NY United States Member #23835 October 16, 2005 3502 Posts Offline

Posted: March 5, 2007, 1:52 am - IP Logged

Quote: Originally posted by justxploring on March 4, 2007

Yes, you make sense. But you are proving my point. You can't possibly be cutting the odds in half by buying 2 tickets for MM. Seriously, once I had a discussion about grammar with a boss when I refused to type a letter the wrong way. I faxed it to Harvard (this is a true story) and it was presented to the English Literature department. I got a response, which was very surprising. I was correct. So maybe I'll send a letter to M.I.T.

In this case I am not 100% sure I am correct. However, I'm thrilled that Coin Toss and I are on the same page, since I began to doubt my logic. So far nobody has convinced me I am wrong since it doesn't make sense. I mean, take the old "odds of being hit by lightning" example. If a man is standing in his yard, there is a possibility that lighting will strike (odds vary by area, weather and other factors of course) So now his neighbor comes over to chat. You have 2 men (or tickets) standing in the same place. Does it now double the chances that lighting will strike? Does it cut the odds in half when one of them walks away? No. That's because the possibility that lightning will strike is not related to the men standing there. The odds of winning a MM or PB game are set odds. If a game has odds of 1 in 140M, then every time a person buys another ticket, the possibility of winning is increased by 1 for that player, not 100%, the same as the odds aren't reduced by 50% if he only buys 1 ticket.

As JAG already wrote, this is going nowhere.

Another decent analogy, but your reasoning is faulty because, the lightning will strike. The question is will it strike a yard with somebody standing in it?

Imagine there are 140 million yards. Some of the yards have a person standing in them and some don't. The yards are all the same size and they all have the same chance of being hit by lightning. There's a storm every Wednesday and Saturday night, and lightning always hits 1 of the 140 million yards. Does it make sense that the chances of the lightning hitting a yard with somebody standing in it are directly proportional to the number of yards with somebody standing in them? On the first Wednesday of the month there's only 1 yard with somebody standing in it. The following Saturday two yards have somebody standing in them. Isn't it twice as likely that the lightning will hit one of th eyards with somebody in it?

Let's go back to Wednesday with only 1 person in 1 yard. That yard happens to be yard # 5-6-7-8-9 / 15, and the lightning bolt comes streaking down right towards the corner where 4 yards meet. It's a toss up. Maybe it will go north and hit yard # 5-6-7-8-9 / 15, but maybe it will go south and hit yard # 5-6-7-8-9 / 16. It could go east and hit yard # 5-6-7-8-9 / 17, or it could go west and hit yard # 5-6-7-8-9 / 18. Now imagine the same scenario, but there's also somebody standing in yard # 5-6-7-8-9 / 16. Is there twice as much chance that it will strike in one of the yards that somebody is standing in?

Zeta Reticuli Star System United States Member #30470 January 17, 2006 10389 Posts Offline

Posted: March 5, 2007, 2:12 am - IP Logged

KY Floyd

"The following Saturday two yards have somebody standing in them. Isn't it twice as likely that the lightning will hit one of the yards with somebody in it?"

"Twice as likely" in a world of 140,000,000 to one is a wee bit optimistic. Actually, when the second person stands in their yard what has happened is the chances of lightning hitting an occupied yard have only been reduced by 1/1400,000,000.

We started out with no one in a yard. the first occupied yard becomes 1 occupies yard our of 140,000,000. the second occupies uyard becomes another one out of 140,000,000.

So we have:

occupied yards: 2

Unoccupied yards: 139,999,998.

Each occupied yard has a 1 in 140,000,000 chance of being hit.

But the way the "cut the odds in half" school of thought portrays this is when the second person goes out in their yeard, they (one person), are occupying half of the remaining 139,999,999 yards.

1 person in 1 yard being followed by another person in their yard being called "twice as likely" may be brilliant in marketing but it's not going to put any shekels in anyone's pocket in lotto.

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.

Wandering Aimlessly United States Member #25360 November 5, 2005 4461 Posts Offline

Posted: March 5, 2007, 2:19 am - IP Logged

KY Floyd, this is interesting. First everyone is telling me that chances aren't the same as odds. Actually, I agree on that to a point, but now you are using the above as an example to prove that the odds are cut in half if you purchase 2 tickets.

