Welcome Guest
You last visited January 19, 2017, 8:41 pm
All times shown are
Eastern Time (GMT-5:00)

# How many sets??

Topic closed. 13 replies. Last post 3 years ago by Greenfox.

 Page 1 of 1
Burnsville
United States
Member #107244
March 4, 2011
853 Posts
Offline
 Posted: August 2, 2013, 3:23 am - IP Logged

Something hit me earlier today as to how to do something with some of what I've been working on. I'm working it out on a sheet myself right now, but was wanting to get some opinions from others on here to make sure I'm right on it.

For a draw that has 6 number positions. 6 balls, or however you want to call them.

If you laid them out in numeric order from smallest to largest, and had say, 4 choices for each position, how many sets would that make to play?

So that would be four digits for the first/smallest. Four for second position. Four for the third and so on.

There would be no need to take one of the first positions choices and switch it to say the third or fifth or whatever. Just four numbers for each position. How many sets total would it be?

Right now, I'm figuring it with:

1,2,3,4 for the first.

2,3,4,5 for the second.

3,4,5,6 for the third.

4,5,6,7 for the fourth.

5,6,7,8 for the fifth.

6,7,8,9 for the sixth.

I know that's not the choices, but just what I'm using to figure out how many.

I'm just wandering for a little test is all and wanted a second or more account to ensure I'm right on it.

Thanks a bunch!!!

You can't steal second and keep your foot on FIRST!!!

When you go through hardships and decide not to surrender, that is strength”.

-Arnold (Ahnald) Schwarzenegger-

Burnsville
United States
Member #107244
March 4, 2011
853 Posts
Offline
 Posted: August 2, 2013, 3:32 am - IP Logged

That's not counting a game with a bonus ball. It's just any game that has 6 ball positions. Doesn't matter which one.

We don't even have a 6 ball draw here, but was just curious and making sure I get it right so I know what to do next.

You can't steal second and keep your foot on FIRST!!!

When you go through hardships and decide not to surrender, that is strength”.

-Arnold (Ahnald) Schwarzenegger-

United States
Member #41846
June 23, 2006
460 Posts
Offline
 Posted: August 2, 2013, 4:04 am - IP Logged

If I understand what you are asking   4x4x4x4x4x4 4096  if some of the numbers in different groups are duplicated and you remove duplicates that will change the answer.  if your question really is how many 6 digits combinations can you make from 24 different #'s that is a different answer 134596

Economy class
Belgium
Member #123700
February 27, 2012
4035 Posts
Offline
 Posted: August 2, 2013, 4:22 am - IP Logged

You should find in Excel help, look for permutation, combination etc.
The functions are lego pieces for further combinations.
We rewrite factorial in VBA, you find the code in a thread.

Krypton
United States
Member #140102
March 11, 2013
904 Posts
Online
 Posted: August 4, 2013, 8:55 am - IP Logged

That's not counting a game with a bonus ball. It's just any game that has 6 ball positions. Doesn't matter which one.

We don't even have a 6 ball draw here, but was just curious and making sure I get it right so I know what to do next.

Hey there Fox. See if this is what your thinking or looking for
their is a formula to work this out, its called combin, below will work out you combinations for you which will be 15
The first part is the is number of items, the second is the number of combination required
=COMBIN(6,4)
Regards
Pup

Economy class
Belgium
Member #123700
February 27, 2012
4035 Posts
Offline
 Posted: August 4, 2013, 12:15 pm - IP Logged

Something hit me earlier today as to how to do something with some of what I've been working on. I'm working it out on a sheet myself right now, but was wanting to get some opinions from others on here to make sure I'm right on it.

For a draw that has 6 number positions. 6 balls, or however you want to call them.

6 number positions as pick 6, ...

If you laid them out in numeric order from smallest to largest, and had say, 4 choices for each position, how many sets would that make to play?
... in ascending order ...
So that would be four digits for the first/smallest. Four for second position. Four for the third and so on.

With totally distinct sets, where the numbers of the next set are all bigger than the numbers of the previous set: 4^6=1296

 N1 N2 N3 N4 N5 N6 1 5 9 13 17 21 1 5 9 13 17 22 1 5 9 13 17 23 1 5 9 13 17 24 1 5 9 13 18 21 1 5 9 13 18 22

...

