Is there any method or system that can an predict the first digit that will fall in Pick 3Prev TopicNext Topic

• Tucker Black

  

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  Feb 20, 2026, 8:49 pm

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  In response to hypersoniq

  Quote: Originally posted by hypersoniq on Feb 20, 2026

  That is where the concept of favorable outcomes to possible outcomes is expressed as a ratio, but is NOT divisible... buy one pick 3 ticket, you have 1:1000, or ONE favorable outcome to 1000 possible outcomes...

  Buy 2 tickets and you have TWO favorable outcomes to 1000 possible outcomes.... the 1000 possible outcomes NEVER goes away. Therefore with ticket purchases 2:1,000 != 1:500, because 500 possible outcomes do NOT vanish because you played two tickets... in pick 3, the 1,000 is not reducible by division, it is a constant.

  2/1000 is equal to 1/500. Or you could write it as 0.002 in digital form, or 0.2% expressed in percentages. They are all the same number.

  Another way to state this is that you have a 99.8% chance of losing. Or 499 in 500 (meaning, on average, out of 500 tries, you'll lose 499 times and win once).
• Tucker Black

  

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  In response to Coin Toss

  Quote: Originally posted by Coin Toss on Feb 20, 2026

  Pure baloney. If playing a second ticket cuts the odds in half why stop with the second ticket, why not keep buying more tickets until you are pretty sure of a win. MM and PB odds can only be reduced by half if you play half of all the combinations.

  That's one way to play, keep buying tickets until you have a sufficiently large chance of winning.

  This is what professionals do when a jackpot is really big, to the point that the expected value of a ticket is greater than its face value.

  But under normal circumstances, it wouldn't make any sense to buy up all the combinations when there are millions of them. You'll spend more on tickets than the jackpot.

  Take Pick 3 for example. It's not difficult for one person to buy 1,000 tickets for the next drawing. But that's a silly thing to do, because you would spend $1,000 and only win $500 of that back (or play the 50 cent version, and spend $500 to win $250 back; same ratio).

  "MM and PB odds can only be reduced by half if you play half of all the combinations"

  If you play half of all the combinations, your odds are not reduced by half, your odds ARE HALF, meaning you have a 50% chance of winning (a far larger chance than 1 in 146 million). It's a coin toss! LOL
• Tucker Black

  

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  Here are the odds of winning the Powerball jackpot for number of unique tickets bought from 1 to 100 (no QP, to have a zero chance of a duplicate)

  num tickets	odds 1 in
  1	292,201,338.00
  2	146,100,669.00
  3	97,400,446.00
  4	73,050,334.50
  5	58,440,267.60
  6	48,700,223.00
  7	41,743,048.29
  8	36,525,167.25
  9	32,466,815.33
  10	29,220,133.80
  11	26,563,758.00
  12	24,350,111.50
  13	22,477,026.00
  14	20,871,524.14
  15	19,480,089.20
  16	18,262,583.63
  17	17,188,314.00
  18	16,233,407.67
  19	15,379,017.79
  20	14,610,066.90
  21	13,914,349.43
  22	13,281,879.00
  23	12,704,406.00
  24	12,175,055.75
  25	11,688,053.52
  26	11,238,513.00
  27	10,822,271.78
  28	10,435,762.07
  29	10,075,908.21
  30	9,740,044.60
  31	9,425,849.61
  32	9,131,291.81
  33	8,854,586.00
  34	8,594,157.00
  35	8,348,609.66
  36	8,116,703.83
  37	7,897,333.46
  38	7,689,508.89
  39	7,492,342.00
  40	7,305,033.45
  41	7,126,861.90
  42	6,957,174.71
  43	6,795,379.95
  44	6,640,939.50
  45	6,493,363.07
  46	6,352,203.00
  47	6,217,049.74
  48	6,087,527.88
  49	5,963,292.61
  50	5,844,026.76
  51	5,729,438.00
  52	5,619,256.50
  53	5,513,232.79
  54	5,411,135.89
  55	5,312,751.60
  56	5,217,881.04
  57	5,126,339.26
  58	5,037,954.10
  59	4,952,565.05
  60	4,870,022.30
  61	4,790,185.87
  62	4,712,924.81
  63	4,638,116.48
  64	4,565,645.91
  65	4,495,405.20
  66	4,427,293.00
  67	4,361,214.00
  68	4,297,078.50
  69	4,234,802.00
  70	4,174,304.83
  71	4,115,511.80
  72	4,058,351.92
  73	4,002,758.05
  74	3,948,666.73
  75	3,896,017.84
  76	3,844,754.45
  77	3,794,822.57
  78	3,746,171.00
  79	3,698,751.11
  80	3,652,516.73
  81	3,607,423.93
  82	3,563,430.95
  83	3,520,498.05
  84	3,478,587.36
  85	3,437,662.80
  86	3,397,689.98
  87	3,358,636.07
  88	3,320,469.75
  89	3,283,161.10
  90	3,246,681.53
  91	3,211,003.71
  92	3,176,101.50
  93	3,141,949.87
  94	3,108,524.87
  95	3,075,803.56
  96	3,043,763.94
  97	3,012,384.93
  98	2,981,646.31
  99	2,951,528.67
  100	2,922,013.38

  As you can see, the odds of winning do not drop by half when you go from n tickets to n+1 tickets, except at the very beginning. Going from 1 ticket to 2 tickets cuts the odds in half because you are buying twice as many tickets.

