The primitive lottery

The primitive lottery

In the game of the primitive lottery we must choose 6 numbers between 1 and 49. My brother says that in the majority of draws at least two numbers appear that are consecutive and I, just to keep the opposite, tell him that they usually do not appear Two consecutive numbers. Since we are very competitive, we decided to play a euro to see who is right.

Who has the reason?


The simplest solution is to calculate the number of combinations in which there are no consecutive numbers and subtract it after the total.

Consider the following two sets:

C1 = Combinations of 6 numbers from 1 to 49 such that there are no two consecutive between them.

C2 = Any combinations of numbers from 1 to 44.

Both sets have the same number of elements, since there is between them the following biunivocal correspondence: (a, b, c, d, e, f) (a, b-1, c-2, d-3, e-4, f-5)

where a, b, c, d, e, f are numbers between 1 and 49 such that there are not two consecutive between them. For example, the combination (1,5,7,20,35,49) of C1 would correspond to that (1,4,5,17,31,44) of C2. The number of C2 is the combinational of 44 over 6 = 7,059,052. Since C1 and C2 have the same number of elements Card (C1) = Card (C2) = 7.059.052.

The total number of combinations of the Primitive Lottery is the combinational of 49 over 6 = 13,983,816.

Then, the number of such combinations in which there are no two consecutive elements is the difference: 13,983,816 - 7,059,052 = 6,924,764

The probability that any two consecutive numbers will appear in a draw is therefore 49.52%, so I was right and In most draws, two consecutive numbers do not appear.
(although for very little)