 4 8 12 16 19 24 4 8 12 16 20 21 4 8 12 16 20 22 4 8 12 16 20 23 4 8 12 16 20 24

If the sets overlap, the case is different. If you have more than 24 numbers, you get more combinations.

Questions:
Are your number sets totally different?
How big is the ball set?

For all or nothing you would have 24!/(24-6)!/6! combinations (134596).
For any condition you add to that, you shrink the count of combinations.

There would be no need to take one of the first positions choices and switch it to say the third or fifth or whatever. Just four numbers for each position. How many sets total would it be?

Right now, I'm figuring it with:

1,2,3,4 for the first.

2,3,4,5 for the second.

3,4,5,6 for the third.

4,5,6,7 for the fourth.

5,6,7,8 for the fifth.

6,7,8,9 for the sixth.

If you have nine numbers with sets 1-4, 2-5, 3-6, 4-7, 5-8, 6-9 then you get 84 combinations.
Note that the sets are overlapping, in set 1 and set 2 you find {2,3,4}.

 1-2-3-4-5-6 1-2-3-4-5-7 1-2-3-4-5-8 1-2-3-4-5-9 1-2-3-4-6-7 1-2-3-4-6-8 1-2-3-4-6-9 1-2-3-4-7-8 1-2-3-4-7-9 1-2-3-4-8-9 1-2-3-5-6-7 1-2-3-5-6-8 1-2-3-5-6-9 1-2-3-5-7-8 1-2-3-5-7-9 1-2-3-5-8-9 1-2-3-6-7-8 1-2-3-6-7-9 1-2-3-6-8-9 1-2-3-7-8-9 1-2-4-5-6-7 1-2-4-5-6-8 1-2-4-5-6-9 1-2-4-5-7-8 1-2-4-5-7-9 1-2-4-5-8-9 1-2-4-6-7-8 1-2-4-6-7-9 1-2-4-6-8-9 1-2-4-7-8-9 1-2-5-6-7-8 1-2-5-6-7-9 1-2-5-6-8-9 1-2-5-7-8-9 1-2-6-7-8-9 1-3-4-5-6-7 1-3-4-5-6-8 1-3-4-5-6-9 1-3-4-5-7-8 1-3-4-5-7-9 1-3-4-5-8-9 1-3-4-6-7-8 1-3-4-6-7-9 1-3-4-6-8-9 1-3-4-7-8-9 1-3-5-6-7-8 1-3-5-6-7-9 1-3-5-6-8-9 1-3-5-7-8-9 1-3-6-7-8-9 1-4-5-6-7-8 1-4-5-6-7-9 1-4-5-6-8-9 1-4-5-7-8-9 1-4-6-7-8-9 1-5-6-7-8-9 2-3-4-5-6-7 2-3-4-5-6-8 2-3-4-5-6-9 2-3-4-5-7-8 2-3-4-5-7-9 2-3-4-5-8-9 2-3-4-6-7-8 2-3-4-6-7-9 2-3-4-6-8-9 2-3-4-7-8-9 2-3-5-6-7-8 2-3-5-6-7-9 2-3-5-6-8-9 2-3-5-7-8-9 2-3-6-7-8-9 2-4-5-6-7-8 2-4-5-6-7-9 2-4-5-6-8-9 2-4-5-7-8-9 2-4-6-7-8-9 2-5-6-7-8-9 3-4-5-6-7-8 3-4-5-6-7-9 3-4-5-6-8-9 3-4-5-7-8-9 3-4-6-7-8-9 3-5-6-7-8-9 4-5-6-7-8-9

I know that's not the choices, but just what I'm using to figure out how many.

If you have developed a model, than you can make a function. Test the function afterwards.
Say you use n numbers, you pick x numbers, then your function is like f(n,x)=... math.

The function needs to get an appropriate name, you have to write a comment what it does and what it returns.
The paramaters also should be defined and get restrictions. Special cases are often zero.
There you may find oddities in mathematics, that require "if".

I'm just wandering for a little test is all and wanted a second or more account to ensure I'm right on it.

The hard thing is to define the function in mathematical writing. You already may find it difficult to formulate it in simple words.

Thanks a bunch!!!

The one man bunch.

Post office scriptum: If you can formulate your question properly, the answer is easier to give.

