Lottery - Odds of winning (1 ticket and 2 tickets)
andymarbles
settled in my head to regain my sanity and hopefully win a bet with a
friend.
The UK lottery has 13,983,816 combinations or a 1 in 13,983,816 chance
if you buy one ticket. Therefore a 2 in 13,983,816 chance if you buy
two tickets. But, can this be written as a 1 in 6,991,908 chance??
I understand that a second ticket will double the chance of winning,
but I don't believe it halves the odds. Help, please. I am sure I
read/heard this somewhere with quite a convincing argument.
Thanks!
Riccibuster
"andymarbles" <andym...@googlemail.com> wrote in message
news:6c1f190e-d930-4802...@a35g2000prf.googlegroups.com...
>I guess this has been argued to death here, but I really need this
> settled in my head to regain my sanity and hopefully win a bet with a
> friend.
>
> The UK lottery has 13,983,816 combinations or a 1 in 13,983,816 chance
> if you buy one ticket. Therefore a 2 in 13,983,816 chance if you buy
> two tickets. But, can this be written as a 1 in 6,991,908 chance??
it may seem intuitive at first glance, but expressing it this way is a naive
oversimplification. each single ticket has an *independent* chance of 1 in
13,983,816.
let's say that you and i each have 1 ticket. those 2 tickets are totally
unrelated to each other and each has an equal but separate chance. this must
be true as i see no alternative to this logic.
now let's say we somehow meet each other and decide to pool our tickets
together. it is absurd to say that these previously independent tickets have
suddenly and magically acquired the property to reduce the overall odds by a
full half. they still retain their independence.
if we apply the same concept to other games of chance, the rule should hold
up. take the game of bridge for instance.
there is a 1 in 158,753,389,900 chance that any one of the 4 players will be
dealt a perfect hand of 13 spades. can we then say that all 4 players
together have an aggregate chance of 1 in 39,688,347,475? the answer is no
because if any one of them is successful, that precludes the chance of any
other hand being successful. they each retain an equal but separate chance.
it's a tough subject to tackle but that's the way i see it anyway. others
will no doubt see it differently.
Gerry
"andymarbles" <andym...@googlemail.com> wrote in message
news:6c1f190e-d930-4802...@a35g2000prf.googlegroups.com...
Doubling your chance and halving the odds are the same thing.
2 in 13,983,816 = 1 in 6,991,908
Washington State Lottery sells 2 tickets for $1. Their website states that
a $1 purchase gives odds of 1 in 6,991,908. Lotteries in the US are
required by law to publish the correct odds.
http://www.walottery.com/sections/LotteryGames/Lotto.aspx?Page=PrizeLevels
LottoHack Jack Ricci
"andymarbles" <andym...@googlemail.com> wrote in message
news:6c1f190e-d930-4802...@a35g2000prf.googlegroups.com...
I play lots of different lotteries, and I buy lots of tickets as a hobby,
( instead of drinking, or chasing after women, or watching TV ...but any
good-looking woman can forward a tasty photo of themselves in their juiciest
outfit to ri...@mnsi.net ). I like to keep things in true perspective,
regardless of what the math seems to indicate. I feel that even the math
language and the many different math expressions can or will lose meaning in
a translation.
According to the figures you gave, and in my opinion only, buying 1
ticket in a 6/49 game buys you 1/13,983,816 th share or possible outcome of
the draw if all possible tickets are sold and no one else buys the same
ticket combo as yours, whether it wins a prize or not. When you buy 2
tickets for the same 6/49 game, you are buying 2/13,983,816 ths shares or
possible outcomes of the draw if all tickets are sold and no one else bought
the same combos as yours.
The 2 tickets you hold may win you a prize or not, but before and after
the draw, you will still be holding 2/13,983,816 ths ...that's 2 of the
13,983,816 total available tickets, or chances at a jackpot, purchased to
try to beat the draw outcome for the big money. In practicality, before a
draw, if you purchase two tickets, you will have 2 real tickets in your
possession for 2 real chances at the jackpot within an enormous
13,983,816-combo field of play, and not just the 1 ticket within only a
half-sized field of 6,991,908 possible combos.
