  # What are the odds Are the lines you play giving you the best bang for your dollar?

Note!
The cNr calculations are done using the standard windows calculator set to scientific mode.I have gone over this several times to reduce errors but some may still exist. My left handtypes faster than my right which causes hard to find errors in the number values. With thatbeing said, here we go.

I will keep this simple as possible so that everyone regardless of you math level will be ableto understand it. I will show why optimizing our sets will increase our wins by increasing thenumber of overall chances. Line optimizing is a very valuable tool for anyone who plays morethan one line per game.

Jade posted something similar to this a year or so back and I think this gives a little moreinsight into the question he posed at that time. If I remember correctly he asked which was abetter bet, playing 1 line for 10 games or playing 10 lines in a single game.  I will show howthat playing 1 line for 10 games has better overall chances of winning a prize than playing 10lines for a single game. This is not the reason for this post but it helps to understand it but  first a little math.

The figures here apply to a pick 5 of 39 game but hold true for any doubledigits lottery game you just need to plug in the correct variabels. .

Below are a few problems for which the nCr function can be used.  These are all lottery relatedand cover various aspects of calculating combinations.

Below are a couple methods one could use to calculate the number of times any one number appearswithin a 5-39 matrix.

Option #1. 575757*5/39 = 73815

Option #2  (39-1)=38!/(5!*34!)= 501942 and 575757-501942=73815

The first option is very simple but what if we need to calculate the number of times any 2 numbersappear together. If we multiply 73,815 by 2 we get 73815*2=147624 which is incorrect.

To calculates pairs (39-2)= 37!/(3!*34!)=7770
To calculates trays (39-3)= 36!/(2!*34!)=630

Here I show how to calculate the odds using the nCr function and some simple math.

To do this we first need to know a couple things. Again for this example I will use a 5-39 but ifworking on a different matrix, just plug in your variables. The list below shows the number ofcombinations of 2, 3, 4 etc.. within pick 5 and 6 lottery games.

To find these values using the nCr function follow the example -> 5!/(2!,3!)=10 which shows the total2 number combos within a set of 5 numbers.

combinations of 2 in 5 = 10
combinations of 3 in 5 = 10
combinations of 4 in 5 = 5
combinations of 5 in 5 = 1

combinations of 2 in 6 = 15
combinations of 3 in 6 = 20
combinations of 4 in 6 = 15
combinations of 5 in 6 = 6
combinations of 6 in 6 = 1

This list can be easily memorized which will cut down on the number of calculations needed to formost of the problems here.

First we will calculate the odds for matching 2 of 5 numbers.  To do this we have to plug in thecorrect variables. Look at this example. 34!/(3!*31!)=5984  Notice the (n) value is 34.  This isgotten by subtracting 5 from 39.  The next variable (r) is 3. To get this value we subtract theprize number match from the total numbers in single line, 5-2=3, The third variable is gotten bysubtracting (r) from (n), 34-3=31

Here are the three variables for calculating 2,3,4 of 5 numbers matches.

2of5 are #1, 39-5=(34), #2=5-2=(3) and #3=34-3=(31)

3of5 are #1, 39-5=(34), #2=5-3=(2) and #3=34-2=(32)

4of5 are #1, 39-5=(34), #2=5-4=(1) and #3=34-1=(33)

34!/(3!*31!)=5984
34!/(2!*32!)=561
34!/(1!*33!)=34

Next we multiply the results by the number of combinations within the a set of 5 numbers.
combinations of 2 in 5 = 10
combinations of 3 in 5 = 10
combinations of 4 in 5 = 5
combinations of 5 in 5 = 1

34!/(3!*31!)=5984*10 =59840
34!/(2!*32!)= 561*10 = 5610
34!/(1!*33!)=  34* 5 =  170
5 of 5          1* 1 =    1
___________________________
Total                =65621

This gives us the number of lines within the entire matrix that if drawn will win a prize.

To calculate the individual odds for matching a prize we divide the total number of sets within the matrixby the total of each possible number match

Odds of matching 2 numbers = 575757/59840=1 in 9.62
Odds of matching 3 numbers = 575757/5610 =1 in 102.63
Odds of matching 4 numbers = 575757/170  =1 in 3386.80
Odds of matching 5 numbers = 575757/1    =1 in 575757

To calculate the overall odds we add all the possible prizes then divide the total lines by thisamount. 575757/65821=8.77

The overall odds are 1 in 8.77 for winning a prize for a 5-39 game.  This applies only to single lineplays regardless of the numbers played. What this means is that there are 65,621 possible lines thatif any one is drawn we will win a prize.  I know this is a bit redundant but we need to understand it.

We now know exactly what our overall odds of winning a prize when playing one ticket but whatif we play more than one line.

