sim \mathsf{Binom}(n = 2, p = 1/3).$ Then $P(X = k) = {2\choose k}(1/3)^k(2/3)^{2-k},$ for $k = 0, 1, 2.$ You can make the table of this distribution in R (ignore row numbers in brackets [ ]).

k = 0:2;  PDF = dbinom(k,2,1/3)
cbind(k, PDF)
     k       PDF
[1,] 0 0.4444444   # 4/9
[2,] 1 0.4444444   # 4/9
[3,] 2 0.1111111   # 1/9

Also, $E(X) = np = 2/3 = 0(4/9) + 1(4/9) + 2(1/9).$

There are useful instances in which a Poisson distribution with the same mean as a binomial distribution gives a close approximation to the binomial distribution. But this is not such an instance.

If $Y \sim \mathsf{Pois}(\lambda=2/3),$ then the distribution table for $Y$ is shown below. Its probabilities are quite different from those in the distribution of $X$ above.

k=0:8;  PDF = round(dpois(k, 2/3), 5)
cbind(k, PDF)
      k     PDF
 [1,] 0 0.51342
 [2,] 1 0.34228
 [3,] 2 0.11409
 [4,] 3 0.02535
 [5,] 4 0.00423
 [6,] 5 0.00056
 [7,] 6 0.00006
 [8,] 7 0.00001
 [9,] 8 0.00000  # subsequent probabilities
                 # smaller than 0.000005

Along with @saulspatz, I wonder what a Poisson distribution has to do with this. Was there another question beyond computing the probability distribution of $X?$

Addendum per Comments: on Poisson approximation to Binomial distributions. Especially if $n$ is large and $p$ is small, Poisson approximations to Binomial distributions may be useful: Consider $\mathsf{Binom}(n=200, p=.02)$ with mean 4, and $\mathsf{Pois}(\lambda =4).$

x = 0:10
pdf.b = round(dbinom(x,200,.02), 4)
pdf.p = round(dpois(x, 4), 4)
cbind(x, pdf.b, pdf.p)
       x  pdf.b  pdf.p
 [1,]  0 0.0176 0.0183
 [2,]  1 0.0718 0.0733
 [3,]  2 0.1458 0.1465
 [4,]  3 0.1963 0.1954
 [5,]  4 0.1973 0.1954
 [6,]  5 0.1579 0.1563
 [7,]  6 0.1047 0.1042
 [8,]  7 0.0592 0.0595
 [9,]  8 0.0292 0.0298
[10,]  9 0.0127 0.0132
[11,] 10 0.0049 0.0053
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