How often P3 digit repeats on top next draw?Prev Topic Next Topic

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Edibs3

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Feb 10, 2016, 1:55 pm

Little help showing the correct P3 Math average for one digit to repeat on top of itself from the last drawn digit .

Include doubles and then just single to single only as well. Thanks

1 n 10 odds for each draw that the same digit will repeat on top of itself correct ? 1/10 x1/10 seems too much.
Edibs3

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Feb 11, 2016, 2:30 pm

Quote: Originally posted by Edibs3 on Feb 10, 2016
Little help showing the correct P3 Math average for one digit to repeat on top of itself from the last drawn digit .

Include doubles and then just single to single only as well. Thanks

1 n 10 odds for each draw that the same digit will repeat on top of itself correct ? 1/10 x1/10 seems too much.

Come on and give a fella' a break Math Masters! Will work for Math Tips.

What are the correct P3 odds for a like digit to hit directly on top of yesterday's draw? How about 3 in a row? Thanks for any help around the return of like digits and the odds.

986

912

993
Wisconsin3054

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Feb 11, 2016, 2:59 pm

It really depends on your point of reference. If you are saying predicting before the first draw that a particular number will appear in the same position on the next two draws, then the odds are 1 in (10X10) 100. 3 draws 1 in (10X10X10) 1000. If you are predicting after the first draw that in the next draw a particular number will appear in the same position as the last draw it would be 1 in 10. The chances are even better that one of the three numbers will repeat: 1-(.9X.9X.9)=27.1% chance, so a little over 1 in 4.
Edibs3

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Feb 11, 2016, 7:35 pm

Quote: Originally posted by Wisconsin3054 on Feb 11, 2016
It really depends on your point of reference. If you are saying predicting before the first draw that a particular number will appear in the same position on the next two draws, then the odds are 1 in (10X10) 100. 3 draws 1 in (10X10X10) 1000. If you are predicting after the first draw that in the next draw a particular number will appear in the same position as the last draw it would be 1 in 10. The chances are even better that one of the three numbers will repeat: 1-(.9X.9X.9)=27.1% chance, so a little over 1 in 4.

Thanks Wisconsin3054! A lot times the stats just don't seem to sync up with current reality.

So much so, that you start to question long-held beliefs. We need to revisit these every once in awhile just to knock down the cobwebs.

These repeating digits have always seemed sort of a flimsy trend but they may be underrated after all.

The longer I'm around the more POSITIONAL repeat digits seem to hold interest. Playing more straights lately has done it.
Wisconsin3054

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Feb 11, 2016, 11:18 pm

Yes, I think when you see the repeating digits they stand out, but the reality of the numbers as I showed is that in about one in four draws there will be a repeating digit in one of the lines. Turns out its just the math and I don't think it makes for a pattern that would help with predictions.
KY Floyd

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Feb 12, 2016, 2:56 am

Quote: Originally posted by Wisconsin3054 on Feb 11, 2016
It really depends on your point of reference. If you are saying predicting before the first draw that a particular number will appear in the same position on the next two draws, then the odds are 1 in (10X10) 100. 3 draws 1 in (10X10X10) 1000. If you are predicting after the first draw that in the next draw a particular number will appear in the same position as the last draw it would be 1 in 10. The chances are even better that one of the three numbers will repeat: 1-(.9X.9X.9)=27.1% chance, so a little over 1 in 4.

"The chances are even better that one of the three numbers will repeat: 1-(.9X.9X.9)=27.1% chance, so a little over 1 in 4."

Just to expand on that a bit, and add either a correction or a clarification: Figuring out the probability of something happening often requires figuring out the probability of the alternative not happening. The chance of not having the first digit match is 9 in 10, so the probability is .9. The probability of not matching the first or 2nd is .9 * .9 = .81. For also not matching the 3rd it's .9 * .9 * .9 = .729. That's the probability for missing all three. That leaves 1 - .729 = .271 as the probability of repeating one or more of the digits.

For repeating exactly one digit you need to figure the chances of matching that digit but not matching either of the others. The probability of matching the 1st digit is .1, and the probability of missing both the 2nd and 3rd is .81, so the probability of matching only the first digit is .081. The same is true for matching only the 2nd, and for matching only the 3rd. Those probabilities are additive, so .081 * 3 = .243 probability of matching just one of the three numbers.

For matching 2 it would be .1 * .1 * .9 = .009 for each of 3 possible two digit matches (AB, AC, BC). Once again those probabilities are additive, so .009 * 3 = .027 for any of the two digit matches. The chance of matching all 3 is (of course) 1 in 1000, so probability of .001.

Adding them all together we get:

.001 chance of all three
.027 chance of any two
.243 chance of exactly one

.271 chance of 1 or more
Wisconsin3054

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Feb 12, 2016, 11:46 am

Quote: Originally posted by KY Floyd on Feb 12, 2016
"The chances are even better that one of the three numbers will repeat: 1-(.9X.9X.9)=27.1% chance, so a little over 1 in 4."

