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Box odds questionReply
ScoobyDue ... yes people seem to get fixated on the overall odds which are for a straight hit, 1 in 10,0000. But when you are playing box the odds of 1 in 715 are correct. Your odds of hitting boxed, while we understand that a single boxed number has 24 combinations, in Maryland for a $1 box you get $200. Suffice it to say if you are playing two different boxed combinations 2 of 715 or .02797% is still a low chance. I have used this strategy and have hit boxed. Still have not hit
Jul 19, 2023, 10:25 am - kdsjeter - Mathematics Forum

Approximating chaotic systemsReply
I don't know if this will help or not, but in both MM and PB they do two test draws that you don't get to see. So your numbers could have come up in the test draws. The winning numbers that come up are in past draws even the powerball or mega number. So pick 25-30 numbers and 5-10 bonus numbers from the past draws and pick from that pool of numbers and you have a better chance of winning, or you can just Quick Pick.
Oct 10, 2023, 4:02 pm - Down - Mathematics Forum

My fascination -The first digitReply
Please explain what this contest means. You're picking a single white ball from 1 to 69, correct? (That presumably corresponds to the Powerball lotto.) Then, your pick is compared to actual (5) lotto winning number results. Correct? Do you win if your pick is any of the 5 actual drawn numbers or only the first one? You were a 12-time winner in 2022, and you are an 8-time winner this year so far, is that correct? How many draws is that in both cases? Let's get the parameters down
Aug 19, 2023, 10:32 am - Orange71 - Mathematics Forum

Arithmetic Complexity in Texas All or NothingReply
I took the liberty of inferring probability, as arithmetic complexity is a fairly imprecise term. Your Groups A, B and C have nothing to do with why All or Nothing is an unfair game. You can pick any groups of 9 numbers, 10 numbers and 5 numbers and say the same thing. They don't have to have 0, 1 or 2 as a first digit. The reason All or Nothing is terrible is the price structure payouts versus the probabilities of hitting the price tiers. I provided an example for the top prize tier in my origi
Jun 17, 2023, 11:50 am - Orange71 - Mathematics Forum

Shuffling and Mixing
Lotto ball machines work well to randomize ball picks because of their mixing properties. A closely related problem is how many shuffles are sufficient to randomize a deck of cards. The following curve is pulled from an AMS (American Mathematic Society) post on this topic. A link to the post is given below. https://www.ams.org/publicoutreach/feature-column/fcarc-shuffle d is a measure of how close the pdf (probability distribution function) of the card ordering is from uni
Dec 24, 2023, 5:01 pm - Wavepack - 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