Buying more tickets = reduced odds

Reply #100

Dowser — Tue, 18 May 2004 00:41:01 GMT

Quote: Originally posted by hypersoniq on May 03, 2004 ...still exist before draw time, they don't magically disappear... (except for here in PA where they had worked wonders with some latex paint and a syringe ;-) Latex paint...whats that all about

Reply #99

ayenowitall — Mon, 17 May 2004 09:24:37 GMT

Oh, what a tangled web we weave.... Colin,You've had more than ample opportunity to come clean on this. Anyone with a shred of decency and integrity would have done so by now. Instead, you change your story every time you're confronted with the facts. You should salvage your honor by admitting you're wrong and apologizing to this entire forum.The truth is that the numbers come out quite differently when the correct values are plugged in. The percentages change greatly and are not nearly so impre... [ More ]

Reply #98

Colin F — Mon, 17 May 2004 02:53:45 GMT

Quote: Originally posted by Colin F on May 12, 2004 My program in this case is just being the medium to run BobP 164 lines against the actual Wisconsin 6/49 History and calculate and record the wins. (It takes just a few minutes.) In the run referred to I started it from the beginning and it scored 24%. Starting it 300 Draws later it scored 21% with 2 Fives.Cost: 500 x 164 x 50c = $41,000Prizes: 3 ; $2 4 ; $40 5 ; $1,500Expected per Probability: 3's - 1447; 4's - 77; 5's - 1 (calc gives 1.5)Pe... [ More ]

Reply #97

ayenowitall — Mon, 17 May 2004 00:20:19 GMT

Quote: Originally posted by Colin F on May 12, 2004In the run referred to I started it from the begining and it scored 24%. Starting it 300 Draws later it scorred 21% with 2 Fives.Cost: 500 x 164 x 50c = $41,000Prizes: 3 ; $2 4 ; $40 5 ; $1,500Expected per Probability: 3's - 1447; 4's - 77; 5's - 1 (calc gives 1.5)Percentage Return: 18% of Cost (7474/41000)Wisconsin:Run 1: 3's - 1455; 4's - 84; 5's - 3 Percentage Return: 24% of CostRun 2: 3's - 1456; 4's - 69; 5's - 2 Percentage Return: 21... [ More ]

Reply #96

Colin F — Sun, 16 May 2004 00:43:41 GMT

Posted: February 08, 2004 at 8:36pm - IP LoggedChief Bottle WasherLottery Post AdministratorNew JerseyUnited StatesMember #1November 29, 20013368 PostsOfflineIMT,Yes, one of the stats that we'll keep track of is the play/win ratio, so if you win with fewer picks, then you will have a higher ratio. That ratio, along with the dollars spent/won ratio is pretty much the Holy Grail of lottery predicting. (Which is one of the main reasons for this poll.)Thanks for the kind words!_________________Goo... [ More ]

Reply #95

BobP — Sat, 15 May 2004 17:26:46 GMT

I think the term dishonest is a little harsh. More a problem in phrasing. I agree recovering 25% of one's investment is still a big loss, but better then nothing. If the odds say you should only recover say 5% then recovering 25% would be beating the odds though it is owed to the reader to explain what this all means.There is also a utility value to these small wins as they form the foundation for winning larger prizes. If you routinely get 3,4, 5 numbers correct you only needed to get 3, 2... [ More ]

Reply #94

ayenowitall — Sat, 15 May 2004 10:16:59 GMT

Quote: Originally posted by BobP on May 13, 2004The idea is to reduce the cost to play with lower tier prizes while waiting to win the jackpot which should pay back all reasonable playing expenses and then some.Of course at this point we're just talking about putting that kind of money into play. Certainly smarter to have a pretty good idea what's going to happen if we do, then to have no idea what-so-ever. BobPBobP,I understand the objective to reduce costs. I also acknowledge that hitting th... [ More ]

Reply #93

Todd — Thu, 13 May 2004 12:34:32 GMT

Colin,Of course you're correct. I'm sure someone will post some irrelevant remarks about how those odds are still pretty terrible or something. Hopefully they can show some restraint.

Reply #92

Colin F — Thu, 13 May 2004 12:30:47 GMT

Quote: Originally posted by keystonechas on May 05, 2004How people come up with some of their odds is beyond me.In a 6/49 game the odds to win a 6 number prize is one in 13,983,816To win a 5 number prize one in 54,201To win a 4 number prize one in 1,032To win a 3 number prize one in 57ChasAnd so if you buy 10 tickets (games)6 1 in 1,398,3815 1 in 5,4204 1 in 1033 1 in 6... and that is a fact that no would be pontificating expert can change.Colin

Reply #91

ayenowitall — Thu, 13 May 2004 10:37:45 GMT

Colin,Did you merely derive those numbers you supplied to me from a run on your computer model? Seriously, I suggest you work them out with a pencil and paper or even a simple calculator. They really don't add up.It's interesting that you won't address the fact that payoff odds are also effectively reduced as more and more tickets are played. You really are not beating any odds.

