Consecutives

Reply #9

cps10 — Thu, 04 May 2006 14:28:54 GMT

Rick

That is very interesting. Even one adjacent pair would be 220,000+ combos, like Ricky suggested. Oh well, back to the drawing board. I was thinking that maybe playing (albeit HUGE money to do so) a consecutive every 5th drawing or so would work out to some big money.

Reply #8

Rick G — Thu, 04 May 2006 13:14:26 GMT

cps10,

In a 5/39 game the breakdown for adjacent numbers drawn is approximately:

0 adjacent pairs...53%

1 " " ....40%

2 " " .....7%

3 or 4 " " .....one half of one percent.

1-2-3-x-x would be considered 2 adjacent pairs.

Good luck

Reply #7

paurths — Thu, 04 May 2006 05:40:23 GMT

You did 50000 by hand???

lol

anyway, there are alot more than 7700 combo's,

i have excluded these 1-2-3-8-9 (where the numbers form all consecutive-groups), 1-2-3-4-9 and 1-2-3-4-5

valid: 1-2-3-8-10, 1-2-4-5-8

Result: 221408 combo's

surely i must have done something wrong...

Reply #6

cps10 — Wed, 03 May 2006 19:51:06 GMT

John

Thank you for your assistance. Actually when I was doing the workout last night by hand (I got tired after 50,000 combos), I found it much more than 7700, because you could still have this situation:

1-2-4-5-6

or

1-2-5-18-19

There could be multiple consecutives.

Now, are you saying that if there was only 1 consecutive in the string, that there would be 7,770 possibilities, as in the 1-2 would be the only consecutive, yielding a

1-2-4-13-25 for example

or

1-10-14... [ More ]

Reply #5

johnph77 — Wed, 03 May 2006 18:58:15 GMT

In a 5/39 matrix there are 575,757 possibilities. Picking two numbers to repeat - ANY two numbers - will yield a matrix which is the same as a 3/37 lottery, giving you a total of 7,770 possibilities.

gl

j

Reply #4

cps10 — Tue, 02 May 2006 20:52:19 GMT

Thanks time...yeah, that is a little above my knowledge, but you have it right. Although the 01-02-03-04 would not be a possibility based on the rules that I set.

But you could have a 01-02-03-07-08

Reply #3

time*treat — Tue, 02 May 2006 20:44:03 GMT

Well, it would be more like adding a series of decreasing numbers (a summation equation)

As far as brute force, your PC can do it in a few mins and will let you decide if you want to consider such consecutives as 1-2-8-14-15; two deltas of one. So you would have your 5 main numbers, and your 4 delta numbers, and your code examines the deltas, and counts and/or prints what you want.

Once you have the code built, the cost of exploring other possibilities drops considerably.

numbersdelt... [ More ]

Reply #2

cps10 — Tue, 02 May 2006 19:24:56 GMT

Thanks time*treat

You are right...it could be brute force, which will stink. However, I think that if you plug together the first set of criteria, you could multiply by the number of two number consecutive combinations, right

Reply #1

time*treat — Tue, 02 May 2006 08:51:18 GMT

I don't know that there is a simple single formula that you could use to find this. It may have to be done by brute force, or in pieces i.e. how many cases of 1-2-a-b-c, a-2-3-b-c and so on. In either case, for an X choose 5 game, the final answer will depend on X . If you already have a method for creating all combinations, then brute force counting is the simpler way to go.

Consecutives

cps10 — Tue, 02 May 2006 00:20:28 GMT

I am trying to find out the number of combinations in a Pick-5 game if you have a consecutive, and no more than 3 consecutives in a row.

In other words, you could have the following:

1-2-8-14-30 OR

1-2-3-14-30

But, you could never have something like this:

1-2-3-4-30

Does this make sense?

Can anyone tell me the formula for something like this for the number of combinations possible given the parameters above for any given number of balls?

For instance, SOUTH CAROLINA has