Yes, if you have 2 people standing outside during an electrical storm there's a greater chance that someone will get hit than if nobody is outside at all. I mean, if nobody is outside then it won't happen at all. Buying 2 tickets does give you twice the chances, but it doesn't cut the odds in half. 2 different animals!

Using your pick-4 example with odds of 1 in 10,000 (those are the odds for the Florida Play-4 game for a straight) then according to your theory, buying 2 tickets would make the odds 1 in 5,000 (half) and 3 would make the odds 1 in 2,500 and 4 would make them 1 in 1,250 and 5 would be 1 in 625 and 6 would be 1 in 312 and 7 would be 1 in 156 and 8 would be 1 in 78 and 9 would be 1 in 39 and 10 would be 1 in 19 and 11 would be 1 in 10 and 12 would be 1 in 5 and 13 would be 1 in 2.5 and then you'd have to win!! So, since Play-4 pays $5,000 I better go out tomorrow and purchase 14 tickets for Play-4 because if it's true that the odds just keep splitting every time I buy a ticket, I'll win $5,000 by only spending $14 or $15. Wow!

Zeta Reticuli Star System United States Member #30470 January 17, 2006 10389 Posts Offline

Posted: March 5, 2007, 2:27 am - IP Logged

Yeah, it's so easy to just keep reducing things by half is amazing more people don;t hit. How can the Pick 4 keep making money like this? All you need is $14.

I'm going to call it a night. Think tonight I'll go to bed one hour earlier, in order to get twice as much sleep!

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.

Michigan United States Member #22395 September 24, 2005 1583 Posts Offline

Posted: March 5, 2007, 6:04 am - IP Logged

quote justx

Using your pick-4 example with odds of 1 in 10,000 (those are the odds for the Florida Play-4 game for a straight) then according to your theory, buying 2 tickets would make the odds 1 in 5,000 (half) and 3 would make the odds 1 in 2,500 and 4 would make them 1 in 1,250 and 5 would be 1 in 625 and 6 would be 1 in 312 and 7 would be 1 in 156 and 8 would be 1 in 78 and 9 would be 1 in 39 and 10 would be 1 in 19 and 11 would be 1 in 10 and 12 would be 1 in 5 and 13 would be 1 in 2.5 and then you'd have to win!! So, since Play-4 pays $5,000 I better go out tomorrow and purchase 14 tickets for Play-4 because if it's true that the odds just keep splitting every time I buy a ticket, I'll win $5,000 by only spending $14 or $15. Wow!

Is this what you meant? Divide 10,000 by the amount of tickets to arrive at:

1 ticket = 1 in 10,000 2 tickets = 1 in 5,000 3 tickets = 1 in 3,333 4 tickets = 1 in 2,500 5 tickets = 1 in 2,000 6 tickets = 1 in 1,666 7 tickets = 1 in 1,428 8 tickets = 1 in 1,250 9 tickets = 1 in 1,111 10 tickets = 1 in 1,000 11 tickets = 1 in 909 12 tickets = 1 in 833 13 tickets = 1 in 769 14 tickets = 1 in 714 15 tickets = 1 in 666 5,000 tickets = 1 in 2 10,000 tickets = 1 in 1

Greenwich, CT United States Member #4793 May 24, 2004 1822 Posts Offline

Posted: March 5, 2007, 9:12 am - IP Logged

Yes truecritic that's it! Each extra ticket reduces the odds by a known percentage. The formula for percent decrease is [(new - old)/old]. Using your example:

The second ticket reduces the remaining odds by 1/2. (5,000 - 10,000)/10,000 = -1/2

The third ticket reduces the remaining odds by 1/3. (3,333-5,000)/5,000 = -1/3

The fourth ticket reduces the remaining odds by 1/4. (2,500-3,333)/3,333 = -1/4

See a pattern? :0)

Of course you're going to have better odds on four distinct tickets together, than on just one. Each individual ticket still retains its odds of 1 in 10,000.

Greenwich, CT United States Member #4793 May 24, 2004 1822 Posts Offline

Posted: March 5, 2007, 9:20 am - IP Logged

Quote: Originally posted by Coin Toss on March 5, 2007

Oh, the same place a lot of these go I guess. A few of us try to convince others that each line of numbers is but one combination up against over 176,000,000 possible combinations and that's all there is to it.

But they insist one more combination they play is going to reduce those combinations by 85,000,000 all at once, not just by one more.

