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Topic closed. 184 replies. Last post 5 years ago by THRIFTY.

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Zeta Reticuli Star System
United States
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January 17, 2006
10388 Posts
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 Posted: March 5, 2012, 6:17 pm - IP Logged

Stack47,

So you believe that the 2nd, 3rd, 4th, and 5th lines of numbers have taken the 175,000,000 to 1 odds down to 35,000,000 to 1?

Wow, that's amazing, just 4 more lines of numbers has eliminated 140,000,000 combinations. Holy Cow! Each one adittional line eliminates 14,000,000 combinations! Holy Lotto Tickets Batman!

In this thread we have this, from mediabrat:

It's just that those odds are so infinitesimal (roughly six ten-millionths of one percent on a single play) as to not make a difference."

and this, from Guru101:

If buying more lines doesn't improve your odds, then WHY would anyone do it? Come on, simplifying a fraction is learned in grade school.

5/20 = 1/4.

Same thing applies when buying 5 lines in the lottery:

5/175,000,000 = 1/35,000,000.

It amazes me the number of people on here that don't know how to simplify a fraction.

Considering the infrequency of jackpots hit, the frequency of rollovers, and the number of lines played per drawing (if 100,000 people play one line, that's still 100,000 lines )- where are these fraction triggered jackpots?

Of the two theories above mediabrat's proves itself over and over.

If the fraction rtheory worked all we would need would be one massive LP pool to hit jackpots left and right.

You're right about the 5 and 9 on craps tables but in casino parlance \$2, \$4, etc is a 'proper bet." If people wanted to bet \$3 or \$5, etc., the casino loves it because there's an unwrittten rule, If it's to the advantage of the house the player can do it. If they bet \$3 odds on a 5 or 9 and won, they should gbt paid \$4.50 but on a dollar game there are no 'birds' (25 cent cheques) so the way the house sees it it is making 50 cents by not paying it.

It's also kind of incredible that you would use the number of combinations on a set of dice to try and prove the fraction argument for a lottery game with 175,000,000 combinations.

As a note on the dice odds, this is how place bets came into being. A floorman in New Orleans told a player if he just wanted to place a number (bet it without the dice putting it on a nnumber) if he would add \$1 to the bet the house would add \$1 to the payoff (or units, a \$5 bet would be one unit).

Thus pass line odds. 4 and 10:

2:1

or

8:4

+1 +1

========

9:5

Same with the 6 & 8 and 5 & 9, 7:6 and 7:5.

Talking about if it's to the advantage of the house it's ok- in some casinos in the South they book "put bets", the player "puts" it on the number (tells the dealer to) instead of placing it. Hello, these bets lose on a winner 7, place bets are off on a winner 7. Yup, if it's to the advantage of the house it's ok.

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.

United States
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May 30, 2004
5142 Posts
Offline
 Posted: March 5, 2012, 6:56 pm - IP Logged

u>>>>>>>>>>>>>>>>>>>>>>>>>>

psykomo

Kentucky
United States
Member #32652
February 14, 2006
7340 Posts
Offline
 Posted: March 5, 2012, 11:16 pm - IP Logged

Stack47,

So you believe that the 2nd, 3rd, 4th, and 5th lines of numbers have taken the 175,000,000 to 1 odds down to 35,000,000 to 1?

Wow, that's amazing, just 4 more lines of numbers has eliminated 140,000,000 combinations. Holy Cow! Each one adittional line eliminates 14,000,000 combinations! Holy Lotto Tickets Batman!

In this thread we have this, from mediabrat:

It's just that those odds are so infinitesimal (roughly six ten-millionths of one percent on a single play) as to not make a difference."

and this, from Guru101:

If buying more lines doesn't improve your odds, then WHY would anyone do it? Come on, simplifying a fraction is learned in grade school.

5/20 = 1/4.

Same thing applies when buying 5 lines in the lottery:

5/175,000,000 = 1/35,000,000.

It amazes me the number of people on here that don't know how to simplify a fraction.

Considering the infrequency of jackpots hit, the frequency of rollovers, and the number of lines played per drawing (if 100,000 people play one line, that's still 100,000 lines )- where are these fraction triggered jackpots?

Of the two theories above mediabrat's proves itself over and over.

If the fraction rtheory worked all we would need would be one massive LP pool to hit jackpots left and right.

