OT: OzLotto Randomness

"David L. Jerkoff"

** Forget all that probability stuff from your HSC Maths ??

OK :

Imagin the seven balls being withdrawn one at a time out of 45 numbered balls.

Ball 1 can be any number.

Ball 2 must be within a group of 15 numbers of which ball 1 is a member.

So, there are 14 candidates for ball 2 in the group.

The chance of picking one of them is 14/44 or 0.318

Similar arguement goes for each of the remaining 5 balls.

So P = 14/44 x 13/43 x 12/42 x 11/41 x 10/40 x 9/39

therefore P = 0.0004254

or 1 chance in 2350

With one draw a week, it means around 45 years until you expect to see such a close grouping.

..... Phil

No, there are up to 28 candidates - 14 either side of the first ball - except that there are fewer if the first ball is less than 15, or more than 31.

After the second ball has been chosen, the number of remaining candidates is determined by how far apart the first ball and second ball are; if the second ball is at the other end of a 15 grouping, then there are only 13 candidates remaining. On the other hand, if the second ball is adjacent to the first, then there are 26 candidates remaining - 13 either side of the pair, but again subject to end effects.

So this probability calculation is nothing like as simple as you're suggesting.

Sylvia.

** My calc was a bit simplified.

But there is a group of 15 consecutive numbers at the end of the picking and the 5 balls chosen are all members of that group.

Do it your way and tell us the answer.

Bet you have to write a program for it.

.... Phil

Which is OK if you say so, and provide an argument that shows that it represents either an upper or a lower limit on the probability. As it stands, it's just a calculation of a probability for a different problem with no indiciation of how the result relates to the question asked.

That seems the obvious approach anyway. The figure I get is exactly 15 in 7052, or about 1 in 470.

For one draw a week, that's once every 9 years.

Sylvia.

On Mon, 29 Jun 2009 13:14:40 +1000, "David L. Jones" put finger to keyboard and composed:

You first need to count the total number of ways that 7 balls can be arranged in a group of 15 where there is a ball at each endpoint.

Ball 1 - 5 balls in 13 positions - Ball 15 Balll 2 - 5 balls in 13 positions - Ball 16 .................................................... Ball 31 - 5 balls in 13 positions - Ball 45

Then you need to do the same for groups of 14, 13, 12, ... 7.

The total number of possible "clumps" is ...

(31 x 13C5) + (32 x 12C5) + (33 x 11C5) + ... + (39 x 5C5)

The total number of ways you can select 7 balls from 45 is 45C7.

So the chance of a clump of 15 or less is ...

(31 x 13! / 8! + 32 x 12! / 7! + 33 x 11! / 6! + 34 x 10! / 5! + 35 x

9! / 4! + 36 x 8! / 3! + 37 x 7! / 2! + 38 x 6! / 1! + 39 x 5! / 0!) x 42 / (45 x 44 x 43 x 42 x 41 x 40 x 39)

= 0.00212705615

= 0.2%

= 1 in 500

- Franc Zabkar

--
Please remove one \'i\' from my address when replying by email.

I wrote one. However...

There are:

39 ways in which the first and last balls extend over 7 numbers.

38 ways in which the first and last balls extend over 8 numbers. The middle 5 balls can be laid out in 6!/5!/1! different ways.

37 ways in which the first and last balls extend over 9 numbers. The middle 5 balls can be laid out in 7!/5!/2! different ways.

36 ways in which the first and last balls extend over 10 numbers. The middle 5 balls can be laid out 8!/5!/3! different ways.

And so on until

31 ways in which the first and last balls extend over 15 numbers. The middle 5 balls can be laid out 13!/5!/8! different ways.

So the total number of ways of laying out 7 balls such that they extend no more than 15 numbers is

(39 + 38 * 6!/1! + 37 * 7!/2! + 36 * 8!/3! .... + 31 * 13!/8!) / 5!

I can't see any way of simplifying it, but it comes to 96525.

The possible ways of laying out 7 balls in 45 positions is 45379620, so the odds are 1 in 45379620 / 96525, which is indeed 15 in 7052 or about

1 in 470.

Sylvia.

That's around about the figure I would have expected by gut feel. So if you take say a dozen games a week (fairly common) you'd expect to see something like that in under a year. Then there's Lotto and Powerball too.

Thanks Dave.

--
---------------------------------------------
Check out my Electronics Engineering Video Blog & Podcast:
http://www.alternatezone.com/eevblog/

Is there more than one OzLotto draw a week?

Sylvia.

Forget that, I misread your original posting.

Sylvia.

If you feel that that doesn't happen, then it suggests that using radom machine picked numbers is a mistake, and you should instead be using obviously non-random sequences.

The point being that it doesn't change your chance of winning, but does change the chance that you'll have to share the jackpot with someone else.

Sylvia.

Computer generated random numbers are NOT truly random.

I've noticed when playing a lot of online games with randomly generated loot drop, that at certain times of the day, the drop is consistently better. I suspect all games rank game drop from lowest to highest value and the underlying bias moves up and down the scale according to a random number affected by time of day. N.B. tested over 1,000 drops, so do not waste your money on Lotto (different number generation mechanism).

True, you cannot generate truly random numbers in software the pattern will always repeat eventually. If you know the period of the algorithm though, you can change the seed before anything repeats.

If you want true random, then you need to do it with hardware, the yanks used radioactive decay to produce one time tapes for the hot line, and the poms at least used to use gas discharge tubes for their premium bonds.

The chances of your numbers being consecutive are the same as them not being consecutive.

The only reason you dont recall seing it is because you probably haven't. I have spend a long time in the gaming and wagering industry and I can assure you that I have seen numbers being drawn consecutivley, from a hardware RNG and a ball cage.

Er, hardly.

The chance of any given sequence, whether consecutive or not, is the same as the chance of any other given sequence. But that's a different thing entirely.

Sylvia

what are the other forms of sequence?

--
Great advances in Debian Linux; post a bug report and get spam in three 
days.

** Anything you like - does not have to follow any rule like these do. 2,4,6,8,10,12 .... 1,3,5,7,9,11,13 ... 1,3,5,11,13,17 ....

.... Phil

To answer your question:

The "truth" is out there, but Australian lotteries commissions are tight lipped with the answers.

"Normally" the numbers for a "quick pick" are selected by a Pseudo random number generator, the seed for which is periodically updated using a clock or timer input. The most commonly used random number generator for this purpose is a type called a "Multiply With Carry" generator.

Imperfections come about when the generator does not have enough "state" and so not all combinations are possible. It should be possible to check this with the lottery commissions as if it has been implemented properly then it makes no difference if everyone knows how it was done. If it was done badly it may be possible to prove that some historically winning combinations were *never* possible to pick with the PRNG, and hence, every one who had a quick pick in those draws has a valid claim to compensation.

My experience with Queensland golden casket is that no information is forthcoming without an FOI application and even then they will obstruct the process and stall beyond what is supposed to be legal.

Actaully, the PRNG is typically determined by the regulator. In Queensland, that was Knuth, now its Mersenne-Twister. Visit the OLGR website, its all documented there somewhere.

All I said was MWC was the most used PRNG - which it is. In this regard I said nothing about specific regulators or any choice they may have made. I ruled out no other possibilities, so please don't put words in my mouth.