Reply #90

Colin F — Thu, 13 May 2004 08:42:20 GMT

GuruDid you read all the previous Posts in this Thread?Colin

Reply #89

Guru — Thu, 13 May 2004 08:23:00 GMT

Quote: Originally posted by Jake649 on May 03, 2004This topic came up in another thread but the thread went off in a tangent so I decided to start a new thread devoted to the following questions.Q: Can lottery odds be reduced by buying more tickets?A: YesQ: How are the new, reduced odds calculated?A: Say that a lottery has X number of unique outcomes. Then the odds of hitting the jackpot is 1 in X. If one buys N number of tickets where no two tickets have exactly the same set of numbers, then th... [ More ]

Reply #88

Colin F — Thu, 13 May 2004 06:52:27 GMT

AyeNoWitAllAs they say A leopard can't change it's spots .1. Lotto is a negative return expectation game.2. If you can't subtract $7,474 from $10,770 and realize you're better off by $3,296 then that's you're problem.4. To plan on winning Lotto with no plan is equivalent to saying I intend to live for 10,000 years.5. The offer you make is not a lottery as it is a one dvent with a 1st prize less than the cost of participation; it's a scam. That you offer it does not surprise me. To... [ More ]

Reply #87

BobP — Thu, 13 May 2004 06:08:18 GMT

Quote: Originally posted by ayenowitall on May 13, 2004Colin,I did go back and read the entire SUMS thread. It was interesting reading. Perhaps the most enlightening aspect of the thread was your statement that your enjoyment of the lottery comes from just being able to improve on the odds ever so slightly. I suppose that's why I failed to recognize the validity of your argument. You see, I thought we were all trying to actually win money.Well, I'm thinking that my computer, my pocket calculator... [ More ]

Reply #86

BobP — Thu, 13 May 2004 05:58:45 GMT

Quote: Originally posted by Colin F on May 12, 2004ayenowitallMay I be the first to congratulate you on your more conciliatry approach. I try not to be drawn too much on commenting on an individuals chances of winning first prize. Dreams are what keep a lot of people going and coping with life.My program in this case is just being the medium to run BobP 164 lines against the actual Wisconsin 6/49 History and caculate and record the wins. (It takes just a few minutes.) In the run referred to I st... [ More ]

Reply #85

ayenowitall — Thu, 13 May 2004 05:49:44 GMT

Colin,I did go back and read the entire SUMS thread. It was interesting reading. Perhaps the most enlightening aspect of the thread was your statement that your enjoyment of the lottery comes from just being able to improve on the odds ever so slightly. I suppose that's why I failed to recognize the validity of your argument. You see, I thought we were all trying to actually win money.Well, I'm thinking that my computer, my pocket calculator, and all of my acquired math skills must be failing me... [ More ]

Reply #84

Colin F — Wed, 12 May 2004 11:01:34 GMT

ayenowitallMay I be the first to congratulate you on your more conciliatry approach. I try not to be drawn too much on commenting on an individuals chances of winning first prize. Dreams are what keep a lot of people going and coping with life.My program in this case is just being the medium to run BobP 164 lines against the actual Wisconsin 6/49 History and caculate and record the wins. (It takes just a few minutes.) In the run referred to I started it from the begining and it scored 24%. Start... [ More ]

Reply #83

ayenowitall — Wed, 12 May 2004 08:24:09 GMT

Quote: Originally posted by Colin F on May 11, 2004So playing BobP's 164 Lines over 500 Draws in a 6/49 Lotto with a random selection of 28 tickets per draw the expected return is 15%; playing 164 the expected return is 18% because a 5 is expected. For Wisconsin he beat the odds and got 3 fives up giving a return around 24%.Colin,Was this a workout based on draw history, a computer model, or actual play? My hat goes off to anyone who can truly beat the odds, but I'm concerned with reality and ac... [ More ]

Reply #82

Colin F — Tue, 11 May 2004 07:43:55 GMT

ayenowitallPlaying random selections will usually give you a return in accordance with the advertised odds. This means as you increase the number of tickets purchased over all the draws your chances of getting the higher prizes kicks in.So playing BobP's 164 Lines over 500 Draws in a 6/49 Lotto with a random selection of 28 tickets per draw the expected return is 15%; playing 164 the expected return is 18% because a 5 is expected. For Wisconsin he beat the odds and got 3 fives up giving a return... [ More ]

Reply #81

ayenowitall — Tue, 11 May 2004 06:26:39 GMT

Quote: Originally posted by Jake649 on May 09, 2004In summary, buying 2 tickets doubles the probability of winning. Buying 3 tickets triples the probability of winning. Buying 4 tickets quadruples the probability of winning, etc.The purpose of this post is to prove that probabilities of winning increases with the number of tickets purchased. This is by no means meant to convince anyone that they should buy more tickets. Spend only what you can afford to lose.Jake,Your math is right, but I wouldn... [ More ]

Reply #80

ayenowitall — Sun, 09 May 2004 19:05:24 GMT

Quote: Originally posted by Colin F on May 07, 2004I don't plan to hit the big one. I'm concerned about increasing the 3's, 4's and 5's wins in for example a 6/49 Lotto for a given number of tickets... For the best systems it hinges on getting the 5's up more consistently and often. For the realists amongst us that is the challenge and enjoyment.The crunch is if there was a contest of 5 x 500 draw runs and the winner was the one who made the most dollars from the 3's, 4's and 5's then with you... [ More ]

