Probability - Consecutive Numbers?

1,2,3,4,5,6 has the same probability as any other *specific* (individual) combo.

Likewise, any specific consecutive combo has the same probability as any individual combo.

But certainly there are more non-consecutive combos than consecutive combos, so the probability of getting "any non-specific non-consecutive combo" is greater than the probability of getting "any non-specific consecutive combo".

It's obvious once you think about it. At first it can be deceptive because we humans see significance in 1,2,3,4,5,6 but not in 3,18,7,40,35,15. We call the 2nd one "random" when really the 1st one is just as random. If we were some alien species, maybe we'd think the 1st one looked random and the 2nd one looked orderly.

If you start watching the drawings hoping to see 3,18,7,40,35,15 you'll notice that you're never seeing it. That's because it's as rare as 1,2,3,4,5,6. It's not rare to see a non-consecutive number, but it's rare to see a specific one that you're paying attention for.

1, 2, 3, 4, 5, 6 is just one of millions of combinations. 3, 15, 28, 41, 47, 49 is also just one of many combinations.

If the numbers can repeat, I don't know if they do, then the number of combinations is 53^6.

P( 1 , 2 , 3 , 4 , 5 , 6 ) = 1 / 53^6

P( 3 , 15 , 28, 41 , 47, 49 ) = 1 / 53^6

P( 1 , 2 , 8 , 16 , 27 , 53 ) = 1 / 53^6

The point is each combination has once chance of occuring, and therefore and equal chance.

PSR2

There are 53C6 ways of picking 6 numbers. The probability of ANY 6 specific number being choice is 1 in 53C6, whether they are in sequence or not. That's 1/ 22,957,480.

How many ways can you have 6 consecutive numbers in a row? They can start with 1, 2, ... 48 (48 49 50 51 52 53 is the highest run of 6).

So the probability of some 6 in a row is

48/ 22,957,480 = 1/ 478,280.83