Although the math appears to make sense, I refuse to acknowledge that
the expression 2/13,983,816 in a 6/49 lottery field is the equivalent of the
expression 1/6,991,908 in the same 6/49 lottery field. I also feel that it
is easier for a real player to hit a jackpot if he/she plays a smaller
lottery field and buys 1 ticket/chance out of a field of only 6,991,908
chances, than it is for that same player to hit a jackpot with 2 tickets
purchased within a much larger field of 13,983,816 possible
combos/chances/tickets. I view the scenario as the " mixing apples and
oranges story " goes.
Different views...different perceptions.
LottoHackJack
LottoHack Jack Ricci
"LottoHack Jack Ricci" <ri...@mnsi.net> wrote in message
news:fivr5...@enews2.newsguy.com...
...Actually, I made a slight miscalculation in my reasoning last post
that might add more fuel to a fire...If the ticket-vending machines would
kick out only quick-picks, and never sell a duplicate combo to any player,
and would then close-out and not sell any more tickets after the total field
of play is purchased and played by the lotto public, then the finite field
produces the following scenario as tickets are systematically purchased...
1/13,983,816...buying 1 ticket would buy you this 1 possible chance/fraction
of the possible draw outcome for a jackpot
2 /13,983,815...buying 2 tickets would buy you 2 of these remaining possible
chances/fractions of the possible draw outcome for a jackpot
3 /13,983,814...buying 3 tickets would buy you 3 of these remaining possible
chances/fractions of the possible draw outcome for a jackpot
4/13,983,813...buying 4 tickets would buy you 4 of these remaining possible
chances/fractions of the possible draw outcome for a jackpot
5/13,983,812...buying 5 tickets would buy you 5 of these remaining possible
chances/fractions of the possible draw outcome for a jackpot
6/13,983,811...buying 6 tickets would buy you 6 of these remaining possible
chances/fractions of the possible draw outcome for a jackpot
7/13,983,810...buying 7 tickets would buy you 7 of these remaining possible
chances/fractions of the possible draw outcome for a jackpot
8/13,983,809...buying 8 tickets would buy you 8 of these remaining possible
chances/fractions of the possible draw outcome for a jackpot
9/13,983,808...buying 9 tickets would buy you 9 of these remaining possible
chances/fractions of the possible draw outcome for a jackpot
10/13,983,807...buying 10 tickets would buy you 10 of these remaining
possible chances/fractions of the possible draw outcome for a jackpot
The next buyer would be purchasing a ticket from the field with chance
expressed as 11/13,983,806, so does every random ticket purchased in a
finite field have the same chance? Does buying 2 tickets double the chances
for that one player, or even maybe for the next two players if they could
meet and pool the tickets together?
The more tickets you or the lotto-playing public buys in a
non-duplicated, finite field, the larger the fraction gets as the remaining
combos available will dwindle towards 0.
Therefore, buying 100 tickets does increase your chances of hitting a
jackpot, but not by the same increment as some would like you to believe, in
my opinion.
In the real world, where we have a non-finite playing field where
duplicate combos can be randomly purchased, 12 million players could buy the
very same combo, while 1,983,816 combos aren't even played at all...So what
are the chances or odds of hitting the jackpot if all 12 million players
play the very same combo and it hits in next draw? How should the odds
before a draw be calculated if all of the punters played the same combo and
it doesn't hit in next draw?
I feel that the calculation of odds before a draw in a non-finite game is
very misleading simply because there can be more than one jackpot winner
having duplicate combos, and as well, the lower tier prize pay-outs may
mushroom out of proportion in certain draws as well. It's just not fair...
LottoHackJack
LottoHack Jack Ricci
"LottoHack Jack Ricci" <ri...@mnsi.net> wrote in message
news:fivus...@enews5.newsguy.com...
...Actually...again...I need to revise one other small thing in the
reasoning from my last previous post...