Reply #79

Colin F — Sat, 08 May 2004 03:28:29 GMT

Probability of getting winning six for x games played in 6/49 Lotto.1 = 1/13,983,8162 = 1/6,991,9084 = 1/3,495,9548 = 1/1,747,97716 = 1/873,98832 = 1/436,99464 = 1/218,497128 = 1/109,248256 = 1/54,624512 = 1/27,3121024 = 1/13,6562048 = 1/6,8284096 = 1/34148192 = 1/1,70716384 = .00117132768 = .00234365536 = .004686131072 = .009373262144 = .018746524288 = .0374921048576 = .0749842097152 = .1499694194304 = .2999398388608 = .59987913983816 = 1This is fun. The things we do and don't get paid for.Rega... [ More ]

Reply #78

JohnC — Sat, 08 May 2004 02:42:38 GMT

I've been following this thread with interest and my take on it is this:If a given draw comprises of 1,000 possible combinations and I buy one ticket, that means that I win if that combination comes up and lose if it doesn't. The term odds is made up of two things... Odds for and odds against and in this case, my odds for = 1 and my odds against = 999. The odds therefore are 1 in 1000 or 999 to 1 against.If I buy two tickets, that means that I win if either of those combinations come... [ More ]

Reply #77

Colin F — Sat, 08 May 2004 02:37:11 GMT

ayenowitallI don't plan to hit the big one. I'm concerned about increasing the 3's, 4's and 5's wins in for example a 6/49 Lotto for a given number of tickets. Now, if you had looked up some of what I've posted before you would have found that I acknowledge that as a difficult task. For the best systems it hinges on getting the 5's up more consistently and often. For the realists amongst us that is the challenge and enjoyment.Some like to play the Pick3 and Pick4 games where I would not discount... [ More ]

Reply #76

BobP — Sat, 08 May 2004 01:20:02 GMT

Let's see now, 6/49 lotto with odds of 1 in 13,983,816Buying X Unique Tickets1 = 13,983,8162 = 6,991,9084 = 3,495,9548 = 1,747,97716 = 873,988.532 = 436,994.2564 = 218,497.125128 = 109,248.5625256 = 54,624.28125512 = 27,312.1406251024 = 13,656.07031252048 = 6,828.035156254096 = 3,414.0175781258192 = 1,707.008789062516384 = 853.5043945312532768 = 426.75219726562565536 = 213.3760986328125131072 = 106.68804931640625262144 = 53.344024658203125524288 = 26.67201232910156251048576 = 13.3360061645507812... [ More ]

Reply #75

ayenowitall — Fri, 07 May 2004 22:31:04 GMT

Colin,You miss my point entirely, but I think there is precious little I can say to help you comprehend. I will say that the concept of odds is not exactly rocket science. What you consider to be the universally accepted meaning is correct as far as it goes, but to stop there without giving consideration to odds as the ratio between the amounts put up by the parties to a bet with respect to the expected probability either way leaves a huge blind spot in your argument. It also leaves the door wid... [ More ]

Reply #74

Todd — Fri, 07 May 2004 20:35:29 GMT

Quote: Originally posted by Jake649 on May 07, 2004For example, it has been stated that if one buys 500 unique pick 3 tickets, then the odds of winning is 1 in 2 (or 50%). But then, the responses deal with the question of whether it is wise to buy 500 tickets. That is not the point of this thread.Jake,Thanks for boiling it down. You are 100% correct: your initial question keeps getting misconstrued. I simply don't understand why this thread keeps getting hijacked. Will those who care about th... [ More ]

Reply #73

Jake649 — Fri, 07 May 2004 20:26:39 GMT

This has been a very interesting discusion.One problem I see is that in discussing the odds, the idea of whether or not one should even buy that many tickets is also presented. That only confuses the issue and is not germain to the discussion.For example, it has been stated that if one buys 500 unique pick 3 tickets, then the odds of winning is 1 in 2 (or 50%). But then, the responses deal with the question of whether it is wise to buy 500 tickets. That is not the point of this thread.I have als... [ More ]

Reply #72

wiseone2 — Fri, 07 May 2004 19:59:04 GMT

Quote: Originally posted by BobP on May 07, 2004If you're playing a game of dodge ball and down to two players on your side and ten on the other the odds are 2 in 10 or 1 in 5. If an army has 10,000 guys and you have 1,000 how many must each of your men cover? Same with lottey tickets, each unique ticket covers a range of possibilities spliting the field among them for coverage. BobPThis whole issue is a matter of one's preferred perspective, rather than an issue of mathematical probability.... [ More ]

Reply #71

LANTERN — Fri, 07 May 2004 18:47:37 GMT

What about this?100 combinations that are not going to win are just as bador worse, because you spend more money, than one combination that is not going to win.What am I talking about?Not about the same thing that you are talking about.I only say, that one bad combination is just as bad as another bad combination.In other words, if you want to improve your chances, buying many tikets is not the way to go, but better number selection is where it's at.I know that this is not at all what you are ta... [ More ]

Reply #70

Todd — Fri, 07 May 2004 17:38:13 GMT

Thank you BobP. One more example proving the odds calculations. Hopefully with this latest example, this odds calculation issue will be put to bed

Reply #69

BobP — Fri, 07 May 2004 17:34:34 GMT

If you're playing a game of dodge ball and down to two players on your side and ten on the other the odds are 2 in 10 or 1 in 5. If an army has 10,000 guys and you have 1,000 how many must each of your men cover? Same with lottey tickets, each unique ticket covers a range of possibilities spliting the field among them for coverage. BobP