You're right about the 5 and 9 on craps tables but in casino parlance \$2, \$4, etc is a 'proper bet." If people wanted to bet \$3 or \$5, etc., the casino loves it because there's an unwrittten rule, If it's to the advantage of the house the player can do it. If they bet \$3 odds on a 5 or 9 and won, they should gbt paid \$4.50 but on a dollar game there are no 'birds' (25 cent cheques) so the way the house sees it it is making 50 cents by not paying it.

It's also kind of incredible that you would use the number of combinations on a set of dice to try and prove the fraction argument for a lottery game with 175,000,000 combinations.

As a note on the dice odds, this is how place bets came into being. A floorman in New Orleans told a player if he just wanted to place a number (bet it without the dice putting it on a nnumber) if he would add \$1 to the bet the house would add \$1 to the payoff (or units, a \$5 bet would be one unit).

Thus pass line odds. 4 and 10:

2:1

or

8:4

+1 +1

========

9:5

Same with the 6 & 8 and 5 & 9, 7:6 and 7:5.

Talking about if it's to the advantage of the house it's ok- in some casinos in the South they book "put bets", the player "puts" it on the number (tells the dealer to) instead of placing it. Hello, these bets lose on a winner 7, place bets are off on a winner 7. Yup, if it's to the advantage of the house it's ok.

"So you believe that the 2nd, 3rd, 4th, and 5th lines of numbers have taken the 175,000,000 to 1 odds down to 35,000,000 to 1?"

Nope and I said there are FIVE 35 million to 1 chances. It means for every 35 million to 1 winning chance, there are 34,999,999 losing chances. Multiple that by five because there are four other 35 million to 1 chances and you'll have the same number of losing chances you began with.

It's like in the 500,000 ticket raffle games with four chances to win \$1 million where the odds are expressed as 125,000 to 1. We all know there are 499,996 tickets that won't win \$1 million just like we know when we buy 5 MM tickets, there are 175,711,531 combo that can be drawn preventing us from winning the jackpot. Would it help if I said I have 1 chance for every 125,000 chances in a raffle game or 1 chance for every 35,142,306 chances in a MM drawing?

"It's also kind of incredible that you would use the number of combinations on a set of dice to try and prove the fraction argument for a lottery game with 175,000,000 combinations."

The odds against a "4" or "10" is 6 to 3 and 6 to 4 with the "5" and "9" and those payoffs are reduced to the lowest common denominator; 2 to 1 and 3 to 2. In MM the number of ways we can lose is determined by the number of chances we buy subtracted from the total number of chances. For simplicity  we'll say there are 175 million chances and minus the 5 bought chances equals 174,999,995.

Ways to lose: 174,999,995. Ways to win: 5. 174,999,995 to 5 equals 34,999,999 to 1.

Ways to lose when the point is "5": 6. Ways to to win: 4. 6 to 4 equals 3 to 2.

If they can reduce the real crap odds, why can't we reduce the real MM odds to the lowest common denominator?

I can't see why there's an argument against expressing the odds against 5 chances as 35 million to 1 when everybody knows it takes 1,757,116 tickets just to get a 1% chance of winning the jackpot.

Australia
Member #37136
April 11, 2006
3315 Posts
Offline
 Posted: March 5, 2012, 11:42 pm - IP Logged

next ill be told buying ea week wont increase my odds of winning!

2014 = -1016; 2015= -1409; 2016  = -1171; 2017 = ?  TOT =  -3596

keno historic = -2291 ; 2015= -603; 2016= -424; 2017 = ? TOT = - 3318

Taunton, Ma
United States
Member #123005
February 11, 2012
136 Posts
Offline
 Posted: March 6, 2012, 12:28 am - IP Logged

u>>>>>>>>>>>>>>>>>>>>>>>>>>

psykomo

Really ? Have you even read this thread? I've been going back and forth from the start.  The emoticon was meant to express my exasperation with this argument. Next time I guess I'll have to spell it out or psykomo the forum police will come and give me the business. Good grief, don't be a troll please.

Taunton, Ma
United States
Member #123005
February 11, 2012
136 Posts
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 Posted: March 6, 2012, 12:52 am - IP Logged

Stack47,

So you believe that the 2nd, 3rd, 4th, and 5th lines of numbers have taken the 175,000,000 to 1 odds down to 35,000,000 to 1?

Wow, that's amazing, just 4 more lines of numbers has eliminated 140,000,000 combinations. Holy Cow! Each one adittional line eliminates 14,000,000 combinations! Holy Lotto Tickets Batman!