...If the playing field were finite, and everyone could buy only one ticket
until all 13,983,816 combos were used up, then the first person to buy a
combo would buy 1 of 13,983,816 possible combos or outcomes, or
1/13,983,816...The second player would be buying 1 of 13,983,815 combos or
possible outcomes in the remaining field, or 1/13,983,815...The third would
buy 1 of 13,983,814 combos or possible outcomes in the remaining field, or
1/13,983,814 of the possible outcomes in the remaining field which has yet
to be purchased, ...and so on, the fraction of purchased tickets gets larger
and larger until the last person buys 1/1 left to buy.
Now if we change the scenario a bit, and we allow just one player to buy
himself a jackpot, and he/she were to buy every possible 6/49 combo, one at
a time over the next few hours :), he would start with the first ticket,
which is 1/13,983,816 of the possible outcomes, and eventually arrives at
buying a whole half of all available combos, when he/she would be buying
that 1 of 6,991,908 available combos left in the remaining playing field!
Please note that he/she did not buy only the 2/13,983,816 combos, which some
would say is the equivalent of doubling your chances, and might be expressed
as the fraction 1/6,991,908, which mathematically equals 2/13,983,816...
To arrive at buying combo number 1/6,991,908, he/she had to purchase a
whole 6,991,907 combos before it to take possession of that 1/6,991,908
combo!
I would say that it is at this point, buying half of the total tickets
in the playing field, that the player is actually doubling his chances of
hitting the jackpot, if he is lucky, and not after buying merely drops in a
bucket... 2 tickets in the entire field of 13,983,816 combos...right...
LottoHackJack
Robert Perkis
What's fixed is the total number of lottery combinations.
If there are 13.9 million combinations we like to say the
odds are 1 in 13.9 million because one dollar buys one
combination giving the purchaser one chance to win.
Fly to a state where you get two combinations for a dollar
the total number of combinations haven't changed, but the
odds of 1 in 13.9 million are 1 in 6.9 million because now
you have two combinations in hand for one/dollar.
In the same sense, buying two tickets for two dollars also
cuts the odds in half for what each ticket faces, but does
not change the odds for one/dollar.
Of course it's still just two combinations out of 13.9
million, but it is the same division you would use if you
were in charge of any kind of team facing a larger team,
you would divide your numbers into theirs and come up with
a figure for how many each of your team would be facing.
To be precise, the total number of lottery combinations
never change. The odds of winning do change for you
depending on how you play, but generally can't be computed
until after the draw.
For example, the decision to put ten numbers into play
rather then six reduces the odds from 1 in 13.9 million
to one in 66,589.60 of having all six among the ten as
opposed to all six among six.
To fully cover 10 numbers takes 210 combinations and 210
times 66,589.60 gives us 13.9 million so in the background
the odds didn't change overall, but they certainly did for
your chance of having the numbers among yours. Actually
even those odds didn't change, they were the odds you took
or chose to accept by putting 10 numbers into play.
So the odds don't change, but there are odds for everything
possible within the lottery and you can choose to play at
one of those odds positions and by doing so be playing at
better or worse odds of winning then the person ahead and
behind you in line.
Robert Perkis / http://www.lotto-logix.com/
Chief Thunderthud
Good grief
Got any more revisions for all that fucked up nonsense ?
Evil Nigel
> In
> practicality, before a draw, if you purchase two tickets, you will have
> 2 real tickets in your possession for 2 real chances at the jackpot
> within an enormous 13,983,816-combo field of play, and not just the 1
> ticket within only a half-sized field of 6,991,908 possible combos.
If you could buy 2 50c different random tickets in a 7 million combo
lottery or 1 $1 ticket random ticket in a 14 million combo lottery,
assuming parity of prize structures, and if you add the consideration
that only an unshared jackpot win would be adequately life-changing,
which would be better and why?
I've just done a quick'n'dirty experiment which seems to show that where
there is a low total number of tickets bought, the 7 million combo lotto
offers the best chance of not sharing, but where a larger total of
tickets have been bought, the 14 million combo lotto is better.
Perhaps a real math-head could either confirm or refute this, and
explain where the tipping-point is.