"... I suggest that you get up earlier!" - Arnold

Burnsville
United States
Member #107244
March 4, 2011
853 Posts
Offline
 Posted: August 8, 2013, 3:56 pm - IP Logged

If I understand what you are asking   4x4x4x4x4x4 4096  if some of the numbers in different groups are duplicated and you remove duplicates that will change the answer.  if your question really is how many 6 digits combinations can you make from 24 different #'s that is a different answer 134596

I must have worded what I was looking for wrong. I was looking for the base amounts of sets that it would take with four digits for the first ball, four for the second ball and so on. Starting with 1-4 for the first, 2-5 for the second and so on. Once you get to the fourth largest digit and it being the 4, you could only have 4-5-6-7-8-9 as a set. I know there would be a few more with some distance between each for real play, but I was looking for a minimum to get an idea.

It wasn't how many combinations could be made for 24 numbers total. Keeping them in numeric order with all the smallest of the four picks for the first number staying as the first number then taking a number from the second numbers choices and moving forward.

I appreciate your time and help!!!

You can't steal second and keep your foot on FIRST!!!

When you go through hardships and decide not to surrender, that is strength”.

-Arnold (Ahnald) Schwarzenegger-

Burnsville
United States
Member #107244
March 4, 2011
853 Posts
Offline
 Posted: August 8, 2013, 3:58 pm - IP Logged

You should find in Excel help, look for permutation, combination etc.
The functions are lego pieces for further combinations.
We rewrite factorial in VBA, you find the code in a thread.

I'll look into that and see what I can figure out. Still learning here and appreciate all help in doing so!!!

You can't steal second and keep your foot on FIRST!!!

When you go through hardships and decide not to surrender, that is strength”.

-Arnold (Ahnald) Schwarzenegger-

Burnsville
United States
Member #107244
March 4, 2011
853 Posts
Offline
 Posted: August 8, 2013, 4:42 pm - IP Logged

Something hit me earlier today as to how to do something with some of what I've been working on. I'm working it out on a sheet myself right now, but was wanting to get some opinions from others on here to make sure I'm right on it.

For a draw that has 6 number positions. 6 balls, or however you want to call them.

6 number positions as pick 6, ...

If you laid them out in numeric order from smallest to largest, and had say, 4 choices for each position, how many sets would that make to play?
... in ascending order ...
So that would be four digits for the first/smallest. Four for second position. Four for the third and so on.

With totally distinct sets, where the numbers of the next set are all bigger than the numbers of the previous set: 4^6=1296

 N1 N2 N3 N4 N5 N6 1 5 9 13 17 21 1 5 9 13 17 22 1 5 9 13 17 23 1 5 9 13 17 24 1 5 9 13 18 21 1 5 9 13 18 22

...

 4 8 12 16 19 24 4 8 12 16 20 21 4 8 12 16 20 22 4 8 12 16 20 23 4 8 12 16 20 24

If the sets overlap, the case is different. If you have more than 24 numbers, you get more combinations.

Questions:
Are your number sets totally different?
How big is the ball set?

For all or nothing you would have 24!/(24-6)!/6! combinations (134596).
For any condition you add to that, you shrink the count of combinations.

There would be no need to take one of the first positions choices and switch it to say the third or fifth or whatever. Just four numbers for each position. How many sets total would it be?

Right now, I'm figuring it with:

1,2,3,4 for the first.

2,3,4,5 for the second.

3,4,5,6 for the third.

4,5,6,7 for the fourth.

5,6,7,8 for the fifth.

6,7,8,9 for the sixth.

If you have nine numbers with sets 1-4, 2-5, 3-6, 4-7, 5-8, 6-9 then you get 84 combinations.
Note that the sets are overlapping, in set 1 and set 2 you find {2,3,4}.