In this thread we have this, from mediabrat:

It's just that those odds are so infinitesimal (roughly six ten-millionths of one percent on a single play) as to not make a difference."

and this, from Guru101:

If buying more lines doesn't improve your odds, then WHY would anyone do it? Come on, simplifying a fraction is learned in grade school.

5/20 = 1/4.

Same thing applies when buying 5 lines in the lottery:

5/175,000,000 = 1/35,000,000.

It amazes me the number of people on here that don't know how to simplify a fraction.

Considering the infrequency of jackpots hit, the frequency of rollovers, and the number of lines played per drawing (if 100,000 people play one line, that's still 100,000 lines )- where are these fraction triggered jackpots?

Of the two theories above mediabrat's proves itself over and over.

If the fraction rtheory worked all we would need would be one massive LP pool to hit jackpots left and right.

You're right about the 5 and 9 on craps tables but in casino parlance \$2, \$4, etc is a 'proper bet." If people wanted to bet \$3 or \$5, etc., the casino loves it because there's an unwrittten rule, If it's to the advantage of the house the player can do it. If they bet \$3 odds on a 5 or 9 and won, they should gbt paid \$4.50 but on a dollar game there are no 'birds' (25 cent cheques) so the way the house sees it it is making 50 cents by not paying it.

It's also kind of incredible that you would use the number of combinations on a set of dice to try and prove the fraction argument for a lottery game with 175,000,000 combinations.

As a note on the dice odds, this is how place bets came into being. A floorman in New Orleans told a player if he just wanted to place a number (bet it without the dice putting it on a nnumber) if he would add \$1 to the bet the house would add \$1 to the payoff (or units, a \$5 bet would be one unit).

Thus pass line odds. 4 and 10:

2:1

or

8:4

+1 +1

========

9:5

Same with the 6 & 8 and 5 & 9, 7:6 and 7:5.

Talking about if it's to the advantage of the house it's ok- in some casinos in the South they book "put bets", the player "puts" it on the number (tells the dealer to) instead of placing it. Hello, these bets lose on a winner 7, place bets are off on a winner 7. Yup, if it's to the advantage of the house it's ok.

You don't get it do you? We're not saying that 140,000,000 combo's were eliminated, We are only saying that your odds can be expressed as a fraction and they improve upon each combination played. Just the same as 20-1 can be expressed as 10-1, Do you beleive that we're saying that 10 of the 20 just disapeared? No, We are simplifying a fraction. I don't understand at all how you and thrifty can make this argument. It's that we're dealing with such mind boggling numbers that they are twisting your view.

175,000,000 combinations

2 tickets will make those odds 87.5mill to one

20 combinations

2 tickets will make those odds 10-1

If you're going to claim that by simplifying the larger number is making the combinations disappear then to be consistant in your argument, you have to say the same with the smaller number. So I ask you, By simplifying 20-2 down to 10-1 , am I making 10 combinations disapear? or am I merely simplifying a fraction to recalculate the odds?

And this statement "If the fraction theory worked all we would need would be one massive LP pool to hit jackpots left and right" does zero for your argument. It's completely ludicrous for a couple reasons. 1-a massive pool is not a garantee, in fact it's still a loooooooong shot and 2-The payoff would not match the investment needed even if we could cover all the combinations. 3-simplifying odds as fractions does not equal covering all combinations even if it's hundreds or thousands, It remains a long shot. Oh and fractions aren't a theory nor some magic formula. Odds can be expressed as fractions and therefore simplified as such.

Whiskey Island
United States
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April 24, 2010
12808 Posts
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 Posted: March 6, 2012, 1:00 am - IP Logged

Your right their is no need to buy large amount of tickets . When you can do it with 1 or 19 set combinations .All that matters is if you have all the right digts being played to win ...

Zeta Reticuli Star System
United States
Member #30470
January 17, 2006
10388 Posts
Offline
 Posted: March 6, 2012, 1:26 am - IP Logged

"We are only saying that your odds can be expressed as a fraction and they improve upon each combination played"

and mediabrat explained how much they are improved:

"It's just that those odds are so infinitesimal (roughly six ten-millionths of one percent on a single play) as to not make a difference."

"2 tickets will make those odds 87.5mill to one"

This is the same as saying the second ticket eliminates 87,500,000 combinations. Saying it reduces the odds by half is smoke and mirrors by the lotteries, ticket sellers, and system sellers.

