Welcome Guest
You last visited January 17, 2017, 5:54 am
All times shown are
Eastern Time (GMT-5:00)

# Pick 22: Combinations and Permutations

Topic closed. 17 replies. Last post 10 years ago by JADELottery.

 Page 1 of 2
Findlay, Ohio
United States
Member #4855
May 28, 2004
400 Posts
Offline
 Posted: November 18, 2006, 3:51 am - IP Logged

Original Quote by JADELottery: "However, when trying to do the work out for Pick 11 there was a computational problem. It had to do with taking 10 numbers 11 at a time. If you think about this, you can see this is impossible. After a few hours, a few beers, and a few nights rest, I realized the solution. The problem has to do with repetition."

I also noticed this problem when I first began listing the patterns for the Pick 22.  The solution I came up with was to first not allow duplicates of any pattern to exist within the list of all possible patterns: AABBC is the same as AABCC and ABBCC because all three patterns each contain two sets of repeating digits and one digit that does not repeat at all.  So when listing all of the possible patterns, I only included AABBC on the list.  As Pick 5 patterns, three of the possible combinations that fit the AABBC pattern are: 55776, 66577 or 75566.  Each of these combinations is included in the total amount of combos for the pattern, but only one arrangement of each is actually counted.  Those three combos may seem similar, but they are actually different because of the digits that are repeating in each.  The second part of the solution was to adjust the formulas to take into consideration equal amounts of repeating digits found within the pattern.  This method produced 100% accurate results, but it was also a very tedious task.  The formula for each pattern has to be modified for the amount of sets of repeated digits it contains.

So even after all ten digits are used up, finding the combinations, their arrangements and total permutations can easily be found-as long as one is willing to list out the patterns and employ the correct formulas.

~~~~~~~

THE COMBINATIONS AND PERMUTIONS OF PICK 22
Pick a digit, any digit...0 to 9, 22 times in a row!

AAAAAAAAAAAAAABCDEFGHI is only one of 807 total outcome patterns for a Pick 22 scenario.  For anyone wondering what an outcome pattern is, AAB represents a typical outcome in Pick 3 game.  AAB represents a double-digit combo with letters instead of actual digits.  The no-match outcome is ABC and the triple is AAA.

The Pick 22 pattern above represents a situation where 22 digits are randomly selected and each of the selections can be any one of the ten digits 0 through 9.  In this particular pattern, the same digit is drawn 14 times and eight other digits are each drawn once.  There are a total of nine different digits in this particular outcome.  I should mention that there isn't a real Pick 22 game that you can actually play as a lottery game, unless of course you are observing the next 22 Pick 3 or Pick 4 games by position!!!  With that in mind, how many unique combinations (without regard to order) can fit the Pick 22 pattern above if each letter represents a different digit?  One might be tempted to try and use the formula (10!)/((10-9)!)/(9!)=10 just because the pattern has 9 different digits from the ten that are possible, but this formula wont work for this situation because the letters in the pattern are not in equal amounts.

Calculating the combinations for outcome patterns with different amounts of repeating letters requires that the formula be done in sections.  The sections are based upon how many letters repeated and in what amounts they repeated in.

For this outcome pattern, we know that the "A" can represent any of the ten different digits.  So, we start with 10 as the first section of the formula.  Obviously, there are ten ways to choose one object from 10.

(10!)÷((10-1)!)÷(1!) = 10

Since the 8 remaining digits, which are all in equal amounts to each other (represented by B, C, D, E, F, G, H, and I) all represent different digits (none of which are the same as each other or the same as that of A), we simply need to know how many combinations there are when choosing 8 of 9 different digits.  We choose 8 from 9 because 10 digits are possible, but the first of the ten was chosen for A, which leaves only 9 digits remaining to choose the eight from. Remember: the outcome pattern contains nine different digits, but any letter can represented any of the ten digits possible.