 1-2-3-4-5-6 1-2-3-4-5-7 1-2-3-4-5-8 1-2-3-4-5-9 1-2-3-4-6-7 1-2-3-4-6-8 1-2-3-4-6-9 1-2-3-4-7-8 1-2-3-4-7-9 1-2-3-4-8-9 1-2-3-5-6-7 1-2-3-5-6-8 1-2-3-5-6-9 1-2-3-5-7-8 1-2-3-5-7-9 1-2-3-5-8-9 1-2-3-6-7-8 1-2-3-6-7-9 1-2-3-6-8-9 1-2-3-7-8-9 1-2-4-5-6-7 1-2-4-5-6-8 1-2-4-5-6-9 1-2-4-5-7-8 1-2-4-5-7-9 1-2-4-5-8-9 1-2-4-6-7-8 1-2-4-6-7-9 1-2-4-6-8-9 1-2-4-7-8-9 1-2-5-6-7-8 1-2-5-6-7-9 1-2-5-6-8-9 1-2-5-7-8-9 1-2-6-7-8-9 1-3-4-5-6-7 1-3-4-5-6-8 1-3-4-5-6-9 1-3-4-5-7-8 1-3-4-5-7-9 1-3-4-5-8-9 1-3-4-6-7-8 1-3-4-6-7-9 1-3-4-6-8-9 1-3-4-7-8-9 1-3-5-6-7-8 1-3-5-6-7-9 1-3-5-6-8-9 1-3-5-7-8-9 1-3-6-7-8-9 1-4-5-6-7-8 1-4-5-6-7-9 1-4-5-6-8-9 1-4-5-7-8-9 1-4-6-7-8-9 1-5-6-7-8-9 2-3-4-5-6-7 2-3-4-5-6-8 2-3-4-5-6-9 2-3-4-5-7-8 2-3-4-5-7-9 2-3-4-5-8-9 2-3-4-6-7-8 2-3-4-6-7-9 2-3-4-6-8-9 2-3-4-7-8-9 2-3-5-6-7-8 2-3-5-6-7-9 2-3-5-6-8-9 2-3-5-7-8-9 2-3-6-7-8-9 2-4-5-6-7-8 2-4-5-6-7-9 2-4-5-6-8-9 2-4-5-7-8-9 2-4-6-7-8-9 2-5-6-7-8-9 3-4-5-6-7-8 3-4-5-6-7-9 3-4-5-6-8-9 3-4-5-7-8-9 3-4-6-7-8-9 3-5-6-7-8-9 4-5-6-7-8-9

I know that's not the choices, but just what I'm using to figure out how many.

If you have developed a model, than you can make a function. Test the function afterwards.
Say you use n numbers, you pick x numbers, then your function is like f(n,x)=... math.

The function needs to get an appropriate name, you have to write a comment what it does and what it returns.
The paramaters also should be defined and get restrictions. Special cases are often zero.
There you may find oddities in mathematics, that require "if".

I'm just wandering for a little test is all and wanted a second or more account to ensure I'm right on it.

The hard thing is to define the function in mathematical writing. You already may find it difficult to formulate it in simple words.

Thanks a bunch!!!

The one man bunch.

Post office scriptum: If you can formulate your question properly, the answer is easier to give.

"... I suggest that you get up earlier!" - Arnold

Thanks again SergeM!!!

That is what I was asking for. In ascending order yes. LOL.

There would only be 24 numbers in total. Four for each position only. Just as you have it with the 4 - 8 - 12  and so on, there would be variations of course. I was only asking with the 1-9 to get a base idea .

There would only be four distinct numbers per position. They would only be per draw and not for future draws. So if I had 1-4 for the first ball, 2-5 for the second and so on, that's the only numbers per position to choose from and/or play.

There would be overlapping. If you have 1-2-3-4-5-6, but 7 was a choice for the fifth digit and none higher than 7 for the sixth, it just wouldn't be used for a combination. Sometimes there will be less than four per set also, and I know that drops the combos down, but looking for a base ground.

The ball set would vary per game. I was going to test this on the Texas lotto and a couple more, so around 55 to 60 balls in total to start with.

I just about have the charts finished that will leave what to play (I hope). I'm still very new to excel and no where near capable to write any VBA or anything like that, but I get done what I want. Learning as I go is about all I can do. And when I have time, so I'll just have to play around with getting it to actually lay out the sets thru formulation. I'll figure it out though.

LOL. Words to me are like pistols to Quigley. "Didn't say I don't know how to use them, just don't like to." Good thing I have patience when it comes to working on this is all I can say. I'll re-write it till it does what I want it to.

I thought it was the one man gang? I thought he passed away, but guess he's still kicking in SC.

If you go all night, your pretty much up early aren't ya? Wasn't Arnold so much better before the politics? Or was he just able to hide it better?

Thanks SergeM for taking the time to reply and any help!!!

You can't steal second and keep your foot on FIRST!!!

When you go through hardships and decide not to surrender, that is strength”.