Then the lottery commission lurkers and "guests" that monitor this board poke thier co-workers and go, "Told ya. Poor bastards, poor dumb bastards." Then in a while the matrix expands again, but hey, that's ok because that "magic dollar" will cut those odds by half, too.

And people will continue to tell stories about the fabulous time they had at the 1994 World Series.

Yeesh indeed.

I don't think that anyone is arguing that a second or third or tenth ticket is going to guarantee a jackpot win. Of course it's still highly unlikely. If every single human on Earth were to buy one ticket for this Mega Millions drawing there would only be around 3 dozen jackpot winners.

That's a weird point. Because....36 people would win the jackpot...that seems like a lot. I think it's tough to grasp how many people live on this planet! Yet you would not personally know any of the winners.

These games seem so easy...just pick 6 numbers.

36 more hours until someone could become very, very wealthy.

New Jersey United States Member #2376 September 25, 2003 582 Posts Offline

Posted: March 5, 2007, 9:47 am - IP Logged

Quote: Originally posted by Coin Toss on March 5, 2007

Oh, the same place a lot of these go I guess. A few of us try to convince others that each line of numbers is but one combination up against over 176,000,000 possible combinations and that's all there is to it.

But they insist one more combination they play is going to reduce those combinations by 85,000,000 all at once, not just by one more.

Then the lottery commission lurkers and "guests" that monitor this board poke thier co-workers and go, "Told ya. Poor bastards, poor dumb bastards." Then in a while the matrix expands again, but hey, that's ok because that "magic dollar" will cut those odds by half, too.

And people will continue to tell stories about the fabulous time they had at the 1994 World Series.

Yeesh indeed.

The other combinations are not just disappearing into thin air. When we say buying 2 tickets will give you 1 in 87,855,768 odds of winning, it doesnt mean the other 87,855,768 combinations just disappeared. Think of it this way, if you buy two tickets, split the total combinations in half, then you will have one ticket give you odds of 1 in 87,855,768 of winning and the other ticket will give you another 1 in 87,855,768 odds of winning. So all the combinations are there, just the combinations have been split amongst different tickets (thats the most crude way to put it but hopefully you will understand).

Zeta Reticuli Star System United States Member #30470 January 17, 2006 10389 Posts Offline

Posted: March 5, 2007, 10:29 am - IP Logged

Quote: Originally posted by twisted on March 5, 2007

The other combinations are not just disappearing into thin air. When we say buying 2 tickets will give you 1 in 87,855,768 odds of winning, it doesnt mean the other 87,855,768 combinations just disappeared. Think of it this way, if you buy two tickets, split the total combinations in half, then you will have one ticket give you odds of 1 in 87,855,768 of winning and the other ticket will give you another 1 in 87,855,768 odds of winning. So all the combinations are there, just the combinations have been split amongst different tickets (thats the most crude way to put it but hopefully you will understand).

It doesn't work like that.

Each indivicual ticket (ticket meaning one line of numbers, $1) is up against the same odds of 175,711, 536 to 1.

What's really intersting here is that the people saying the second ticket cuts the odds in half seem to be stopping there. When we take it to the point like justxploring did, of $14 reducing pick 4 odds down to being too tempting to pass up for the player, it's "Oh no, that's not it."

No matter how many tickets are played only one combination can win, and of 175,711,536 combinations only one can be drawn.

If Bob plays one ticket and Debbie plays 101 tickets, Debbie may think she's got 100 more chances of hitting than Bob, but in a world of 175,000,000 + 100 is still infintesimal.

Let's say you're at a beach and there are 175,711,536 grains of sane on this one spot. Oneo of those is dyed a different color than regular sand, but you're blindfolded. You're on a game show and rthe idea is to pick out the colored grain of sand, blindfolded. But you answer a trivia question and to give you a "break", they color one additional grain of sand.

No one is going to say, "I can't believe I didn't pick a colored grain of sand since my chances doubled."

This should be a mantra for everyone on this board:

"Each individaul ticket is up against the same odds."

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.

Greenwich, CT United States Member #4793 May 24, 2004 1822 Posts Offline

Posted: March 5, 2007, 10:38 am - IP Logged

Coin Toss,

No one on this side is disputing your point. Each individual ticket is up against the same odds. We are all in agreement.