(9!)÷((9-8)!)÷(8!) = 9

There are 9 combinations when choosing 8 different digits from 9 without regard to order.  Regardless of which of the 10 digits is actually selected for A, there will always be nine other possibilities left to choose the eight from.  So to get the total amount of combinations, we only need to multiply the ten possibilities for the duplicated digit of A by the number of ways to choose eight from the remaining nine:

(10!)÷((10-1)!)÷(1!) × (9!)÷((9-8)!)÷(8!) = 90

There are a total of 90 combinations that fit the AAAAAAAAAAAAAABCDEFGHI Pick 22 pattern.  Now we need to figure out how many unique ways that each of those 90 combinations can be arranged so we can find the total amount of permutations.  To do this, we start with the factorial of 22 (since the pattern contains 22 individual digits, spaces or trials) then divide it by the corresponding factorials for the amounts of each of the ten different digits found within the outcome.  The pattern in mention contains 14 A's, 1 B, 1 C, 1 D, 1 E, 1 F, 1 G, 1 H, 1 I and 0 J's.  I do know that continually dividing the preceding number by factorial 1 is redundant and unnecessary, but this method can help to create a "code" or method of sorts when you enter it into spreadsheet to do the calculations.  I like to call it "pattern partitioning."

Letter count within the pattern = partition:

A B C D E F G H I J
14 1 1 1 1 1 1 1 1 0

The formula using the partition equates to the following:

(22!)/(14!)/(1!)/(1!)/(1!)/(1!)/(1!)/(1!)/(1!)/(1!)/(0!)

When the formula is properly worked out, you should arrive at exactly 12,893,126,400 unique ways that each one of those 90 combinations can be arranged!!!  When the 90 combinations are multiplied by the 12,893,126,400 different ways that each can be arranged, you will get the total number of permutations that are possible:

90 × 12,893,126,400 = 1,160,381,376,000 Permutations

That's a pretty good-sized number!  One trillion, one hundred-sixty billion, three hundred eighty-one million, three hundred and seventy-six thousand total permutations!  It seems like an awful lot, but its nothing in comparison to the astronomical total of possibilities that encompass the entire Pick 22 scenario, which is ten sextillion, written as 10^22 or 10,000,000,000,000,000,000,000.

A trickier Pick 22 Pattern to calculate is AAAABBBCCCDDDEEFFGGHIJ.  In actual play, this pattern might look something like this: 4009585373960697385120.  In this pattern, all ten digits were selected.  One digit was selected four times, three digits were each selected three times, three other digits were each selected two times and finally, the last three digits were each selected one time.

Sectioning and partitioning outcome patterns such as this one will make the formulas much easier to put together so you can find the correct totals.  The formula used to find the combinations for this pattern can be broken down into four sections.  The sections and the logic behind them are shown below:

1.) We know that there are four A's in the combo.  Since A is the only letter that appears in the quantity of four, we have to treat it as its own section.  Obviously, there are ten possible digits that could represent A.  The formula for this section is a no-brainer; you can use the formula if you want or just write 10.

(10!)÷((10-1)!)÷(1!) = 10

2.) There are three letters that each appear in the quantity of three: BBBCCCDDD.  Because each of these three letters appears in equal amounts, they can be put into the same section.  The logic is as follows:  Since one of the ten digit was used for A, how many ways (without regard to order) can three different digits be selected from the remaining nine?

(9!)÷((9-3)!)÷(3!) = 84

3.) There are three letters that each appear in the quantity of two: EEFFGG.  Again, these three letters are in equal amounts to each other so they are put into their own section.  The logic behind this is as follows:  Four of the ten digits were used for the letters ABCD, how many ways (without regard to order) can three digits be drawn from the remaining six?

(6!)÷((6-3)!)÷(3!) = 20

4.) We have now used up seven of the ten possible digits for the letters ABCDEFG.  We have three different letters remaining (HIJ) and we also know that there are only 3 possible digits left out of the ten that are possible.  The letters HIJ appear in equal amounts of one of each, so they get put into the last section.  No matter which of the seven digits were used to represent the letters ABCDEFG, there are exactly three digits remaining to choose three from.

(3!)÷((3-3)!)÷(3!) = 1

All we have to do to get the total amount of combinations (without regard to order) is multiply the four sections together:

((10!)/((10-1)!)/(1!)) × ((9!)/((9-3)!)/(3!)) × ((6!)/((6-3)!)/(3!)) × ((3!)/((3-3)!)/(3!))

