Megaball prediction

Reply #19

Wavepack — Thu, 11 May 2023 00:49:55 GMT

Yeah, there's also the scratcher games like tic-tac-toe and bingo where you can look at how the boards are filled out, look for singletons that line up on a board, and with 60% probability, know whether a scratcher card is a winner. So if you could look at a stack of scratchers before buying, you could pick out the most likely winners (without scratching anything off). Here's an article about that.

https://www.wired.com/2011/01/cracking-the-scratch-lottery-code/

1 - alpha = confidence i... [ More ]

Reply #18

Orange71 — Thu, 11 May 2023 00:10:41 GMT

alpha is basically 1 - upper confidence level (decimalized) in this case. So alpha=0.05 is 1 - upper confidence level. Therefore, upper confidence level in this case is 0.95 or 95% of the distribution.

Reply #17

Orange71 — Wed, 10 May 2023 23:59:24 GMT

I might add, in jest, that if you do find a lottery with a prize structure flaw, don't post it on lotterypost.com. That would be seriously bad for your return

Reply #16

Orange71 — Wed, 10 May 2023 23:55:30 GMT

Analyzing numbers is fun, but there is hardly any reason to believe that lotteries, at least government-run ones in democratic countries, should be anything other than random.

If you want to find an edge in lotteries, look for flaws in the prize structures, not the number draws. There's the famous story of the Midwest couple Jerry and Marge Selbee who did just that. They found a lottery with a prize structure such that jackpots that were not won were distributed down to lower prize categori... [ More ]

Reply #15

Wavepack — Wed, 10 May 2023 22:07:08 GMT

It seems like your result is saying that there is an 89% (or ~ 95%) probability that the sample is from a non-uniform pdf. Also, not sure what your alpha is. The Chi-square test that I use has no alpha.

My white ball Chi-square test result is 75% probability that the sample is from a non-uniform pdf.

Our red ball Chi-squared test results seem to match.

Reply #14

Wavepack — Wed, 10 May 2023 21:55:57 GMT

About a year ago, I did the Chi-square test on Mega Millions and Powerball, both white and non-white balls. I was going from memory that each had too high probability estimates for a non-uniform pdf test.

Those test results from a year ago is what motivated me to see if there was any pdf non-uniformity that could be used to improve predictions. It's been a fun side project every since that I work on every now and then. The project drives home probability and statistics fundamentals, along w... [ More ]

Reply #13

Orange71 — Wed, 10 May 2023 21:47:39 GMT

Here is the analysis for the same 577 draws for the white balls (1-70). The pre-factor modifier for critical 2 due to sampling without replacement here is (N-1)/(N-k) where N=70 and k=5. Same conclusion, we fail to reject the Null Hypothesis of randomness for alpha=0.05 . Test statistic = 71. Modified critical Chi-square, [(N-1)/(N-k)] 2 , is 94.89.

Ball Observed Expected (O-E) 2 / E

1 43 41.2143 0.0774

2 40 41.2143 0.0358

3 50 41.2143 1.8729

4 49 41.2143 1.4708

5 3... [ More ]

Reply #12

Orange71 — Wed, 10 May 2023 20:04:18 GMT

Wait, I thought you were talking about Megaball (red) non-randomness in the original post. You're saying now the white balls (main pool) is where the apparent non-randomness is. Correct?

Your post:

I've found a method, using only past draw information, to correctly predict the next Megaball draw at 8.07% (N = Number of tests = 322). The expected prediction match with uniform random guessing is 4%.

...

I have found no method that predicts Mega Millions white balls over expected w... [ More ]

Reply #11

Wavepack — Wed, 10 May 2023 19:33:05 GMT

Test the white balls, and you will see something may be amiss.

ChiSquaredW = 75.084428

Prob. white draws from non-uniform pdf = 0.712206 for # draws = 574

ChiSquaredR = 21.121951

Prob. red draws from non-uniform pdf = 0.368479 for # draws = 574

Sampling without replacement on the white balls complicates the Chi-square test formula a bit. Below is the Matlab code for it.

ChiSquaredW = NumW * (NumW - 1) * (sum(DistW .* DistW) - (NumWBalls^2 * N^2 / NumW)) ...

Reply #10

Orange71 — Wed, 10 May 2023 18:23:47 GMT

Indeed, the data is posted here for 577 draws since 2017. The Chi-square Goodness of Fit results analysis are shown below using the usual alpha = 0.05. Nothing appears amiss to me. Null Hypothesis of randomness is not rejected.

https://www.lottonumbers.com/mega-millions/number-frequency

Reply #9

Wavepack — Wed, 10 May 2023 15:51:20 GMT

As stated in the initial post, the predictions were made using only past draw data, and there was no training of the method with data at all (other than adjusting two time window length parameters)

Maybe 2 to 3 percent prediction accuracy over expected was due to chance, but unlikely 4% over expected. I've got good statistical significance. Over three sigma.

