d6 Poisson encounters

The Imperial March trumpets in the background as you party moves silently through the star destroyer. You dive into cover as a group of storm troopers turns the corner and comes your way. It's the second group you've encountered in the last hour. They must be looking for something, are they aware of your presence in the ship?

The Poisson distribution named after the french Siméon Denis Poisson allows us to calculate the odds of events occurring within a set period of time, area or volume. For example on the average a group of storm troopers is encountered every half hour, how many will the party encounter in the next half hour? According to the following graph, following the orange dots we have 37% chance of having 0 encounters, 37% chance of having 1, 18% of having 2, 6% of having 1 and so on.

The cumulative distribution is as follows:

The numbers are as follows

	Prob. Mass	Cumulative
# events	lambda = 1	lambda = 1
0	36.79%	36.79%
1	36.79%	73.58%
2	18.39%	91.97%
3	6.13%	98.10%
4	1.53%	99.63%
5	0.31%	99.94%
6	0.05%	99.99%
7	0.01%	100.00%
8	0.00%	100.00%
9	0.00%	100.00%
10	0.00%	100.00%

What's really cool, and those d6 fans out there will love it, is that the 5d6 - 5d6 probability curve looks just like the Poisson distribution for lamda = 1. So you don't need to remember those odds.

Roll 5d6 subtract another 5d6, divide by 3, round, and subtract one.

2 + 2 + 2 + 4 + 6 = 16

3 + 5 + 4 + 5 + 5 = 22

Difference = 6

Divide by 3 = 2

Subtract 1 = 1 encounter in the following half hour.

2 + 2 + 3 +1 + 6 = 14

6 + 4 + 4 + 4+ 1 = 19

Difference =  5

Divide by 3 = 1 2/3, rounded = 2

Subtract 1 = 1 encounter in the following half hour.

5 + 5 + 6 + 3 + 1 = 20

1 + 2 + 2 + 3 + 4 = 12

Difference = 8

Divide by 3 = 2 2/3, rounded = 3

Subtract 1 = 2 encounters in the following half hour