Now that we have calculated the total chances one set has for a single drawing we can calculate thetotal chances for playing one line for 10 games. 65621*10=656210. Playing one line each game givesus the best cost/chance ratio we can get. The CC ratio for a one line play is calculated by dividingthe cost by the total chances. \$1.00/65621=\$0.000015239.

You may be a bit confused by this so let's dig a little deeper.  What if we play 2 lines? Letslook at the two examples below and I will introduce what I call "prize overlap lines", POL forshort.

Player #1
01-02-03-04-05
06-07-08-09-10

Player #2
01-02-03-04-05
01-02-03-04-06

Above we have two players that show the extreme ranges for betting on the lottery. Player #1plays 10 different numbers on his/her 2 lines and player #2 plays 6 total numbers. There areno repeating numbers in the lines of player #1 but player 2's lines share 4 numbers.

To calculate the number of POL's I use a algorithm that won't be posted here but will use theresults gotten from it.

I need to back up a bit and explain what POL's are.  Remember above where I calculated the number oflines within a matrix that if drawn could win a prize, these totals are based on a single play. Whenwe play more that one line the overall chances will change depending on the numbers played.

A PLO is any line in the matrix that shares a win with more than one line played.

Example.
Let's say that the set 01-04-08-10-24 is drawn. Both lines of player #1 match 2 numbers,  01-04 and08-10.  The set drawn is a POL, "Prize Overlap Line"

Player #1's share 3,100 lines/POL's.
Player #2's share 43,797 lines/POL's.

How does this effect the overall chances?

We can see from above what the overall chances are for a single bet game but when playing more than oneline we have to make adjustments when we calculate the overall.

Player #1's first line has 65621 chances of winning a prize and so does the second but some of the linesare POL's.

01-02-03-04-05 = 65621 total chances
06-07-08-09-10 = 65621 total chances

There are 3100 lines within the matrix that share prizes with both lines so 65621+65621=131242-3100=128142
This is the overall chances these two lines have of winning a prize in the next draw and 57575/128142=4.493

Now lets look at the overall chances for player #2. These lines both have the same overall chances as player#1 but the POL's are much higher. 65621+65621=131242-43797=87445 so overall chances are 575757/87445=6.584.

We can see from this that player #1 will on average win a prize for every \$4.39 he bets while player #2will win one prize for every \$6.58. For my 5-39 a match-2 breaks even and 3 and up makes money.

The CC ratio gives us the cost per chance so \$2.00/128142=.000015608 vs \$2.00/87445=.000022872.

Playing one line per game will always have the lowest cost to chance ratio.  The down side is that our oneline will never give better odds than 1 in 8.77.

RL RL,

Thank you for doing this, very informative. I have a question, in a 5/36 game how many lines would I have to play to ensure each digit is covered at least once. If this takes you away from the direction you are going or headed, forget it. Carbob

A round figure would be 36/5=7.2.

Also note that the statement below is incorrect.

We can see from this that player #1 will on average win a prize for every \$4.39 he bets while player #2will win one prize for every \$6.58. For my 5-39 a match-2 breaks even and 3 and up makes money.

The dollar value is doubled because we are playing two lines. You can't drop

below the overall odds for a dollar bet. The first part proves this is true.

Each line we purchase will never have the same overall chances as the previous

line because of overlaps.  Our chances can however be increased but it comes

at a price. I should have said for each combined bet and left the dollar sign

off. Sorry for the mistake, it was due to a edit mistake. The statement was to

show the differences in the number of chances based on the numbers we put into

play.

RL Bob

I reread your question and I am working on a digit version of this that shows why I play digits.  We don't have

to cover all the digits the lines in the matrix take care of that.  If we put 5 digits into play, lets say 1-2-3-5-7 then

we are playing 19 lines in a 5-36.

01-02-03-05-07

11-12-13-15-17

21-22-23-25-27

31-32-33-35

This is over half the numbers in the matrix but let's say that we only play sets that have all 5 digits in each line.

For my 5-39 I can only build 3 lines before I have to reuse a number.  I will get into detail about this and show

why it works.  Five digits, 3-base + 2 non-base will reduce the matrix to less than 6000 lines. If I say that each

non-base digit can only show once in a set then it drops to 3000 and change.  When I first read your reply I

missed the word digit and thought you were talking numbers, sorry again for yet another mistake.

RL In a nutshell.

Unless a person can predict the exact numbers then the worst possible bet one could make is to

play many lines using the same numbers for each set.  The overall chances would not increase at

all.  The first ticket we buy has the best CC ratio as every additional ticket we purchase drops in

value.   When I play I don't consider my bet in lines played but the number of overall chances my

tickets give me.  If I can pay \$10.00 for  250,000 chances or \$10.00 for 145,000 chances, which one

is the better bet.  It's not a \$1.00 per ticket bet it's a \$10.00 fee for 250,000 chances.

