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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 6/47 Probability QuestionReply
The probability of getting 5 out of 6 balls on a single ticket is (6 choose 5)*(41 choose 1)/(47 choose 6) = 246/10737573. So if you've got 20 tickets, no two of which have the same set of 5 numbers, then your chance of winning the second place prize in a single drawing is 20*246/10737573 = 4920/10737573. To solve your problem, you just put that new fraction in the place of where you have 20/10737573 and keep the rest of the equation the same. The whole thing works out to about a 1 in 21.48 c
May 20, 2020, 5:38 pm - cottoneyedjoe - Mathematics Forum

Statistics around the balance of even/odd and small/big numbersReply
As everyone has stated, you have 20 odd and 19 even. The odd has a slight advantage in this case. The confusion is the numbers. Any NUMBER have equal chance since there is only one of each in the pool. However....THROW THE BALLS AWAY! Instead let's use 20 Black Beans, those we'll call odd. And 19 White Beans, which we call even. That changes our perception a little, eh? We have a 20 in 39 chance of drawing a Black Bean (odd) on the first draw. Let's say we do. We now have an e
Jan 3, 2011, 6:47 pm - garyo1954 - Mathematics Forum

I Believe hitting lottery is all mathematics-- I can prove it!Reply
That reminds me of the systems that add or subtract something from the previous drawing; add 1 to the first digit, subtract 3 from the second, and add 7 to the third. These systems usually come from reverse engineering when someone notices that by adding or subtracting something there would be 3 straight hits and maybe some boxed hits over a drawing period of say 200 drawings. If the previous drawing was 278 the play would be 345 so the assumption is that 345 will always follow 278 and we kno
Oct 27, 2009, 11:26 am - Stack47 - Mathematics Forum

Mathematics and the LotteryReply
Koycerin states that only the very first bet at the beginning of a cycle will truly be 50/50. If we take the coin example as head or tails, and the first toss was heads, second was heads, it will always tends to correct itself by striving to blance itself out. Regardless of whether it's 10 times in a row, what would the bettor choose, heads or tails? You can have all the theories in the world but the when you factor in the human factor into the equation, the most logical bet would be tails. S
Aug 7, 2011, 2:53 pm - RJOh - Mathematics Forum

Calculating a percentageReply
In MegaMillions, any particular number has a 5 in 56 chance of being drawn. It doesn't make any difference in what order it is drawn, as long as it is drawn. The only thing this sort of thing will accomplish is to add suspense to the draw, if you're into that sort of thing. But, yes, any given number has a 1 in 56 chance of being drawn first. If that number isn't drawn, it has a 1 in 55 chance of being drawn second, and so on. The odds are lower and the percentages are higher on any particula
May 7, 2009, 12:53 am - johnph77 - Mathematics Forum

At what level of information does chaos appear?Reply
There is a flaw in your logic, 0 and 1 can't both equal randomness. One has to equate to yes and the other has to be no. In order to prove your theory you would have to flip a coin 5 times at least 1000 times in order to compile a result set that had any meaningful data. Basically in flipping a coin 5 times there are 32 possible outcomes. The majority of those outcomes will result in 3 of the flips even splitting 1 and 0 in various combinations. The first 5-10 times may possible result in all
Feb 28, 2017, 2:52 pm - Novan60 - Mathematics Forum

Preferred lottery prediction methodsReply
Second the data of 1 in 3,800,000 to match all 5 numbers is filled with a lot of near impossible combinations like....1,2,3,4,5,....2,3,4,5,6.......3,4,5,6,7.....3,4,5,7,8..etc. So really the realistic odds are much much less. I wonder why people think those combinations are nearly impossible . Even Gail Howard in her Lottery Masters Guide claims that the combo 1,2,3,4,5,6 will probably never fall because its too far out of balance. The drawing machines and the balls they contain knows
Mar 21, 2006, 12:17 pm - Thoth - Mathematics Forum

Can math and logic improve chances of winning a jackpot?Reply
I pasted the correctly calculated odds above. The odds are posted on the Texas Lottery website and only you know why it was necessary to calculate and show them here. Jackpots are paid for the extremes, so there are 2 jackpots. If there really are two jackpots, someone is ignoring the fact one ticket can only win one jackpot and believe it or not none of the other prizes with two chances to win can't win twice either. The odds are correct on the website (1 in 2,704,156) because once
Aug 7, 2013, 1:24 am - Stack47 - Mathematics Forum

Calculating a percentageReply
The fallacy in this argument is that, if one assumes there will be five numbers drawn, that would equal 100% of the numbers. That would only be true if the five numbers to be drawn were already known or predetermined, or if there were only five balls to be drawn. Then your argument is valid, and the odds for drawing a specific number from that draw would be: 1::5, or 20%, on the first ball. 1::4, or 25%, on the second ball. 1::3, or 33%, on the third ball. 1::2, or 50%, on the fourth
May 7, 2009, 8:40 pm - johnph77 - Mathematics Forum