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 my knowledge of probabilities goes only as far as joint distribution and conditional probability.

  • 2
    $\begingroup$ Lotteries are generally done such that each individual number's chance of being drawn is as equally likely as that of any other number's chance. As such, each lotto ticket is equally likely to win as any other lotto ticket (regardless of what numbers are actually on the ticket, assuming the number selection was valid). For intuitive proof: the machine that selects the balls for numbers cannot read and has no way of knowing what number is on each ball. $\endgroup$
    – JMoravitz
    Mar 4, 2015 at 4:33
  • 1
    $\begingroup$ Biff’s odds of matching the drawn numbers is the same as George’s, but if the jackpot prize for selecting all the drawn numbers is shared, Biff may have a better chance than George of not sharing the prize, if he wins, and perhaps more so if he chooses consecutive numbers unlikely to be lucky number or birthdates. Under the shared jackpot scenario, a ticket containing numbers less likely to have been chosen by other players has a greater expected return. Which tickets fall into this category is hard to say, but perhaps consecutive numbers in the 30’s or 40’s would. $\endgroup$
    – Steve Kass
    Mar 4, 2015 at 5:25

1 Answer 1


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).


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