Is the probability of drawing six consecutive (for example, eleven, twelve, thirteen, …) lotto numbers different than the odds of getting six randomly chosen numbers? Say Biff has a ticket with the numbers
$$7, 8, 9, 10, 11, 12$$
and George has six randomly chosen numbers
$$1, 13, 15, 29, 49, 67$$
on his ticket. Are Biff's chances of winning less than or equal to George's? I'd love to see a proof, but bear in mind that my probabilities knowledge goes only as far as joint distribution and conditional probability.
Best Answer
We would intuit that the probability of drawing a tight sequence of numbers is less than than a wide spread.
This is indeed true, but that is simply because there are more possible wide spreads than possible tight sequences.
What you have is a particular tight sequence, and a particular wide spread. In a fair draw, no particular spread of numbers is more likely to result than any other.
Lotto balls do not magically repel closer numbers more than distant numbers.
Select any array of $6$ different, valid lotto numbers by whatever method you choose, as long as it is completely independent of knowing what numbers will be drawn.
The Lotto draw then selects $6$ balls from $45$ (here; may vary by locality). Compare the first of your numbers with the results. Whatever number it is, there is a $6/45$ probability that it is on one of the balls drawn. When given that, there is a $5/44$ conditional probability that the second number is on one of the remaining numbers drawn. And so on, et cetera.
Thus there is a $\frac{6!\;39!}{45!}$ probability that all six of your numbers will be drawn, however you chose them; including by using Biff's method, or by using George's method.
$$P = \frac 1{8145060}$$
So Biff's method and George's method are both equally unlikely to win.
Edit: I will note that very few Lottos use number higher than 60, and thus George may be certain to lose (should 67 be an invalid number).