Evil Nigel
Evil Nigel
Evil Nigel wrote:
> LottoHack Jack Ricci wrote:
>
>> In practicality, before a draw, if you purchase two tickets, you will
>> have 2 real tickets in your possession for 2 real chances at the
>> jackpot within an enormous 13,983,816-combo field of play, and not
>> just the 1 ticket within only a half-sized field of 6,991,908 possible
>> combos.
>
>
> If you could buy 2 50c different random tickets in a 7 million combo
> lottery or 1 $1 ticket random ticket in a 14 million combo lottery,
> assuming parity of prize structures, and if you add the consideration
> that only an unshared jackpot win would be adequately life-changing,
> which would be better and why?
Make that 1 ticket in the 7M lotto and 2 tickets in the 14M lotto.
Chief Thunderthud
"Evil Nigel" <use...@nospam.com> wrote in message
news:13l8fv1...@corp.supernews.com...
> Grrrrrrrrrrr!
I agree 100%
Riccibuster
"Riccibuster" <ricci...@byenow.com> wrote in message
news:fivaoi$ik0$1...@aioe.org...
i think what i was trying to argue here is that every single combo or line
played always has the same fixed odds no matter how many lines are played.
on second reading i realize that i was addressing the question incorrectly.
a *block* of lines taken as a single entity or unit of any size from 2 to
the max does indeed give the player a proportional improvement of the odds
in his favor. reducing a large fraction to its lowest common denominator is
the proper thing to do in this case.
gerry's answer is correct out of the gate. RP raises some interesting points
about conditional odds. wacky whacky jacky is typically in way over his head
and lost at sea. i wouldn't want to ask that sucker for driving directions.
Lottoking
Play 2 sets of multi tickets of 7 numbers,generated from 16
numbers.Total cost of 14*0.75 Euro
Chances are high to win a X-1 hit when 16 numbers have X numbers.
It just depends on your 16-numbers hitting chance of 3 hit,4 hit,5 hit
or 6 hit.
Nan
Riccibuster
"Lottoking" <lotteryp...@gmail.com> wrote in message
news:7d00e59f-0b39-4e0d...@e67g2000hsc.googlegroups.com...
------------------------------------------
fucking nonsense that has nothing to do with the topic.
Lottoking
> "Lottoking" <lotterypredict...@gmail.com> wrote in message
>
> news:7d00e59f-0b39-4e0d...@e67g2000hsc.googlegroups.com...
> On 12ÔÂ2ÈÕ, ÏÂÎç8ʱ38·Ö, andymarbles <andymarb...@googlemail.com> wrote:
>
> > I guess this has been argued to death here, but I really need this
> > settled in my head to regain my sanity and hopefully win a bet with a
> > friend.
>
> > The UK lottery has 13,983,816 combinations or a 1 in 13,983,816 chance
> > if you buy one ticket. Therefore a 2 in 13,983,816 chance if you buy
> > two tickets. But, can this be written as a 1 in 6,991,908 chance??
>
> > I understand that a second ticket will double the chance of winning,
> > but I don't believe it halves the odds. Help, please. I am sure I
> > read/heard this somewhere with quite a convincing argument.
>
> > Thanks!
>
> Play 2 sets of multi tickets of 7 numbers,generated from 16
> numbers.Total cost of 14*0.75 Euro
> Chances are high to win a X-1 hit when 16 numbers have X numbers.
> It just depends on your 16-numbers hitting chance of 3 hit,4 hit,5 hit
> or 6 hit.
> Nan
> ------------------------------------------
>
> fucking nonsense that has nothing to do with the topic.
Yes,it does,as it improves the chance of winning a higher prize ratio
Riccibuster
On 12?4?, ??10?22?, "Riccibuster" <riccibus...@byenow.com> wrote:
> "Lottoking" <lotterypredict...@gmail.com> wrote in message
>
> news:7d00e59f-0b39-4e0d...@e67g2000hsc.googlegroups.com...
>
> > I guess this has been argued to death here, but I really need this
> > settled in my head to regain my sanity and hopefully win a bet with a
> > friend.