Reply #68

Colin F — Fri, 07 May 2004 08:59:12 GMT

AyenowitallI agree with those who have wondered about the comprehension skills of some members. (I have to be true to my honesty and say that in a face to face situation I would just walk away as I have always held the view that you cannot have a worthwhile arguement with a stupid person.) Far be from me to call anyone stupid here. Some little exercises I have done were to produce biased histories and see how they fared. So for the 6/49 example I gave above it means that for every 49 Draws each

Reply #67

keystonechas — Fri, 07 May 2004 08:20:11 GMT

Quote: Originally posted by RJOh on May 06, 2004wiseone2,I'm starting to get the idea that you can't comprehend a lot of things, but to each his own. Good luck to you.RJOhI agree!!This thread is like the movie GOUNDHOG DAY, with the same scene over and over and over.....Chas

Reply #66

ayenowitall — Fri, 07 May 2004 08:09:01 GMT

Colin,I think we all know the difference between giving odds and taking odds. They're just two ways of saying the same thing, and in an honest game there's no advantage of one over the other.I think Professor Croucher is right to some extent because the odds of the game don't change no matter how many tickets a player buys. The odds of the game are also unaffected by the most brilliant strategy or exotic computer program.Of course, every different ticket that any individual buys is another chanc... [ More ]

Reply #65

Colin F — Fri, 07 May 2004 04:10:36 GMT

Let's cut to the chase.Buying more tickets and increasing your chances of winning orreducing your chances of losing should be a no-brainer (I love thatAmericanism) for this Forum; our most ardent critics do not have a problem with this so why some members here do strains my powers of comprehension.The real question is, Can the odds be reduced other than bybuying more Random Selection Tickets? I say YES, DA, JA, SI, OUI, HAI and whatever other language you care to use.What does my opposition say... [ More ]

Reply #64

Todd — Fri, 07 May 2004 01:58:12 GMT

AAAAHHHHHHHHHHH!!!!!That's me screaming, because I can't believe you're going to take us down this same road again!No one said that buying a second ticket makes 60,000,000 combinations disappear.It reduces your odds to 1 in 60,000,000.Also, buying 11 lines can contain all 53 numbers... play them mixed up 4 times... 44 lines total does not increase your odds of winning the jackpot (over 44 Quick Picks). It would only ensure that you win at least a lower-tier prize.

Reply #63

hypersoniq — Fri, 07 May 2004 01:51:10 GMT

so then buying a second ticket makes 60,000,000 combinations disappear then?think of it how you like, but the fact remains that the number of possible outcomes NEVER changes... Besides, playing 500:1000 in pick3 in PA which pays 500:1 would be a wasted exercise IF you won...there are still 500 uncovered combos... think of it as 1:2 if it helps justify placing 500 bets, but the possible combinations will always be the same...Rather than bulk tickets, I would lean towards a full coverage wheel...

Reply #62

RJOh — Fri, 07 May 2004 01:22:46 GMT

wiseone2,I'm starting to get the idea that you can't comprehend a lot of things, but to each his own. Good luck to you.RJOh

Reply #61

wiseone2 — Fri, 07 May 2004 01:16:34 GMT

sorry RJOh, but i can't comprehend whatever you 're saying. Nothing in reality changes the fact that winning a lottery jackpot is more likely for one person over another, except for the person that buys a ticket versus one that doesn't, not buying many tickets, not using math, nothing.

Reply #60

RJOh — Fri, 07 May 2004 01:03:09 GMT

Quote: Originally posted by wiseone2 on May 06, 2004I only attempted to use math (however inaccurately) to mainly save a lot of people some money. I personally don't think math and the lottery go together. It that were the case, the greatest mathematicians in the world would be winning the jackpots all the time.The greatest mathematicians in the world would know their odds of winning or losing a lottery. They would tell you if you told a person buying a ticket for a pick3 game that he would n... [ More ]

Reply #59

wiseone2 — Fri, 07 May 2004 01:01:49 GMT

it's not me trying to relate to others that by using math you can guarantee with certainty a win. nor am i remotely or indirectly trying to imply that.

Reply #58

Todd — Fri, 07 May 2004 00:53:39 GMT

Of course, that's how ALL gambling works, not just the lottery. When's the last time you walked into a casino or track and had a probablility of winning? If you want certainty of income, get a second job.

Reply #57

wiseone2 — Fri, 07 May 2004 00:50:38 GMT

everything the lottery does is geared towards prdventling anyone from having an edge . that is real.