RL RL-RANDOMLOGIC,

COST PER CHANCE

"We can see from this that player #1 will on average win a prize for every \$4.39 he bets while player #2 will win one prize for every \$6.58. For my 5-39 a match-2 breaks even and 3 and up makes money.

The CC ratio gives us the cost per chance so \$2.00/128142=.000015608 vs \$2.00/87445=.000022872."

I haven't checked your math but your results are consistent with what is widely known, and that is that the "DISTRIBUTION" of your winnings can be manipulated by choosing betting patterns like those you suggest.  I discussed this in 2010.

Unfortunately, "COST PER CHANCE" is NOT "EXPECTED VALUE," which is really what "winning \$" is all about.  BTW, are you sure JadeLottery showed there was any difference over the long haul in the 10/1 vs 1/10 question?

See my signature line.

--Jimmy4164 Jimbo

I was wondering if you could stay away, I guess not.  I also did not make any reference to what Jade's post said

one way or the other.  As far as I remember he just posted the question for us to ponder, I could be wrong about

that as I don't remember visiting the post but a few times.   I did pose a very similar question in your last topic but

you never replied.  Maybe you caught on to my little play on math, maybe not.  However I must say I am glad to see

you post without using the word innumeracy.  The first statement above was retracted as being incorrect and the \$

sign should have not been used.  The figures are for a 2 line bet without regard to the price of the tickets.  The CC

ratio I believe to be correct but my haste in getting this up has already yielded a few mistakes.

RL Part 2

First let me say that I will be posting this in several parts due to time restraints.  I have about 10
projects I am working on and to stay on tract I decided to devote no more than one hour a day to
lottery.  This includes all my lottery related stuff so it may be a few days between post.

Now I will talk a little about the digits and why I use them.  All of us are familiar with the base 10
numerical system.  Using digits is really nothing more than going back to the basics we were taught
in grade school.  The number 37 represents 3*10 + 7*1 so working with digits is really nothing new.

Back in the early days of computers, PC's were slow, I mean really slow. Back when I first started
witting lottery stuff I was limited to using brute force to prove certain calculations which could
take days or even weeks.  I started looking for shortcuts to prove/solve problems even though I had
never even heard of a algorithm.  My first algorithm reduced over a 500 lines of code to less than 5
and took a few minutes to solve a problem that before took days.  I would never have been able to
figure out how to write that algorithm without reducing numbers to digits and have been using them
since that time.

Digits and the lottery.
The digits show at a greater frequency then the numbers they represent. The greater the frequency the
easier it is to approximate when something will show.  I can't predict the numbers but I can make a
few good approximations of when certain digits will show.  Digit tracking provides more information
that can be used to analyzing data.  Not all digits have equal overall chances of showing in the next
draw. Digits alone are not enough to increase your winnings but they are a good place to start.

System players mostly look for biases which in turn gives them reasons to choose one value over another.
Without a bias we are left with using methods that are not mathematical. IMHO there are two types of
tools that can be used for picking numbers.

Predictors and system.
Predictors attempt to predict the best numbers to play with little or no user input and reliey on historic
data analysis. Systems provide the user with many data types which should show biases for one value over
another. Predictors make these choices for you systems don't.

Most here know my stance on the lottery and random but I will go over it again.  A lottery has a finite
number of combinations which can be drawn.  Each number has a equal chance of being drawn but over time
the sets or lines drawn will mimic the overall matrix in population and distribution.  This applies not
only to the numbers but almost any observation we can make. The only random attribute of a lottery is
the order in which the numbers are drawn.  Analysis of secondary data such as digits can show us a bias
that is not a product of random but a product of the matrix.  The digit system is really just a lottery
profiling tool very similar to those used in law enforcement and other disciplines.

Below is a list of the last 10 drawings for the MO. cash 5-39 game

DAY  DATE       NUMBERS        Total Digits   Digits
=================================================================
SUN  04/27/14   04 09 11 18 20      6         1-2-4-8-9-0
SAT  04/26/14   19 20 26 27 33      7         1-2-3-6-7-9-0
FRI  04/25/14   05 19 21 29 35      5         1-2-3-5-9
THU  04/24/14   03 23 33 36 39      4         2-3-6-9
WED  04/23/14   07 10 11 37 38      5         1-3-7-8-0
TUE  04/22/14   04 11 32 38 39      6         1-2-3-4-8-9
MON  04/21/14   01 02 07 30 35      6         1-2-3-5-7-0
SUN  04/20/14   02 03 05 27 30      5         2-3-5-7-0
SAT  04/19/14   05 16 32 36 39      6         1-2-3-5-6-9
FRI  04/18/14   01 11 25 30 31      5         1-2-3-5-0

The digit system divides the digits into two groups called [Base=1-2-3] and [Non-base=4-5-6-7-8-9-0]
Base digits are digits that can take up either the left or right position within a number as in 32.