= 10 × 84 × 20 × 1 = 16,800 Combinations

The number of unique ways that each of those 16,800 combinations can be arranged is calculated by using the patterns partition:

A B C D E F G H I J
4 3 3 3 2 2 2 1 1 1

(22!)/(4!)/(3!)/(3!)/(3!)/(2!)/(2!)/(2!)/(1!)/(1!)/(1!)

This equation works out to be exactly 27,102,641,005,440,000.  No, that's not an error! There are over well over twenty-seven quadrillion ways that each of the 16,800 combinations can be arranged!  ...And multiplying those two figures together to give us the total amount of permutations produces an even larger number.

16,800 × 27,102,641,005,440,000 = 455,324,368,891,392,000,000

There are exactly 455,324,368,891,392,000,000 unique outcomes that fit the AAAABBBCCCDDDEEFFGGHIJ pattern.  That number is 455 quintillion and then some!

The combined amount of permutation for the two outcome patterns is nothing short of colossal.  Even still, they only represent a small fraction of total 10^22 that are possible.  Remember, these are only two of 807 total possible outcome patterns!

When I originally embarked on my mission to calculate the combinations, arrangements and permutations of a Pick 22 scenario, I realized that the biggest obstacle would be the finding of the Patterns themselves.  I have heard of a mathematical formula called the "partition function", but this was of no use beyond anything past Pick 10.  I had to actually generate and list each pattern that could possibly occur off the top of my head.  During this mind-numbing process, I also had to be careful not to create any duplicated patterns.  For example, AAAABBBCCCDDDEEFFGGHIJ is exactly the same pattern as that of AAABBBCCCDDDDEEFFGGHIJ.  I found that using a very specific and orderly approach made this job a little easier.  I always started with the highest amounts of repeating digits to the left of the pattern and kept the lowest amount of repeating digits to the right.  The very first outcome pattern on the list is AAAAAAAAAAAAAAAAAAAAAA and the last pattern is AAABBBCCDDEEFFGGHHIIJJ.  I started the project with the single goal of determining the TRUE probability of digits to become Hot or Cold within 22 consecutive games for the specific positions as observed in Pick 3 and Pick 4.

After about a month of work (in my spare time), I had the entire scenario worked out and decided to check my work.  I knew that there had to be a few errors somewhere because the sum of all permutations needed to equal exactly 10^22 or 10,000,000,000,000,000,000,000.  My total was a several million larger.  After a few days of double-checking the patterns and another few days of verifying the 807 un-automated formulas that calculate the combos for each pattern, I discovered that I had made three small errors.  I corrected the errors, but to my dismay, the total was still 979,200 too much!  I went through the whole process of double-checking everything yet another time but no matter what I did, I could not find any more errors whatsoever. The answer finally dawned on me: Excel cannot calculate with accuracy beyond 15 significant digits!  Excel was rounding the largest of the numbers up in many of the patterns.  The only remedy for this situation was to buy a special Excel Add-In program...one that allowed it to perform high precision calculations with up to 32,767 significant digits.  Once I installed the Add-In, the total was a perfect 10,000,000,000,000,000,000,000.  Of course I wanted to be sure that the numbers didn't somehow round out and equal 10^22 just by chance, so I used the Add-In to recalculate the combos and arrangements, as well as the permutations and their sum.  Everything worked out perfectly and I knew the totals were now 100% accurate.

Now that I had the whole scenario accurately worked out, I started over and adjusted the formulas for the combinations to pertain to a specific digit.  For example, the pattern AAAABBBCCCDDDEEFFGGHIJ represents any outcome in which there are ten different digits drawn, one of which is selected four times and three other different digits are each selected three times and three more digits are each selected two times and yet another three digits are each selected one time.  I know that there are 16,800 different combinations and that each one of them can be arranged 27,102,641,005,440,000 unique ways.  But suppose I want to know how many combos there are where the four A's are actually 7's.  To figure this out, I had to change the formula for the combos.  This was also a long drawn out task because depending on which letter in the pattern you want to represent as a particular digit, the formulas have to change to account for the amount of the specific letter or repetitions that each letter represents.  To make a long story short, different formulas are used depending on which letter in the pattern you want to find the combos for.  Since the combos are not what's really important as far as the probability is concerned, I won't go into the details of this process.  What is important is the total number of permutations...and there is an easier way to get these totals rather than adjusting the combo formulas just so you can multiply them by the number of arrangements to get permutations. This shortcut is the percent method.  If you take AAAABBBCCCDDDEEFFGGHIJ and break it down into sections based on the amounts of letters it contains, you can easily find the percent and/or total of all permutations that contain a specific digit in specific amounts. AAAABBBCCCDDDEEFFGGHIJ breaks down as follows:

AAAA = 10% because there is 1 group of 4

BBBCCCDDD = 30% because there are 3 groups of 3

EEFFGG = 30% because there are 3 groups of 2

HIJ = 30% because there are 3 "groups" of 1

Since we know that there are 455,324,368,891,392,000,000 total permutations that fit the pattern, you simply find those percents out of that total to get arrive at your number.  For example, in the outcome pattern AAAABBBCCCDDDEEFFGGHIJ, how many permutations of contain exactly four 7's?

