thecatlover
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- Jul 15, 2017
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Hi,
Caution: long post ahead.
My question is a bit tricky but deals in probability. I'm trying to determine what a few different numbers should be on average. My assumption is that the iterations are not truly random and not chosen by a human. Rather, my assumption is that the iterations are determined by an algorithm and that the algorithm does not choose randomly but only gives the impression of being random.
Let me explain.
There are six slots from left to right (see the bold numbers below).
In theory, in each slot a number between 1 and 49 can occur. Each day a new drawing takes place in the very same way.
In each slot from left to right, the number is always higher than in the previous slot. In slot one, it is assumed therefore that 49 cannot be a number because the rest of the slots must be filled with higher numbers and the highest possible number is 49. This begs the question that the first slot is chosen first.
In slot one the number is in practice nearly always inferior to ten. Nearly always? - Out of 30 iterations thus far, 26 of the numbers are inferior to ten.
In slot one, it is assumed that the number can be as high as possible given the rest of the following slots but in practice none have gone beyond the number 20.
In slot six, out of 30 iterations, 21 are between 41 and 49.
In slot six, the number has not been below thirty but is very often in the forties. - Knowing the nature of these two terminal slots may help to determine the future outcome of the inner slots (slots 2, 3, 4, and 6. See below for further reasoning).
Horizontal Averages:
In slot two, 12 iterations are between the numbers 11 and 20. In slot two, 10 of the iterations are between 1 and 10. This means that over the course of thirty different drawings, slot two has most often been filled with a number between eleven and twenty. It might be the number fifteen, the number twenty, the number twelve, but the there have been more occurrences between those two numbers than any other in slot two.
In slot three, 13 iterations are between 21 and 30. In slot three, 12 iterations are between 11 and 20.
In slot four, 15 iterations are between 21 and 30. And 12 iterations in slot four are between 31 and 40.
Inductions:
If in slot six the number is 42 or greater, out of 21 iterations, 14 iterations in slot four are 40 or above. That is, if in slot six the number is 42 or higher, then in slot five, the number is 40 or higher fourteen times out of twenty-one, hence the chance that the number in slot four will be 40 or higher when in slot five the number is 42 or higher is significant. How significant? Well, 21 divided by 14 is a 1.5 percent chance of it happening. That means that every three iterations, when slot six is 42 or higher, slot five will be 40 or higher. It is slightly less than that because in total there are 30 iterations and not 21. But since 21 is very high out of 30 and since 14 is considerable when to 21, then it should stand to reason that it is a consistent, predictable pattern that will therefore repeat in future iterations. Hence, it is a probable outcome and not random.
Between each slot, the average difference of numbers varies between 6 and 8. That is, if in slot three the number is 26, on average, the number in the next slot (slot four) is no more nor less than six to eight numbers higher. It can be more or less than six to eight numbers higher but on average it is not. This is important because it can be combined together with the averages of each slot to help determine the most probable outcome of each slot in each new iteration, I assume.
Between slots three and four, there is a peculiar incidence whereby out of 30 iterations, there are 7 iterations whose difference is only 1 or 2. That is, if in slot three the number is 29, the next number (in slot four) in such a case will be 30. This incidence happens between no other slots more than once and is therefore assumed truly random otherwise but is assumed not random here (between slots three and four). In slots four and five it also happens 5 times but five out of 30 is not enough to be useful when we know that the relationship between slot 4 and slot 5 is likely one where both will be in the forties already - although this assumption may be unmerited? Hence, in talking about the incidence in slots three and four; if 30 is divided by 7 that means that about every fifth iteration will have the same pattern. Again, this is of course just an average out of thirty, but possibly somewhat useful in predicting future outcomes. In short however, the relationship between slots three and slots four does not seem as random as one may superficially assume in first looking at the numbers and it is plainly obvious that in slots four and five there is a clear pattern whereby the numbers are in the forties.
Question:
Therefore, my question is, if in future iterations, in slot one, the number is six (because as you can see it has not yet occurred where all other numbers between 1 and 10 have occurred in the present 30 iterations) and if in slot six the number is above 42 so as to take advantage of the pattern that sometimes occurs between slots five and six, and if the other averages are taken into consideration, what then will be the most probable outcomes of slots two, three, four, and five in your calculations?
