1 in 13,983,816

[therichster]

hey guys,
I'm very new to the forum and I have some questions about the lottery. As my presence on this forum might have already given away a little about my positive outlook on lotteries and optimism about being able to improve your odds by picking "smarter" number, I wanted to know about the odds one might have playing the lottery.
As mathematics and the lottery board shows us that the odds of winning the grand-prize (6/6) playing 6/49 are 1 in 13,983,816, I wanted to know how one would calculate the odds of winning with each ticket purchased.
This might seem like very simply fraction calculations, but if you purchase only one ticket the odds of winning 6/6 are 1 in 13,983,816. If you purchase 2 tickets, do they become 1 in 6,991,908? And if you purchase 3 tickets, do they become 1 in 4,661,272?
If this thinking is correct and people say that getting hit by lightning has much better odds (1 in 240,000), wouldn't one only have to purchase 100 tickets ( odds become 1 in 139,839 to win 6/6 ) to even out the playing field and make it theoretically more probable of winning the jackpot twice before even getting hit by lightning once?

Also on a different note, lets say that you purchase only one ticket ( making your odds again 1 in 13,983,816 of winning ) but you automatically eliminate the numbers from the previous draw ( because it is possible of the same set of numbers being picked consecutively two or three weeks...it would be safe to say that its more unlikely to happen). Would it be optimistic of me to say that your odds would automatically improve to that of having purchased 2 tickets ( one in 6,991,908 ) ? Although ur actually not playing that set of numbers from the last draw, its like knowing ahead of time that they won't be picked but yet playing them anyway? Just as if you would have played them nonetheless?

I hope my questions are clear enough and that someone has an answer, even if the questions seems a little simple minded.
Thank you

 

[blitzed]

hello therichster,

nope, 2 tickets = 2 chances in 13,983,816...

only way to sufficiently reduce the odds, is to successfully reduce the number playfield.

ok, take your 6/49 with 13,983,816 possibilities. now ifya can successfully reduce 49 numbers down to 42 numbers, now only have 5,245,786 possibilities according to johnph77's calculations for a 6/42 game:
http://www.johnph77.com/math/stlot.html#phll642

cheers!
blitzed

 

[therichster]

that is what I used to think as well. But when odds are calculated for example for the super 7 grand prize ( even by loto-quebec's calculations ), they usually come up with the 1 in 60,000,000 per line played.
But since you get 3 lines per 2 dollars played, they automatically reduce your odds to 1 in 20,000,000 per 2 dollars played.
If mathematicians and even loto-quebec can do it...how come it doesn't apply with each additional ticket purchased?

 

[blitzed]

say there is a raffle game with one jackpot, 10,000 sequentially numbered tickets, and they are all sold.

if you bought 3 of those 10,000 tickets, it would not suddenly give you a 1 in 3,333 chance to win, what about the people holding the other 9,997 tickets?

each Super7 pick is actually 1 out of 62,891,499 possibilities:
http://www.johnph77.com/math/stlot.html#cndmps7

blitzed

 

[therichster]

The website says 1 in 62 million, but then how come loto-quebec has it listed as 1 in 20,963,833 ?
http://loteries.loto-quebec.com/web/jsp/MainPage.jsp?Params=Y.US.80200.0&section=logo-super-7&objet=image

And back to your example, theoretically you do have 1 in 3,333 chance of winning if you were to buy the 3 tickets. Its odds...and odds don't have to add up to those odds of other players. Depending on how many tickets to other people bought...those would be their odds.
Maybe I'm not getting this right, but lets make the example an extreme one. Lets say you were to buy 5,000 of those 10,000 tickets. Because you have half the tickets your chances are 1 in 2 of actually holding the winning ticket. Aren't odds simply fractions? 5000/10000 is it not the same as 1/2? Please do correct me if I'm misunderstanding something here...