455,324,368,891,392,000,000 × .10 (10%) = 45,532,436,889,139,200,000

Or how many contain three 7's?

455,324,368,891,392,000,000 × .30 (30%) = 136,597,310,667,417,600,000

Here's the outcome pattern AAAABBBCCCDDEEFFGGHHIJ, it breaks down as follows:

AAAA = 10% because there is 1 group of 4

BBBCCC 20% because there are 2 groups of 3

DDEEFFGGHH = 50% because there are 5 groups of 2

IJ = 20% because there are 2 "groups" of 1

Breaking the permutations down in this way also lead to another proof that the totals are 100% accurate.  When the combos were all adjusted to let each letter represent a particular digit being selected in the specified amount within each outcome pattern, the sum of all permutations equaled exactly:

9,015,229,097,816,388,767,119

This number is significant because it is the result of 10^22-9^22, the former being the total number of possible permutations (outcomes) and the later being the total amount of permutations where a particular digit is NOT selected.  Since there are exactly 9^22 possibilities for a digit to not be drawn out of 10^22 possibilities, then 10^22 - 9^22 must equal the number of possibilities that a particular digit will be drawn AT LEAST once...and it is!

As trackers and system players, many of us choose the digits in the Pick 3 and Pick 4 based on odds and probability.  The "odds" are 1 in 10 for any particular digit to be drawn during any individual game (in a specific position).  However, the true probability for the digit to be drawn sometime within 22 games is precisely:

9,015,229,097,816,388,767,119  in  10,000,000,000,000,000,000,000

Using the outcome patterns we can also find the true probability for a digit to be drawn anywhere from exactly 1 time, to as many as exactly 22 times.  Or, we can find the true probabilities for a digit to be drawn AT LEAST X amount of times within 22 games.  Of course, to do this, the list of all outcome patterns has to be sorted and grouped by the quantities of repeated letters they each contain.  Eventually, I may post some of the info that I have derived from the scenario.  For now, here's a tad bit:

Listed below are the top 20 Pick 22 outcome patterns.  These are only 20 of the 807 possible, but yet their combined total probability (as a percent) is 52.59%!  Take a look at the last 22 position-one, position-two or position three digits in your states game and see how they hold up.

AAAABBBBCCCDDDEEFFGGHI

AAAABBBCCCDDDEEFFGGHIJ

AAAAABBBBCCCDDDEEFFGHI

AAAABBBCCCDDDEEEFFGGHI

AAAABBBCCCDDDEEFFGGHHI

AAAAABBBBCCCDDEEFFGGHI

AAAABBBCCCDDEEFFGGHHIJ

AAAAABBBCCCDDDEEFFGGHI

AAAABBBBCCCDDDEEFFGHIJ

AAAABBBBCCCDDEEFFGGHIJ

AAAABBBBCCCDDDEEEFFGHI

AAAAABBBBCCCDDDEEFFGGH

AAAAABBBBCCCDDEEFFGHIJ

AAAAABBBCCCDDEEFFGGHIJ

AAAABBBBCCCDDDEEEFFGGH

AAAABBBBCCCCDDDEEFFGHI

AAABBBCCCDDDEEFFGGHHIJ

AAAAABBBBCCCCDDDEEFFGH

AAAABBBBCCCDDEEFFGGHHI

AAAAABBBBCCCDDDEEEFFGH

~Probability=Odds in Motion~

United States
Member #5344
June 30, 2004
23641 Posts
Offline
 Posted: November 18, 2006, 10:02 am - IP Logged

That was Beautiful.. Thanks..  Lots of work too...  Appreciated.