Reply #56

wiseone2 — Fri, 07 May 2004 00:35:34 GMT

ok, i'm wrong with the math calculations, but i don't feel i'm wrong perspectively about the realization of winning a jackpot based on increasing the amount of money you play for a drawing. i'm sure mathematical probability was used to create the lottery and and ensure no one, no where, no how, would have an advantage over anyone in winning the jackpot. just because on paper the odds look reduced using math, doesn't mean in reality they are reduced. anyone selling lottery strategy software or... [ More ]

Reply #55

Todd — Fri, 07 May 2004 00:08:53 GMT

wiseone,First you posted a bunch of messages in which you claimed to mathematically prove that buying two Powerball tickets didn't lower your odds to 1 in 60,000,000. And now, after being faced with indisputable facts showing that you were wrong, you claim that you personally don't think math and the lottery go together. You even proved yourself wrong by quoting the mathematician who said Buying two powerball tickets comes extremely close to doubling your chance of winning the jackpot. (It... [ More ]

Reply #54

wiseone2 — Thu, 06 May 2004 21:42:03 GMT

You can argue if it is effective reducing of the odds, meaning whether or not you are much closer to winning a jackpot, but you are being intellectually dishonest if you refuse to admit that you have reduced the odds by buying additional tickets.Arguing whether using math to give oneself a sense that purchasing more tickets to win a pb jackpot or any lottery game is effective is exactly my point. I only attempted to use math (however inaccurately) to mainly save a lot of people some money. I p... [ More ]

Reply #53

Todd — Thu, 06 May 2004 17:48:01 GMT

RJOh

Reply #52

RJOh — Thu, 06 May 2004 16:57:36 GMT

I plan to increase my chances of winning something in the PowerBall drawing this weekend by purchasing several combinations. I know there are 120,526,770 possible combinations and 3,342,046 of them will pay $3 or more and I hope that some of my combinations are among them. I'll be looking at past history and hoping to pick only numbers that are most likely to be drawn to reduce my selections to about 20 lines which in well within my budget. My chances aren't as good as being in a pool of pla... [ More ]

Reply #51

Colin F — Thu, 06 May 2004 13:46:49 GMT

ODDS10 little boxes; one of them has a valuable diamond in it.Cost you $100 to look in a box and keep what's in it.Buy 1 box:Odds for or probability (favourable/ all outcomes): 1 in 10 or 1/10 or .1 or 10%.Odds against (bookies way) : 9 chances of losing: 1 chance of winning or 9:1 or 9/1 (but the / does not mean division).Buy 5 boxes:Odds for or probability: 5 in 10 or 5/10 or 1/2 or .5 or 50%.Odds against: 5 of losing: 5 of winning or evens.Odds against = (1/Probability) - 1Probability of 1/4... [ More ]

Reply #50

ayenowitall — Thu, 06 May 2004 06:13:13 GMT

On a jackpot game, you can enhance your chances of winning (albeit ever so slightly) by buying more sets of numbers, but the odds of the game remain the same. The odds of any one set of numbers being drawn are unaffected.

Reply #49

keystonechas — Thu, 06 May 2004 04:17:38 GMT

Welcome dragon,Relax, no one is going to run you out. Todd's KISS (Keep It Simple Stupid) method to explain all this is much better than using numbers in the millions and thousands.Chas

Reply #48

dragon — Thu, 06 May 2004 02:14:22 GMT

Todd:I am glad you said the above, not me, even though I felt like it throughout this post. If I had said that, I'd be chased out of the forum by now for being a goose.I think wiseone is splitting hairs. If he ( or she ) believed what he is saying, it would not make sense for him to play the lottery. Just play one ticket, that's all you need. Given that that is absurd, he must just be playing a game of semantics. The Saliu paragraphs have no bearing on his argument.

Reply #47

Todd — Wed, 05 May 2004 21:31:16 GMT

Quote: Originally posted by wiseone2 on May 05, 2004I said I wasn't going to post anything else on this topic, but I must clarify the issue. Each number combination you purchase has a 1:120,000,000 chance of winning or the odds against you are 120,000,000 to 1. The purchase of 2 number combinations does not change the odds against you to 60,000,000 to 1. Probability is figured by calculating total probable cases. Comparing number of chances to probability is like apples to oranges. The numb... [ More ]

Reply #46

wiseone2 — Wed, 05 May 2004 21:29:56 GMT

For everyone that thinks they know how to figure odds:There are 1000 total possible cases in pick-3 lottery, the straight bet being considered, from 0 (000, in fact) to 999. Only one 3-digit combination is drawn, therefore the winning probability is 1 / 1000 (0.001). The lottery commission says the odds are 1 in 1000, or 1000 to 1.There are 10000 total possible cases in pick-4 lottery, the straight bet being considered, from 0 (0000, in fact) to 9999. Only one 4-digit combination is drawn, ther... [ More ]

Reply #45

Colin F — Wed, 05 May 2004 21:16:23 GMT

KeyStoneChasSpot on.BobP1 in 8,145,06012 in 8,145,060 = 12/8,145,060 = 1/678,7554,072,530 in 8,145,060 = 1/2 or .58,145,059 in 6,145,060 ie 99% sure but not certain.8,145,059 in 8,145,060 = 1 ie certain but may have to share with other winners.RegardsColin

Reply #44

wiseone2 — Wed, 05 May 2004 21:14:35 GMT

My statement of odds is concerning Powerball only.

