Odds of a win with 4 picks per option in a 5 pick game?Prev TopicNext Topic

• Greenfox

  

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  Dec 2, 2019, 8:05 pm

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  Here in NC we have the 5/43 Cash 5. If choosing options smallest to largest, you would have balls 1 thru 39 as options for your smallest pick to play. 2 thru 40 as your second to the smallest to play. 3 thru 41 as your next option. 4 thru 42 as your fourth option and 5 thru 43 as your fifth option. Smallest to largest, that would be the only plausible options per. You will never see ball 2 picked as the third option this way. On a 43 ball set, that means you have 39 options at most without per pick without any narrowing down. Once you make your first pick, your options for each additional pick go down depending on where you start.

  My question is, if you picked 4 options as your smallest, 4 options as your second pick, 4 as your third and so on, only keeping those options in each position how many sets would you have to play? This is NOT figuring in that if you had 10 as your smallest option and had 10 as your second option. Of course you can't play the same number in two positions on one line and win, but I really am just looking at the overall lines with 4 options per. I'm not interested in if there are duplicates for each position for this is what I'm trying to say and I'm not looking for 39-43 or 1-5 in the equation. Just an overall guestimate with 4 options per 5 positions.

  With that said, playing a $1.00 per line, if you had all 5 winning numbers in each of those sets of 4 options, how much you have to spend to win the $100,000, smallest jackpot? Would the total bet cost be more than the jackpot even at it's smallest. 

  Just out of curiosity. Of course.

  Thanks guys and gals.

  You can't steal second and keep your foot on FIRST!!!

  “Strength does not come from winning. Your struggles develop your strengths.
  When you go through hardships and decide not to surrender, that is strength”.

  -Arnold (Ahnald) Schwarzenegger-
• Greenfox

  

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  Dec 4, 2019, 4:03 am

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  Went ahead and worked it all out and got my numbers.

  In case anyone didn't know and is even curious about the answer:

  It would take 1,024 tickets. So $1,024.00 to win the jackpot. That's provided you had one of the winning numbers in each position for numbers drawn 1 thru 5.

  I ran a set of stats using 1-4 as the first pick, 5-8 as the second, 9-12 as the third, 13-16 as the fourth and 17-20 as the fifth. Choosing any number out of that set of options as the winning number in that position as the number that was drawn.  I could change the winning numbers for the smallest  from 1 to 4, second to 5 to 8 and so on and it made no difference in the stats.

  Provided you used the four numbers for each position, it would take 1,024 tickets to win the jackpot. And the winning numbers was in each of your 4 sets of course.

  With what I ran using those numbers you would have 1 jackpot win, so say, $100,000.00 minimum jackpot.

  You'd have 15 four of fives, and the pari-mutual here averages around $200 to $250, I just ran it as $200 and that was $3,000.00

  There would be 90 three of fives at $5.00 a whack, is $450.00

  And there would be 270 two of fives at $1.00 a line for $270.00.

  Then you'd have 405 tickets with one of five and 243 with zero hits for 648 losing tickets where you got nothing.

  So, with this test, had I had 4 options per pick, each one of the 5 sets had that positions winning number in it, it would have cost me $1,024.00 to win $103,720.00.

  That would be 101.29 times your bet for a $102,696.00 profit.

  You can't steal second and keep your foot on FIRST!!!

  “Strength does not come from winning. Your struggles develop your strengths.
  When you go through hardships and decide not to surrender, that is strength”.

  -Arnold (Ahnald) Schwarzenegger-
• cottoneyedjoe

  

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  Dec 6, 2019, 9:24 pm

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  Interesting analysis. You got me thinking about how many tickets you would need if you allowed some of the 5 sets to have elements in common, so I made a big Jupyter notebook to analyze various cases of shared elements. 

  • if 2 sets have 1 number in common, you'd need only 960 tickets
  • if 3 sets have 1 number in common, you'd need only 864 tickets
  • if 2 sets have 1 number in common and 2 other sets have a different number in common, you'd need 900 tickets

  I won't bore you with more of these, I did about 40 or so cases. Some of them get really hairy when sets share different things with different sets and share more than one element.
• Greenfox

  

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  Dec 8, 2019, 3:38 am

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  In response to cottoneyedjoe

  Quote: Originally posted by cottoneyedjoe on Dec 6, 2019

  Interesting analysis. You got me thinking about how many tickets you would need if you allowed some of the 5 sets to have elements in common, so I made a big Jupyter notebook to analyze various cases of shared elements. 

  • if 2 sets have 1 number in common, you'd need only 960 tickets
  • if 3 sets have 1 number in common, you'd need only 864 tickets
  • if 2 sets have 1 number in common and 2 other sets have a different number in common, you'd need 900 tickets

  I won't bore you with more of these, I did about 40 or so cases. Some of them get really hairy when sets share different things with different sets and share more than one element.

  Hello cotton,

  Just starting off, I had to go look up what that Jupyter notebook is. Not hearing of it, I needed to know. Does it not leave your work out there for easy access for someone to get into? Never heard of it and was just curious. I'm sure it's not just an open source or anything. I'm also just guessing it's something like cloud computing, where you can work off a server, do large projects off of a smaller running PC without it slowing down? Just looked at it slightly to see.

  So your saying that with the data I posted, if one number showed up in two of the 5 sets, it would cut you down from 1,024 lines played to 960 lines for the win? Provided you had all winning numbers to choose from to start with. If so, that's 64 less tickets. 3 sets would cut you down another 96 tickets. I'm guessing that's also keeping 4 options per pick and still having one of those 4 in 2 or 3 of the sets? Gonna say that it would have to be that way if your cutting down lines to play by that much.

  With what I've got, it cuts out the possibility of any number needed in more than one position is why I ask. What your saying also helps with determining to keep a number as an option in more than one position.

  Where I stand with what I've been working on, I have 11 options for my smallest number drawn. 15 for the second. 18 for the third. 21 for the fourth and 15 for the fifth. That's my options per position, smallest to largest, before doing any filtering down per game. Each position is broken down into 4 sections where the numbers layout. The winning number for all 5 picks show up in those sections, the way I have it tuned at the moment, on average right at 10 games per month with another 10 games per month showing 4 of the 5 winning numbers. Some sections may have only one option, some may have half the options. Depending on the math. But on average, about 4 to 5 options per section. I can tune each position to give me more or less options, depending on how I want to play it. Safe or risky. Obviously, I need to tune down the 3rd and 4th option and run tests to see how it affects the monthly average starting output and hopefully not decrease the average. Will do so soon. Tuning those down, and keeping my monthly average up, would put me about 3 options per section for each pick and help out more than anyone would know. The way I have this thing setup, with the 4 large sections per pick is why I wanted to find out the possibility of just picking a number out of those and getting a win. Provided I chose right out of those of course.

  It's also funny that I'm reading this whilst watching old 60's sci-fi about other planets and such. That made me think about this whole journey I've been on with this. If I can get to Jupiter and back, out west shouldn't be that big of a deal.

  Anyway, thank you for the reply and the further research.

  You can't steal second and keep your foot on FIRST!!!

  “Strength does not come from winning. Your struggles develop your strengths.
  When you go through hardships and decide not to surrender, that is strength”.

  -Arnold (Ahnald) Schwarzenegger-
• cottoneyedjoe

  

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  March 28, 2019
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  Dec 9, 2019, 6:48 pm

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  You can run Jupyter notebooks locally and totally offline if you have Sage Math installed. It's basically a free and opensource version of Mathematica or Matlab.