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I'm actually not sure if this question has a concrete answer or even makes sense. If you were to analyze a particular filter [from whatever lottery software] and say its Due percentage is high, and the max interval is around the highest based on a large number of draws, is there a particular formula you could use that might suggest what its chance of working are? In other words, if the filter says it has a 13 percent chance after back testing, is there a formula that might give a different percentage based on these variables?
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Quote: Originally posted by Lottofoolinn on Nov 16, 2018
I'm actually not sure if this question has a concrete answer or even makes sense. If you were to analyze a particular filter [from whatever lottery software] and say its Due percentage is high, and the max interval is around the highest based on a large number of draws, is there a particular formula you could use that might suggest what its chance of working are? In other words, if the filter says it has a 13 percent chance after back testing, is there a formula that might give a different percentage based on these variables?
If the probability of an outcome is P (where P is between 0 and 1) then the probability the outcome will occur at least once within the next N drawings is
1 - (1-P)^N
This assumes the drawings are independent. I don't know if that's what you're looking for because I don't use lottery software, and I couldn't tell you what formulas various software authors use to give you probabilities like 13% chance that such and such will hit.
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Quote: Originally posted by oate on Nov 17, 2018
If the probability of an outcome is P (where P is between 0 and 1) then the probability the outcome will occur at least once within the next N drawings is
1 - (1-P)^N
This assumes the drawings are independent. I don't know if that's what you're looking for because I don't use lottery software, and I couldn't tell you what formulas various software authors use to give you probabilities like 13% chance that such and such will hit.
Thanks for the reply. You probably don't have to have lottery software to do what i did. The 13 percent was jus an example. If you backtest the filter, and say it passed 2/13 times, the percentage would be around 13, so its just a percentage based on how often it has passed given a specified range of draws [i usually like to anaylze the whole history if possible]
So, if you backtested a larger range, say 5000 draws and found out the max amount of draws between when the filter passed was 70 [so it failed 70 in a row once], you'd have the max interval. And if you also had the information that its average interval is 10, and its recently failed 65 times in a row [close to the max] I was wondering if the somehow make a difference in probabilty. It would likely be a good educated guess to say this filter has a better chance than a different one at 13 percent no where near its max interval, since the filters essentially will pass again at some point, but I was wondering if there was some way to quantify this potential difference. Sorry, I'm not good with statistics at all, so I don't know if the formula you gave really does answer the question or not.
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Ok, I think I see part what you're asking. Yes, you can get a probability from an average interval. Let's say that on average an outcome happens once every N draws. Then the probability of that event happening is 1/N. The probability that that particular event doesn't occur in K consecutive draws is
(1 - (1/N))^K
An example: The probability of an Even-Even-Even combo in Pick 3 is 1/8 = 0.125 = 12.5%. You would expect the lottery to draw an Even-Even-Even combo once every 8 draws on average. The probability of Even-Even-Even not happening 60 times in a row is
(1 - (1/8))^60 = (7/8)^60 = 0.0003315
This example uses an event whose theoretical probability is known and could be worked out easily if you didn't know it. It sounds like you don't know the theoretical probability of an event and are trying to estimate it from the data. In which case I would suggest using all the back data you have. If you have data for D drawings and your event has occurred E times, then the estimated probability is E/D.
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Quote: Originally posted by Lottofoolinn on Nov 17, 2018
I get what your saying. I think this pretty much covers what I was asking. Thanks.
This is a Pick 3 Tutorial from Lottoman from a number of years ago. This has some very good info regarding probability for Playing the Pick 3 game but can be applied to other games: