New Jersey United States Member #1 May 31, 2000 23259 Posts Offline

Posted: January 4, 2011, 7:29 am - IP Logged

Quote: Originally posted by Coin Toss on January 4, 2011

Nope.

Every line of numbers you play is up against the same odds, there is only one set of winning numbers drawn.

barbos is right.

If the fractionalizing theory worked people would just form pools and bet enough to "make sure" they won. Ha ha ha, the lottery loves that. They have this all figured out.

Take a pick 5 game where the odds are 575,757 to one. $5000 in tickets, according to your thoey, would bring the odds down to 115 to one. $20,000 would make the odds 28 to one. If it worked people would be taking out loans and winning $100,000 or more a pop.

And if you're theory worked on Mega Millions it would never have rolled over 15 times to get to a $330,000,000 jackpot.

You have $13 worth of tickets, each one up against odds of 175,711,536.

You are up against 175,711,536: 1 - 13 times.

That would only be true if a different person holds each ticket. We are talking about one person holding all the tickets, so COLLECTIVELY the odds of 13 tickets are 13 in 175,711,536. There is really no way you can argue with that. You have covered 13 of the 175,711,536 possible combinations.

So reduce the fraction. You do know that odds are simply a fraction, right?

The Ville, FL United States Member #95879 August 19, 2010 1708 Posts Offline

Posted: January 4, 2011, 9:06 am - IP Logged

Quote: Originally posted by barbos on January 3, 2011

You might be right but buying 87,855,768 tickets does not reduce the odds to 1 in 2 - there are still almost 88 millions possible combinations left over.

barbos - that will give you a 50/50 chance persay...That is exactly what it does. Increases your odds if you bought half of the amount of possible combinations.

Problem with doing that: if someone else wins also, then you just spent87,855,768 and wont come out ahead because you would be splitting itwith someone else.

** With your logic you are basically saying that if a person buys every combination available, then they still only have a 1 in 175,711,536 chance. Which is not true. If a person bought every single combination in Mega Millions that means they would have a 175,711,536 in 175,711,536 chance (or 100%) chance of winning....

Let look at this at another angle. The pick 3 games have 1000 possibilities. If you spent $500 on pick 3, that would also give you a 50/50 chance or a 1 in 2 chance of winning.

Zeta Reticuli Star System United States Member #30470 January 17, 2006 10344 Posts Offline

Posted: January 4, 2011, 10:25 am - IP Logged

Quote: Originally posted by Todd on January 4, 2011

That would only be true if a different person holds each ticket. We are talking about one person holding all the tickets, so COLLECTIVELY the odds of 13 tickets are 13 in 175,711,536. There is really no way you can argue with that. You have covered 13 of the 175,711,536 possible combinations.

So reduce the fraction. You do know that odds are simply a fraction, right?

Todd,

With 13 tickets you have covered 13 sets of a posssible 175,711,536 sets of numbers, leaving 175,711, 523 sets not covered.

From GURU101,

"if you buy a $5 quick pick, your odds are 1 in 35,142,307(175,711,536 / 5). I happen to have $13 worth of tickets, so my odds would be 1 in 13,516,272, which is about the same as a single line in a Pick 6/49 Lotto game."

If the "magic fraction odd reduction" worked, those 13 tickets would have covered some 40,000,000 + combinations. They don't, they only cover 13.

According to you guys, those 13 tickets have covered 162,195,264 (175,711,536 minus 13,516,272 stated by having 13 tickets) sets of numbers! THE "MAGIC BULLET" THEORY FROM THE WARREN COMMISSION ON THE JFK ASSASSINATION MEETS THE MEGA MILLIONS ODDS!

Once again, if it worked, with the over 100,000 members you have here on LP, we could all go $5 or $10 apiece and have a "lock". Once upon a time.....

Good Luck everyone!

Those who run the lotteries love it when players look for consistency in something that's designed not to have any.

There is one and only one 'proven' system, and that is to book the action. No matter the game, let the players pick their own losers.

The Ville, FL United States Member #95879 August 19, 2010 1708 Posts Offline

Posted: January 4, 2011, 10:34 am - IP Logged

Quote: Originally posted by Fla386 on January 4, 2011

i have 330 million reasons for a road trip to Ga !!!! Good Luck Everyone!!!

Yea man Im glad I live so close to GA I went twice last week over my vacation. Gave me and my family something to do while we were off ....Less than an hour drive. Went up and bought some lotto tix, got lunch and headed back.

Over and done in 2 hours tops.