>
> > The UK lottery has 13,983,816 combinations or a 1 in 13,983,816 chance
> > if you buy one ticket. Therefore a 2 in 13,983,816 chance if you buy
> > two tickets. But, can this be written as a 1 in 6,991,908 chance??
>
> > I understand that a second ticket will double the chance of winning,
> > but I don't believe it halves the odds. Help, please. I am sure I
> > read/heard this somewhere with quite a convincing argument.
>
> > Thanks!
>
> Play 2 sets of multi tickets of 7 numbers,generated from 16
> numbers.Total cost of 14*0.75 Euro
> Chances are high to win a X-1 hit when 16 numbers have X numbers.
> It just depends on your 16-numbers hitting chance of 3 hit,4 hit,5 hit
> or 6 hit.
> Nan
> ------------------------------------------
>
> fucking nonsense that has nothing to do with the topic.
Yes,it does,as it improves the chance of winning a higher prize ratio
---------------------------------------------------
"any of this gettin' through to you son?"...as foghorn leghorn would say.
why don't you follow your own advice then? just about everything you post
here uses some new and even more fucked-up wheeling method than the day
before...in some cases the hour or minute before. fer chrissakes man! you're
all over the goddamn place like a frog on a hot griddle.
winn...@gmail.com
> "Lottoking" <lotterypredict...@gmail.com> wrote in message
>
> news:f8cb6a86-d940-43eb...@e4g2000hsg.googlegroups.com...
> On 12?4?, ??10?22?, "Riccibuster" <riccibus...@byenow.com> wrote:
>
>
>
>
>
> > "Lottoking" <lotterypredict...@gmail.com> wrote in message
>
> >news:7d00e59f-0b39-4e0d...@e67g2000hsc.googlegroups.com...
>
> - 显示引用的文字 -
Then lassen mir in ruhig!
simo...@gmail.com
I choose 6 numbers from a possible of 49 (1,2,3,4,5,6,) which gives me the odds of 1 in 13,983,816
When I buy a second ticket (1,2,3,4,5,7) there are now only 13,983,815 combinations left therefore the odds are reduced to 1 in 13,983,815. Furthermore the odds are now the same on the first ticket BUT this is NOT the simplistic math 2 in 13,983,815, the way to view it is that I am a single (1) competitor and MY odds are now 1(me) in 13,983,815
This works perfectly with all the extremes of scenario which conflict with each other.
ie. I buy half the tickets (6,991,908) then 13,983,816 minus 6,991,908 equals 6,991,908 so I have the odds of 1 in 6,991,908
just another thought if you were to buy 13,983,815 tickets you would have a 1 in 13,983,816 chance of loosing hmmm.... I hear you say "well you are not going to loose are you! well then you are not going to win either are you! ...no wonder I have always nicknamed it The Irrational fluttery!
I still have the problem of is it better to buy 52 tickets once a year or buy 1 ticket each week for a year
nigel
>
> I still have the problem of is it better to buy 52 tickets once a
> year or buy 1 ticket each week for a year
However if you buy 52 tickets once a year you get a slightly better
chance of winning a jackpot, but if you buy one ticket each week for a
year, although you have a slightly reduced chance of winning a jackpot,
that is counterbalanced by a miniscule chance of winning multiple jackpots.
Evil Nigel
fitzyb...@gmail.com
fitzyb...@gmail.com
novembe...@gmail.com
The lottery is now pick 6 in 59 so the
odds should be updated ==> 1 in 45,057,474
What the OP should have asked is : If I buy 2 tickets
a) with different numbers
b) with the same numbers
do the odds change to 2 in 45M ? Seems a) yes and b) no
ions...@gmail.com
I.
novembe...@gmail.com:
“The lottery is now pick 6 in 59…”
In your neck of the woods! There are multiple 6/N lotteries in the world.
“What the OP should have asked is : If I buy 2 tickets
a) with different numbers
b) with the same numbers “
A good point, usually overlooked. But if a weirdo buys 2 tickets with the exact same numbers, the probability does NOT change. Two tickets in that case equal 1 FAVORABLE case. In case only the weirdo hits the 6 winners, he would share the jackpot… with himself.