Reply #43

wiseone2 — Wed, 05 May 2004 21:09:29 GMT

I said I wasn't going to post anything else on this topic, but I must clarify the issue. Each number combination you purchase has a 1:120,000,000 chance of winning or the odds against you are 120,000,000 to 1. The purchase of 2 number combinations does not change the odds against you to 60,000,000 to 1. Probability is figured by calculating total probable cases. Comparing number of chances to probability is like apples to oranges. The number (1) in the stated odds is represented as (1) beca... [ More ]

Reply #42

keystonechas — Wed, 05 May 2004 21:07:36 GMT

How people come up with some of their odds is beyond me.In a 6/49 game the odds to win a 6 number prize is one in 13,983,816To win a 5 number prize one in 54,201To win a 4 number prize one in 1,032To win a 3 number prize one in 57Chas

Reply #41

RJOh — Wed, 05 May 2004 20:52:26 GMT

wiseone2,Are you self taught? Where did you get all these mathematical rules? Each possible combination has a chance to win, if you buy ten of them then you have 10 chances to win, that is ten times the chances of winning with one combination.RJOh

Reply #40

wiseone2 — Wed, 05 May 2004 20:06:04 GMT

This is the last I'm going to post on this topic: My point is that the odds stated for a lottery game compare two quantities of the same kind; 1:120,000,000 is a comparison of one ticket combination out of 120,000,000 ticket combinations. The odds are expressed in a ratio. The mathematical rule for a ratio is that it is expressed in its lowest terms. So we are mistaken when we express the purchase of 2 tickets as changing the odds to 2:120,000,000. To do so violates the mathematical rule, t... [ More ]

Reply #39

Todd — Wed, 05 May 2004 18:33:06 GMT

Quote: O

Reply #38

BobP — Wed, 05 May 2004 18:23:40 GMT

Quote: Originally posted by Colin F on May 05, 2004ToddYour right of course. Where it starts to get interesting is if you play 1,991,908. ie 50/50 1/2 etc. Now, some of you may remember I think in the SUMS thread where I referred to Kerry Packer Lotto after a benevolant Media Magnate in Australia who is also known to win and lose millions at the Blackjack high roller tables. After I got my Computer Model operational I would amuse myself by playing 200,000 or 300,000 tickets. The thing that ama... [ More ]

Reply #37

wiseone2 — Wed, 05 May 2004 18:21:39 GMT

I think the reason why some of us don't accept that as truly reducing odds is because to say 1:60,000,000 in the case of 2 pb tickets, is saying your one ticket that was once 1:120,000,000 is now 1:60,000,000, when in fact it's still 1:120,000,000 for each ticket. If they drew two sets of winning numbers, then looking at the odds that way would make more sense. The odds being expressed in the number of possible combinations against your ticket(s).

Reply #36

Todd — Wed, 05 May 2004 18:01:09 GMT

If you want to reduce your chances of hitting Powerball to 1 in 1 million, buy 120 tickets. MUSL is correct that it is not an effective way to reduce your odds, but 1 in 1 million is better than 1 in 120 million no matter how you slice it.

Reply #35

Todd — Wed, 05 May 2004 17:58:18 GMT

MUSL's answer was correct, but they did not answer your question.For some reason that I can't comprehend, people seem unable to come to grips that the odds become 1 in 60 million when you buy 2 Powerball tickets. Their answer should have been simply yes instead of all the stuff about thinking about it as numbers.

Reply #34

wiseone2 — Wed, 05 May 2004 17:19:23 GMT

I sent in this question to Powerball.com and they provided the following answer:Q: CAn a person effectively reduce the odds of winning a powerball jackpot by usingthis formula?; 1:120,526,770/N, where N is the number of tickets purchased. Inother words, can 2:120,526,770 be reduced to 1:60,263,385, where 2 is the numberof tickets purchased?A: You cannot purchase enough tickets to effectively reduce your odds in thisgame. Instead of thinking about it as odds, think about it as numbers.There a... [ More ]

Reply #33

Colin F — Wed, 05 May 2004 07:37:08 GMT

Correction:Replace 1,991,908 with 6,991,908.(One should always check mentally the result from your Computer Calculator. They are such fickle things when clicking in numbers.)Colin

Reply #32

Colin F — Wed, 05 May 2004 07:31:46 GMT

ToddYour right of course. Where it starts to get interesting is if you play 1,991,908. ie 50/50 1/2 etc. Now, some of you may remember I think in the SUMS thread where I referred to Kerry Packer Lotto after a benevolant Media Magnate in Australia who is also known to win and lose millions at the Blackjack high roller tables. After I got my Computer Model operational I would amuse myself by playing 200,000 or 300,000 tickets. The thing that amazed me was that playing such a high number of combi... [ More ]

Reply #31

Colin F — Wed, 05 May 2004 07:11:16 GMT

Further clarification:Replace 13,983,816 1 duplicates don't countWith 13,983,817 1 duplicates don't count.Colin

Reply #30

Todd — Wed, 05 May 2004 07:09:24 GMT

Colin,Your line with 2 tickets can also be expressed as 1/6,991,808.