  If you go from 50 tickets to 100 tickets, you again cut the odds in half, from 1 in 5,844,027 to 1 in 2,922,013. But if you go from 50 tickets to 51 tickets, your odds of winning improve only slightly, from about 1 in 5.8 million to about 1 in 5.7 million. The odds improve by 2% because you're buying 2% more combinations.
• Coin Toss

  

  Zeta Reticuli Star System
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  You are really entertaining those who manage powerball, your reasoning is really off. So you are saying that the first ticket is up against 292,201,338 odds and the second ticket eliminates 146,100,669 combinations. IF ONLY!   Start with 292,201,338 and go down the list by subtracting 1 each time until you have subtracted 100 from 292,201,338 which is 292,201, 238.  That's how it works, not voo doo math.
• hypersoniq

  

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  Feb 20, 2026, 11:36 pm

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  Apply the following correction...

  Number of tickets | Possible Combos (constant)

                                 1 | 292,201,338

                                     ...

                             100 | 292,201,338

  Nothing you do can reduce that total number of possible combos... it is not a pure math problem, so the Possible combos is simply not divisible by the number of tickets you buy, unless you buy them ALL, then you are 

  292,201,338 | 292,201,338 <== STILL not reduced.

  1 ticket gets you in the game, any more than that is just lining the pockets of the MUSL.

  Have an EXCELlent day!
• Tucker Black

  

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  In response to hypersoniq

  Quote: Originally posted by hypersoniq on Feb 20, 2026

  Apply the following correction...

  Number of tickets | Possible Combos (constant)

                                 1 | 292,201,338

                                     ...

                             100 | 292,201,338

  Nothing you do can reduce that total number of possible combos... it is not a pure math problem, so the Possible combos is simply not divisible by the number of tickets you buy, unless you buy them ALL, then you are 

  292,201,338 | 292,201,338 <== STILL not reduced.

  1 ticket gets you in the game, any more than that is just lining the pockets of the MUSL.

  OK, forget about mathematics for a minute. Just use some intuition -- if you buy ALL 292,201,338 combinations, you're saying that you are not guaranteed to win? How would the lottery choose a combination you did NOT buy?
• Coin Toss

  

  Zeta Reticuli Star System
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  How long do you think it would take to fill out the playslips to play all the combinations! By the time you had them all filled out the game may have changed!
• Tucker Black

  

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  In response to Coin Toss

  Quote: Originally posted by Coin Toss on Feb 20, 2026

  You are really entertaining those who manage powerball, your reasoning is really off. So you are saying that the first ticket is up against 292,201,338 odds and the second ticket eliminates 146,100,669 combinations. IF ONLY!   Start with 292,201,338 and go down the list by subtracting 1 each time until you have subtracted 100 from 292,201,338 which is 292,201, 238.  That's how it works, not voo doo math.

  I'm not eliminating millions of combinations. This is introductory probability theory: probability of winning = number of ways to win divided by total number of ways. It's 2 divided by 292 million. This is not voodoo math. This is real math.

  Your reasoning is nonsensical, because you're implying that if you were to buy 292,201,337 different combinations (1 less than all of them), you still have only a 50% chance of winning? That's clearly ridiculous, as at that point, your chances of losing are 1 in 292 million. Winning is virtually guaranteed by buying all but just one combination.

  Let's try this one: suppose you have a six sided die. You can bet $1 and pick a number between 1 and 6. Now roll the die. If it lands on your number, you win $5 (making the expected house win 1/6 = 16%).

  Suppose you put down $2 and bet on two numbers. It should be obvious that your chances of winning are 1 in 3, because you've bet on a third of the numbers. While you didn't reduce the number of possible outcomes from 6 to 3, by betting on 2 of them, your chances of winning are 2 in 6, which equals 1/3.

  Now put down $5 and bet on 5 of the possible numbers. Say that you bet on 1,2,3,4 and 5. You win $5 if the die lands on anything other than 6. What is the probability of a six sided die landing on 1,2,3,4 or 5? It's 5/6, or 83.333%. It is NOT 1 in 2.
• Tucker Black

  

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  In response to Coin Toss

  Quote: Originally posted by Coin Toss on Feb 21, 2026

  How long do you think it would take to fill out the playslips to play all the combinations! By the time you had them all filled out the game may have changed!