A post like this shouldn’t trigger a debate. Every probability book everywhere in the world defines PROBABILITY very clearly:
PROBABILITY = (FAVORABLE CASES) / (TOTAL POSSIBLE CASES)
Examples:
~ 1 ticket in 6/49 lotto: 1/13983816 = 0.000000071 = 0.000007151%
~ 2 tickets with at least one different number: 2/13983816 = 1/6991908 = 0.000000143 = 0.000014302%
The confusion is also “influenced” by interchanging the correct term PROBABILITY with an Old English term — ODDS. Yes, I still use the ‘odds’ term more often than I should, especially when analyzing lottery:
http://saliu.com/oddslotto.html
II.
Doollow (aka nigel):
“However if you buy 52 tickets once a year you get a slightly better chance of winning a jackpot…”
That phenomenon is called ‘Ion Saliu’s Paradox’. In the 1980s, a group of Austreyelian investors brought a number of tickets equal to total number of combosnations in the Virginia 6/49 lotto. The jackpot was over $100 million.
Of course, they couldn’t fill out all those 13+ million tickets. So, they bought computer-generated combinations. Little did they know that they had covered only 63% of all possible 6/49 outcomes (that’s a law of randomness):
http://saliu.com/theory-of-probability.html
(Theory of Probability, Ion Saliu’s Paradox)
Ion Saliu,
Founder of Lottery Mathematics,
Founder of Lottery Paradox,
Etc.
ions...@gmail.com
1) Coin-tossing is the most convincing. There are 2 possibilities (outcomes): ‘Heads’ and ‘Tails’. The probability to get one possibility is p = 1/2. If we play the 2 possibilities, the chance is p = 2/2 = 100% or ‘1 in 1’. But if we play 1 outcome in 2 trials, the chance is only 75% or ‘1 in 3.3’.
2) Roll one die: 6 possibilities. The probability is p = 1/6. If we play the 3 possibilities, the chance is p = 3/6 = 50% or ‘1 in 2’. But if we play 1 outcome in 3 trials, the chance is 34.7% or ‘1 in 2.88’.
3) Let’s do now calculations for 6/49 lotto, for we got great software to do such calculations: SuperFormula.EXE.
So, play 52 tickets (distinct 6-number combinations) once a year or one ticket a week (52 weeks, right?) If we play the 52 possibilities, the chance is p = 52/13983816 = 0.000371858% or ‘1 in 268919.5’. But if we play 1 combonation in 52 weeks, the chance is 0.0004% or ‘1 in 268920.5’.
Not a big difference here. But how about 10 years? Play 520 tickets (distinct 6-number combinations) ONCE or one ticket a week (520 weeks, right?) If we play the 520 possibilities, the chance is p = 520/13983816 = 0.00372% or ‘1 in 26892’. But if we play 1 combonation in 520 weeks, the chance is 0.0037% or ‘1 in 26893’.
The more combination played, the wider the gap between the two methods. And there is a limit in all this matter, as calculated by the ‘Ion Saliu Paradox’. If we play all 13983816 combosnations at once (as the Austreyelian investors did), the chance to hit the jackpot is 100%. If we play 1 ticket for the next 13983816 drawings (good luck with that!) the chance to hit the jackpot is only 63.2%.
That’s how randomness works for all intents and purposes.
We talkin here hitting the jackpot…
http://saliu.com/formula.html
(SuperFormula: The Best Software for Statistics Mathematics Probability)
Ion Saliu,
Founder of Probability Theory of Life,
Founder of Randomness Philosophy,
Etc.
nikc...@gmail.com
In the UK 6/59 draw..
If you buy ONE ticket you have ONE chance in 45,057,474 of winning the Jackpot. You now have 45,057,473 combinations AGAINST you.
If you buy TWO tickets you have TWO chances in 45,057,474 of winning the Jackpot. You now have 45,057,472 combinations AGAINST you.