OLD/Vtrac

The Quantum Master
West Concord, MN
United States
Member #21
December 7, 2001
3680 Posts
Offline
 Posted: November 20, 2006, 9:37 am - IP Logged

I have most of the logic worked out, however, I need to work it into a single formula or combination of formulae.

Here's the Pick 22 sheet I've come up with all the possible permutations for a Pick 22 permutation having N numbers the same.

 N Permutation Count 3 1,185,137,049,600,000 4 2,583,137,881,497,600,000 5 1,410,819,743,342,820,000,000 6 3,554,460,203,427,300,000,000 7 3,214,703,533,042,230,000,000 8 1,371,270,936,447,220,000,000 9 365,057,858,936,212,000,000 10 69,605,360,580,772,800,000 11 10,188,597,297,213,000,000 12 1,187,675,594,368,960,000 13 112,454,310,917,340,000 14 8,728,162,536,240,000 15 555,762,131,640,000 16 28,835,890,272,000 17 1,201,585,833,000 18 39,267,585,000 19 969,570,000 20 17,010,000 21 189,000 22 1,000 Total 10,000,000,000,000,000,000,000

Presented 'AS IS' and for Entertainment Purposes Only.
Any gain or loss is your responsibility.

Order is a Subset of Chaos
Knowledge is Beyond Belief
Wisdom is Not Censored
Douglas Paul Smallish
Jehocifer

The Quantum Master
West Concord, MN
United States
Member #21
December 7, 2001
3680 Posts
Offline
 Posted: November 20, 2006, 9:57 am - IP Logged

Here's the basic permutational form count:

 N Base Form Count 3 1 4 177 5 8,290 6 40,928 7 89,517 8 114,837 9 102,525 10 72,875 11 44,889 12 25,185 13 13,262 14 6,662 15 3,110 16 1,288 17 487 18 173 19 58 20 18 21 5 22 1 Total 524,288

Presented 'AS IS' and for Entertainment Purposes Only.
Any gain or loss is your responsibility.

Order is a Subset of Chaos
Knowledge is Beyond Belief
Wisdom is Not Censored
Douglas Paul Smallish
Jehocifer

The Quantum Master
West Concord, MN
United States
Member #21
December 7, 2001
3680 Posts
Offline
 Posted: November 20, 2006, 10:22 am - IP Logged

Nevermind, I'm spending too much time on this.

My mind is turning in to permutational mush.

Presented 'AS IS' and for Entertainment Purposes Only.
Any gain or loss is your responsibility.

Order is a Subset of Chaos
Knowledge is Beyond Belief
Wisdom is Not Censored
Douglas Paul Smallish
Jehocifer

Findlay, Ohio
United States
Member #4855
May 28, 2004
400 Posts
Offline
 Posted: November 20, 2006, 12:12 pm - IP Logged

Nevermind, I'm spending too much time on this.

My mind is turning in to permutational mush.

LOL!!! Yea, I think I fried a few circuits of my own.

Its pandora's Box...inside another box...inside a box...inside a another box...inside a box...inside another  box ...inside..

~Probability=Odds in Motion~

The Quantum Master
West Concord, MN
United States
Member #21
December 7, 2001
3680 Posts
Offline
 Posted: November 20, 2006, 4:01 pm - IP Logged

LOL!!! Yea, I think I fried a few circuits of my own.

Its pandora's Box...inside another box...inside a box...inside a another box...inside a box...inside another  box ...inside..

I think I'm relatively close though.

I have to rework the logic in deducing the counts.

Presented 'AS IS' and for Entertainment Purposes Only.
Any gain or loss is your responsibility.

Order is a Subset of Chaos
Knowledge is Beyond Belief
Wisdom is Not Censored
Douglas Paul Smallish
Jehocifer

Findlay, Ohio
United States
Member #4855
May 28, 2004
400 Posts
Offline
 Posted: November 20, 2006, 9:10 pm - IP Logged

I think I'm relatively close though.