  That's a deflection. The difficult logistics of playing all the combinations in Powerball do not change the math behind it.

  If MUSL allowed it, a computer could generate as many combinations as you want and simply enter those tickets electronically, and skip the paper printing. Less than a second.
• Stack47

  

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  In response to Tucker Black

  Quote: Originally posted by Tucker Black on Feb 20, 2026

  Let's try this with the simple game of Pick 3.

  In Pick 3, there are 1,000 combinations, from 000 to 999, all of which have the same probability of being the winning combination.

  Let's say you purchase one ticket. You pick 567 and bet $1. If the winning combination is 567, you win $500. Your chances of winning is 1 in 1000, because you have bet on one combination out of 1,000 equally likely combinations.

  Now let's say you decide to purchase a second ticket for the same drawing. You pick 234 and bet another $1. If the winning combination is 567 OR 234, you win $500. You cannot win $500 twice, because your tickets have different numbers on them.

  The probability that you, the holder of two tickets, win $500 is 2 divided by 1,000. 2/1000 = 1/500. You have doubled your chances of winning.

  The odds that the second ticket is a winner is not 1 in 999. It's still 1 in 1000. But you have TWO tickets.

  ..............

  If the game is structured differently, it can be the case that a second ticket has odds of winning 1 in 999. In this version, you buy one ticket. You are only allowed to buy one ticket. Then, the lottery picks one combination of three digits out of a hat, without replacement. If your ticket matches the winning numbers, you win $500. But if your ticket didn't win, then you can buy a second ticket for the second and final drawing. That second ticket has a 1 in 999 chance of being a winner because, indeed, there are 999 combinations remaining in the hat.

  In this two-step version of Pick 3, the probability that you win $500 is 1/1000 + 999/1000*1/999. The initial 1/1000 is the first ticket & drawing; the 999/1000*1/999 is the event [first ticket loses] AND [second ticket wins]. The two can be added together because they are mutually exclusive (i.e. you cannot win both drawings due to the rules of this game).

  Doing some easy math here, 1/1000 + 999/1000*1/999 = 1/1000 + 1/1000 = 2/1000 = 1/500. So even in this bizarre version of Pick 3, you still have exactly a 1 in 500 chance of winning. However, it would be slightly better for the player, the difference being that in this version, you might not have to buy two tickets. You could win on the first ticket, and if you need to buy a second ticket, your chances of winning are slightly better.

  In this version of Pick 3, the average number of tickets you need to buy is 1.999, and the game is worth 0.05% more due to this structure.

  But of course, that is not how two-ticket Pick 3 works! If you want to buy two tickets for the same drawing, you are forced to purchase both in advance of the drawing. Both tickets have exactly the same chances of winning, 1 in 1000.

  Had this type of "the odds are" debates many times and usually both sides are saying the same thing but expressing it differently. Some divide the number of outcomes by the number of chances and others divide the number chances by the number of outcomes and get a percentage. In the first example 1000 divided by 2 equals 500 to 1. In the second, 2 divided by 1000 equal .002 which is the inverse of 500 to 1. 

  Most standard calculators have a inverse "1/x" button. If we put in 500 and hit the "1/x" button we get .002 and then by hitting the "1/x" again, it shows 500. Looks weird but odds of 500 to 1 have a 0.2% chance of winning so two different pick 3 straight combos have a 0.2% chance of winning at odds of 500 to 1.

  Apparently fashohits is looking for a way to beat 10 to 1 odds and probably make a $5 or $10 front pair bet or something like that and another way to express odds. The odds against picking the first digit are 10 to 1, but since there is no "one digit" bet on lottery terminals, 10 front or outside pair bets must be made. The odds of winning now become 100 to 1 because at least 10 bets must be made unless there is a follow up "is there a method or system that can predict the first two digits?" thread. 

  Let's not forget 10 $1 front pair bets is wagering $10 to win $40 and not much room for error.

  "Some people have something to say while other people feel the need to say something"
• Stack47

  

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  In response to Tucker Black

  Quote: Originally posted by Tucker Black on Feb 21, 2026

  I'm not eliminating millions of combinations. This is introductory probability theory: probability of winning = number of ways to win divided by total number of ways. It's 2 divided by 292 million. This is not voodoo math. This is real math.

  Your reasoning is nonsensical, because you're implying that if you were to buy 292,201,337 different combinations (1 less than all of them), you still have only a 50% chance of winning? That's clearly ridiculous, as at that point, your chances of losing are 1 in 292 million. Winning is virtually guaranteed by buying all but just one combination.

  Let's try this one: suppose you have a six sided die. You can bet $1 and pick a number between 1 and 6. Now roll the die. If it lands on your number, you win $5 (making the expected house win 1/6 = 16%).

