Is it worthwhile to find gats with high 1-hits

[Sooz]

Just wondering. This may seem like a silly question. But you could ask WG to pick just one number from 6 different tables.
Another way to ask similar question: Probability-wise, is it better to pick the best 3-hitters from 2 gat tables of 6 selected numbers each? OR to pick the best 2-hitters from 3 gat tables of 6 selected numbers each? OR to pick the best 1-hitters from 6 gat tables of 6 selected numbers. (As you may know, I've been using 2 gat good-3-hitting tables of 6 numbers each and asking WG for 20 blocks.)
I suppose it depends entirely on what expected percentage for each. Well, the average good 3-hitter table has about 14 3-hits per 100 draws. The average good 2-hitter table has - I didn't keep track of the results but I think it was aroun 21%. The average good 1-hitter table has about 85 1-hits (augmentative) per 100 draws.
Can you help me figure out how to calculate which method has best probability?

Sooz

[lottoarchitect]

Code: Select all

Total numbers 49, Drawn numbers 6, Selected numbers 6
Probabilities to match correct numbers.
Correct | Probability % | once in N draws  
      0 | 43,5964975512 |            2,29
      1 | 41,3019450485 |            2,42
      2 | 13,2378029002 |            7,55
      3 |  1,7650403867 |           56,66
      4 |  0,0968619724 |         1032,40
      5 |  0,0018449900 |        54200,84
      6 |  0,0000071511 |     13983816,00

Natural probability suggests
- best 3-hitters on 2 GAT tables : 1.765% ^ 2 = 0.031%
- best 2-hitters on 3 GAT tables : 13,238% ^ 3 = 0.232%
- best 1-hitter on 6 GAT tables : 41,302% ^ 6 = 0.496%

However, picking 6 numbers on 2 GATs produces 12 numbers total, 6 numbers on 3 GATs GATs produces 18 numbers and 6 numbers on 6 GATs produces 36 numbers. Obviously these can't get compared like that.

The correct estimation would be:
3-hitters on 2 GATs of 6 numbers each (12 picked numbers)
2-hitters on 3 GATs of 4 numbers each (12 picked numbers)
1-hitter on 6 GATs of 2 numbers each (12 picked numbers)

Using GAT's odd calculator

Code: Select all

Correct | Probability % | once in N draws  
      3 |  1,7650403867 |           56,66 for a 3-hitter in one GAT of 6 numbers
      2 |  6,3928901810 |           15,64 for a 2-hitter in one GAT of 4 numbers
      1 | 21,9387755102 |            4,56 for a 1-hitter in one GAT of 2 numbers

Thus
- best 3-hitters on 2 GAT tables : 1.765% ^ 2 = 0.031%
- best 2-hitters on 3 GAT tables : 6,393% ^ 3 = 0.026%
- best 1-hitter on 6 GAT tables : 21,939% ^ 6 = 0.011%

Based on natural probability, your best approach is to pick 3-hitters from 2 different GATs.
Since GATs performance does not corellate to natural probability (it is a predictor after all), your only means of computing these probabilities is to use the indicated hit ratio at the multiplications needed to get the overall hit ratio.

[Sooz]

Thanks Anastasios - I'm trying to digest this. So for the best 3-hitters on 2 GAT tables - if GAT shows a Ratio % of 12%, then should I use 12 instead of 1.7650? Or I could use (improvement ratio * 1.7650) which in this case is (6.798711 * 1.7650%) which also equals 12%.
Thus best 3-hitters on 2 GAT tables: 12% ^ 2 = 1.44% Is that right?

That never occurred to me - of course! for it to be a proper comparison, the number of picked numbers must stay constant at 12. So in order to compare, I'll now have to run GAT again requesting best 2-hitters from only 4 selected numbers, and also requesting best 1-hitters from only 2 numbers. This will show the proper ratio percent to plunk into the equation.

Sooz

[lottoarchitect]

Both are equivalent really. The Imp. ratio shown by GAT is actually the Ratio % achieved by GAT / the % of natural probability. For simplicity, just use the ratio %.

Thus best 3-hitters on 2 GAT tables: 12% ^ 2 = 1.44% Is that right?

Right, provided the selected GATs, both will continue their illustrated performance on 3-hitters. Since we never have a GAT do exactly the same, treat the outcome as a rough estimation of what to expect. Also, these equations compute the chance to have all the x-hitters show up at the same time in your GATs.

[Sooz]

Yes, I understand. These odds are actually amazing!
Sooz