I have to rework the logic in deducing the counts.

I think I craked it...can't believe an answer so simple eluded me for so long!

I think this should work, provided that N^0=1 and 0!=1.

To get the precise number of permutations that contain exactly X amount of a specified digit:

n = trials or spaces (i.e. Pick 3 = 3, Pick 4 = 4, Pick 22 =22)

r = exact amount of repeated digit (two digits 4,s or three digit 4,s etc...)

(n!)/(r!)/((n-r)!)*(9^(n-r))

~Probability=Odds in Motion~

Findlay, Ohio
United States
Member #4855
May 28, 2004
400 Posts
Offline
 Posted: November 21, 2006, 2:45 am - IP Logged

Everything adds up perfect!  It will be interesting to see what you can come up with for the "outcome patterns" or Basic Forms.

(n!)/(r!)/((n-r)!)*(9^(n-r))

 REPEATS PERMUTATIONS PROBABILITY PERCENTAGE 22 1 0.0000000000000000000001 0.00000000000000000001 21 198 0.0000000000000000000198 0.00000000000000000198 20 18,711 0.0000000000000000018711 0.00000000000000018711 19 1,122,660 0.000000000000000112266 0.000000000000011227 18 47,993,715 0.0000000000000047993715 0.00000000000047994 17 1,554,996,366 0.0000000000001554996366 0.00000000001555 16 39,652,407,333 0.0000000000039652407333 0.00000000039652 15 815,706,665,136 0.0000000000815706665136 0.0000000081571 14 13,765,049,974,170 0.000000001376504997417 0.00000013765 13 192,710,699,638,380 0.000000019271069963838 0.0000019271 12 2,254,715,185,769,046 0.0000002254715185769046 0.000022547 11 22,137,203,642,096,088 0.0000022137203642096088 0.00022137 10 182,631,930,047,292,726 0.0000182631930047292726 0.0018263 9 1,264,374,900,327,411,180 0.000126437490032741118 0.012644 8 7,315,311,923,322,878,970 0.000731531192332287897 0.073153 7 35,113,497,231,949,819,056 0.0035113497231949819056 0.35113 6 138,259,395,350,802,412,533 0.0138259395350802412533 1.3826 5 439,176,902,879,019,428,046 0.0439176902879019428046 4.3918 4 1,097,942,257,197,548,570,115 0.1097942257197548570115 10.979 3 2,080,311,645,216,407,817,060 0.208031164521640781706 20.803 2 2,808,420,721,042,150,553,031 0.2808420721042150553031 28.084 1 2,407,217,760,893,271,902,598 0.2407217760893271902598 24.072 0 984,770,902,183,611,232,881 0.0984770902183611232881 9.8477 CHECK PROOF SUM of Repeats 1 through 22: 9,015,229,097,816,388,767,119 10^22-9^22: 9,015,229,097,816,388,767,119 Digit Not Drawn (0): 984,770,902,183,611,232,881 9^22: 984,770,902,183,611,232,881

~Probability=Odds in Motion~

The Quantum Master
West Concord, MN
United States
Member #21
December 7, 2001
3680 Posts
Offline
 Posted: November 21, 2006, 2:52 am - IP Logged

Everything adds up perfect!  It will be interesting to see what you can come up with for the "outcome patterns" or Basic Forms.

