Lotto numbers in a run of six

No doubt the recent results will raise some eyebrows. Apparently the winning numbers of 5,6,7,8,9 &10 were predicted by 20 people. What are the chances of a run of six falling out of 49 balls? Let’s compare answers.

ETA: I got (1)(5/48)(4/47)(3/46)(2/45)(1/43)

Rare …

Ok, a friend just disagreed with my solution, and this is why:

Say for instance the first ball to drop is number 10. Then the second ball can be either one of 5,6,7,8,9 or 11,12,13,14,15 to allow for a run of six to be possible. So the second ball, depending on its value, should contribute anywhere from a maximum factor of (10/48) and a minimum factor of (5/48). Not a constant (5/48) as I proposed above.

Stuff it, I’ll go monte carlo the answer on an idle raspberry. :-[

This :

import random random.seed(a=None, version=2) draws = 10000000 runs = 0 for draw in range(draws): draw_set = [ball for ball in range(1,50)] drop_set = [] for drop in range(6): n = random.randint(0,len(draw_set)-1) drop_set.append(draw_set.pop(n)) drop_set.sort() if drop_set[5] - drop_set[0] == 5: runs += 1 if draw % 1000000 == 0: print(f'{draw} draws checked with {runs} runs so far') print(f'{runs} draws out of {draws} were runs') print(f'The chance of a run is {runs/draws}')

Yielded this:

>>> %Run 0 draws checked with 0 runs so far 1000000 draws checked with 3 runs so far 2000000 draws checked with 7 runs so far 3000000 draws checked with 13 runs so far 4000000 draws checked with 15 runs so far 5000000 draws checked with 18 runs so far 6000000 draws checked with 24 runs so far 7000000 draws checked with 26 runs so far 8000000 draws checked with 32 runs so far 9000000 draws checked with 35 runs so far 37 draws out of 10000000 were runs The chance of a run is 3.7e-06

So looks like we can expect a run every 270 000 draws.

All the sound and fury this Lotto draw has unleashed again reveals the pervasive ignorance of randomness and statistics that encumbers the self-accorded “clever ones” in society.

Monte Carlo is fine as far as it goes, but with today’s computing power, you could brute force an exact answer because there are 10,068,347,520 different unordered sequences of six numbers drawn from 49 (10,068,347,520 = 49×48×47×46×45×44), making it a 34-bit problem (log₂10,068,347,520 = 33.229…). This is well within brute-force feasibility. Think nested loops six levels deep, and with some thought and ingenuity, you can reduce it further to 69,919,080 (why?), making it a 27-bit problem.

Except that I wouldn’t do it using Python because I don’t want to wait days for the result.

That said, I think you’re neglecting the issue of the PowerBall: The highest number of the sequence must also be drawn last, but the other five can be drawn in any order.


There is a nice way of illustrating to the number conspiracy theorist (or is it number theory conspirator?) that the numbers on the balls are only nominal, and only for purposes of identification. Ask him to imagine, if he will - which he probably won’t - that he replaces each of the 49 numbers on the balls with a picture of, say, a different breed of dog. This will still allow a perfectly viable Lotto draw, since the 49 balls are still separately identifiable. So a run of 5,6,7,8,9 and10 may now manifest as collie, chihuahua, great Dane, pekingese, Jack Russel and standard poodle. The theorist will be forced to admit that there is nothing special about the canine arrangement, and by extension, its numerical equivalent. Or am I hoping for too much?

Not sure I follow - is there something about the rules of the game I’m missing?

Yup, that’s exactly the point of the numbers on the Lotto balls, namely to distinguish them from one another, so that any set of 49 unique identifiers could be used. Numbers just happen to be compact and convenient, and easy for people to understand (well, most of them, anyway).

But I fear the committed conspiracist will simply argue that the use of numbers is a deliberate part of the plot to dupe the great unwashed. Such tends to be their unsinkable rubber duck resilience that any discordant bit of evidence or reason that challenges the supposed subversive shenanigans is deftly woven into the narrative as part of the plot’s cleverness.

I have pointed out to relatives and acquaintances who blew the “Lotto Deception!” whistle following the sequential draw that there are far more subtle and effective ways of skimming cash from our NLC (as very likely is already the case) and that such blatant manipulation as they hypothesise would be flatly stupid.

Fat lot of good that did.

As for the Lotto rules, it’s my understanding that the Bonus ball is always the one drawn last. That is, for the draw 5, 6, 7, 8, 9 with 10 as the PowerBall, the 10 must necessarily have been the last ball drawn after the other five, but those five can be drawn in any order. So, if instead the 8 had been the last ball to be drawn, the draw would have been reported as 5, 6, 7, 9, 10 with 8 as the Bonus ball.

With that in mind, I wrote some code that systematically exhausts the draw space, counting the total draws, the number of times sequential numbers are drawn (in any order), and the number of times sequential numbers are drawn where the last ball drawn is also the highest number in the sequence. The code took about 16 minutes to run on my dodgy eight-year-old laptop and produced these results:
Total draws = 10 068 347 520
Sequential numbers draws (any order) = 31 680
Sequential numbers draws (last is highest) = 5 280

The above numbers mean that on average a sequential numbers draw (any order) occurs once in every 317 814 draws (probability ≈ 0.000 003 146 5), and a sequential numbers draw where the last ball is the one with the highest number in the sequence will occur once in every 1 906 884 draws (probability ≈ 0.000 000 524 42).


;D ;D ;D

Ok, got it.

Nice work. But I’d keep such odds a secret from above mentioned family and acquaintances … it can only add fuel to the fire. ???

Incidentally, what is wrong with this solution (ignoring the PowerBall complication):

Total possible number of runs = 44 ( [1,2,3,4,5,6], [2,3,4,5,6,7], [3,4,5,6,7,8] … [44,45,46,47,48,49] )

Total possible combinations = 49!/43! = 10 068 347 520

Therefore, probability of drawing a run = 44/10 068 347 520
≈ 4.4 × 10⁻⁹ which is admittedly orders of magnitude too small.

The problem with it is that it implicitly assumes each (n, n+1, n+2n+5) set can be drawn in only one way when in fact each can be drawn in 6! = 720 different ways (pigeonhole principle) because the order in which they are drawn doesn’t matter. Now, it’s no coincidence that 44×720 = 31 680. It’s also no coincidence that 31 680 = 6×5 280.


It is when you realize that in six random numbers, the draw 1, 2, 3, 4, 5, 6 is exactly as likely as, say, 4, 6, 7, 3, 9, 1, that you realize how small your chances really are. :slight_smile:

Fantastic. Then 6! X 44 / (49!/43!) is the answer. And with no further appeals to overclocking processors at that.


I reckon this means we won’t be seeing more runs of consecutive numbers any time soon… ::slight_smile:


Or, erm, any numbers. Except maybe number of years in prison. :slight_smile: