by **sh2187** » Tue Aug 16, 2022 10:09 am

Yes it's normal to have 0.61% if we didn't apply filters which is 10,000/1,623,160.

Regarding the 100% it's not 100% of the 1,623,160 tickets, it's 100% of the filtered tickets which means we suppose that these 10,000 filtered tickets will contain at least one match 6 if 6 drawn.

But also we have 10 drawn instead of 6, so we can reduce the tickets more, I will explain in detail what I mean:

The nature of KENO style makes the cover optimization difficult, in normal lotto we can reduce 35/6 to 35 6 5 6 for example, but in KENO we need to reduce 35/6 to 35 6 6 10 which I don't know if it's possible or not.

We need to play 35 total balls, ticket has 6 numbers, match 6 required, drawn balls = 10 (i.e. 35 6 6 10)

So we first make a full wheel 35/6 = 1,623,160 tickets (because the ticket contains only 6 numbers not 10).

Then we apply filters to reduce the full wheel, let's suppose the result of filters is 10,000 filtered tickets.

We suppose that one of these 10,000 tickets will win (until now the calculations didn't consider 10 balls drawn).

Then now because it's a KENO style, the numbers of drawn balls is more than the numbers in ticket, we have 10 balls drawn, it means the 10,000 tickets can be reduced according to that because:

Mathematically:

(35 6) if no filters applied = 1,623,160 tickets

(35 6 6 10) if no filters applied = 7730 tickets (the 10 balls drawn reduced the tickets without any filter but of course it's not enough because the prize is much less than 7730)

So 1,623,160 unfiltered tickets reduced to unfiltered 7730 tickets

then 10,000 filtered tickets will be reduced to x filtered tickets

That's why I requested from EL to reduce the 10,000 tickets according to 35 6 6 10 but I need to tell EL to consider that these 10,000 contains at least one 6 match, this is the point that I think not possible in EL.

Also I am not sure if EL can work with cases that have number of drawn balls more than the numbers in the ticket.