 

[t.lawler]

The website says 1 in 62 million, but then how come loto-quebec has it listed as 1 in 20,963,833 ?
http://loteries.loto-quebec.com/web/jsp/MainPage.jsp?Params=Y.US.80200.0&section=logo-super-7&objet=image

And back to your example, theoretically you do have 1 in 3,333 chance of winning if you were to buy the 3 tickets. Its odds...and odds don't have to add up to those odds of other players. Depending on how many tickets to other people bought...those would be their odds.
Maybe I'm not getting this right, but lets make the example an extreme one. Lets say you were to buy 5,000 of those 10,000 tickets. Because you have half the tickets your chances are 1 in 2 of actually holding the winning ticket. Aren't odds simply fractions? 5000/10000 is it not the same as 1/2? Please do correct me if I'm misunderstanding something here...

3 in 10,000 does in fact = 1 in 3,333
stop and think for moment
if you triple your chances 3 in 10,000 vs 1 in 10,000 then you have reduced
your odds accordingly

 

[t.lawler]

The website says 1 in 62 million, but then how come loto-quebec has it listed as 1 in 20,963,833 ?
http://loteries.loto-quebec.com/web/jsp/MainPage.jsp?Params=Y.US.80200.0&section=logo-super-7&objet=image

And back to your example, theoretically you do have 1 in 3,333 chance of winning if you were to buy the 3 tickets. Its odds...and odds don't have to add up to those odds of other players. Depending on how many tickets to other people bought...those would be their odds.
Maybe I'm not getting this right, but lets make the example an extreme one. Lets say you were to buy 5,000 of those 10,000 tickets. Because you have half the tickets your chances are 1 in 2 of actually holding the winning ticket. Aren't odds simply fractions? 5000/10000 is it not the same as 1/2? Please do correct me if I'm misunderstanding something here...

hi to the richster
yes you are right
3 in 10,000 does in fact = 1 in 3,333

if you triple your chances 3 in 10,000 vs 1 in 10,000 then you have reduced
your odds accordingly

 

[t.lawler]

The website says 1 in 62 million, but then how come loto-quebec has it listed as 1 in 20,963,833 ?
http://loteries.loto-quebec.com/web/jsp/MainPage.jsp?Params=Y.US.80200.0&section=logo-super-7&objet=image

And back to your example, theoretically you do have 1 in 3,333 chance of winning if you were to buy the 3 tickets. Its odds...and odds don't have to add up to those odds of other players. Depending on how many tickets to other people bought...those would be their odds.
Maybe I'm not getting this right, but lets make the example an extreme one. Lets say you were to buy 5,000 of those 10,000 tickets. Because you have half the tickets your chances are 1 in 2 of actually holding the winning ticket. Aren't odds simply fractions? 5000/10000 is it not the same as 1/2? Please do correct me if I'm misunderstanding something here...

hi to the richster
yes you are right
3 in 10,000 does in fact = 1 in 3,333

if you triple your chances 3 in 10,000 vs 1 in 10,000 then you have reduced
your odds accordingly

 

[t.lawler]

sorry about the repeated post . i got distracted and lost track of what i was doing.

 

[johnph77]

t.lawler is correct. Odds are a ratio. If odds are stated as 1::13,983,816 for one ticket and you buy two tickets, your odds of winning just became 2::13,983,816, reduceable to 1::6,991,908. Odds expressions are like algebraic formulas - if you reduce or increase both sides of an expression or formula by a like amount, the expression or formula stays the same.

In the example of the cited Quebec lottery, my website gives the odds of winning at 1::62,891,499. And this is correct - per game. Since one gets three plays per two-dollar ticket, the odds of a win on one ticket then become 3::62,891,499 - or 1::20,963,833, as stated on the Quebec lottery site.

 

[daleks]

