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

Reply #4

cottoneyedjoe — Mon, 09 Dec 2019 23:48:36 GMT

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.

Reply #3

Greenfox — Sun, 08 Dec 2019 08:38:32 GMT

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 y... [ More ]

Reply #2

cottoneyedjoe — Sat, 07 Dec 2019 02:24:14 GMT

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,... [ More ]

Reply #1

Greenfox — Wed, 04 Dec 2019 09:03:32 GMT

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 positi... [ More ]

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

Greenfox — Tue, 03 Dec 2019 01:05:56 GMT

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 p... [ More ]