Reply #29

Colin F — Wed, 05 May 2004 07:07:28 GMT

Clarification:13,983, 816 different tickets etcColin

Reply #28

Colin F — Wed, 05 May 2004 07:03:58 GMT

Consider a 6/49 Lotto Game:Possible Combinations of 6 Integers: 13,983,816 (only)Tickets purchased Probability of having Winning Ticket 0 0 ie impossible 1 1/13,983,816 2 2/13,983,616 13,983,816 1 ie certain (also qualify as a fool because of House Margin.) 13,983,816

Reply #27

BobP — Wed, 05 May 2004 04:45:25 GMT

What's fixed is the total number of lottery combinations.If there are 13.9 million combinations we like to say theodds are 1 in 13.9 million because one dollar buys onecombination giving the purchaser one chance to win.Fly to a state where you get two combinations for a dollarthe total number of combinations haven't changed, but theodds of 1 in 13.9 million are 1 in 6.9 million because nowyou have two combinations in hand for one/dollar.In the same sense, buying two tickets for two dollars alsoc... [ More ]

Reply #26

RJOh — Wed, 05 May 2004 03:18:38 GMT

Quote: Originally posted by wiseone2 on May 04, 2004So RJOh, how much was the jackpot they won? One of my points in the scenario is that anyone that purchased that many tickets can still lose. And if they won....would they make a profit after taxes?Several poster have mentioned the story. As I remembered the value was high enough that they made a profit. I don't know if Virgina had a cash option at that time, the jackpot was over $25M. They didn't just play any lottery as a spokesman for th... [ More ]

Reply #25

prob987 — Wed, 05 May 2004 03:13:17 GMT

Reply #24

keystonechas — Wed, 05 May 2004 00:13:12 GMT

Quote: Originally posted by keystonechas on May 04, 2004Unless your talking pick 3 or pick 4 games where payouts are set, any calculations at buying more tickets to reduce the odds is useless. As wiseone2 stated your dealing with a parimutuel game. Number of winning tickets sold divided by the jackpot.So why bother with ticket reduction odds?ChasCan anyone answer my orginal question? Why bother with ticket reduction odds if your dealing with a parimutuel game?Chas

Reply #23

wiseone2 — Tue, 04 May 2004 23:38:45 GMT

So RJOh, how much was the jackpot they won? One of my points in the scenario is that anyone that purchased that many tickets can still lose. And if they won....would they make a profit after taxes

Reply #22

RJOh — Tue, 04 May 2004 23:27:26 GMT

Quote: Originally posted by wiseone2 on May 04, 2004Even if you used one person in this scenario purchasing $60,000,000 in powerball tickets, the break even odds are the same. Some years back, an Australia group tried to play all the possible combinations in a Virgina lottery which took about 11M tickets. They took control of several SevenEleven terminals for 3 days and the most they could play were 8M combinations. They won the lottery but the point is that there wasn't enough time to play... [ More ]

Reply #21

wiseone2 — Tue, 04 May 2004 20:58:45 GMT

The only way odds are ever reduced by purchasing multiple tickets is in a drawing where more than one winning combination is drawn. Think about it

Reply #20

wiseone2 — Tue, 04 May 2004 20:43:33 GMT

the only example that clearly reflects reducing odds with the number of tickets purchased formula, is when you purchase half of the possible outcomes of a drawing. You have a 50% chance of winning. But in the case of the pick 3 straight combo paying $500, would investing $500 to reduce your odds in half, and having one of your 500 numbers drawn winning ? No, it's just a break even prize. Odds are normally figured with the inclusion of break even prizes. In another scenario, assume ever... [ More ]

Reply #19

LANTERN — Tue, 04 May 2004 17:13:30 GMT

It's not how many tickets you buy, but which numbers you buy.How you get your numbers is what will determine your odds and not so much how many tickets or numbers you buy.

Reply #18

RJOh — Tue, 04 May 2004 16:28:05 GMT

So you are saying the thread should read:Buy more tickets = Increase chances to win with same odds**same amount of winning/losing combinations regardless of the number of tickets soldRJOh

Reply #17

Todd — Tue, 04 May 2004 16:14:48 GMT

Quote: Originally posted by hypersoniq on May 04, 2004Quote: Originally posted by Jake649 on May 03, 2004hypersonig,To use your example of the pick 3 game. Say you played 500 tickets. My formula says your new odds are 1 in 2 (50% chance). Your formula says you have only odds of 1 in 500 becuase there are 500 numbers not covered.I am willing to bet my paycheck that I am right.Jakeno, I would say it would be 500:1000the 1000 is constant...Saying that it's 500:1000 is THE SAME THING as saying 1 in... [ More ]

Reply #16

wiseone2 — Tue, 04 May 2004 14:24:23 GMT

I don't think so...the topic of the discussion is whether buying more tickets reduces odds . No, it doesn't reduce odds, it increases chances. In Jake's formula....1:x/n, it decreases the number of possible outcomes. So in a 1:1000 game, if you bought 5 tickets, the formula says you have a 1:200 chance of winning as opposed to 5:1000. The number of possible outcomes is not reduced to 200 by buying 5 tickets. More tickets only amounts to more chances, without affecting the odds. The odds... [ More ]

Reply #15

dragon — Tue, 04 May 2004 12:43:22 GMT

Quote: Originally posted by hypersoniq on May 04, 2004no, I would say it would be 500:1000the 1000 is constant...Are maybe all of us saying the same thing -- just with different examples? The comment above in blue comes full circle to the beginning of this thread, doesn't it

Reply #14

hypersoniq — Tue, 04 May 2004 08:19:02 GMT

Quote: Originally posted by Jake649 on May 03, 2004hypersonig,To use your example of the pick 3 game. Say you played 500 tickets. My formula says your new odds are 1 in 2 (50% chance). Your formula says you have only odds of 1 in 500 becuase there are 500 numbers not covered.I am willing to bet my paycheck that I am right.Jakeno, I would say it would be 500:1000the 1000 is constant...

