$300.00 worth of QP'sPrev TopicNext Topic

• Sleepingdogslie

  

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  August 4, 2003
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  May 6, 2004, 9:00 pm

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  If one were going to s
• RJOh

  

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  mid-Ohio
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  May 6, 2004, 9:16 pm

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  If I was going to buy $300.00 worth of tickets, I would pick my own numbers and make sure I didn't repeat any combination of 5 that had won or try to avoid repeating any combinations of fours within the group.  That's just me, I read of several people who won the jackpot by just buying $5-$100 worth of tickets at one terminal. 

  RJOh

   * you don't need to buy every combination, just the winning ones * 

         
  
• Jake649

  

  Canada
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  May 6, 2004, 9:49 pm

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  I would buy all the tickets at one terminal. Your probability of winning does not change by varying the number of different terminals used to buy your tickets.

  Good luck,
  Jake
• ohiopick3

  

  Ohio
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  May 6, 2004, 10:24 pm

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  Quote: Originally posted by RJOh on May 06, 2004

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  I would pick my own numbers and make sure I didn't repeat any combination of 5 that had won or try to avoid repeating any combinations of fours within the group.  That's just me, RJOh
  
  
  
  

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  RJOh just hit the nail right on the head! That's the way to go!
• weshar75

  

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  May 7, 2004, 1:34 am

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  If I had the opportunity to buy $300 dollars worth of Powerball lottery I would go to the lottery retailer you prefer and go early in the morning or late at night before the drawing.  You could get 60-5 dollar tickets without powerplay or 30-10 dollar tickets with powerplay.  Your not going to hit the jackpot on every ticket so powerplay option will maximize your prizes that don't get all six numbers.  Your going to have repeated numbers so you might take the extra time and fill out play slips with different number combinations to minimize the repeats. 
• BobP

  

  Dump Water Florida
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  Quote: O
• JohnC

  

  Melbourne
  Australia
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  May 7, 2004, 10:55 pm

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  There has been a lot of debate over the years about this and my thinking on this one is that it doesn't really matter......everything depends on your partner (Lady Luck).  Over the years, there have been endless times when there's been a big draw and I've lost count of the number of times when big syndicates have staked large sums of money and come up empty but some guy somewhere has bought a minimum-stakes ticket and hit the jackpot.

  About three years ago we had a $30 million Powerball draw here in Australia (which is as big as it gets so far) and they sold umpteen tickets for it.....there were syndicates formed all over the country.  As things turned out, the jackpot was won by one guy in Perth who bought one "Powerplay Six" ticket for $3.25, which is the smallest ticket you can buy.

  If Lady Luck is in your corner, you'll win.
• twisted

  

  New Jersey
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  May 7, 2004, 11:40 pm

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  Quote: O
• BobP

  

  Dump Water Florida
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  May 8, 2004, 12:13 am

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  Quote: Originally posted by JohnC on May 07, 2004
  

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  There has been a lot of debate over the years about this and my thinking on this one is that it doesn't really matter......everything depends on your partner (Lady Luck).  Over the years, there have been endless times when there's been a big draw and I've lost count of the number of times when big syndicates have staked large sums of money and come up empty but some guy somewhere has bought a minimum-stakes ticket and hit the jackpot.

  About three years ago we had a $30 million Powerball draw here in Australia (which is as big as it gets so far) and they sold umpteen tickets for it.....there were syndicates formed all over the country.  As things turned out, the jackpot was won by one guy in Perth who bought one "Powerplay Six" ticket for $3.25, which is the smallest ticket you can buy.

  If Lady Luck is in your corner, you'll win.

  
  

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  Yes, I too agree one needs a huge dose of luck to win a jackpot.  It has also been proven hundreds of thousands of Quick Picks can be purchased with no guarantee of any win, while permutations / wheeling / cover designes can guarantee lower tier prize wins, providing a win you need only one or two more winning numbers to jackpot.  Say a game is 5/52 and you have guaranteed a 4if5 prize by wheeling, now that ticket has a one in 47 chance of the five number jackpot. 
  