If you buy THREE tickets you have THREE chances in 45,057,474 of winning the Jackpot. You now have 45,057,471 combinations AGAINST you.
Couple of friends of mine and I share and play 16 lines in the UK 6/59. Therefore we have 16 chances in 45,057,474 of winning the Jackpot. The fact remains, we have 45,057,458 combinations AGAINST us.
Playing 1, 2 or 16 combinations; the chances of winning the Jackpot are.. ZILCH! Well, as near as dammit!
No need for a mathematician here.
Unless??...
...................................................
On 26/11/2016 22:36, fitzyb...@gmail.com wrote:
> On Saturday, December 1, 2012 at 6:16:27 PM UTC-5, simo...@gmail.com wrote:
>> Just to throw a spanner in the works! I came on here to find out how buying 2 lotto tickets changes the odds....I got all the answers I had already thought of ,,,,,,,THEN it dawned on me! its so obvious really.
>>
>> I choose 6 numbers from a possible of 49 (1,2,3,4,5,6,) which gives me the odds of 1 in 13,983,816
>> When I buy a second ticket (1,2,3,4,5,7) there are now only 13,983,815 combinations left therefore the odds are reduced to 1 in 13,983,815. Furthermore the odds are now the same on the first ticket BUT this is NOT the simplistic math 2 in 13,983,815, the way to view it is that I am a single (1) competitor and MY odds are now 1(me) in 13,983,815
>>
>> This works perfectly with all the extremes of scenario which conflict with each other.
>> ie. I buy half the tickets (6,991,908) then 13,983,816 minus 6,991,908 equals 6,991,908 so I have the odds of 1 in 6,991,908
>>
>> just another thought if you were to buy 13,983,815 tickets you would have a 1 in 13,983,816 chance of loosing hmmm.... I hear you say "well you are not going to loose are you! well then you are not going to win either are you! ...no wonder I have always nicknamed it The Irrational fluttery!
>>
>> I still have the problem of is it better to buy 52 tickets once a year or buy 1 ticket each week for a year
>
ions...@gmail.com
I.
nikc...@gmail.com:
“I ain't no mathematician…”
You WAS a mathematician, though! You expressed a mathematical formula in words, instead of symbols/numbers:
You sez:
“TWO tickets you have TWO chances in 45,057,474 of winning the Jackpot”
It comes from mathematics:
When buying 2 tickets, the probability to win the jackpot is:
p = 2/45057474.
II.
“ZILCH! Well, as near as dammit” — you can improve the chance to win the jackpot by playing more tickets less frequently. Say, after a win, you do not play for the next 5 drawings. Then, “play 16 * 5 = 80 lines”.
“God hates mathematics, Einstein hates gambling.”
http://saliu.com/formula.htm
ions...@gmail.com
> On Tuesday, November 29, 2016 at 8:48:55 PM UTC-5, nikc...@gmail.com wrote:
>
> I.
> nikc...@gmail.com:
> “I ain't no mathematician…”
>
> You WAS a mathematician, though! You expressed a mathematical formula in words, instead of symbols/numbers:
>
> You sez:
> “TWO tickets you have TWO chances in 45,057,474 of winning the Jackpot”
>
> It comes from mathematics:
>
> When buying 2 tickets, the probability to win the jackpot is:
> p = 2/45057474.
>
> II.
> “ZILCH! Well, as near as dammit” — you can improve the chance to win the jackpot by playing more tickets less frequently. Say, after a win, you do not play for the next 5 drawings. Then, “play 16 * 5 = 80 lines”.
>
> http://saliu.com/formula.htm
I.
“Say, after a win, you do not play for the next 5 drawings. Then, “play 16 * 5 = 80 lines”.”
That is valid for RANDOM play only. It is unadulterated mathematics.
II.