-Arnold (Ahnald) Schwarzenegger-

Burnsville
United States
Member #107244
March 4, 2011
853 Posts
Offline
 Posted: August 8, 2013, 4:50 pm - IP Logged

Hey there Fox. See if this is what your thinking or looking for
their is a formula to work this out, its called combin, below will work out you combinations for you which will be 15
The first part is the is number of items, the second is the number of combination required
=COMBIN(6,4)
Regards
Pup

Sup Sky?

I'm just about there with getting to that point. If it does what I hope it will, that part will make it easier for sure. I'll figure out something though.

I'll check in with you later....

Thanks for the reply and info!!!

You can't steal second and keep your foot on FIRST!!!

When you go through hardships and decide not to surrender, that is strength”.

-Arnold (Ahnald) Schwarzenegger-

Economy class
Belgium
Member #123700
February 27, 2012
4035 Posts
Offline
 Posted: August 8, 2013, 6:02 pm - IP Logged

If you use covermaster, you can copy a full wheel to the clipboard, and paste it where you want to have it, or just save as text file, later import the text file to Excel or other spreadsheet program.

When you have the wheel in Excel, activate the filters, mark the numbers you want to see displayed. When you like the result, select the data and paste it on the next sheet.

When it doesn't fit on one sheet, you can split the data on several sheets.

Another option is to filter a conditioned wheel to the point that will fit on the sheet. I tried it, and can say that it may come out very efficient.

If you want to try AON, you might use Jade's wheel machine, put the wheel matrix in Excel and link your numbers. After that use your filters.

Burnsville
United States
Member #107244
March 4, 2011
853 Posts
Offline
 Posted: August 9, 2013, 3:42 am - IP Logged

If you use covermaster, you can copy a full wheel to the clipboard, and paste it where you want to have it, or just save as text file, later import the text file to Excel or other spreadsheet program.

When you have the wheel in Excel, activate the filters, mark the numbers you want to see displayed. When you like the result, select the data and paste it on the next sheet.

When it doesn't fit on one sheet, you can split the data on several sheets.

Another option is to filter a conditioned wheel to the point that will fit on the sheet. I tried it, and can say that it may come out very efficient.

If you want to try AON, you might use Jade's wheel machine, put the wheel matrix in Excel and link your numbers. After that use your filters.

Thank you for the information SergeM!!! I'm not sure what covermaster is, but I'll check into it. I've never really used much other than what my peabrain has come up with on this stuff.

I've read some about wheels on here, but have never used any as of yet. Been thinking about trying them for some time also, just never got around to it. I've always went at them with a direct win or no win with as little as invested monetarily-wise as possible. Not that worried about the 'ole mighty dollar, just for personal interest of wanting to do it that way. No more sets than I've ever played at once, I could match them up fairly quickly by hand on the slip. What I'm doing now, a wheel would come in handy dandy. I'm thinking about an even/odd, up/down from previous draw only to test with. That would leave me four digits per position, unless the data said It should be less.

I've got this copied to a notepad so I can get to it and learn how. When I work on my charts and stuff, I disconnect from the net all together and that way I will have it at hand.

I don't have anything set up for the AON game just yet, but guess I could. With what I do, it would take a farily large file to make one. Shouldn't be real complicated to do so, just time consuming for a draw with 12 positions. Right at finished with what I was doing for the 2 Step, 6 draws and 5 draws, so will probably start one for that next.

Again, I really appreciate the information and help!!!

You can't steal second and keep your foot on FIRST!!!

When you go through hardships and decide not to surrender, that is strength”.

-Arnold (Ahnald) Schwarzenegger-

Economy class
Belgium
Member #123700
February 27, 2012
4035 Posts
Offline
 Posted: August 9, 2013, 4:06 am - IP Logged

You can't compare to pick 3.

1-4, pick 3:

 N1 N2 N3 1 1 1 1 1 2 1 1 3 1 1 4 1 2 1 1 2 2 1 2 3 1 2 4 1 3 1 1 3 2 1 3 3 1 3 4 1 4 1 1 4 2 1 4 3 1 4 4 2 1 1 2 1 2 2 1 3 2 1 4 2 2 1 2 2 2 2 2 3 2 2 4 2 3 1 2 3 2 2 3 3 2 3 4 2 4 1 2 4 2 2 4 3 2 4 4 3 1 1 3 1 2 3 1 3 3 1 4 3 2 1 3 2 2 3 2 3 3 2 4 3 3 1 3 3 2 3 3 3 3 3 4 3 4 1 3 4 2 3 4 3 3 4 4 4 1 1 4 1 2 4 1 3 4 1 4 4 2 1 4 2 2 4 2 3 4 2 4 4 3 1 4 3 2 4 3 3 4 3 4 4 4 1 4 4 2 4 4 3 4 4 4

