Hi,

    First of all, thanks to everyone for their comments. Especially, RJOh for submitting a suggestion.

KY Floyd

    It’s good to see your comments even if we end up agreeing to disagree. *L*  Let me start by giving you a perspective from my point of view.

1)      Because the lottery is random, the only sure fire way to win it is to bet all the combinations.

2)      Because the lottery is random, there is currently no fool proof way to predict what the next numbers picked will be.

3)      Any attempt to eliminate individual lottery numbers or combinations is an assumption. The lottery Post is a place where those assumptions are submitted, scrutinized, and measured for accuracy.

    With that out of the way, lets go back to my initial suggestion.

    Would you agree that if you select one number out of 39 different numbers that there is a 3% chance that you are either right or wrong? ( I’ll let you choose 1 different number every game for the next 100 draws and let you decide whether you where right 97% of the time or wrong 97% of the time) *L*

    The best suggestion I have for you is not to think of beginnings of this system in the usual terms. It is and will be a little different than you are used to. Be patient, weigh it on its own merits, and you might possibly have a new tool to work with.   And in the end, if your disagree and think it’s all hogwash, that’s fine too.    

    RJOh

    I really like the way your data is organized. Very cool. I’m not sure the best way to quantify the amount of combinations eliminated versus error though.

In the process to eliminate one number from 39, I have been looking at the number that has not been drawn for awhile, or, to put it another way, the number furthest out.

What I am seeing is that this number, currently the number 5 in Tennessee's 5/39 game, has gone 27 draws without showing. 

Be eliminating the number furthest out, we will be right more often than wrong.

I can also consider the number following the 5, in this case 6.  So I could concievably eliminate two numbers. 

Impressive research, RJOH.  I feel humbled by your work.

I see that you divide the 39 numbers into three groups in decending order.  Have you compiled the three groups using a different standard?  By that I mean something like taking the 13 most frequently drawn numbers over the life of the drawing to be in the "A" group, and the 13 least drawn numbers over the life of the crawing to be in the "C" group, and allowing the remaining numbers fall into the "B" group?

@JKING: I've been working on this kill-one-number angle for a while. I think it has merit.

@GASMETERGUY: Farthest out is a good place to start.

I knock out five numbers from the grid 80% of the time that will not be drawn when I play my choices on the one line I play when I do play.

What state would anyone like 5 numbers eliminated in?

I could get real lucky though and hit five of five.

Would you agree that if you select one number out of 39 different numbers that there is a 3% chance that you are either right or wrong?

No. If only one number out of 39 was drawn then there would be a 1 in 39 chance of of correctly guessing that number. Since there are 5 numbers that are drawn there is a 5 in 39 chance. When 5 of 39 numbers are drawn then 5/39ths of the numbers are drawn. 5/39ths is 12.82%, so there is a 12.82% chance that picking one number randomly will result in picking one of the 5 numbers that are drawn. The flip side is that if you decide that a particular number won't be drawn the chance that you will be correct is 87.18%, not 97%. You aren't eliminating 12.82% of the combinations with a 3% chance of choosing incorrectly. You're eliminating 12.82% of the combinations with a 12.82% chance of choosing incorrectly, so it's a wash.

As BobP notes, you can increase your chances of winning if a certain result occurs in exchange for an reduced chance of winning if that result doesn't occur. For a simple analogy imagine playing pick 3 with only even numbers. If the winning number is even you have a 1 in 500 chance of winning instread of 1 in 1000. Since there is only a 50% chance that the winning number will be even your overall chance of winning is .50 * 1 in 500, which is 1 in 1000. There is no net gain.

I did write a routine that calculates the average and median rate of occurances of the numbers in the winning combinations and compare that to the number of time they hit in a most recent period and color them accordinly, 0=blue, 2=red, 3 or more=yellow and everything else=green.  When ever all of the winning numbers are one color and I calculate all the numbers that could be in that group, they're 80% of the total number pool so there's no real advantage.  I keep checking hoping that one time there will be a group small enough that I could cover a winning combination in ten or twenty lines.

Cough !!! You might want to take a second look at playing with only
five even digits and how many straight numbers one can make with
any five digits.  When the winning number is even and you are playing
an even number you have a 1 in 65 chance of winning $500.

1: 000
2: 002
3: 004
4: 006
5: 008
6: 020
7: 022
8: 040
9: 044
10: 060
11: 066
12: 080
13: 088
14: 200
15: 202
16: 220
17: 222
18: 224
19: 226
20: 228
21: 242
22: 244
23: 262
24: 266
25: 282
26: 288
27: 400
28: 404
29: 422
30: 424
31: 440
32: 442
33: 444
34: 446
35: 448
36: 464
37: 466
38: 484
39: 488
40: 600
41: 606
42: 622
43: 626
44: 644
45: 646
46: 660
47: 662
48: 664
49: 666
50: 668
51: 686
52: 688
53: 800
54: 808
55: 822
56: 828
57: 844
58: 848
59: 866
60: 868
61: 880
62: 882
63: 884
64: 886
65: 888

BobP

Hi,

   Below is a listing for how many combinations there are for different amounts of lottery number pools. I believe the math to be correct, but wouldn't mind a double check. Like my initial example, you'll find 575757 combinations for a 5/39 game. If you eliminate one specific number from all the combinations you'll find 501942 combinations,  for what becomes a 5/38 game. So, if you eliminate approximately 50% of the numbers accurately, then you would be playing say a 5/20 game with 15504 combinations or a 5/19 game with 11628 combinations.

   I would have you note that this method reduces combination on a different scale than using combinations to reduce from the total amount combinations. I guess you could say that reduction methods using combinations is linear, while reduction methods using individual numbers is non-linear.

  Sounds fairly good huh. But, KY Floyd is 100% correct when he says "There is no such thing as a free lunch." If you elect to elimate one number that ends up being picked, then every subsequent combination you bet will be unable to win the 5 out of 5 prize. To make matters worse, look at Maddogs challenges. Everyone, including myself, is having a tough time getting enough numbers right for a payoff.

  The upside is, when you do select correclty there is a dramatic reduction in combinations.

  My suggestion is to use this system first. Only eliminate those numbers which you deem to have the lowest probabilty of occuring. Then use the usual combination reductions methods to obtain your final picks.

  Anyway, this completes what I want to say about the method I started this thread with.

  Are ther any others methods anyone would like to suggest?