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649 formulaReply
RJOh wrote: ``Had I played and picked 31 as my key number, my odds of picking the other five winning numbers would have been 1:1,712,304``. Yes, if you could know a priori that 31 would be picked. But looking forward, you cannot know that. So your true odds are the probability of matching a combination with 31 (1/1,712,304) times the probability that one of those combinations is drawn (1,712,304/13,983,816). That is 1:13,983,816, the same probability as matching all 6 randomly. So your odds a
Jun 13, 2013, 2:27 pm - mathhead - Mathematics Forum

How are odds calculated?Reply
Where does those odds come from? 2 of 5 has 1:9.6 odds. 3 of 5 has 1:103 odds Where do they come from? === They come from the multiple ways of having some of the 5 winning numbers. As mentioned above, the order isnt important, but there's only 1 chance to have all 5 numbers, so there's only 1 winning combination out of the 575,757 possible combinations. There are multiple way to have some, but not all of the winning numbers, so that gives you more than 1 chance in 575,757 of winni
Oct 1, 2009, 2:09 pm - KY Floyd - Mathematics Forum

The odds the Jackpot reached $390 million are...
What are the odds of winning the MEGA MILLIONS Jackpot? The answer is 1 in 175,711,536. Since all the parameters are known, i.e. - the odds of winning the MEGA MILLIONS Jackpot: 1 in 176 million; - the price for each bet - the distribution of the revenue from those bets, in particular which percent goes to the jackpot; - how the jackpot increased week after week until reaching the record of $390 million ... it should be peanuts for you to calculate what are the odds that the Jackpot c
Mar 19, 2007, 5:38 pm - MattMatt - Mathematics Forum

Pick 3 Paradox? Question of probability
The standard (mechanical) way that a Pick 3 digit is selected is to draw one ball from a machine that contains a pool of ten balls. This is done 3 separate times (there are 3 machines that each contain ten balls) to give us our winning Pick 3 combination. Since there are ten balls in each machine we can calculate the total possible combinations by multiplying each of the machines ten balls by the other machines ten balls and so on. When this is done we get 10x10x10, which equals 1000 possib
Nov 10, 2005, 1:12 pm - Thoth - Mathematics Forum

Lotto 7/35Reply
The odds don't change because you change the way the numbers are wheeled / generated. While your playing from a reduced matrix, the actual game still plays from the 7-35 pool. The odds that the winning set will fall within the reduced pool is 6,724,520/1,046,520=6.426. On average you can expect the winning set to come from the 1046520 lines around 1 in every 6.426 games. To calculate the lower prizes could be done but they too would be conditional. I have a old tool that calculat
Sep 25, 2018, 11:47 am - RL-RANDOMLOGIC - Mathematics Forum

How to simulate a distribution and use the lottery's own randomness against itself.Reply
Jade has done a lot of hard work and posted it for us to use, if we wish. With all due respect, Tucker Black, from this post, you don't seem to believe in game theory which is pure mathematics. in fact you utter these words: This is the superstition of numbers chosen more frequently in the past are more likely to hit in the future which is just as false as the superstition numbers chosen less frequently in the past are due to hit in the future . I have written lots of programs to simulate
Oct 29, 2016, 5:59 pm - DJTRUMP - Mathematics Forum

Daily 4 Formulas
I am just about finished with a workout in Excel using past numbers (PN) and Constants (C) in formulas for CA Daily 4. I select from 11 Number Columns 4N1 4N4; 4M1 4M4; 4NRS; 4MRS; 4GRS (Root Sums for numbers and mirrors; and the root sum of the game number digits) in a previous game (Game Back). I set a Constant to be used in the formula. I am running the following formulas: (1) PN-PN, (2) PN-C, (3) PN^2-PN, (4) (PN^2+PN^2)^.5. I have generated thousands of results for each formula going back 5
Oct 5, 2015, 1:33 pm - AllenB - Mathematics Forum

Some basic math formulas applied to pick3/pick4Reply
I understand the idea behind this and if the trend you said later about the k digit being USUALLY 0-4 could be beneficial. However I'm curious if you have tried this theory by moving the mirror so to speak. Technically a mirror is assigned arbitrarily. Perhaps there is a universal idea of what a mirror is and how its constituted, in reality a mirror is just a form of a cypher and moving the digits from one place to another since the numbers still run consecutively. In which case you could move t
May 29, 2013, 6:42 am - AlgorithmGuru - Mathematics Forum

Backtesting and Simulating Lottery SystemsReply
Jimmy You've got to be kidding! Are you saying that picking so many digits to play needs some advanced math to calculate what the odds for matching 3,4,5 or more of them. Looks like another side step to me. RL
Apr 12, 2011, 12:45 am - RL-RANDOMLOGIC - Mathematics Forum

Check my math, please?? :)Reply
So 1 in 11.2 chances of 1 particular number being chosen, That's fine for whether or not a particular number will be selected in any given drawing, but you're looking for the chances that any of the 5 numbers from a drawing will repeat in the next drawing. Here's how that works. 5 of the 56 numbers were in the last drawing and 51 of 56 were not. When the first number is selected there's a 51/56 chance it will not be any of the 5. If that happens there will be 55 numbers remaining, of whi
May 26, 2010, 3:49 am - KY Floyd - Mathematics Forum