(n!)/(r!)/((n-r)!)*(9^(n-r))

 REPEATS PERMUTATIONS PROBABILITY PERCENTAGE 22 1 0.0000000000000000000001 0.00000000000000000001 21 198 0.0000000000000000000198 0.00000000000000000198 20 18,711 0.0000000000000000018711 0.00000000000000018711 19 1,122,660 0.000000000000000112266 0.000000000000011227 18 47,993,715 0.0000000000000047993715 0.00000000000047994 17 1,554,996,366 0.0000000000001554996366 0.00000000001555 16 39,652,407,333 0.0000000000039652407333 0.00000000039652 15 815,706,665,136 0.0000000000815706665136 0.0000000081571 14 13,765,049,974,170 0.000000001376504997417 0.00000013765 13 192,710,699,638,380 0.000000019271069963838 0.0000019271 12 2,254,715,185,769,046 0.0000002254715185769046 0.000022547 11 22,137,203,642,096,088 0.0000022137203642096088 0.00022137 10 182,631,930,047,292,726 0.0000182631930047292726 0.0018263 9 1,264,374,900,327,411,180 0.000126437490032741118 0.012644 8 7,315,311,923,322,878,970 0.000731531192332287897 0.073153 7 35,113,497,231,949,819,056 0.0035113497231949819056 0.35113 6 138,259,395,350,802,412,533 0.0138259395350802412533 1.3826 5 439,176,902,879,019,428,046 0.0439176902879019428046 4.3918 4 1,097,942,257,197,548,570,115 0.1097942257197548570115 10.979 3 2,080,311,645,216,407,817,060 0.208031164521640781706 20.803 2 2,808,420,721,042,150,553,031 0.2808420721042150553031 28.084 1 2,407,217,760,893,271,902,598 0.2407217760893271902598 24.072 0 984,770,902,183,611,232,881 0.0984770902183611232881 9.8477 CHECK PROOF SUM of Repeats 1 through 22: 9,015,229,097,816,388,767,119 10^22-9^22: 9,015,229,097,816,388,767,119 Digit Not Drawn (0): 984,770,902,183,611,232,881 9^22: 984,770,902,183,611,232,881

Something don't seem right. With 22 numbers, you a guaranteed to have 3 repeating numbers, absolutely, positively. There should be no way to have only 2, 1, or 0 repeating numbers because of repetition with only 10 numbers (0 through 9).

Presented 'AS IS' and for Entertainment Purposes Only.
Any gain or loss is your responsibility.

Order is a Subset of Chaos
Knowledge is Beyond Belief
Wisdom is Not Censored
Douglas Paul Smallish
Jehocifer

The Quantum Master
West Concord, MN
United States
Member #21
December 7, 2001
3680 Posts
Offline
 Posted: November 21, 2006, 3:02 am - IP Logged

Here's why. Let's say all the numbers in the first 10 selections are different and just by chance they are:

0123456789

When it comes time to picking the 11th selection, what's left to select?

0123456789?  (? is any number 0 through 9)

No matter what number is selected, there will be at a minimum at least 2 matching numbers and no less.

Now extend this to 20 numbers for the minimum of 2 matching, the first 20 selections are:

01234567890123456789

What happens on the 21st selection?

01234567890123456789?  (? is any number 0 through 9)

No matter what number is selected, there will be at a minimum at least 3 matching numbers and no less.

Presented 'AS IS' and for Entertainment Purposes Only.
Any gain or loss is your responsibility.

Order is a Subset of Chaos
Knowledge is Beyond Belief
Wisdom is Not Censored
Douglas Paul Smallish
Jehocifer

Findlay, Ohio
United States
Member #4855
May 28, 2004
400 Posts
Offline
 Posted: November 21, 2006, 3:40 am - IP Logged

Everything adds up perfect!  It will be interesting to see what you can come up with for the "outcome patterns" or Basic Forms.

(n!)/(r!)/((n-r)!)*(9^(n-r))

 REPEATS PERMUTATIONS PROBABILITY PERCENTAGE 22 1 0.0000000000000000000001 0.00000000000000000001 21 198 0.0000000000000000000198 0.00000000000000000198 20 18,711 0.0000000000000000018711 0.00000000000000018711 19 1,122,660 0.000000000000000112266 0.000000000000011227 18 47,993,715 0.0000000000000047993715 0.00000000000047994 17 1,554,996,366 0.0000000000001554996366 0.00000000001555 16 39,652,407,333 0.0000000000039652407333 0.00000000039652 15 815,706,665,136 0.0000000000815706665136 0.0000000081571 14 13,765,049,974,170 0.000000001376504997417 0.00000013765 13 192,710,699,638,380 0.000000019271069963838 0.0000019271 12 2,254,715,185,769,046 0.0000002254715185769046 0.000022547 11 22,137,203,642,096,088 0.0000022137203642096088 0.00022137 10 182,631,930,047,292,726 0.0000182631930047292726 0.0018263 9 1,264,374,900,327,411,180 0.000126437490032741118 0.012644 8 7,315,311,923,322,878,970 0.000731531192332287897 0.073153 7 35,113,497,231,949,819,056 0.0035113497231949819056 0.35113 6 138,259,395,350,802,412,533 0.0138259395350802412533 1.3826 5 439,176,902,879,019,428,046 0.0439176902879019428046 4.3918 4 1,097,942,257,197,548,570,115 0.1097942257197548570115 10.979 3 2,080,311,645,216,407,817,060 0.208031164521640781706 20.803 2 2,808,420,721,042,150,553,031 0.2808420721042150553031 28.084 1 2,407,217,760,893,271,902,598 0.2407217760893271902598 24.072 0 984,770,902,183,611,232,881 0.0984770902183611232881 9.8477 CHECK PROOF SUM of Repeats 1 through 22: 9,015,229,097,816,388,767,119 10^22-9^22: 9,015,229,097,816,388,767,119 Digit Not Drawn (0): 984,770,902,183,611,232,881 9^22: 984,770,902,183,611,232,881

Note to readers, the table is digit specific. so these percentages apply to all digits by themselves.  For example, the probability to see exactly five digit 7's drawn in position-one within the next 22 pick 3 or pick 4 games is 4.39%.  The probability to see exactly six digit 7's is 1.38%, etc.

