Again, if you don't like my wording, I can't do anything about it. I clearly stated in my post that you CANNOT increase the overall chances. If you continue to choose
to purposely interpret my post in the wrong way, I cannot help you.
However, your third paragraph makes a point. I may not have been clear enough. For my first point, I meant you should not buy the same ticket more than once
in the SAME DRAW. Of course the chances of 4-13-21-35-43-47 has an equal chance of appearing in any draw. There's no difference in playing that every draw
than to play something different every draw. What IS different, however, is if you're buying the same ticket more than once for the same draw. Instead of increasing
your chances to win the jackpot, you're increasing the number of times you will win, if you win.
If you want to see math, then I'll show you some. Let's take a really simple lottery game, where there is only a jackpot and no other prizes, because that's what
most players are after and care about, and is what I addressed in my points. Let's say the goal is to hit 2 numbers, with replacement, from 1-10.
Now, Let's say you bought 10^2, or 100 tickets for the same draw, each one being different. You covered all the possibilities, so you have a 100% chance of winning
Now suppose you bought 2 tickets per draw for 50 draws. You're still buying 100 tickets in total, but clearly there's a chance you might not win the jackpot at all.
But to compensate, you might win the jackpot twice, or 3 times or 50 times.
Now suppose again, that you bought 100 tickets for the same draw, but you bought the SAME numbers 100 times. The odds of you winning are still 1%. It's just
that if you win, you win with all 100 tickets.
My claim was that if you disregard winning the jackpot many times, i.e treat it the same way as winning once, then buying different tickets for the same time
increases your odds of winning the jackpot.
Put mathematically, if your odds of winning the jackpot with a single ticket is 1/x, then:
Odds of winning once or more with y unique tickets for a single draw is:
Odds of winning once or more with y tickets but only y - z unique tickets for a single draw is:
(y - z)/x
Odds of winning once or more with 1 ticket for y draws would be:
(1 - 1/x)(1/x) + <--- this is because 1/x of the time, the condition has already been satisfied, so you only look at the remaining cases.
(1 - 1/x - (1 - 1/x)(1/x))(1/x) + <--- again, the condition has been satisfied by some more cases so you only look at the remaining ones.
(1 - 1/x - (1 - 1/x)(1/x)) - (1 - 1/x - (1 - 1/x)(1/x))(1/x))(1/x) +
^--- each time you have to subtract the cases that satisfies the condition from above and multiply by (1/x), which is the chance of winning in that draw
So, if you call each of these cases Cy where Cy denotes the probability of having won the jackpot for the first time at draw y, then the total probability comes to
1(1/x) + (1 - C1)(1/x) + (1 - C1 - C2)(1/x) + ... + (1 - C1 - C2 - ... - Cy-1)(1/x) = (y - (y - 1)C1 - ... - Cy-1)/x
Since Cy > 0 for all y > 0, (y - (y - 1)C1 - ... - Cy-1) < y.
Thus (y - (y - 1)C1 - ... - Cy-1)/x < y/x, as is the desired result.
Also, (y - z)/x < y/x when z > 0, i.e there are duplicate tickets.
If you have anymore concerns, feel free to ask. I'm sorry I didn't resort to name calling and insults, it's just not my style. If I realized I was wrong, as I often do,
I may be slightly embarassed but I will admit my error.
Edit: added cases for duplicate tickets