Just to expand on that a bit, and add either a correction or a clarification: Figuring out the probability of something happening often requires figuring out the probability of the alternative not happening. The chance of not having the first digit match is 9 in 10, so the probability is .9. The probability of not matching the first or 2nd is .9 * .9 = .81. For also not matching the 3rd it's .9 * .9 * .9 = .729. That's the probability for missing all three. That leaves 1 - .729 = .271 as the probability of repeating one or more of the digits.

For repeating exactly one digit you need to figure the chances of matching that digit but not matching either of the others. The probability of matching the 1st digit is .1, and the probability of missing both the 2nd and 3rd is .81, so the probability of matching only the first digit is .081. The same is true for matching only the 2nd, and for matching only the 3rd. Those probabilities are additive, so .081 * 3 = .243 probability of matching just one of the three numbers.

For matching 2 it would be .1 * .1 * .9 = .009 for each of 3 possible two digit matches (AB, AC, BC). Once again those probabilities are additive, so .009 * 3 = .027 for any of the two digit matches. The chance of matching all 3 is (of course) 1 in 1000, so probability of .001.

Adding them all together we get:

.001 chance of all three
.027 chance of any two
.243 chance of exactly one

.271 chance of 1 or more

Good catch. I was thinking of there being at least one repeat as opposed to only one. Better to be precise though when we are talking math!
Edibs3

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Feb 12, 2016, 1:39 pm

Seemed as if we were getting an average of around THREE 1st position repeats per month for evening draws.

It also looked like the same for the other 2 positions in general.

What would that look like using the correct math averages for each position per month?

Secondly, what would 3 like digits in a row look like odds wise? Ever seen 4 in a row?

Sure glad we have good help on these odds/probability questions here and willing to share it with the earthlings. Thanks
Wisconsin3054

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Feb 12, 2016, 2:05 pm

Quote: Originally posted by Edibs3 on Feb 12, 2016
Seemed as if we were getting an average of around THREE 1st position repeats per month for evening draws.

It also looked like the same for the other 2 positions in general.

What would that look like using the correct math averages for each position per month?

Secondly, what would 3 like digits in a row look like odds wise? Ever seen 4 in a row?

Sure glad we have good help on these odds/probability questions here and willing to share it with the earthlings. Thanks

Repeats in a particular position are 1 in 10 odds so 3 per month per position should be considered normal. Odds of a 3-peat in a particular position would be 1 in 100. So on average 3-4 times a year would be expected. The 4-peat goes to 1 in 1000 odds so expected once every 3 years or so I would say.

Remember these are odds for each line though so the odds that it occurs in one of the three lines is better.
Lucky Loser

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Feb 13, 2016, 2:04 am

Hey guys, here's a link to a discussion on this very same topic about nine months ago. I tried my best to establish the validity of using a system based on this approach but, the OP of the topic felt otherwise. Since then, this same method has continued to prove out...and always will. The topic is called 'Indicators' and I hope it helps.

https://www.lotterypost.com/thread/289701

Small games, frequent wins, and regular payouts 'cause.....

There are seven days in the week...'Someday' isn't one of them.

#lotto-4-a-living
Wisconsin3054

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Feb 13, 2016, 10:32 am

Quote: Originally posted by Lucky Loser on Feb 13, 2016
Hey guys, here's a link to a discussion on this very same topic about nine months ago. I tried my best to establish the validity of using a system based on this approach but, the OP of the topic felt otherwise. Since then, this same method has continued to prove out...and always will. The topic is called 'Indicators' and I hope it helps.

https://www.lotterypost.com/thread/289701

Do you have long term results to show that it proves out? What we have shown above is that when you are only dealing with 10 digits, the chance that one of those digits, especially if you are looking at all positions, will repeat, that is just math not a pattern. There is no way to predict which number or numbers will repeat nor what position they will repeat in. I may not be great at this, but I believe the math looks like this:

1-(.7 X .7 X .7) = 65.7% chance of at least one repeat looking at any position.

So over time, more often than not you will see one repeat digit each draw. Again, its not a pattern its just the math. Also, this fact does not help you predict when the 65.7% chance of repeat will occur vs. the 34.3% chance that it does not occur; it does not help you predict if one or more than one number will repeat; it does not help you predict which digit will repeat; and it does not help you predict on what position it will repeat.

I also notice someone on that thread was surprised/excited to see that in pick 4 the method works even better. Not really a surprise when you are adding another potential repeating digit on each side of the equation:

1-(.6 X .6 X .6 X .6) = 87.04% chance of at least one repeat looking at any position. Again, doesn't help you predict any of the things noted above.
Wisconsin3054

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Feb 13, 2016, 10:38 am

Quote: Originally posted by Wisconsin3054 on Feb 13, 2016
Do you have long term results to show that it proves out? What we have shown above is that when you are only dealing with 10 digits, the chance that one of those digits, especially if you are looking at all positions, will repeat, that is just math not a pattern. There is no way to predict which number or numbers will repeat nor what position they will repeat in. I may not be great at this, but I believe the math looks like this:

1-(.7 X .7 X .7) = 65.7% chance of at least one repeat looking at any position.