If you are coming from South FL get off at exit 3 approximately 2 miles past the border...there are 3 gas stations that sell tix right there. You would be amazed at how many scratchoff books these dudes have. Like 8-10 of each one and the walls are plastered with past winners. Its big business there because they are the first big exit past the border

New Jersey United States Member #1 May 31, 2000 23259 Posts Offline

Posted: January 4, 2011, 10:34 am - IP Logged

Quote: Originally posted by Coin Toss on January 4, 2011

Todd,

With 13 tickets you have covered 13 sets of a posssible 175,711,536 sets of numbers, leaving 175,711, 523 sets not covered.

From GURU101,

"if you buy a $5 quick pick, your odds are 1 in 35,142,307(175,711,536 / 5). I happen to have $13 worth of tickets, so my odds would be 1 in 13,516,272, which is about the same as a single line in a Pick 6/49 Lotto game."

If the "magic fraction odd reduction" worked, those 13 tickets would have covered some 40,000,000 + combinations. They don't, they only cover 13.

According to you guys, those 13 tickets have covered 162,195,264 (175,711,536 minus 13,516,272 stated by having 13 tickets) sets of numbers! THE "MAGIC BULLET" THEORY FROM THE WARREN COMMISSION ON THE JFK ASSASSINATION MEETS THE MEGA MILLIONS ODDS!

Once again, if it worked, with the over 100,000 members you have here on LP, we could all go $5 or $10 apiece and have a "lock". Once upon a time.....

Good Luck everyone!

OMG, odds calculations do no express the NUMBER of combinations you've covered. The express the FRACTION of the combinations you've covered!

Zeta Reticuli Star System United States Member #30470 January 17, 2006 10344 Posts Offline

Posted: January 4, 2011, 10:43 am - IP Logged

The number of possible combinations is how the odds are established to begin with. Every line of numbers played (one combination) reduces those odds by one combination. There are no fractions that magically take one more combination and reduce the odds by millions.

The belief that there is sure sells a lot of tickets and systems though!

But I do think I might use numbers extracted from 13, 516, 272 tonight!

13 51 6 27 2 .....

1 35 16 27 2 .... darn still need a Mega number!

Those who run the lotteries love it when players look for consistency in something that's designed not to have any.

There is one and only one 'proven' system, and that is to book the action. No matter the game, let the players pick their own losers.

New Jersey United States Member #1 May 31, 2000 23259 Posts Offline

Posted: January 4, 2011, 10:48 am - IP Logged

Quote: Originally posted by Coin Toss on January 4, 2011

The number of possible combinations is how the odds are established to begin with. Every line of numbers played (one combination) reduces those odds by one combination. There are no fractions that magically take one more combination and reduce the odds by millions.

The belief that there is sure sells a lot of tickets and systems though!

But I do think I might use numbers extracted from 13, 516, 272 tonight!

13 51 6 27 2 .....

1 35 16 27 2 .... darn still need a Mega number!

OK, you are reluctant to have a discussion that expresses odds in a mathematical format, so let's try this a different way.

I would assume that you know that all odds can be expressed as "1 chance in _____". If we can't agree on that, then I'm not sure we can have any discussion at all. So please just validate that you agree with this statement.

Then, as step two, please fill in the blank: After I buy 13 tickets, my odds of winning tonight are 1 in ______.

Zeta Reticuli Star System United States Member #30470 January 17, 2006 10344 Posts Offline

Posted: January 4, 2011, 10:55 am - IP Logged

Each ticklet is facing odds of 175,711, 536 :1.

You have those odds against you 13 times over. You've reduced the odds by 13 combinations and that's as good as it gets, you havent's divided those original odds by 13. 13/175,711,536 is hype, that's all.

Thinko of it ie thsi, there are 46 Mega numbers (bottom matrix). The odds against one number hitting are 46:1 (46 to one, or 1 in 45).

So you play two Mega numbers. If you say you've just cut those odds to 23:1, you are implying that the seond Mega number has eliminated not one more but 23 more of the remaining numbers, 2/46 as opposed to 1/46.

Those who run the lotteries love it when players look for consistency in something that's designed not to have any.

There is one and only one 'proven' system, and that is to book the action. No matter the game, let the players pick their own losers.

The Ville, FL United States Member #95879 August 19, 2010 1708 Posts Offline

Posted: January 4, 2011, 11:07 am - IP Logged

Quote: Originally posted by Coin Toss on January 4, 2011

Each ticklet is facing odds of 175,711, 536 :1.

You have those odds against you 13 times over. You've reduced the odds by 13 combinations and that's as good as it gets, you havent's divided those original odds by 13. 13/175,711,536 is hype, that's all.

Thinko of it ie thsi, there are 46 Mega numbers (bottom matrix). The odds against one number hitting are 46:1 (46 to one, or 1 in 45).