...ODDS do not change for each combination of numbers purchased in any particular draw.......the ODDS are found as a simple nfactorial....for example, in the 6/49 games, they are calculated as.....(49x48x47x46x45x44) / (1x2x3x4x5x6).....that gives the ODDS for each and every single possible line purchased/generated......once someone purchases two tickets, the ODDS remain the same but the CHANCES to win are cut in half, because you have now bought two tickest....you will find that if you were to purchase one set of numbers, the ODDS and CHANCES are both identical....with the purchase of two sets of numbers, the ODDS stay the same for each individual set, but the CHANCES of winning have now been reduced by half...if you follow that through, and then buy 2,3,4,5 .....13,983,815 tickets, the CHANCES will be halved each time, but the ODDS will remain the same for each set purchased, and the ticket remaining, ticket number 13,983,816, has the same ODDS of winning as any of the others....if you divide the 13-odd million possibilities in half each time, you will find it will take you 13-odd million dollars (at a dollar a game) to buy ALL the CHANCES, and make you a sure winner...the CHANCES of winning are now reduced to 1:1, yet the ODDS remain at 13-odd million for each and every combination....a lot of the lottery corporations will publish the CHANCES of winning as being accurate because you get two or three tickets for a dollar....so since when has the price of a bet ever affected the possible outcome of a chance generation of numbers????.....in the example of the Quebec lottery posted just above this post, where the odds are given as 1:20-odd million because you bought 3 tickets, they are simply LYING to you....and, no, an nfactorial is NOT a simple algebraic expression....sheesh.....

 

[johnph77]

you will find that if you were to purchase one set of numbers, the ODDS and CHANCES are both identical....with the purchase of two sets of numbers, the ODDS stay the same for each individual set, but the CHANCES of winning have now been reduced by half...if you follow that through, and then buy 2,3,4,5 .....13,983,815 tickets, the CHANCES will be halved each time, but the ODDS will remain the same for each set purchased, and the ticket remaining, ticket number 13,983,816, has the same ODDS of winning as any of the others

If you have one chance to win - i.e. buying one ticket - the odds of winning in your cited lottery are 1::13,983,816. If you have two tickets, you have two chances to win (2::13,983,816), but your odds of winning that particular draw with those two tickets are 1::6,991,908, a reduction of that ratio. If you have three tickets, you have three chances to win (3::13,983,816), your odds of winning are 1::4,661,272, again a reduction of that ratio. Your odds of winning are not halved each time by buying an additional ticket save for the second one.

 

[t.lawler]

hey guys,
I'm very new to the forum and I have some questions about the lottery. As my presence on this forum might have already given away a little about my positive outlook on lotteries and optimism about being able to improve your odds by picking "smarter" number, I wanted to know about the odds one might have playing the lottery.
As mathematics and the lottery board shows us that the odds of winning the grand-prize (6/6) playing 6/49 are 1 in 13,983,816, I wanted to know how one would calculate the odds of winning with each ticket purchased.
This might seem like very simply fraction calculations, but if you purchase only one ticket the odds of winning 6/6 are 1 in 13,983,816. If you purchase 2 tickets, do they become 1 in 6,991,908? And if you purchase 3 tickets, do they become 1 in 4,661,272?
If this thinking is correct and people say that getting hit by lightning has much better odds (1 in 240,000), wouldn't one only have to purchase 100 tickets ( odds become 1 in 139,839 to win 6/6 ) to even out the playing field and make it theoretically more probable of winning the jackpot twice before even getting hit by lightning once?

Also on a different note, lets say that you purchase only one ticket ( making your odds again 1 in 13,983,816 of winning ) but you automatically eliminate the numbers from the previous draw ( because it is possible of the same set of numbers being picked consecutively two or three weeks...it would be safe to say that its more unlikely to happen). Would it be optimistic of me to say that your odds would automatically improve to that of having purchased 2 tickets ( one in 6,991,908 ) ? Although ur actually not playing that set of numbers from the last draw, its like knowing ahead of time that they won't be picked but yet playing them anyway? Just as if you would have played them nonetheless?

I hope my questions are clear enough and that someone has an answer, even if the questions seems a little simple minded.
Thank you

hi therichster
As i have said befor the only true way do increase one's odds of winning anything is to bet more. If you bet 2 then you have twice the chaance of winning. if you bet 100 the you have 100 times of the chance of winning.

However you asked so many questions in the second paragraph that I lost track.

If you elimanate 6 numbers and are right then of course you have improved your odds. In a 6/49 lottery with 14 meg chances then if you throw out 6 numbers then it becomes a 6 meg lotto. Even though I am rounding the odds
I am trying to show the point that if you throw out the last 6 numbers drawn then it is not any better than throwing out any other 6 numbers.

TIP OF THE DAY the second way to improve your odds of winning is to SUCCESSEVLY reduce the pool.

"life is like a box of chocolates you never know what you are going to get "
forrest gump