  If the lucky fellow you mentioned paid $3.25, as I understand it Au. Powerball tickets are only $1.10 each $.80 cents from U.S. so maybe he was buying a 8 or 9 number wheel in three combinations.  Heck, even three tickets covering 15 different numbers among them is a 2if4of5 wheel in 3 lines with a high percentage chance of a 3# win.
  
  http://www.ozlotteries.com/free_account.php?aff_id=AE83-C230-052A-313D
  
  See the link for odds and prices of Au games.  I wish we had a game like the OZ game here, million dollar tax free prize, 6/45 with two bonus balls and two plays for a dollar playable online, sounds good to me. BobP
• JohnC

  

  Melbourne
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  May 8, 2004, 6:44 pm

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  Hi BobP,

  In Australia, Powerball tickets cost 50 cents a game.  A Powerplay 6 is a ticket with six numbers from which the first five came (any-five-from-six) plus one specific number for the Powerball.

  e.g 10, 14, 17, 23, 31, 40  (from which the first five numbers must come), plus

        say 37 (which must be the Powerball).

  The odds of winning Powerball over here are just under 55 million to 1 and as this guy bought six games, the odds against him winning were around 9 million to 1.  With a bit of help from Lady Luck though, he made it.
• twisted

  

  New Jersey
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  May 8, 2004, 8:34 pm

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  Quote: Originally posted by JohnC on May 08, 2004
  

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  Hi BobP,

  In Australia, Powerball tickets cost 50 cents a game.  A Powerplay 6 is a ticket with six numbers from which the first five came (any-five-from-six) plus one specific number for the Powerball.

  e.g 10, 14, 17, 23, 31, 40  (from which the first five numbers must come), plus

        say 37 (which must be the Powerball).

  The odds of winning Powerball over here are just under 55 million to 1 and as this guy bought six games, the odds against him winning were around 9 million to 1.  With a bit of help from Lady Luck though, he made it.

  
  

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  Are you saying that if he would have played 300 games he would have cut his odds of winning to 1 in 183333???  Doesnt make sense to me.
• BobP

  

  Dump Water Florida
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  May 9, 2004, 1:44 am

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  Quote: Originally posted by twisted on May 08, 2004
  

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  Quote: Originally posted by JohnC on May 08, 2004
  

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  Hi BobP,

  In Australia, Powerball tickets cost 50 cents a game.  A Powerplay 6 is a ticket with six numbers from which the first five came (any-five-from-six) plus one specific number for the Powerball.

  e.g 10, 14, 17, 23, 31, 40  (from which the first five numbers must come), plus

        say 37 (which must be the Powerball).

  The odds of winning Powerball over here are just under 55 million to 1 and as this guy bought six games, the odds against him winning were around 9 million to 1.  With a bit of help from Lady Luck though, he made it.

  
  

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  Are you saying that if he would have played 300 games he would have cut his odds of winning to 1 in 183333???  Doesnt make sense to me.
  

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  Not exactly as it depends on whether we're talking about 300 games or $300. buying 600 games and I don't know what kind of coverage you can get without generating games for it.
  
  In any regard, lets pretend we're only trying to win the 5if5 part of the game with the powerball portion excluded.  What are the odds of say getting two numbers correct on a combination you play?  Maybe 1 in 20?
  
  Let's say with 300/600 I can get multiple 3if5 coverage, each of those tickets then have a 1 in 20 chance of picking up the remaining two numbers for a 5 number win.  You'd already have 3# wins guaranteed, you just have to get two more numbers the bulk of the work is already done.
  
  Well ok, since three numbers out of five are already busy giving us the 3# wins we're really looking at a 2if2 coverage requirement for picking up the two additional numbers, still it more interesting starting with 3#'s correct.
  
  If we could afford a 4if5 wheel, we'd have 4 of the 5 numbers guaranteed which would mean only one of the game's numbers are left to be picked up by our ticket out of the numbers not making up the four.  If there are only say 40 numbers left for that fifth number to come from, our odds of winning the 5# prize would be 1 in 40 not gazillians to one.  I'm afraid a 4if5 wheel for all the numbers would be thousands of dollars, but not out of reach for a syndicate.  BobP
• JohnC

  

  Melbourne
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  May 9, 2004, 3:24 am

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  No...............55 million divided by 300 is 182,333.33.
• BobP

  

  Dump Water Florida
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  May 9, 2004, 5:52 pm

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  Quote: Originally posted by JohnC on May 09, 2004
  

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  No...............55 million divided by 300 is 182,333.33.
  

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  No............you're missing the point.  The girl has just called out four of the five winning numbers and you have all four of them, what are the odds you will have the fifth winning number in your combination?  BobP
• JohnC

  

  Melbourne
  Australia
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  May 11, 2004, 9:17 am

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  Sorry, BobP....this is all a bit complicatd for me.