I think this is a good idea, and I reckon it's been suggested before.
I do think that the prizing you've put up is a weird considering:
Putting 6 numbers of 0-40 is a total of 41 numbers, would lead to the mathematics being 41^6 which equals to 1 chance of 4750104241 (unless I'm totally wrong) - which obviously is never going to happen.
For the second one, 41^3 equals to 1 chance of 68921.
And for the third, that's 1 chance of 41 (logically).
Say the jackpot is $1,000,000 (1 million USD).
The chance of winning the first bet would then equal to, indeed, 1 of 4750104241. In other words, assuming you win the 4,750,104,241th time - you have wasted a total of $950,020,848,200.
The second bet then gives 1/3 of the jackpot ($333,333) - and as you win once out of 68921, you will have wasted and average of $27,568,400.
The third bet which oddly gives you 3/4 of the jackpot (being $750,000) - happens once out of 41 times. And then you have only wasted $8200.
In other words, the lottery prizes needs a little rethinking.
Please correct me if I'm wrong.
I do think that the prizing you've put up is a weird considering:
Putting 6 numbers of 0-40 is a total of 41 numbers, would lead to the mathematics being 41^6 which equals to 1 chance of 4750104241 (unless I'm totally wrong) - which obviously is never going to happen.
For the second one, 41^3 equals to 1 chance of 68921.
And for the third, that's 1 chance of 41 (logically).
Say the jackpot is $1,000,000 (1 million USD).
The chance of winning the first bet would then equal to, indeed, 1 of 4750104241. In other words, assuming you win the 4,750,104,241th time - you have wasted a total of $950,020,848,200.
The second bet then gives 1/3 of the jackpot ($333,333) - and as you win once out of 68921, you will have wasted and average of $27,568,400.
The third bet which oddly gives you 3/4 of the jackpot (being $750,000) - happens once out of 41 times. And then you have only wasted $8200.
In other words, the lottery prizes needs a little rethinking.
Please correct me if I'm wrong.