Nick Koutras
Member
I usually do not post such strategies
but I'm willing to share this one since
it does not affect the pay-out of the game and now
most of you do have LottoSelectorXL.
I sincerely admit that this strategy has make allot
of money for me.
It applies to Keno games in Ont/Que and using LottoSelectorXL
Step1. Use LottoSelectorXL and the 3-Bins method
with the parameters for Next Draw
but set the slider to 1 <-- very critical
The other parameters do not matter.
Step2. Mark the First 6 Numbers of Row 6:6 and Last 6 numbers of the
same row.
Now you have the two Groups of Numbers to use.
Now apply the following Lotto Design the each of the two groups
1 6
1 3
5 6
2 4
3 5
2 4
The above is the design LD(6,2,2,4,L=2)=6
So play every combination of Group1 x Group2=6x6=36 tickets of 4 numbers
If from each Group we have 4 correct we will have 4 out of 4 correct tickets
for a payoff of 400$
This system will be a loser if within 11 trials our hypotheses
will be wrong ie: 4 out of 6 correct on both of our groups.
A more adventurer player can use the next system,
for 64$ but for payoff of 900$:
This system has a break even point of 900/64=14 trials.
5 6
2 5
2 6
3 4
1 2
1 5
3 4
1 6
Which is the same design but it gives us 3x2 correct
for a total of 9 correct prices minimum!
I have to emphasize the minimum here.
To prove me wrong do some historical testing of the above
and follow the number of hits of the first 6 and last 6 numbers of Row 6:6
but I'm willing to share this one since
it does not affect the pay-out of the game and now
most of you do have LottoSelectorXL.
I sincerely admit that this strategy has make allot
of money for me.
It applies to Keno games in Ont/Que and using LottoSelectorXL
Step1. Use LottoSelectorXL and the 3-Bins method
with the parameters for Next Draw
but set the slider to 1 <-- very critical
The other parameters do not matter.
Step2. Mark the First 6 Numbers of Row 6:6 and Last 6 numbers of the
same row.
Now you have the two Groups of Numbers to use.
Now apply the following Lotto Design the each of the two groups
1 6
1 3
5 6
2 4
3 5
2 4
The above is the design LD(6,2,2,4,L=2)=6
So play every combination of Group1 x Group2=6x6=36 tickets of 4 numbers
If from each Group we have 4 correct we will have 4 out of 4 correct tickets
for a payoff of 400$
This system will be a loser if within 11 trials our hypotheses
will be wrong ie: 4 out of 6 correct on both of our groups.
A more adventurer player can use the next system,
for 64$ but for payoff of 900$:
This system has a break even point of 900/64=14 trials.
5 6
2 5
2 6
3 4
1 2
1 5
3 4
1 6
Which is the same design but it gives us 3x2 correct
for a total of 9 correct prices minimum!
I have to emphasize the minimum here.
To prove me wrong do some historical testing of the above
and follow the number of hits of the first 6 and last 6 numbers of Row 6:6

