I have been studying an interesting pattern of numbers
recently.
It is related to a comment I have heard said repeatedly by
statisticians, that seems like it should not be true.
The comment is: “If you play the lottery why not just pick
the number 1,2,3,4,5 or 2,3,4,5,6? They are just as likely to show up as any
other number.”
However, if they are just as likely to show up why do they
so seldom appear in the random number generator that is the lottery?
Let us see if we can determine why this is the case.
But first we must create some narrow definitions, and a
formula.
The above sequence fits the definition of five consecutive
numbers equally incremented by some number. In this case the number one.
When
defining the odds for winning the lottery the total number of possibilities is
referenced. Your odds of winning are 1/(N choose K). The number you have chosen
must match the 1 set of numbers that come out of the drawing based on the selection
of K numbers form N possibilities.
Do equally incremented sequential numbers appear to behave differently?
To be able to count how many equally incremented sequential number sets of length K could be
selected from a set of size N the mathematical notation becomes:
N-S(K-1)
In human readable terms this yields the total number of
equally incremented sets of K items from the set size N, with a sequential
increment of S.
This expression gives the results it does based on the following:
By the nature of consecutive numbers the maximum first
number in the set produced will be the product of the increment and the length
of the selection set minus one.
Since these are consecutive numbers this maximum first
number is also the total number of all equally incremented consecutive numbers
that can be drawn from the set of size N.
Let us do a small demonstration.
Setting N to 8, and K to 3
the total number of selections we could get out of this combination is 8 choose
3 which is 56. To get the total number of possible equally incremented sequential numbers we must increment count the number numbers produced by each increment of S
The colors represent the counts.
Purple:
8-1(3-1) = 6 {(1,2,3)(2,3,4)(3,4,5)(4,5,6)(5,6,7)(6,7,8)}
Green
: 8-2(3-1) = 4
{(1,3,5)(2,4,6)(3,5,7)(4,6,8)}
Blue : 8-3(3-1) = 2 {(1,4,7)(2,5,8)}
The
summation is:
The sum of these individual calculations is 12.
Now that we have this number the question we want to ask is:
What is the probability that 3 numbers chosen from the set of 8 will be equally
incremented sequential numbers ?
The probability is 12/56.
What is the probability that 3 numbers chosen from the set
of 8 will not be equally
incremented sequential numbers ?
44/56.
Reducing the results, we have 3/14, and 11/14 respectively.
There are 11 to 3 odds against choosing three equally
incremented sequential numbers from a set of 8 numbers.
This is one way of determining the odds of equally incremented sequential numbered sets of numbers coming out of a random number generator
like a lottery drawing.
There are a few other use cases for this formula related to path length calculations for graphs that I will continue to research. More to come on this interesting formula.