  Suppose you put down $2 and bet on two numbers. It should be obvious that your chances of winning are 1 in 3, because you've bet on a third of the numbers. While you didn't reduce the number of possible outcomes from 6 to 3, by betting on 2 of them, your chances of winning are 2 in 6, which equals 1/3.

  Now put down $5 and bet on 5 of the possible numbers. Say that you bet on 1,2,3,4 and 5. You win $5 if the die lands on anything other than 6. What is the probability of a six sided die landing on 1,2,3,4 or 5? It's 5/6, or 83.333%. It is NOT 1 in 2.

  "I'm not eliminating millions of combinations."

  Correct; try expressing it as, "two different combos each have one chance out of every 146,100,669 chances".

  "Some people have something to say while other people feel the need to say something"
• Tucker Black

  

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  In response to Stack47

  Quote: Originally posted by Stack47 on Feb 21, 2026

  "I'm not eliminating millions of combinations."

  Correct; try expressing it as, "two different combos each have one chance out of every 146,100,669 chances".

  That is incorrect. Two different combos each have a one in 292,201,338 chance.
• dr san

  

  bgonçalves
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  hello pick3=

  🎯 Pick3 Mathematical Model

  Total possible combinations:

  U

  = 1000

  U=1000

  Each draw chooses 1 number from 1000.

  We want to know:

  After N draws, how many different numbers have appeared?

  📊 Coverage Formula

  The probability of a number not being drawn in a lottery is:

  999

  1000

  1000
  999

  ​

  After N draws:

  (
  999
  1000
  )
  𝑁
  (
  1000
  999

  ​

  )
  N

  Therefore, the expected number of numbers that have never been drawn is:

  1000

  ×
  (
  999
  1000
  )
  𝑁
  1000×(

  1000
  999

  ​

  )
  N

  Thus, the numbers that have been drawn at least once will be:

  1000
  −

  1000
  ×
  (
  999
  1000
  )
  𝑁
  1000−1000×(
  1000
  999

  ​

  )
  N

  This is the The amount that will be excluded by your filter.

  🎯 Now let's solve for 75%–80% safety.

  You want only 20% to 25% of the universe to be excluded.

  That is:

  𝐸
  𝑥
  𝑐
  𝑙
  𝑢
  𝑠
  𝑎
  ~
  𝑜
  ≈
  200
  𝑎
  250
  𝑛
  𝑢
  ˊ
  𝑚
  𝑒
  𝑟
  𝑜
  𝑠
  Exclus
  a
  ~
  o≈200a250n
  u
  ˊ
  meros

  Let's calculate.

  🔵 Case 1 — 200 numbers excluded (20%)

  1000
  −
  1000
  (
  999
  1000
  )
  𝑁
  =
  200
  1000−1000(
  1000
  999
  ​

  )
  N
  =200

  Solving:

  (
  999
  1000
  )
  𝑁
  =
  0.8
  (
  1000
  999
  ​

  )
  N
  =0.8

  Applying log:

  𝑁
  ≈
  223
  𝑠
  𝑜
  𝑟
  𝑡
  𝑒
  𝑖
  𝑜
  𝑠
  N≈223 draws
  🔵 Case 2 — 250 numbers excluded (25%)
  (
  999
  1000
  )
  𝑁
  =
  0.75
  (
  1000
  999
  ​

  )
  N
  =0.75
  𝑁
  ≈
  288
  𝑠
  𝑜
  𝑟
  𝑡
  𝑒
  𝑖
  𝑜
  𝑠
  N≈288 draws
  ✅ Final Result

  To maintain a 75% to 80% guarantee, the ideal number is:

  🎯 Between 220 and 290 previous draws

  📊 Real Comparison
  Draws % excluded
  200 ~18%
  250 ~22%
  300 ~26%
  400 ~33%
  600 ~45%
  700 ~50%
  🔥 Conclusion Technique

  If you want statistical precision:

  👉 Best ideal point: 250 draws
  👉 Acceptable range: 220 to 290
  👉 Above 350, your coverage starts to decrease significantly.
• dr san

  

  bgonçalves
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  hello=So, in pick3, to use the filters from previous draws with an 80% guarantee, it's from 250 draws, and when you generate the pick3 bet, it cannot be the same as the previous 250.
• Coin Toss

  

  Zeta Reticuli Star System
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  ▲▼
  In response to Tucker Black

  Quote: Originally posted by Tucker Black on Feb 21, 2026

  That's a deflection. The difficult logistics of playing all the combinations in Powerball do not change the math behind it.

  If MUSL allowed it, a computer could generate as many combinations as you want and simply enter those tickets electronically, and skip the paper printing. Less than a second.

  You said it yourself, IF MUSL ALLOWED IT. Real big if there.