United States Member #23787 October 16, 2005 68 Posts Offline

Posted: December 30, 2005, 8:29 pm - IP Logged

Quote: Originally posted by Rick G on December 30, 2005

Rlevins,

When you buy 10 combos at 175 million to one odds of hitting the jackpot you are saying that your odds are lowered to 17.5 million to one if I'm not mistaken.

I understand what you are saying when you extrapolate and buy all 175 million combos. But this concept would have to be accepted on an extremely "sliding scale".

But going back to the original example......what happened to those other 174,999,990 combinations that could have beat you? They don't just disappear (as hypersoniq so eloquently put it).

Odds are odds. If you are playing a one digit out of ten game and you play one digit your odds are 9-1 of getting it right. If you play two numbers your odds are lowered to 8-1 of getting it right. Why? Because your first number had to have been wrong so it must be included in the losing propositions. The other digits remaining are still possibilities and can't be ignored.

If you play all ten digits then you will win...but what is the formula for this "sliding scale"?

PS...I'm referring to odds not probability. In my experience odds and probability are the same thing. I'd love to have someone show me the difference.

You can think this in this way , say you buy 175 tickets out of 175 then you have probability of 1. All 175 combinations have equal chance and the sum of probabilities of all numbers would be equal to 1. So to exaplin your "others", I am not saying that they disappeared. What I am saying is that when you got your 1 ticket, You have a probability of 1/175 and for others it is 174/175. 1/175 + 174/175 = 1. Similarly if you buy 10 tickets your probability of winning becomes 10/175 and for others it becomes 165/175. Here again 10/175 + 165/175 =1 Here is the formula for ODDS=PROBABILITY/(1-PROBABILITY) http://www.colorado.edu/education/DMP/activities/discrete_prob/jdaact04.html "Probability of an event occuring can be calculated as the number of ways that the event can occur divided by the total number of outcomes. The odds of something occuring can be calculated as the ratio of the number of ways that the event can occur to the number of ways in which the event does not occur"

The Carolinas - Charlotte United States Member #21627 September 12, 2005 4138 Posts Offline

Posted: December 30, 2005, 8:43 pm - IP Logged

Quote: Originally posted by Rick G on December 30, 2005

Rlevins,

When you buy 10 combos at 175 million to one odds of hitting the jackpot you are saying that your odds are lowered to 17.5 million to one if I'm not mistaken.

I understand what you are saying when you extrapolate and buy all 175 million combos. But this concept would have to be accepted on an extremely "sliding scale".

But going back to the original example......what happened to those other 174,999,990 combinations that could have beat you? They don't just disappear (as hypersoniq so eloquently put it).

Odds are odds. If you are playing a one digit out of ten game and you play one digit your odds are 9-1 of getting it right. If you play two numbers your odds are lowered to 8-1 of getting it right. Why? Because your first number had to have been wrong so it must be included in the losing propositions. The other digits remaining are still possibilities and can't be ignored.

If you play all ten digits then you will win...but what is the formula for this "sliding scale"?

PS...I'm referring to odds not probability. In my experience odds and probability are the same thing. I'd love to have someone show me the difference.

RickG - like you were stating in your 10 digit example. That is right. If you play 1 number, then you have a 1/10 chance or 10%. Your odds are 1:10 of winning. The percentage is calculated by dividing the number you have by the number of overall chances.

When you play two, then your odds go down to 1/5 or 2/10, meaning you have a 20% chance of winning and an 80% chance of losing. When you are playing the 175 million that you are talking about, and you play 10 sets of numbers, then your chances are 10:175,000,000 or 1:17,500,000 it's just that your odds of losing are now 17,499,999:17,500,000. The odds are completely different than what you are suggesting. It does in fact lower your odds of winning the more numbers you buy.

mid-Ohio United States Member #9 March 24, 2001 19825 Posts Offline

Posted: December 30, 2005, 8:48 pm - IP Logged

Rick G writes: "If there are 175 million combinations and you buy 55 tickets your odds of winning are 175 million MINUS 56 (including the winning ticket), not DIVIDED by 55; (that comes out to 174,999,944 to one, not 3,180,000 to one)."

Put another way, if there are 175 million combinations and you buy 0 tickets your odds of winning are 175 million MINUS 1 (including the winning ticket), not DIVIDED by 0; (that comes out to 174,999,999 to one, not infinity to one or impossible).

* you don't need to buy more tickets, just buy a winning ticket *

FEMA Region V Camp #21 United States Member #520 July 27, 2002 5699 Posts Offline

Posted: December 30, 2005, 9:20 pm - IP Logged

Rlevins and Cps,

I understand what you guys are saying and will check out the link Rlevins provided. But I can't get past the example I provided above.

A one of ten digit game should have the same probabilities (or odds) as any X of N digit game.

Can either of you refute the logic of my one digit out of ten example? If so, please explain. I am totally open to your input and would appreciate your feedback. So far I'm not getting it.

I don't want to sound like a Vulcan but I need some logic here. (And if you could, limit it to a one out of ten digit game proposition so we're all on the same page).

Thanks!

