Re: Three + Bonus and Fours in Pool 45, Pick 6 Lotto game with 2 Bonus integers



Robert Perkis schrieb:
smanf@xxxxxx wrote:

Robert,

if you make wrong assumptions, you will get a wrong argumentation
My answers are embedded.

Regards, Manfred

Persons A and B decide to play their method for one
year and compare results. Each buys 100 tickets per
draw.

Person A plays only combinations randomly selected
from the 43,560 that contain four consecutive numbers.

Person B plays only combinations randomly selected
from the 13,940,256 that contain three or fewer
consecutive numbers.

By the end of the year there have been zero draws with
four or more consecutive numbers.

That's only sure after the year end.
Before the years end all possibilities are open.
So you make a wrong assumption.

Also I could say that at the end of the year there were
2 drawings with 4 consecutive numbers and calculate the
chance for winning the jackpot:

2 times the chance 100 / 43560 = 1 / 435,6 for winning
the jackpot. Is this bad? Also if this happens in only 5 years,
it's a good chance. Better than 100 / 13940256.
The 52 times for 52 weeks don't increase the winning chances.

It's unrealistic to say that after a year there were no drawings
with 4 consecutive numbers, also if the past showed years
where no 4 consecutive numbers were drawn

Because all weeks chances we cannot summarize, person
A has the highest chance for winning the jackpot as soon as
4 consecutive numbers will be drawn..

Example for that we cannot summarize the chances of
different drawings:
If a company exists many years and plays 13,983,816 times
for every drawing with one (same or random) ticket, there is
no guarantee to win the jackpot.

But if the company buys in one week all 13,983,816 tickets,
the jackpot is won.

Therefore lottoplayers could increase their winning chances if they
save every week the money for playing and with this money at
the end of the year they buy all tickets for only one drawing.


For person A there have been zero draws where a jackpot
was possible.

Actually at the last drawing of the year person A has the same chance
to win the jackpot like person B because it is not sure what
the next drawings numbers are.


For person B there have been 104 draws where a jackpot
was possible.

For person A there were 100 draws x 52 weeks = 5200 chances
to win the jackpot like for person B because nobody knows at the
beginning
of the year how the drawings will be.


Of course I understand that if there had been one draw
with four or more consecutive numbers A would have had
just as good a chance of winning in the single draw as
B has in 104 draws.

But because we cannot summarize the chances of different
drawings, A had the better chances over all to win
the jackpot for the assumption that 4 consecutive numbers
will be drawn


My point is the uselessness of sitting out 99+ draws
waiting for that one opportunity of a chance to win
that may not come.

Robert, we cannot sitting out anything at a random drawing.
It is also possible, that we have in the next drawing the same
numbers like at the drawing before. This can happen tomorrow
or never in humans existence, independent of the probability
calculation.


At least B can look forward to each draw knowing there
is a small chance of winning.

For the assumption that 4 consecutive numbers will be drawn,
the chance to win the jackpot for this drawing is for A very much
higher than for B and we saw that we cannot summarize the
chances of different drawings.

So there remains the question:
Is it better to have many times a small chance for winning
the jackpot, or is it better to have many times a very very small
chance to win the jackpot together with a few high chances
to win the jackpot (we speak only about 6 right, not about
3, 4,or 5 right numbers)

If we examine each of
those chances we will find other populations of various
sizes in there that excluded each combination from winning
so the most likely outcome is neither win.

It's kind of like two boxers getting into the ring, one
very outmatched and the other with two broken hands.

Robert Perkis
http://www.lotto-logix.com

Sorry to continue to disagree with you Manfred. If this
were a math class discussing the lottery over time I'd
have to agree.

However, I have to make up tickets to play in the next
draw. I do so knowing four or more consecutive numbers
will only prove to have been the correct answer one time
in 321 draws.

I start with all 13,983,816 combinations, actually this
being Florida 6/53 I start with 22,957,480 combinations.

Like a sculpture, I chip away what I don't think the next
draw will look like.

Seeing four or more consecutive numbers average 1 in 406
draws (6/53) I chip away 56,448 combinations knowing I will
be correct in this choice at least 400+ draws out of 406.

Yes it's random, the next 406 draws could all be four or
more consecutive numbers, but that's not the way to bet.

Outside the discussion is also 3-2-1 consecutive that is
expected 0.5% and 3-3 expected 0.0% I also chip away.

(6/53) 1-1-1-1-1-1 occurs 53.5% and 2-1-1-1-1 occurs 37.5%
and 2-2-1-1 occurs 5.1% and 3-1-1-1 3.4%

To me, the way to formulate this portion of the wager, is
obvious and proven correct draw after draw. Expecting the
others to come up draw after draw is wishful thinking.

Robert Perkis
http://www.lotto-logix.com

Robert,

I assume you do incorporate also past drawings
into your playing strategy for the next drawing and you
don't like combinations with 4 consecutive numbers
because they appear not very often.

I wanted to show that for having 6 right numbers only
the number of tickets is important and the population of
the numbers is irrelevant. Also if we look for tickets
for the next drawing.
The only successful strategy for playing to get 6 right
numbers can be to buy more tickets and nothing else.

I play only max. 12 tickets per drawing.

For this few tickets the chance for getting the jackpot
is nearly cero in my life, but if I don't play the lotto
then the chance is really cero.

My goal for playing is to find a method for maximum
3-number winnings for reducing the playing costs
and contemporaneously having a chance for higher
winnings.

If we want 3-number winnings, the discussion
is total different.

For such a strategy I have two options:

a) Playing with an optimised wheel with lowest redundant
3-number combinations.

b) Playing with random numbers or other wheel
types like combinations of two groups with 3 numbers

In case a) I will have more often a 3-number win and
the playing is less draggy and I get some money back
earlier.

In case b) there are less drawings where I will win,
But if I win I will win much more

But over the time both methods give the same amount
of payouts if the numbers of bought tickets are the
same in case a) and b)

I prefer case a)

Good luck, Manfred
.

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