It's interesting to note that:

• It's almost two and a half times more likely to see your digit drawn exactly one time than not at all.
• The chance that your digit will be drawn exactly two times is better than the chance that it will only be drawn once OR exactly three times
• The chance that your digit will only be drawn once is less than 4% greater than the chance of seeing it drawn exactly three times.
• The chance to not see your digit drawn at all is 9.85%, but to see it drawn exactly 4 times is almost 11% (10.979%).

~Probability=Odds in Motion~

Findlay, Ohio
United States
Member #4855
May 28, 2004
400 Posts
Offline
 Posted: November 21, 2006, 4:04 am - IP Logged

Here's why. Let's say all the numbers in the first 10 selections are different and just by chance they are:

0123456789

When it comes time to picking the 11th selection, what's left to select?

0123456789?  (? is any number 0 through 9)

No matter what number is selected, there will be at a minimum at least 2 matching numbers and no less.

Now extend this to 20 numbers for the minimum of 2 matching, the first 20 selections are:

01234567890123456789

What happens on the 21st selection?

01234567890123456789?  (? is any number 0 through 9)

No matter what number is selected, there will be at a minimum at least 3 matching numbers and no less.

Your picking a number between 0 and 9, (thats ten digits to choose from) but your guessing/picking 22 in a row.  Like Pick 3 (3 times) or Pick 4 (4 times), but for pick 22 it's 22 times.

The digits could be the same all 22 times:

7777777777777777777777

OR the digit doesn't have to come up at all:

1859095435601128890252  (no 7 in there)

See where I'm coming from?

~Probability=Odds in Motion~

The Quantum Master
West Concord, MN
United States
Member #21
December 7, 2001
3680 Posts
Offline
 Posted: November 21, 2006, 4:07 am - IP Logged

Ah, ok... you are referring to consecutive repetition only and not repetition in general.

Hmmm, makes a little more sense.

Presented 'AS IS' and for Entertainment Purposes Only.
Any gain or loss is your responsibility.

Order is a Subset of Chaos
Knowledge is Beyond Belief
Wisdom is Not Censored
Douglas Paul Smallish
Jehocifer

The Quantum Master
West Concord, MN
United States
Member #21
December 7, 2001
3680 Posts
Offline
 Posted: November 21, 2006, 4:16 am - IP Logged

Your picking a number between 0 and 9, (thats ten digits to choose from) but your guessing/picking 22 in a row.  Like Pick 3 (3 times) or Pick 4 (4 times), but for pick 22 it's 22 times.

The digits could be the same all 22 times:

7777777777777777777777

OR the digit doesn't have to come up at all:

1859095435601128890252  (no 7 in there)

See where I'm coming from?

However in the number 1859095435601128890252 there are the following repeated numbers

0 - 3
1 - 3
2 - 3
5 - 4
8 - 3
9 - 3

But they are not all consecutive.

1 and 8 have a consecutive repeat of 11 and 88.

Presented 'AS IS' and for Entertainment Purposes Only.
Any gain or loss is your responsibility.