So over time, more often than not you will see one repeat digit each draw. Again, its not a pattern its just the math. Also, this fact does not help you predict when the 65.7% chance of repeat will occur vs. the 34.3% chance that it does not occur; it does not help you predict if one or more than one number will repeat; it does not help you predict which digit will repeat; and it does not help you predict on what position it will repeat.

I also notice someone on that thread was surprised/excited to see that in pick 4 the method works even better. Not really a surprise when you are adding another potential repeating digit on each side of the equation:

1-(.6 X .6 X .6 X .6) = 87.04% chance of at least one repeat looking at any position. Again, doesn't help you predict any of the things noted above.

I thought of one thing that complicates this a little bit, but I don't have the time at the moment to try to do the math on it.

The above math is based on the preceding draw having unique digits, ie. no doubles, triples, quads.

When you factor those in the overall daily percentage probably goes down a little bit.
Edibs3

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Feb 13, 2016, 12:02 pm

Thinking out loud. In general terms at first ok?

As we are waiting for the next double just about to hit we can prepare to do the following:

$$ We can assume at least ONE of the 3 digits {0,1,2,} will repeat in the next draw today, and it will be a DOUBLE.

The odds of our being right are these 48 of the 90 total boxes below. Playing all/or part.

055, 118, 226, 244, 299, 001, 119, 155, 227, 002, 011, 066, 228, 255, 003, 166, 229, 004, 022, 077, 112, 266, 005, 113, 122, 177, 006, 033, 088, 114, 277, 007, 115, 133, 188, 223, 008, 044, 099, 116, 224, 233, 288, 009, 117, 144, 199, 225

or-- these 144 straights below

001, 002, 003, 004, 005, 006, 007, 008, 009, 010, 011, 020, 022, 030, 033, 040, 044, 050, 055, 060, 066, 070, 077, 080, 088, 090, 099, 100, 101, 110, 112, 113, 114, 115, 116, 117, 118, 119, 121, 122, 131, 133, 141, 144, 151, 155, 161, 166, 171, 177, 181, 188, 191, 199, 200, 202, 211, 212, 220, 221, 223, 224, 225, 226, 227, 228, 229, 232, 233, 242, 244, 252, 255, 262, 266, 272, 277, 282, 288, 292, 299, 300, 303, 311, 313, 322, 323, 330, 331, 332, 400, 404, 411, 414, 422, 424, 440, 441, 442, 500, 505, 511, 515, 522, 525, 550, 551, 552, 600, 606, 611, 616, 622, 626, 660, 661, 662, 700, 707, 711, 717, 722, 727, 770, 771, 772, 800, 808, 811, 818, 822, 828, 880, 881, 882, 900, 909, 911, 919, 922, 929, 990, 991, 992,

=====================================
Edibs3

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Feb 13, 2016, 1:56 pm

Then again, we could bet the other half of the coin. Instead of 48 we could go with the opposite42 boxes below. This other half is a little bit cheaper.

No return digits from the previous draw 0,1,2

#42 box

334, 488, 668, 677, 335, 344, 399, 588, 669, 336, 499, 688, 778, 337, 355, 445, 599, 779, 788, 338, 446, 455, 699, 339, 366, 447, 799, 889, 448, 466, 556, 899, 377, 449, 557, 566, 477, 558, 388, 559, 577, 667

or --these 126 straight versions of those box up there.

334, 335, 336, 337, 338, 339, 343, 344, 353, 355, 363, 366, 373, 377, 383, 388, 393, 399, 433, 434, 443, 445, 446, 447, 448, 449, 454, 455, 464, 466, 474, 477, 484, 488, 494, 499, 533, 535, 544, 545, 553, 554, 556, 557, 558, 559, 565, 566, 575, 577, 585, 588, 595, 599, 633, 636, 644, 646, 655, 656, 663, 664, 665, 667, 668, 669, 676, 677, 686, 688, 696, 699, 733, 737, 744, 747, 755, 757, 766, 767, 773, 774, 775, 776, 778, 779, 787, 788, 797, 799, 833, 838, 844, 848, 855, 858, 866, 868, 877, 878, 883, 884, 885, 886, 887, 889, 898, 899, 933, 939, 944, 949, 955, 959, 966, 969, 977, 979, 988, 989, 993, 994, 995, 996, 997, 998

We do have a better chance hitting STRAIGHT with doubles because of the 1 in 3 odds instead of the 1 in 6 singles.

Playing at futuristic State odds of .25 cents per ticket is only 10.50. Even playing all/part boxes would be a 75.00 payout.

Straights 31.50 at a 225.00 payout. Even playing all straight 126.
Wisconsin3054

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Feb 13, 2016, 5:58 pm

Your first problem is getting to the point where you could assume that one of the numbers will repeat and assume that it will be a double. If you somehow knew that would happen in advance then yes, you have improved your odds with a filter. However, my view of reality is that the past draws do not affect the future draws, so I am not sticking down 42, 48, 126, or 144 plays in a single draw based on the initial assumption.

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