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same exact pick 3 numbers played in the exact same order, for eve and night what are the odds?
So for Tuesday 8/22/23 Georgia pick 3 numbers for the evening draw are 858, and for the night? 858, what are the odds of that happening, I would think it's very high .
Aug 23, 2023, 2:27 am - lottoman626 - Mathematics Forum

Approximating chaotic systemsReply
I had found that time series over a mostly uniform distribution tends to return an average when trying to make a prediction. Had hopes for SARIMAX when studying python time series analysis, but the predictions always came out between 4.8 and 5.2 (for pick 3). Then when you see how predictions based on ARIMA are calculated, you realize it was the wrong tool to use My focus is shifting to classification, though I am not exactly sure what it will look like just yet.
Aug 25, 2023, 8:46 pm - hypersoniq - Mathematics Forum

Approximating chaotic systemsReply
I only recently started reading about PRNGs. I was surprised to learn that some PRNGs that are considered cryptographically secure aren't one large cycle in state space, but instead have some short cycles. That indicated to me that if the PRNG was seeded to start in those short cycles, the PRNG wouldn't be random enough. I'm aware of some satisfiability modulo theory libraries used to solve for the seed given PRNG output. But in those cases, you know the PRNG because it's in the code, and th
Aug 23, 2023, 11:21 am - Wavepack - Mathematics Forum

Arithmetic Complexity in Texas All or NothingReply
It's a really awful game in terms of odds vs returns The odds vs return in most pick 3 games are less than 50% and other games are even worse. The chances of a $2 All or Nothing ticket winning something is 22.22% or 1 out of 4.5 tickets. Though it could be $10, $50, $500 or quarter million, players should expect the minimum return of about $2 for every 5 tickets they purchase. Wouldn't call it an awful game but showing a profit playing the game might require winning the jackpot.
Jul 1, 2023, 2:52 pm - Stack47 - Mathematics Forum

Proof-based math puzzlesReply
1) Since the density of primes decreases as the integers increase, our best chance of proving this existence for all N is to make sure the starting odd number is large enough. We want a starting number such that the next N numbers are factorable/composite. The best way to do this is to make the starting number minus three a composite of factors of offsets from the starting number. Let S = starting number = (2N+1)! + 3 . Candidate set of consecutive odd numbers = {(2N+1)! + 3, (2N+1)! + 5, .
Aug 8, 2023, 7:48 am - Wavepack - Mathematics Forum