1-4, combination lotto type, pick 3

 N1 N2 N3 1 2 3 1 2 4 1 3 4 2 3 4

If you include positional by drawing:

 N1 N2 N3 1 2 3 1 2 4 1 3 2 1 3 4 1 4 2 1 4 3 2 1 3 2 1 4 2 3 1 2 3 4 2 4 1 2 4 3 3 1 2 3 1 4 3 2 1 3 2 4 3 4 1 3 4 2 4 1 2 4 1 3 4 2 1 4 2 3 4 3 1 4 3 2

That doesn't make much sense,
because there is no straight only boxed.
If it can help for selection then it is okay,
but that will probably fool you.

Wheels are reduced: 1 to 6 can be:
1 2 3
4 5 6
If 3/6 come out, you have a 2/3 guarantee for 3/6 drawn,
shortly 2if3.

Burnsville
United States
Member #107244
March 4, 2011
853 Posts
Offline
 Posted: August 9, 2013, 4:55 am - IP Logged

You can't compare to pick 3.

1-4, pick 3:

 N1 N2 N3 1 1 1 1 1 2 1 1 3 1 1 4 1 2 1 1 2 2 1 2 3 1 2 4 1 3 1 1 3 2 1 3 3 1 3 4 1 4 1 1 4 2 1 4 3 1 4 4 2 1 1 2 1 2 2 1 3 2 1 4 2 2 1 2 2 2 2 2 3 2 2 4 2 3 1 2 3 2 2 3 3 2 3 4 2 4 1 2 4 2 2 4 3 2 4 4 3 1 1 3 1 2 3 1 3 3 1 4 3 2 1 3 2 2 3 2 3 3 2 4 3 3 1 3 3 2 3 3 3 3 3 4 3 4 1 3 4 2 3 4 3 3 4 4 4 1 1 4 1 2 4 1 3 4 1 4 4 2 1 4 2 2 4 2 3 4 2 4 4 3 1 4 3 2 4 3 3 4 3 4 4 4 1 4 4 2 4 4 3 4 4 4

1-4, combination lotto type, pick 3

 N1 N2 N3 1 2 3 1 2 4 1 3 4 2 3 4

If you include positional by drawing:

 N1 N2 N3 1 2 3 1 2 4 1 3 2 1 3 4 1 4 2 1 4 3 2 1 3 2 1 4 2 3 1 2 3 4 2 4 1 2 4 3 3 1 2 3 1 4 3 2 1 3 2 4 3 4 1 3 4 2 4 1 2 4 1 3 4 2 1 4 2 3 4 3 1 4 3 2

That doesn't make much sense,
because there is no straight only boxed.
If it can help for selection then it is okay,
but that will probably fool you.

Wheels are reduced: 1 to 6 can be:
1 2 3
4 5 6
If 3/6 come out, you have a 2/3 guarantee for 3/6 drawn,
shortly 2if3.

Hey again SergeM. The 3 and 4 is a total different beast. I tried to set some stuff up on it the same way as I do some of the other games, but the repeating in the same draw just took out any ideas I had for it. When trying to use it for what I do on the others anyway. I set up the tracking charts for them, but what I use for the larger draws wasn't happening.  Not with my limited excel skills anyway.

My main focus was the Powerball. That's where all my work stems from. I just kind of moved over into the 3 and 4 to try to pick up some wins here and there and see what I could do with it. On these games I only play one set of numbers per draw, (minus the couple of times I did pick two sets and showed on here), and it's win or lose. On the PB or jackpot games, it's never more than 5 picks. Not since the first time I tried to pick win anyway.

No, I would never have any reason to use a wheel on the 3 or 4 draws. It really wouldn't make much sense. Especially not the way I play. It would be strictly for any larger games where double digits are concerned. The only thing I really use on the 3 and 4 is what is on my other post. That crazy jumbled up hot mess I've been trying  to explain in hopes it will just hit me on what exactly does what. I can start to understand them better when I'm explaining what I see.

Thank you again for you time and help!!!

You can't steal second and keep your foot on FIRST!!!