Eighty percent of the last 10 games have 5 or 6 total digits. There is one game with 7 and one with 4.
Fifty percent of the draws all share the digits 1-2-3 and fourty percent share digits 1-2-3-9.  Digit
(0) and (9) appear in sixty percent overall.  The digit (0) is not counted when used to pad a single
digit number as in the number 01.

I will keep this simple so that it can be done using pen and paper.

A digit setup is done by first deciding how many digits we want to use in each set. For this example
I will will be using 6.  The six digits I choose are 1-2-3 + 5-7-9.  Every set I build from these 6
digits must contain at least one of each digit.

These six digits repersent the numbers

01-02-03-05-07-09
11-12-13-15-17-19
21-22-23-25-27-29
31-32-33-35-37-39

So we have 24 numbers.  There are 42,504 combinations of 5 numbers in 24 but if we say that every set
must contain at least 1 of every digit then the total combinations are reduced to 5421 a reduction of
37,083 lines.  If we say that the non-base digits can only be used once then the total combinations
are reduced to 3033.

This is however way to many sets to play so how do we select which numbers to actually play.  Lets say
that the set drawn is 01-07-23-25-39

In the 3033 lines there are

680 that match 2of5
182 that match 3of5
24 that match 4of5
1 that match 5of5
____________________
total  = 887 overall chances for winning a prize.

3033 / 887 = 3.42  overall odds of winning.

Above is a best case but what if digit 9 does not show and the set 03-22-27-33-35 is drawn.  This set
has 4 digits 2-3-5-7.

620 that match 2of5
112 that match 3of5
3 that match 4of5
0 that match 5of5
____________________
total = 735 overall chances of winning a prize

3033 / 735 = 4.13 overall odds of winning a prize.

Now lets do one more but this time we will add another digit to the set drawn and again leave out digit
9. Set drawn = 07-15-24-33-38.  We know that the max prize will be a 3of5 because 2 of the numbers 24
and 38 are not in our list of 24 numbers above and the digits are 1-2-3-4-5-7-8 = 7

310 that match 2of5
29 that match 3of5
0 that match 4of5
0 that match 5of5
_____________________
total = 349 overall chances of winning a prize

3033 / 349 = 8.69

Remember the overall odds are 1 in 8.77 overall for winning a prize for this game and all 3 of the simulations
here fall below that number.  If playing more than one line and use the information posted at the start of this
topic then you can see that playing digits can work out very good even when the actual draw takes another
direction.  As with any method we do need to make so many correct selections but we don't have to be spot on.

In part 3 I will dig a little deeper and show how to select the numbers to play from the list.

RL RL, I'm with you, thank you. RL-RANDOMLOGIC,

You said,  "I also did not make any reference to what Jade's post said one way or the other.  As far as I remember he just posted the question for us to ponder, I could be wrong about that as I don't remember visiting the post but a few times."

His simulation supported the theoretical answer in that it resulted in no statistically significant difference between "1 Play for 10 Draws" vs  "10 Plays for 1 Draw."

Your Cost Per Chance [of winning some
thing] is NOT the EXPECTED VALUE of a Play, or set of plays.  Consequently, your methods will produce more HITS, but in the long run, your WINNINGS will be the same as with Quick Picks or any other selection method you choose.  A stack of \$1 and \$2 dollar winning tickets doesn't buy you much, especially when you consider what you spent to accumulate them.  If your goal is to impress the lottery outlet clerks with lots of small wins, you're on the right track! You are not the only player who believes that more 1 and 2 number matches increases your chances of matching 3, 4, and 5/6.  You are wrong.

Your saving grace is that your methods will definitely not decrease your winnings. --Jimmy4164 It was If you had 10 bucks to bet... back in 2011.

We keep our posted topics in our favorites, https://www.lotterypost.com/member/1170/favorites This what we be usin'

Factorial - n! = n · (n -1) · (n - 2) · ... · 3 · 2 · 1 and 0! = 1

Permutation - P(n, r) = n! / (n - r)!

Combination - C(n, r) = P(n, r) / r!

Probability of Win - Wp(n, r, w) = (C(r, w) · C(n - r, r - w)) / C(n, r)

Odds to Win - Wo(n, r, w) is 1 in (C(n, r) - C(r, w) · C(n - r, r - w)) / (C(r, w) · C(n - r, r - w))

n - total numbers in lottery pool
r - pick size from the pool of numbers
w - winning drawn numbers in a played combo RL

It is spring and I am getting the lottery fever too.

Thanks for the information you have posted so far. Combinations(39,5)= 39!/[(39-5)!5!] = 39*38*37*36*35/5/4/3/2/1 I tried to format the combo stuff by keystroke using the windows calculator, I figured some might not be fimilar

with the nCr function and this would make it easier.  Big thanks for sharing.

RL

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