Posted 4/6: IL Pick 3 midday and evening until they hit: 555, 347 (str8).

United States Member #23787 October 16, 2005 68 Posts Offline

Posted: December 30, 2005, 10:56 pm - IP Logged

I did few searches and found something which somewhat clarifies these.

http://mathforum.org/library/drmath/view/56706.html Probability vs. Odds, Explained Probability (Chance) and Odds Are Not Quite the Same Thing Odds and probability contain the same information, but express it differently. Say a rectangle is divided into boxes ( 1 X 4 matrix) and fill in one box. The probability of being colored in is 1/4 or 25% but the "odds" of being colored in is 1:3 or 0.33. For example, when we may say that the odds are 4:1 that the Indiana Pacers will win their basketball game, this means the probability of winning is 80%. The probability of an event is defined as: (Chances for) P(x) = --------------- (Total chances) So, for example, the probability of drawing an ace in a single deck of 52 cards is 4/52 = 1/13 (or about 0.077 = 7.7%). Odds, on the other hand, are given as: (Chances for) : (Chances against) Incidentally, odds of 1:1 would be read as "one TO one", not "one OUT OF one." (The words "out of" seem to imply total chances, which is probability, not odds.) Since (Total chances) = (Chances for) + (Chances against), we can calculate (Chances against) = (Total chances) - (Chances for). The odds of drawing an ace in a deck of cards is 4:(52-4) = 4:48 = 1:12. Notice the difference in the second value; probability uses (Total chances), but odds use (Chances against). This is why the probability (if considered as a ratio) and the odds are different.

If you consider your example of 1 digit out of 10. let us say that winning number is 6 (but you do not know it) So chances of picking 6 is 1/10 (probability) or 1 in 10 odds. but let us say that you are allowed to pick to numbers and then what are the cahnces that you will pick 6? probability will be 1/10 + 1/10. Odds will be 2 in 8 (not 1 in 8) . You are thinking that since there is only 1 number by which you can win,you are thinking that this makes chances to win also as 1. Although there is only 1 way to win,you get 2 chances to pick the number so chances/ways to win are 2 and chances/ways to loose are 8. this is also evident from formula probability/(1-probability) .2/(1-0.2)=2/8

Coastal Georgia United States Member #2653 October 30, 2003 1866 Posts Offline

Posted: December 30, 2005, 11:17 pm - IP Logged

Quote: Originally posted by rlevins on December 30, 2005

I did few searches and found something which somewhat clarifies these.

http://mathforum.org/library/drmath/view/56706.html Probability vs. Odds, Explained Probability (Chance) and Odds Are Not Quite the Same Thing Odds and probability contain the same information, but express it differently. Say a rectangle is divided into boxes ( 1 X 4 matrix) and fill in one box. The probability of being colored in is 1/4 or 25% but the "odds" of being colored in is 1:3 or 0.33. For example, when we may say that the odds are 4:1 that the Indiana Pacers will win their basketball game, this means the probability of winning is 80%. The probability of an event is defined as: (Chances for) P(x) = --------------- (Total chances) So, for example, the probability of drawing an ace in a single deck of 52 cards is 4/52 = 1/13 (or about 0.077 = 7.7%). Odds, on the other hand, are given as: (Chances for) : (Chances against) Incidentally, odds of 1:1 would be read as "one TO one", not "one OUT OF one." (The words "out of" seem to imply total chances, which is probability, not odds.) Since (Total chances) = (Chances for) + (Chances against), we can calculate (Chances against) = (Total chances) - (Chances for). The odds of drawing an ace in a deck of cards is 4:(52-4) = 4:48 = 1:12. Notice the difference in the second value; probability uses (Total chances), but odds use (Chances against). This is why the probability (if considered as a ratio) and the odds are different.

If you consider your example of 1 digit out of 10. let us say that winning number is 6 (but you do not know it) So chances of picking 6 is 1/10 (probability) or 1 in 10 odds. but let us say that you are allowed to pick to numbers and then what are the cahnces that you will pick 6? probability will be 1/10 + 1/10. Odds will be 2 in 8 (not 1 in 8) . You are thinking that since there is only 1 number by which you can win,you are thinking that this makes chances to win also as 1. Although there is only 1 way to win,you get 2 chances to pick the number so chances/ways to win are 2 and chances/ways to loose are 8. this is also evident from formula probability/(1-probability) .2/(1-0.2)=2/8

Owwwwwch !!!! Stop it, I give !!! Information overload......

There is a 100% probability I am signing off and going nite nite.

NY United States Member #23835 October 16, 2005 3474 Posts Offline

Posted: December 31, 2005, 5:00 am - IP Logged

Quote: Originally posted by Rick G on December 30, 2005

KY,

Maybe I read the article wrong, but I thought it specifically said that the winning ticket came from the same block of tickets purchased at the same time, location and number of tickets purchased (in this case quick picks which makes it easier to decide).

The defendant DOES have to refute the prosecution's allegations and therefore PROVE their innocence before a jury without any doubts from the jury before they can be found innocent or guilty.

Let's let 12 of his peers figure it out. We don't know all the facts and are not in the courtroom right now.