When there were only 29 iterations, my calculation was this:
5, 13, 21, 29, 37, and 41.
If you see the most recent iteration, you see that 9 was chosen in the first slot. This is logical. First, in slot one, 9 had never occurred, which I hadn't noticed because my analysis wasn't as complete. Also, 9 is below ten, so the pattern played out as predicted. As just said, I had not noticed that 9 wasn't used and that is why I am now still sticking with 5 for the first slot. If I take into consideration the average between the slots (6 to 8 numbers difference) and if I assume the following arithmetic of the rest of the slots given the predetermined numbers in slots four and five (in the forties), I think that 13 was a bad choice on my part. It is true that it is 8 numbers higher and therefore satisfies the horizontal average. But since 13 has already happened two times in slot 2, it should seem that 13 will not happen before another number happens that has not occurred as frequently and that still satisfies said averages.
A caveat that is perhaps noteworthy is that the first slot is perhaps not chosen first in an algorithm. Perhaps the algorithm starts with slots six and five and goes backwards. I think though that it does start with slot one because of the way human beings think in the West, generally speaking. (From left to right). If such human beings designed said (assumed) algorithm, then it is safe to assume that the first slot is determined before the other slots. Be that as it may, the first slot’s iteration will most likely be between 1 and 10. And the second slot could be between 1 and 10 or between 10 and 20. Since 13 has been iterated twice in slot two already, it should stand to reason that 14 be the better number. While 14 has already occurred once, the average of any other available number defies the mentioned pattern of averages. That is, if I were to pick a number that hasn't been chosen so as to get away from numbers that have, the only ones that have not are several numbers outside the average difference between slots and so this may be a kind of dilemma. Still, as we can see in the 30th iteration, the number was 15. So out of all the apparently random numbers that could have occurred, 15 is not very far away from 13. There are other assumptions that can be made, but there is not enough support from the numbers alone to think that they carry any predictive power.
Caution: long post ahead.
My question is a bit tricky but deals in probability. I'm trying to determine what a few different numbers should be on average. My assumption is that the iterations are not truly random and not chosen by a human. Rather, my assumption is that the iterations are determined by an algorithm and that the algorithm does not choose randomly but only gives the impression of being random.
Let me explain.
There are six slots from left to right (see the bold numbers below).
In theory, in each slot a number between 1 and 49 can occur. Each day a new drawing takes place in the very same way.
In each slot from left to right, the number is always higher than in the previous slot. In slot one, it is assumed therefore that 49 cannot be a number because the rest of the slots must be filled with higher numbers and the highest possible number is 49. This begs the question that the first slot is chosen first.
In slot one the number is in practice nearly always inferior to ten. Nearly always? - Out of 30 iterations thus far, 26 of the numbers are inferior to ten.
In slot one, it is assumed that the number can be as high as possible given the rest of the following slots but in practice none have gone beyond the number 20.
In slot six, out of 30 iterations, 21 are between 41 and 49.
In slot six, the number has not been below thirty but is very often in the forties. - Knowing the nature of these two terminal slots may help to determine the future outcome of the inner slots (slots 2, 3, 4, and 6. See below for further reasoning).
Horizontal Averages:
In slot two, 12 iterations are between the numbers 11 and 20. In slot two, 10 of the iterations are between 1 and 10. This means that over the course of thirty different drawings, slot two has most often been filled with a number between eleven and twenty. It might be the number fifteen, the number twenty, the number twelve, but the there have been more occurrences between those two numbers than any other in slot two.
In slot three, 13 iterations are between 21 and 30. In slot three, 12 iterations are between 11 and 20.
In slot four, 15 iterations are between 21 and 30. And 12 iterations in slot four are between 31 and 40.