Many lottery players, however, apply STRATEGIES or SYSTEMS related to past results or drawings. In such cases, one must play each and every draw required by the lottery strategy. Skipping drawings can lead to losses. Here are but a few real-life cases encountered by this very author:
http://saliu.com/frequency-lottery.html
(Lottery Strategy, Systems, Software Based on Lotto Number Frequency)
http://saliu.com/strategy-gambling-lottery.html
(The Best Strategy for Lottery, Gambling, Sports Betting, Horse Racing, Blackjack, Roulette, Baccarat, Craps)
Ion Saliu,
Founder of Mathematical Axiomaticism
“A good man is an axiomatic man; an axiomatic man is a happy man. Be axiomatic!”
novembe...@gmail.com
ions...@gmail.com
Actually, only ‘Ion Saliu’s Paradox’ does the most precise calculations. It calculates the degree of certainty for ‘AT LEAST one success in N trials’. The binomial distribution formula (‘Probability of EXACTLY n successes in N trials’) still proves that the chance is better when playing more tickets less frequently.
My previous post showed results calculated by the binomial distribution formula. The most precise results, as calculated by the ‘Ion Saliu Paradox’ for the die rolling read:
2.1) Roll one die: 6 possibilities. The probability is p = 1/6. If we play 3 possibilities, the chance is p = 3/6 = 50% or 1 in 2. But if we play 1 outcome in 3 trials, the degree of certainty is 42.1% or 1 in 2.38. The difference is 7.9%
2.2) We can play all 6 point-faces at once; the probability is p = 6/6 = 100% or 1 in 1. If we play 1 point-face in 6 rolls, the degree of certainty is 66.5% or 1 in 1.5. The difference is 33.5%
Read further:
http://saliu.com/oddslotto.html
(‘Multum in parvo: Much in little’ section)
Best of luck in your gambling adventures during this holiday season!
Ion Saliu,
Founder of Gambling Mathematics
“Truth be told, Randomness Almighty plays Cosmic Roulette with the Universe, including our disorderly world.”
http://saliu.com/almighty_number.html
Ion Saliu
Ion Saliu
> On Tuesday, November 29, 2016 at 9:54:01 AM UTC-5, novembe...@gmail.com wrote:
>
> I.
> novembe...@gmail.com:
> “The lottery is now pick 6 in 59…”
>
> In your neck of the woods! There are multiple 6/N lotteries in the world.
>
> “What the OP should have asked is : If I buy 2 tickets
> a) with different numbers
> b) with the same numbers “
>
> A good point, usually overlooked. But if a weirdo buys 2 tickets with the exact same numbers, the probability does NOT change. Two tickets in that case equal 1 FAVORABLE case. In case only the weirdo hits the 6 winners, he would share the jackpot… with himself.
>
> A post like this shouldn’t trigger a debate. Every probability book everywhere in the world defines PROBABILITY very clearly:
>
> PROBABILITY = (FAVORABLE CASES) / (TOTAL POSSIBLE CASES)
>
> Examples:
> ~ 1 ticket in 6/49 lotto: 1/13983816 = 0.000000071 = 0.000007151%
> ~ 2 tickets with at least one different number: 2/13983816 = 1/6991908 = 0.000000143 = 0.000014302%
>
> The confusion is also “influenced” by interchanging the correct term PROBABILITY with an Old English term — ODDS. Yes, I still use the ‘odds’ term more often than I should, especially when analyzing lottery:
>
> II.
> Doollow (aka nigel):
> “However if you buy 52 tickets once a year you get a slightly better chance of winning a jackpot…”
>
> That phenomenon is called ‘Ion Saliu’s Paradox’. In the 1980s, a group of Austreyelian investors brought a number of tickets equal to total number of combosnations in the Virginia 6/49 lotto. The jackpot was over $100 million.
>
> Of course, they couldn’t fill out all those 13+ million tickets. So, they bought computer-generated combinations. Little did they know that they had covered only 63% of all possible 6/49 outcomes (that’s a law of randomness):
>
>
> Ion Saliu,
> Founder of Lottery Mathematics,
> Founder of Lottery Paradox,
• All-Matters Lottery, Lotto: Strategies, Systems, Software
• https://groups.google.com/forum/?hl=en#!topic/rec.gambling.lottery/OR5o1HDcy-U