Maybe a better way for us to phrase the "cutting odds" conundrum is: the second ticket, in conjunction with the first, cuts the odds by 1/2. The third ticket, in conjunction with the second and first, cuts the remaining odds by 1/3. The 567th ticket, in conjunction with the previous 566 tickets, cuts the remaining odds by 1/567th.

United States Member #24380 October 21, 2005 623 Posts Offline

Posted: March 5, 2007, 10:52 am - IP Logged

"Let them laugh," the winner of the MM and PB in the same week said as the monies were invested quite nicely in that special place for those who DARE to Dream. How sweet it is!

Greenwich, CT United States Member #4793 May 24, 2004 1822 Posts Offline

Posted: March 5, 2007, 11:18 am - IP Logged

Quote: Originally posted by mylollipop on March 5, 2007

"Let them laugh," the winner of the MM and PB in the same week said as the monies were invested quite nicely in that special place for those who DARE to Dream. How sweet it is!

Yes, that would be fantastic!

But you would create so much attention, and have so many critics and conspiracy theories and state and federal investigations....if I were in that position, I think I would almost give the smaller jackpot ticket anonymously to a complete stranger.

NY United States Member #23835 October 16, 2005 3502 Posts Offline

Posted: March 5, 2007, 11:44 am - IP Logged

Quote: Originally posted by justxploring on March 5, 2007

KY Floyd, this is interesting. First everyone is telling me that chances aren't the same as odds. Actually, I agree on that to a point, but now you are using the above as an example to prove that the odds are cut in half if you purchase 2 tickets.

Yes, if you have 2 people standing outside during an electrical storm there's a greater chance that someone will get hit than if nobody is outside at all. I mean, if nobody is outside then it won't happen at all. Buying 2 tickets does give you twice the chances, but it doesn't cut the odds in half. 2 different animals!

Using your pick-4 example with odds of 1 in 10,000 (those are the odds for the Florida Play-4 game for a straight) then according to your theory, buying 2 tickets would make the odds 1 in 5,000 (half) and 3 would make the odds 1 in 2,500 and 4 would make them 1 in 1,250 and 5 would be 1 in 625 and 6 would be 1 in 312 and 7 would be 1 in 156 and 8 would be 1 in 78 and 9 would be 1 in 39 and 10 would be 1 in 19 and 11 would be 1 in 10 and 12 would be 1 in 5 and 13 would be 1 in 2.5 and then you'd have to win!! So, since Play-4 pays $5,000 I better go out tomorrow and purchase 14 tickets for Play-4 because if it's true that the odds just keep splitting every time I buy a ticket, I'll win $5,000 by only spending $14 or $15. Wow!

One of the things I think we all agree on is that the terminology can get confusing. Odds are just one way of describing your chances, just like quarts is one way of describing how many pints you have. The odds are the same as chances at the same time that they aren't. For pick 3 the odds are 1 to 999, and the chances of winning are 1 in 1000. You have 1 quart of milk, and I have 2 pints. Different numbers telling the same story.

"Using your pick-4 example with odds of 1 in 10,000 (those are the odds for the Florida Play-4 game for a straight)"

You say that as if the odds might be different in other places. I trust you understand that the odds are determined by math and not the people running the game?

"buying 2 tickets would make the odds 1 in 5,000 (half) and 3 would make the odds 1 in 2,500"

You clearly missed my earlier post or didn't understand it. Nobody has said that's how it works. Several of us have repeatedly said that to halve your odds you need to buy twice as many tickets. 2 is twice as much as 1, so somebody with 2 tickets is twice as likely to win as somebody with 1 ticket. To be twice as likely to win as somebody with 2 tickets you would need to have twice as many tickets as them. How many is twice as much as 2? Several of us say it's 4. What do you think? Several of us say that 3 is three times as much a 1, so buying 3 tickest reduces your odds to 1/3. For pick 4 that's 1 in 3,333.3. The only people who think it works the way you described it are the imaginary characters in coin toss' head.

Let's revisit some earlier posts. You say you're good at logic, so let's start with simple logic. Note that the conclusion could be wrong if any of the conditional statements are wrong, but all we're doing is seeing if the reasoning is correct. If I said elephants are pink, that animal is an elephant, therefore that animal is pink, the reasoning would be correct. The conclusion may be wrong because elephants aren't pink, but it might be right because the animal could be a flamingo. All we're interested in right now is the reasoning process. so, here's my proposal.

