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Pick 3 Paradox? Question of probabilityReply
Good to see someone caught on - Indeed it is all about perception!A front loaded double does in fact have a 10% chance of creating a triple. In just the the same way a back loaded double and a split loaded double also have the same 10% chance, but when and only when they are looked at individually. The problem arises in my so called paradox because of how the numbers were perceived or observed. All the double types (front - back - split) do have the individual 10% chance of creating a triple
Nov 27, 2005, 7:13 pm - Thoth - Mathematics Forum

Testing data for randomness / predictabilityReply
I'm with you, he makes no sense. In a 5 bit binary string there are 32 possible ways to sequence the 0's and 1's. 00000=0 to 11111 = 31. I think he's trying to make the point that just like in a coin toss the next value could be either heads or tails and the outcome has nothing to do with what came before it, Bravo. The lottery is not truly random, it's sufficiently random which is something else. In lottery games random is limited to the order of numbers being drawn. For instance, i
Jan 25, 2017, 3:53 am - RL-RANDOMLOGIC - Mathematics Forum

Can math and logic improve chances of winning a jackpot?Reply
I pasted the correctly calculated odds above. Jackpots are paid for the extremes, so there are 2 jackpots. The odds don't change. When you play AON 1 to 12 is the same then 13 to 24. When you play a regular lotto, it is different, there is only one jackpot, that eventually is split because there are several winners. If you play pick 3, ten different lines, your odds are 10:1000, the notation 1:100 might be seen on a playslip. The scientific notation would be 0.01. Only one combinatio
Aug 5, 2013, 6:42 pm - SergeM - Mathematics Forum

odds of winning
As a new lottery player, I'd like to calculate the odds of winning. Each NYLottery game has posted odds, however, my calculations give different odds. Perhaps I am not doing it right, or they use a different method, but here are my results compared to theirs: If we use #'s 0-9, and pick one number, then odds of winning are 1 in 10. Thus, I just have to buy 10 tickets, play all numbers, and will get a winner 100%. If 3 numbers are picked, then it's 10x10x10, or 000-999, thus 1:1
Jul 21, 2010, 10:29 am - Gomar - Mathematics Forum

Help JADELotteryReply
Thanks for taking your time to help on this one JadeLottery! There is brilliance in your logic and the way you apply those formulas. However, the odds of 1 in 24,939,240,465 for 21 hits in 979 draws seem extraordinarily inflated to me. I was expecting the odds/probability to be more around one in several hundred thousand or possibly as high as 1 in 2,000,000 or so. From a statistacal perspective, the frequency of boxed combos that show 10 and 11 hits within 979 consecutive trials would s
Nov 13, 2006, 1:41 am - Thoth - Mathematics Forum

Pick 3 Repeat ProbabilityReply
hello cottoney, great job on the odds percentages, can you calculate the value of the probabilities of the pick3 positions? example 478 = last draw 726 = next repeated a digit, but not in the same position, so the question for pick3, calculate the probability of pick3 of a digit not repeating in the same position as the last draw? please we could create a filter at the next not repeat the position
Apr 29, 2020, 7:13 am - dr san - Mathematics Forum

quiz pick-4 daily game oddsReply
Jade Based on my test play where the average hit rate is 9 of 12 step values, is it possible that I can calculate my odds for hitting all 12 by calculating the odds for hitting 3 of 3 steps. In my test play I managed 3 attempts with 11 out of 12. The other games ranged 8 to 10 the overall average was 9 of 12. Most of my misses are step-1 values which are always 1 in 3. It seems logical that I could get a good estimate by calculating the odds for hitting 3 or 4 step-1 values.
Jan 25, 2016, 11:02 pm - RL-RANDOMLOGIC - Mathematics Forum

What Does It Take To Win (Mathematically Speaking?)Reply
jimmy Black Jack vs lottery Card counters and lottery self pics. When a player starts playing BJ he first lays down his bet before the first card is drawn from the deck. He may however watch the game for a while before deciding to enter the game waiting for the most oppertune time to begin play. He may calculate the odds for the remaining cards left in the deck. The lottery player watches the game and then decides to give it a try by picking his numbers based on information go
Feb 3, 2011, 2:26 pm - RL-RANDOMLOGIC - Mathematics Forum

Pick 22: Combinations and Permutations
Original Quote by JADELottery: However, when trying to do the work out for Pick 11 there was a computational problem. It had to do with taking 10 numbers 11 at a time. If you think about this, you can see this is impossible. After a few hours, a few beers, and a few nights rest, I realized the solution. The problem has to do with repetition. I also noticed this problem when I first began listing the patterns for the Pick 22. The solution I came up with was to first not allow duplicates of
Nov 18, 2006, 3:51 am - Thoth - Mathematics Forum

What are the oddsReply
mikemmk1990 I don't know of any books but I have never really looked, the idea here is to gain an advantage but that is based on the users ability to select digits. The game is random and as such there is no secure method of selection. The overall odds for a JP win are better for the lines generated from a pool of 24 numbers selected using typical methods but only because there are many more lines that can be generated. If however a person is not put off by winning more l
Oct 23, 2014, 11:02 am - RL-RANDOMLOGIC - Mathematics Forum