The article said the winning ticket came from a block of tickets and wasn't a "separate" ticket. It says nothing about how many tickets were in the block, and it doesn't say whether or not they were quickpicks. It's still there if you want to look at it again.

As a matter of law the defendant doesn't have to prove anything. The burden of proof is on the plaintiff or prosecution. If the prosecution can't prove their claim they shouldn't win. I'd say they "don't win," but juries, and occasionally judges, often make stupid decisions. You can see how many of the "jurors" here decided he's guilty simply because of unsupported claims by people who want a share of the pie. As a matter of practicality, if any of the evidence presented suggests that the claims might be true it would be a very good idea for the defendant to refute the evidence if possible, but there is no requirement to do so. Depending on whether it's a criminal or civil matter the proof must be beyond a reasonable doubt or the preponderance of the evidence must be against the defendant.

In this case, the plaintiffs can all get up and testify that they had whatever agreement and that the ticket was bought for the pool, but that's not proof, and arguably isn't even evidence at all. For a verdict that fairly decides in their favor they need verifiable evidence that the ticket was one of the 55 that should have been bought for the pool by the defendant, that the defendant violated the rules of the pool, or that the defendant (or his wife) didn't buy any tickets for himself. Right now we don't have any information on what their rules were or if they have a way to prove what they were, and it's probably impossible to prove that he didn't buy any tickets for himself. That only leaves the possibility of proving that the ticket really was bought for the pool. Showing that the block of tickets including the winner was 55 tickets would be very good proof and make a reasonable juror conclude that the ticket was bought for the pool. As a practical matter, it would then be up to the defendant to prove that he also bought 55 tickets for himself to have any chance of winning.

As for letting a real jury decide, it should only get that far if the plaintiffs have something more than their allegations.

Euclid United States Member #29327 December 31, 2005 5 Posts Offline

Posted: December 31, 2005, 7:32 am - IP Logged

Okay folks.. try this one. You can get by, but you can never get away.

The problem with group plays rests with the the leader of the group. When I say leader, I mean one with the moral turpitude to do the right thing and that includes being fair. All lotteries have the ability to determine when, where, and what time an online ticket is purchased. This also confirms why there really is no perfect crime, because you don't know what detail is missing that will get you caught.

As for purchasing tickets as an annuity, Mega Millions does not have a cash option feature at the time of purchase, all tickets are sold as annuities. For the record, the claim was paid that's why the plaintiffs request the defendant's assets be frozen.

I find it rather odd that since the lottery proved the ticket came from the block of tickets sold for the group that the man did not just say" My bad, here's your share" or he could have blamed it on "the Mrs" and saved face and friendship.

Another theory is that the wife could have checked the numbers without her husband's knowledge and swapped out the ticket! It is quite possible that he really didn't know a ticket was swapped. If the ticket was purchased at a different location, it should be simple enough to examine which one was swapped out, provided they still have all the other original tickets.

There is nothing wrong with playing with a group, but the guidlines have to be so tight that they squeak when they walk to prevent this type of situation from occuring. One of the main features of a group play should be, any ticket played outside of the group should be done at a different location than that of the group play. But the bottom line is honesty is your best policy!

Dallas, Texas United States Member #3275 January 7, 2004 25 Posts Offline

Posted: December 31, 2005, 7:52 am - IP Logged

“Ohio Couple who won $250,000 playing Mega Million discovered the hard way that money can come between friends” is a stupid statement.

First off Friends don’t steal from you. This couple stole $250,000 from people they were using.

I run two lottery pools, one of my pools just turned 1 year old on Christmas Day. My Mega Million Pool will be two years old this coming February.

My members, who number 18 to 27 people, always get copies of their tickets. I ask everyone who wants to play give me $10.00 for 1 share (they can and do buy more than one share). I give me at least once a week updates (newsletters) concerning their wins, and other lottery news. I live in Texas but I have two members, who have been playing with me for some time, live out of state.

I communicate with my members via e-mail. They get all their tickets with their newsletter via e-mail. I scan all our tickets and include them with their first newsletter and any other tickets we get thru winning; I scan those as well and send them with my updates.

Everyone gets, not only their tickets, but a membership list. I might have 25 members but have sold 35 shares. That tells you that I have a least one member who purchase more shares.

There is never a question about what numbers they have, who is in the club, and how much everyone paid. I don’t want any surprises when we get to Austin to claim our money.

I think most of my members of my club trust me. Most of them have been with me for a long time. Almost all of them have worked with me at some point. Even the ones who leave the company that I work for still keep in touch and either drop by to give me their $10 or $20.00.

It has never occurred to me to try to steal from them. I can’t imagine violating their trust. Most of my members, like me, live from paycheck to paycheck. Above all, I can imagine the punishment from God for doing anything to harm any of members. I am absolutely terrified of Him.

For all of those who made comments like “Stay out pools” and such, I need to tell you there is nothing wrong with pools, if you do it correctly.

The last line says, “Salcone no longer joins the others for breakfast.” Is it any wonder? He is a thief and tried to cheat his friends, he deserves more than No Breakfast.