1. If the outcome of an event is "A" I have a 1 in 5000 chance of winning pick 4.

2. If the outcome of an event is "B" I have a 1 in 5000 chance of winning pick 4.

Those are both conditional statements of the form "if A then B", and for the purpose of reaching a conclusion that is logical we can assume that they are true.

3. The outcome will be either "A" or "B"

4. Therefore I have a 1 in 5000 chance of winning pick 4.

That's the conclusion I get. Do you agree or disagree? Why?

Now back to our lightning storm.

A town has 1000 yards and one of them is about to be hit by one bolt of lightning. 500 yards are 1 acre and 500 yards are 2 acres. The lightning will strike purely at random, so every square foot of the town has an equal chance of being hit. Is it twice as likely that the lightning will hit one of the yards with 2 acres?

NY United States Member #23835 October 16, 2005 3502 Posts Offline

Posted: March 5, 2007, 11:49 am - IP Logged

Quote: Originally posted by Coin Toss on March 5, 2007

It doesn't work like that.

Each indivicual ticket (ticket meaning one line of numbers, $1) is up against the same odds of 175,711, 536 to 1.

What's really intersting here is that the people saying the second ticket cuts the odds in half seem to be stopping there. When we take it to the point like justxploring did, of $14 reducing pick 4 odds down to being too tempting to pass up for the player, it's "Oh no, that's not it."

No matter how many tickets are played only one combination can win, and of 175,711,536 combinations only one can be drawn.

If Bob plays one ticket and Debbie plays 101 tickets, Debbie may think she's got 100 more chances of hitting than Bob, but in a world of 175,000,000 + 100 is still infintesimal.

Let's say you're at a beach and there are 175,711,536 grains of sane on this one spot. Oneo of those is dyed a different color than regular sand, but you're blindfolded. You're on a game show and rthe idea is to pick out the colored grain of sand, blindfolded. But you answer a trivia question and to give you a "break", they color one additional grain of sand.

No one is going to say, "I can't believe I didn't pick a colored grain of sand since my chances doubled."

This should be a mantra for everyone on this board:

"Each individaul ticket is up against the same odds."

"Each indivicual ticket (ticket meaning one line of numbers, $1) is up against the same odds of 175,711, 536 to 1."

Yup. That's true, and I like that you felt the need to emhasize "each individual ticket" but don't understand the significance of it.

"in a world of 175,000,000 + 100 is still infintesimal."

Yup. That's true also.

"No one is going to say, "I can't believe I didn't pick a colored grain of sand since my chances doubled."

Yup. You got that one right, to.

"It doesn't work like that."

Bzzzt. Sorry. You seem to actually see some of the dots, but you just can't manage to connect them. It's shame you can't see how funny it is.

Wandering Aimlessly United States Member #25360 November 5, 2005 4461 Posts Offline

Posted: March 5, 2007, 12:10 pm - IP Logged

No, True Critic, although you are correct when using that formula. I was just showing KY Floyd that his theory about "every time you buy another ticket you cut the odds in half" isn't valid. However, I see your point, because I've seen this many times on math boards. JAG said a few posts back that this isn't getting anywhere, and I agree. The reason isn't because one of us is right or one of us is wrong, but that in a game where the odds are over a 100 million to one, buying 2 or 10 or even 100 tickets makes such a small dent. I've known people who splurged when a jackpot hit $50 million and spent $100 and didn't even get 4 numbers, and our state game has odds of "only" 1 in 22.9 million. Have you ever gone on a state site and used the feature (FL has one) that lets you see if your numbers have ever been drawn in the history of the lottery? Out of curiosity, I once wrote down 100 random combinations and none of them had ever won the 6/53 jackpot. In fact, not one of them had ever won 5 out of 6. The most ever paid was about $75 for 4 numbers. What does that tell you about the "odds?" When you have a date with Lady Luck, you'll win your big jackpot. Buying one more ticket means nothing. As Coin Toss said, when people walk around saying "spend another buck and your odds are cut in half" the casino managers love it.

I shouldn't be spending so much time on this. I need to work! I wanted to answer you, so it wouldn't seem as if I was throwing in the towel, so to speak. I also appreciate it when other members post their opinions, because that's what a forum is all about, right?