THE ODDS ARE BASED ON THE TOTAL NUMBER OF POSSIBLE COMBINATIONS IN ANY GAME. ANY SET OF ADDITIONAL NUMBERs PLAYED REDUCES THOSE COMBINATIONS BY 1, NOT HALF.

THE LOTTERIES LOVE SUCH BELEIFS.

Let's try this - you go to a Vegas sports book before the NFL season starts and you make two futures bets, you bet on two teams to win the Super Bowl. There are 30 teams. The odds are going to vary, but of 30 teams. Only one is going to win it. The odds with a field of 30 are 29:1, 29 to 1.

You bet on two teams. You have two 29 to 1 shots. Neither of those bets is 15 to 1, although you bet two of thirty. There are two separate entities, each 29 to 1. That's as good as it gets.

GOOD LUCK

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.

Taunton, Ma
United States
Member #123005
February 11, 2012
136 Posts
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 Posted: March 6, 2012, 1:54 am - IP Logged

Not saying that an additional ticket reduces the combinations in half merely the odds.

And it wouldn't continue to be cut in half with each ticket merely divided into the total number of combinations.

Your arguing with mathematical facts. Odds are indeed based on total number of combinations VERSUS COMBINATIONS PLAYED.

It is a fact that 2 - 175mill still leaves 174,999,998 but that does not mean that the odds can't be expressed as 1 - 87.5 mill. One fact does not preclude the other. You seem to think the former doesn't allow for the latter but you're wrong.

No one here is naiive enough to believe that an additional ticket or an additional 1,000 tickets gives them favorable odds. Just slightly better odds than 1 ticket alone that is all. If you can't concede any of these points then I fear you have closed your mind to anything that counters your argument. You haven't rebutted any piece of math we've given you merely rehashed the same false statements over and over. You can google fractions, odds and percentages and everything you find will support what we've said. I'm done with this subject. I'm actually getting aggrevated at this point.

Zeta Reticuli Star System
United States
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January 17, 2006
10388 Posts
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 Posted: March 6, 2012, 2:28 am - IP Logged

"Your arguing with mathematical facts. Odds are indeed based on total number of combinations VERSUS COMBINATIONS PLAYED."

Read the back of the playslip. Odds are based on having the one winning combination of the total number of possible combinations.

If you think that's a false statement I suggest from hereon you skip the formality of actually playing your tickets and just send the money that would have been spent on fractions straight to the lottery of your choice.

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.

NY
United States
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October 16, 2005
3502 Posts
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 Posted: March 6, 2012, 2:29 am - IP Logged

You don't get it do you? We're not saying that 140,000,000 combo's were eliminated, We are only saying that your odds can be expressed as a fraction and they improve upon each combination played. Just the same as 20-1 can be expressed as 10-1, Do you beleive that we're saying that 10 of the 20 just disapeared? No, We are simplifying a fraction. I don't understand at all how you and thrifty can make this argument. It's that we're dealing with such mind boggling numbers that they are twisting your view.

175,000,000 combinations

2 tickets will make those odds 87.5mill to one

20 combinations

2 tickets will make those odds 10-1

If you're going to claim that by simplifying the larger number is making the combinations disappear then to be consistant in your argument, you have to say the same with the smaller number. So I ask you, By simplifying 20-2 down to 10-1 , am I making 10 combinations disapear? or am I merely simplifying a fraction to recalculate the odds?

And this statement "If the fraction theory worked all we would need would be one massive LP pool to hit jackpots left and right" does zero for your argument. It's completely ludicrous for a couple reasons. 1-a massive pool is not a garantee, in fact it's still a loooooooong shot and 2-The payoff would not match the investment needed even if we could cover all the combinations. 3-simplifying odds as fractions does not equal covering all combinations even if it's hundreds or thousands, It remains a long shot. Oh and fractions aren't a theory nor some magic formula. Odds can be expressed as fractions and therefore simplified as such.

"And this statement "If the fraction theory worked all we would need would be one massive LP pool to hit jackpots left and right" does zero for your argument."

You're not understanding what it is or how much he doesn't understand. Notice that he said "Each one adittional line eliminates 14,000,000 combinations!" Forget that it doesn't seem to have any realistic relationship to the numbers he started with. Ignore his disagreement for the moment. He doesn't even understand what it is he's being told. He doesn't know whether to add and subtract or to multiply and divide. In the past he's made posts that indicate that he thinks people are claiming that each additional ticket doubles the chances of winning, instead of only the 2nd ticket doing so.