If you do a Chi-square analysis on past Megaball draw data to test the uniform pdf hypothesis, the results suggest there are good o... [ More ]

Reply #8

Orange71 — Wed, 10 May 2023 14:26:56 GMT

The only way that conjecture makes sense is if you had hypothesized that particular ball X would occur 26 or more times before seeing the 322 draw results. As far as I can tell, you didn't, it was after seeing the data. Now trying applying the conjecture about ball X to the next 322 draws, as it only matters what happens in the future. You might get extremely lucky and it'll happen again, i.e. 1 in about 1550 chance. I wouldn't hold out too much hope.

Reply #7

Wavepack — Wed, 10 May 2023 02:37:26 GMT

So using the exact binomial pdf, and not a Gaussian approximation, results in a 1 in 1550 chance that the 8% correct Megaball prediction method was due solely to uniform random chance.

That is statistically significant, which is the main point. Losing less money when playing the Lotto is a good thing.

Reply #6

Orange71 — Wed, 10 May 2023 00:58:20 GMT

The simulation code is correct for how I framed the problem statement. Consider the cumulative Binomial distribution probability that a particular ball will be drawn at least 26 times out of 322. This is the same as one minus the cumulative Binomial probability that up to 25 draws will occur out of 322. We can easily calculate this in Excel with the following formula:

1-BINOM.DIST(25,322,0.04,TRUE) = 0.000644

We might, at first glance, think we can just multiply this probability by 25 t... [ More ]

Reply #5

Wavepack — Tue, 09 May 2023 05:56:46 GMT

Assuming all megaballs are equally likely to be chosen, 1/25 = p = probability that a draw prediction is correct due purely to chance. Let Y = number of correct predictions in n=322 independent draws. The probability distribution of Y = P(Y=y) is binomial with parameters (p, n = 322). Since n is large, P(y = Y) = cumulative distribution function of Y = cdf of Y is very well approximated by the Central Limit Theorem as

P(y = Y) = N[(y - n*p) / sqrt(n*p*(1 - p))],

where N(x) is the cumulat... [ More ]

Reply #4

Orange71 — Tue, 09 May 2023 02:50:10 GMT

I agree that the probabilities of a particular vs. any number are different.

I'm supposing you looked at the data (322 draws) and observed a particular ball that was drawn 26 times. However, there would have been no rational basis, before seeing the data, to predict that specific ball would appear more than any other. That's why my perspective is, if you are looking for evidence of bias, you have consider any ball.

I also agree that coincidences, or unusual occurrences, happen all the ti... [ More ]

Reply #3

Wavepack — Tue, 09 May 2023 01:47:26 GMT

The probability of ANY ball occurring more than 26 times out of 322 tests due to uniform random chance is different than estimating the probability of ONE chosen ball of the same. In other words, your simulation over estimates the probability due to random chance.

This reminds me of the difference between any two people in a group having the same birthday versus any two particular people in the group having the same birthday. Different probabilities.

It also reminds me of coincidences in... [ More ]

Reply #2

Orange71 — Tue, 09 May 2023 00:31:34 GMT

I ran a simulation of 100000 loops of picking 322 random numbers (balls) each from 1 to 25. 1.576% of the time out of 100000 loops (fraction = 0.01576) at least one ball occurred 26 or more times out of 322. I'm not changing my view that the apparent Megaball results are consistent with just random fluctuation.

Python 3.x code:

#code start

import random

def simulation():

results = list()

balls = 25

iterations = 322

for i in range(iterations):

results.append(ra... [ More ]

Reply #1

Orange71 — Mon, 08 May 2023 23:30:24 GMT

How do you figure you can predict a random process? Yes, you will see fluctuations from past results, but that doesn't mean that a hot number from past drawings will carry forward to future draws or that a cold number is overdue . 8.07% out of 322 sounds like you found a number that occurred 26 times. Try running a simulation of say, 1 million iterations, picking 322 random numbers from 1 to 25 on each iteration. Then see what percentage of simulations result in any one or more numbers occurring... [ More ]

Megaball prediction

Wavepack — Fri, 05 May 2023 03:49:36 GMT

Because of the limited draw data set size and high noise, it's very difficult to ascertain whether there is any signal in the past data that can be used to predict future draws.

I've found a method, using only past draw information, to correctly predict the next Megaball draw at 8.07% (N = Number of tests = 322). The expected prediction match with uniform random guessing is 4%.

No training cheating (other than adjusting two time length/window parameters).

I have found no method that... [ More ]