Inductions:
If in slot six the number is 42 or greater, out of 21 iterations, 14 iterations in slot four are 40 or above. That is, if in slot six the number is 42 or higher, then in slot five, the number is 40 or higher fourteen times out of twenty-one, hence the chance that the number in slot four will be 40 or higher when in slot five the number is 42 or higher is significant. How significant? Well, 21 divided by 14 is a 1.5 percent chance of it happening. That means that every three iterations, when slot six is 42 or higher, slot five will be 40 or higher. It is slightly less than that because in total there are 30 iterations and not 21. But since 21 is very high out of 30 and since 14 is considerable when to 21, then it should stand to reason that it is a consistent, predictable pattern that will therefore repeat in future iterations. Hence, it is a probable outcome and not random.
Between each slot, the average difference of numbers varies between 6 and 8. That is, if in slot three the number is 26, on average, the number in the next slot (slot four) is no more nor less than six to eight numbers higher. It can be more or less than six to eight numbers higher but on average it is not. This is important because it can be combined together with the averages of each slot to help determine the most probable outcome of each slot in each new iteration, I assume.
Between slots three and four, there is a peculiar incidence whereby out of 30 iterations, there are 7 iterations whose difference is only 1 or 2. That is, if in slot three the number is 29, the next number (in slot four) in such a case will be 30. This incidence happens between no other slots more than once and is therefore assumed truly random otherwise but is assumed not random here (between slots three and four). In slots four and five it also happens 5 times but five out of 30 is not enough to be useful when we know that the relationship between slot 4 and slot 5 is likely one where both will be in the forties already - although this assumption may be unmerited? Hence, in talking about the incidence in slots three and four; if 30 is divided by 7 that means that about every fifth iteration will have the same pattern. Again, this is of course just an average out of thirty, but possibly somewhat useful in predicting future outcomes. In short however, the relationship between slots three and slots four does not seem as random as one may superficially assume in first looking at the numbers and it is plainly obvious that in slots four and five there is a clear pattern whereby the numbers are in the forties.
Question:
Therefore, my question is, if in future iterations, in slot one, the number is six (because as you can see it has not yet occurred where all other numbers between 1 and 10 have occurred in the present 30 iterations) and if in slot six the number is above 42 so as to take advantage of the pattern that sometimes occurs between slots five and six, and if the other averages are taken into consideration, what then will be the most probable outcomes of slots two, three, four, and five in your calculations?
When there were only 29 iterations, my calculation was this:
5, 13, 21, 29, 37, and 41.
If you see the most recent iteration, you see that 9 was chosen in the first slot. This is logical. First, in slot one, 9 had never occurred, which I hadn't noticed because my analysis wasn't as complete. Also, 9 is below ten, so the pattern played out as predicted. As just said, I had not noticed that 9 wasn't used and that is why I am now still sticking with 5 for the first slot. If I take into consideration the average between the slots (6 to 8 numbers difference) and if I assume the following arithmetic of the rest of the slots given the predetermined numbers in slots four and five (in the forties), I think that 13 was a bad choice on my part. It is true that it is 8 numbers higher and therefore satisfies the horizontal average. But since 13 has already happened two times in slot 2, it should seem that 13 will not happen before another number happens that has not occurred as frequently and that still satisfies said averages.
A caveat that is perhaps noteworthy is that the first slot is perhaps not chosen first in an algorithm. Perhaps the algorithm starts with slots six and five and goes backwards. I think though that it does start with slot one because of the way human beings think in the West, generally speaking. (From left to right). If such human beings designed said (assumed) algorithm, then it is safe to assume that the first slot is determined before the other slots. Be that as it may, the first slot’s iteration will most likely be between 1 and 10. And the second slot could be between 1 and 10 or between 10 and 20. Since 13 has been iterated twice in slot two already, it should stand to reason that 14 be the better number. While 14 has already occurred once, the average of any other available number defies the mentioned pattern of averages. That is, if I were to pick a number that hasn't been chosen so as to get away from numbers that have, the only ones that have not are several numbers outside the average difference between slots and so this may be a kind of dilemma. Still, as we can see in the 30th iteration, the number was 15. So out of all the apparently random numbers that could have occurred, 15 is not very far away from 13. There are other assumptions that can be made, but there is not enough support from the numbers alone to think that they carry any predictive power.
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