So you play two Mega numbers. If you say you've just cut those odds to 23:1, you are implying that the seond Mega number has eliminated not one more but 23 more of the remaining numbers, 2/46 as opposed to 1/46.

coinstoss -

Another way to look at this is to establish the odds if a person spent $46 bux to cover every Mega Ball #.

If that same person played the same 5 set of #'s 46 times with a different MegaBall number (1-46) each time, that persons chances of winning would be 1 in 3,819,816. That would garauntee he/she's chances of winning was 1 in 3,819,816 (on each of those 46 tickets) since they already got the megaball right (they covered all of the possible Megaballs).

Its all math just like Todd is trying to explain. Buying more tickets does in fact give a person better chances.

Go to MS excel and run this formula and play around with the numbers: =combin(56,5) then multiply that # by 46 (# of MegaBalls) ....that will give you the total MegaBall possibilities which is 175,711,536.

If you run this formula =combin(56,5) it will give you 3,819,816. Like I said above, if a person plays the same #'s 46 times with a different MegalBall for each line, that persons chances would be 1 in 3,819,816.

New Jersey United States Member #1 May 31, 2000 23259 Posts Offline

Posted: January 4, 2011, 11:13 am - IP Logged

Quote: Originally posted by Coin Toss on January 4, 2011

Each ticklet is facing odds of 175,711, 536 :1.

You have those odds against you 13 times over. You've reduced the odds by 13 combinations and that's as good as it gets, you havent's divided those original odds by 13. 13/175,711,536 is hype, that's all.

Thinko of it ie thsi, there are 46 Mega numbers (bottom matrix). The odds against one number hitting are 46:1 (46 to one, or 1 in 45).

So you play two Mega numbers. If you say you've just cut those odds to 23:1, you are implying that the seond Mega number has eliminated not one more but 23 more of the remaining numbers, 2/46 as opposed to 1/46.

I understand what you are saying, but you have completely ignored everything I have said. Not worth having a discussion if you can't try to re-phrase the discussion in order to gain consensus. I tried taking your examples and pointing out the flaws, whereas you have ignored what I said. Too bad.

United States Member #13801 April 14, 2005 48 Posts Offline

Posted: January 4, 2011, 11:37 am - IP Logged

Quote: Originally posted by Coin Toss on January 4, 2011

Each ticklet is facing odds of 175,711, 536 :1.

You have those odds against you 13 times over. You've reduced the odds by 13 combinations and that's as good as it gets, you havent's divided those original odds by 13. 13/175,711,536 is hype, that's all.

Thinko of it ie thsi, there are 46 Mega numbers (bottom matrix). The odds against one number hitting are 46:1 (46 to one, or 1 in 45).

So you play two Mega numbers. If you say you've just cut those odds to 23:1, you are implying that the seond Mega number has eliminated not one more but 23 more of the remaining numbers, 2/46 as opposed to 1/46.

2/46 = 4.3%

1/23 = 4.3%

2/46 = 1/23

If you buy 2 numbers in a 46-number game, you have a 4.3% chance of winning.

NEW YORK United States Member #90535 April 29, 2010 11974 Posts Offline

Posted: January 4, 2011, 12:04 pm - IP Logged

Quote: Originally posted by Coin Toss on January 4, 2011

Nope.

Every line of numbers you play is up against the same odds, there is only one set of winning numbers drawn.

barbos is right.

If the fractionalizing theory worked people would just form pools and bet enough to "make sure" they won. Ha ha ha, the lottery loves that. They have this all figured out.

Take a pick 5 game where the odds are 575,757 to one. $5000 in tickets, according to your thoey, would bring the odds down to 115 to one. $20,000 would make the odds 28 to one. If it worked people would be taking out loans and winning $100,000 or more a pop.

And if you're theory worked on Mega Millions it would never have rolled over 15 times to get to a $330,000,000 jackpot.

You have $13 worth of tickets, each one up against odds of 175,711,536.

You are up against 175,711,536: 1 - 13 times.

You are right, whenever someone buys $5 worth of mega millions this does not change the equation to win 1 in 175,711,536. I feel that we are confusing odds and chances. You can increase your chances to win like TODD stated, but you are right COIN TOSS the odds to win never change no matter how many lines we purchase. Each line you play whether 1 or 100 tickets faces the odds of 1 in 175,711,536 to win.Athough the lottery makes you believe otherwise by dividing it 175,711,536/2=87,855,768 . The fact is that you can't cut the odds in half by purchasing 2 tickets like the example implies.I do not believe in purchasing multiple tickets since it takes one set of numbers to win.So when you purchase 2 tickets you have two chances to win out of 175,711,536 each number line facing the odds of 1:175,711,536. So 13 tickets will be 1:175,711,536 , 1:175,711,536 and so on until you reach 13 times, not 13:175,711,536. The equation never changes, it stays the same 1:175,711,536.