That's where the nonsense about a buying a bunch of tickets to make it a sure thing are coming from. His ability to grasp what he's being told seems to have him convinced that people are claiming you can have a 100% chance of winning by buying a relatively modest fraction of the combinations.

Then there's this: "It's also kind of incredible that you would use the number of combinations on a set of dice to try and prove the fraction argument for a lottery game with 175,000,000 combinations."

He's apparently convinced that math works differently when the numbers are bigger. In fact, I'm pretty sure he specifically said that once.

That's the one thing he's extremely good at.  Ask an incredibly simple question where a correct answer goes against his argument and odds are he'll go off in some wildly different direction and avoid the question, or maybe claim that lottery math is different than math about other things.

So let's imagine a simple scenario. I buy 2 tickets for pick 4. One is for 0000 and the other is for 0001. There are 10,000 possible combinations. 5000 of them are odd and 5000 are even. My first ticket gives me a 1 in 5000 chance if the winning number is from the 5000 numbers in the even group. My second ticket gives me a 1 in 5000 chance if the winning number is from the 5000 numbers in the odd group. That means my 2 tickets gives me odds of 1 in 5000 whether the winning number is odd or even, even though there are still 9,998 numbers that I didn't play. That clearly demonstrates that my two 1 in 10,000 chances double my chances to 1 in 5,000.

The exact same concept works by splitting the possible numbers into 10 groups of 1000 each and buying 10 tickets. Buy 0000 to cover the 10% of possible numbers from 0000 to 0999. Buy 1000 to cover the 10% of possible numbers from 1000 to 1999. Buy 2000 to cover the 10% of possible numbers from 2000 to 2999. And soon. Whichever group the winning number is from there's a 1 in 1000 chance it will match the ticket for that group.  10 tickets, 10 in 10,000 chances gives you a 1 in 1000 chance of winning.

We know Coin Toss will see those two scenarios. If I'm wrong, I'm sure he can clearly explain why, so let's see what happens. Of course Thrifty is welcome to play, too.

Zeta Reticuli Star System
United States
Member #30470
January 17, 2006
10388 Posts
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 Posted: March 6, 2012, 2:58 am - IP Logged

Floyd,

Most of this discussion is about the false belief that more tickets increases the player's odds tremendously.

It just isn't the case and the results prove it over and over most drawings.

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.

Australia
Member #37136
April 11, 2006
3315 Posts
Offline
 Posted: March 6, 2012, 6:42 am - IP Logged

Floyd,

Most of this discussion is about the false belief that more tickets increases the player's odds tremendously.

It just isn't the case and the results prove it over and over most drawings.

2/175 000 000 = 1/87 500 000 = 99.99999885714285714% of failure

theyre all the same thing.

2014 = -1016; 2015= -1409; 2016  = -1171; 2017 = ?  TOT =  -3596

keno historic = -2291 ; 2015= -603; 2016= -424; 2017 = ? TOT = - 3318

Taunton, Ma
United States
Member #123005
February 11, 2012
136 Posts
Offline
 Posted: March 6, 2012, 11:24 am - IP Logged

Floyd,

Most of this discussion is about the false belief that more tickets increases the player's odds tremendously.

It just isn't the case and the results prove it over and over most drawings.

SHOW ME ONE SINGLE POST WHERE ANYONE HAS MADE THAT CLAIM !!!!!!! WHO THE HECK HAS SAID THAT ?

Zeta Reticuli Star System
United States
Member #30470
January 17, 2006
10388 Posts
Offline
 Posted: March 6, 2012, 2:40 pm - IP Logged

Well I guess a lot of people just don't realize how odds are expressed.

Since dice have been mentioned here, the payout for hitting a 12 or 2  (not the odds, the payout, follow aliong)

is 30 TO 1. This is expressed as 30:1.

some casinos, to make it sound better, labeled their prop bets 31 FOR 1. leading some players to think they were getting an extra \$1. Not the case, the for includes the bet.

The real odds are 35:1, 36 combinations, one has to hit. 35 TO 1 is expressed as 35:1, i in 36 is expressed as 1:36.

That is how odds are expressed.

Since that is how odds are expressed, 87.5 M to one means one thing and one thing only, 87.5M to one

87,500,000 :1........but it's not valid when winning is based on 175,000,000 to one.

Good luck getting the extra dollars on the prop bets.

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.

 Page 7 of 13