Reply #13

keystonechas — Tue, 04 May 2004 08:10:13 GMT

Unless your talking pick 3 or pick 4 games where payouts are set, any calculations at buying more tickets to reduce the odds is useless. As wiseone2 stated your dealing with a parimutuel game. Number of winning tickets sold divided by the jackpot.So why bother with ticket reduction odds?Chas

Reply #12

wiseone2 — Tue, 04 May 2004 07:54:57 GMT

All that says BobP is that your chances of having the 6 winning numbers amongst your 24 is 1 in 100. But your chances of having the six numbers in one grid are not 1 in 100. I don't know what the odds ratio is for 6/49 lotto, but say it's 1: 20,000,000. Your 24 numbers only add up to 4:20,000,000 chance of winning. Say your numbers are (1-2-3-4-5-6-), (7-8-9-10-11-12), (13-14-15-16-17-18), (19-20-21-22-23-24). Are you saying that if you played those numbers you would win the jackpot at leas... [ More ]

Reply #11

keystonechas — Tue, 04 May 2004 07:45:06 GMT

To wheel 24 numbers in a 6/49 game reduces the odds to one in 134,596Chas

Reply #10

BobP — Tue, 04 May 2004 07:12:32 GMT

Of course the ratio remains the same, you only have two combinations out of a thousand, but each now can be said to face off against 500 lines each. If you keep going doubling the amount of tickets you go from 1 in 1000 to 1000 in 1, full circle. The Hyper Geometric Distribution doesn't go away just because some don't care for how it is described in this case. ;-)Take 6/49 lotto, there is no question playing 24 numbers reduces the odds of having all six winning numbers among your 24 to roughly... [ More ]

Reply #9

wiseone2 — Tue, 04 May 2004 07:09:45 GMT

For those of you that want to reduce the odds in the lottery by purchasing more tickets.....look at it this way.....you can subtract from the number of possible outcomes by the number of tickets you purchase, but you certainly cannot divide. Example.....1:1000, buy 10 tickets with different numbers=1:990. What this means is that you have a 1:990 chance of losing, not winning. Chances of winning would be 10:1000. Hope this helps.

Reply #8

wiseone2 — Tue, 04 May 2004 06:31:52 GMT

Todd, in your example of buying two tickets in a game with odds of 1:1000, you are correct in that if you bought two tickets, it will equal 2:1000, but it does not reduce the odds in half, obviously. Odds are expressed in ratios. The definition of a ratio is a fraction expressed in its lowest terms. That is the rule. So no matter how many tickets you buy, the 1000 portion of the ratio stays the same, does not change. So in a game where the odds ratio is X:1000, X is the number of tickets p... [ More ]

Reply #7

dragon — Tue, 04 May 2004 04:13:31 GMT

That's what it looked like. My company specializes in MS technologies also, SQL 2000 or 2003, we did a lot of work in ASP.NET, but are now completely converted to .NET and C#.

Reply #6

Todd — Tue, 04 May 2004 04:07:04 GMT

JavaScript is only a very small fraction, and only deals with small functions on the Web pages themselves - not with any of the code which runs the site.I use 100% Microsoft technologies, including Windows Server 2003, MS SQL 2000, ASP (on IIS 6.0), and a small amount of ASP.NET. Also several VB.NET and VB 6.0 programs which run some of the automated stuff.I'll dventually publish full documentation of the site technologies. (Too busy working on new features at the moment

Reply #5

dragon — Tue, 04 May 2004 03:55:26 GMT

Todd, I see your point, but I think you can, in this case, deduce this intuitively for the pick-3 game. Buy 1 ticket with one permutation: 1:1000. Buy 500 tickets: 1:2. Buy 1000 different tickets 1:1.Incidentally, I said this before, I am really impressed with your coding (or the end-results of it that I can see). You use Javascript. What's the platform, what's the database

Reply #4

Todd — Tue, 04 May 2004 03:47:44 GMT

dragon,To add Jake's argument to the mix, he is saying that if you buy the 123 and the 124, the odds of winning are essentially cut in half, going from 1 in 1000 to 1 in 500. Basic mathematics would say that you can reduce a fraction from 2/1000 to 1/500, but the question here is does that same reduction principle apply to the odds

Reply #3

Jake649 — Tue, 04 May 2004 01:39:31 GMT

hypersonig,To use your example of the pick 3 game. Say you played 500 tickets. My formula says your new odds are 1 in 2 (50% chance). Your formula says you have only odds of 1 in 500 becuase there are 500 numbers not covered.I am willing to bet my paycheck that I am right.Jake

Reply #2

prob987 — Tue, 04 May 2004 01:32:43 GMT

Reply #1

hypersoniq — Tue, 04 May 2004 00:56:32 GMT

it doesn't work like that...let's use the simplicity of the pick 3... 1,000 possible straight combinations (ignore boxes for the purpose of the discussion...)you buy 1 ticket (1 combo out of 1,000)... you leave 999 uncovered possibilities...you buy 2 tickets (2 combos out of 1,000)... you leave 998 uncovered possibilities... by your formula above it would be a reduction from 1:1000 to 1:500, but you leave 498 combos unaccounted for.....they may express odds as a fraction, but that does not mean... [ More ]