Lotto - Gaps - probability, HELP

From: djura (djura-12_at_gawab.com)
Date: 02/07/05 

Date: Mon, 07 Feb 2005 09:39:19 +0100

Hi all,

(btw, it's a long time I left Lotto, but
this idea is very oold one in my head ;)

example: (Lotto 6 of 36)

A B C D E F
1 4 11 12 24 33

gaps:
B-A, C-B, D-C, ...

A B C D E
3 7 1 12 9

That means, I'm not interested in exact numbers (all play).

Let's assume, statistically,
I want these gaps for my system ;)

I can arrange the gaps in
5! = 120 ways.

For 6 of 36, there are 36-(A+B+C+D+E) =
= 36 - 32 = 4 real (sub)combinations.
So, in total 4x120=480 real Lotto
combinations for full system.

Example (I'll take 1 gap combination):

1 3 7 9 12
----------------------------------------------------
1, 2, 5, 12, 21, 33
2, 3, 6, 13, 22, 34
3, 4, 7, 14, 23, 35
4, 5, 8, 15, 24, 36
---------------------------------------------------

What's for sure:

If I guess N gaps, the garantie is (N+1)
Lotto numbers.

Question:

For N=5, there is one winning comb. (6),
? (5)'s,
? (4)'s
? (3)'s

thank You.

ps. The number of gaps can be reduced
      by statistical review, let's say ;)

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