I have a math question

Reply #22

dr san — Sat, 02 Jan 2016 10:52:47 GMT

Jaden, you can create a macro on the following criteria

Most = 02 = 01 Less Equal = 00 Always the last draw for the next Example

15 23 35 45 02 52 02 code

05 18 29 38 48 59

already is 01 (less)

01 13 20 32 42 51 or other smaller or the larger numbers also

Reply #21

SergeM — Fri, 01 Jan 2016 19:28:58 GMT

Systems extended

Reply #20

lakerben — Fri, 01 Jan 2016 00:02:08 GMT

Impressive!

Happy new year Jade

Reply #19

JADELottery — Thu, 31 Dec 2015 21:52:58 GMT

Yep.

Hence, the files we made.

The files can be found here:

Excel 2007 - ftp://www.jadexcode.com/Excel/CombinatorialIndex.xlsm

Excel 2003 - ftp://www.jadexcode.com/Excel/CombinatorialIndex.xls

These have function description and usage in the Function Information tab.

=Fact(N)

Returns -

- An integer

- The factorial of N, mathematically written as ' N! '

- N!

- N (N - 1) (N - 2) 3 2 1

- 0! = 1

=Perm(N, R)

Returns -

- An integer... [ More ]

Reply #18

A649KING2B — Thu, 31 Dec 2015 21:12:46 GMT

LOL, that looks more complicated than Chinese to me.

Im still counting with my fingers.

Reply #17

SergeM — Thu, 31 Dec 2015 20:41:55 GMT

I think that ranges are just too easy.

Do a mix such as:

1 3 5 7 8

6 7 8 9 10

9 15 16 17 20

30 34 38

This pure mixing, not by indeces and you can leave out the rotten apples!

I make this on LINK with groups:

Input: [[1, 3, 5, 7, 8], [6, 7, 8, 9, 10], [9, 15, 16, 17, 20], [30, 34, 38]].

Count of combinations: 315. Limit display: 3000.

1 10 15 30

1 10 15 34

1 10 15 38

1 10 16 30

1 10 16 34

1 10 16 38

1 10 17 30

1 10 17 34

1 10 17 38

1 10 20

Reply #16

JADELottery — Thu, 31 Dec 2015 20:22:45 GMT

Thanks.

Here's the Excel VB code we used to generate the combinations.

____________________________________________________________________________________________________

Function Fact(ByVal n As Integer) As Double

On Error GoTo exitfunction

If (n = 1) Then

Fact = 1

Else

Fact = n * Fact(n - 1)

End If

Exit Function

exitfunction:

Fact = -1

End Function

Function Perm(ByVal n As Integer, ByVal R As Integer) As Double

On Error GoTo exitfunction

Dim... [ More ]

Reply #15

A649KING2B — Thu, 31 Dec 2015 19:34:45 GMT

Hey, ya'll, Thanks for helping me out. I figured it out by cutting and pasting with excel, took me 6 and a half hours, which is WHY I had to ask for help.

Looks like JADELottery had the correct answer to my question. = 39,389,

later on when I am done with my little project, ill let you in on what i figured out and Ill ask for your opinions on my findings, Thanks again

.

Reply #14

winsumloosesum — Wed, 30 Dec 2015 17:30:09 GMT

Here is an Excel file with the 4/38 game.

https://copy.com/SWZQstJViQOUPHns

You can filter by position by entering your numbers in the Dashboard2 worksheet and clicking the toolbar.

I would think the number of combinations left is based on what numbers you play by position

There are 2 Excel files

Create a new folder and unzip the contents into your new folder. You can name the folder anything you want.

You will need Excel version 2007 and newer for this to work since there are

Reply #13

JADELottery — Wed, 30 Dec 2015 13:50:13 GMT

'4 separate pools.' Meaning no balls are common between pools.

pool 1 - 11 numbered balls for first position, Balls: 01, 02, 03, 04, 05, 06, 07, 08, 09, 10, 11

pool 2 - 10 numbered balls for second position, Balls: 12, 13, 14, 15, 16, 17, 18, 19, 20, 21

pool 3- 23 numbered balls for third position, Balls: 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44

pool 4- 19 numbered ball for forth position, Balls: 45, 46, 47, 48, 49, 50, 51, 52, 53, 54... [ More ]

Reply #12

DCTreybil — Wed, 30 Dec 2015 13:24:20 GMT

If you're drawing numbers in these ranges from 4 different pools, the total number of possible combinations is yielded by simple multiplication:

11 X 11 X 22 X 16 = 42,592

If you draw 4 numbers in these ranges from a single pool with a maximum value of 38, the total number of possible combinations becomes 2,910.

Dim count As Double

Dim position_1 As Integer

Dim position_2 As Integer

Dim position_3 As Integer

Dim position_4 As Integer

' 1 to 11 for fir... [ More ]

Reply #11

A649KING2B — Wed, 30 Dec 2015 02:58:51 GMT

So not to confuse more than I already have, Ill rephrase the question. Even though this is not the actual scenario, So, say 4 separate pools.

pool 1 - 11 numbered balls for first position

pool 2 - 10 numbered balls for second position

pool 3- 23 numbered balls for third position

pool 4- 19 numbered ball for forth position

= how many combinations? , Thanks

.

.

Reply #10

SergeM — Tue, 29 Dec 2015 18:24:41 GMT

Use indexes:

high index - low index + 1

You can do this with a calculator. You may need pen and paper too.

Reply #9

DCTreybil — Tue, 29 Dec 2015 00:12:19 GMT

I typed some code into the Form_Load event of a Visual Basic program.

The first set of nested loops (without indents here) double checks the 4 elements with a maximum value of 38 result. Result is verified by this code and my combinations calculator app.

The second set of nested loops includes a set of nested if-then checks for the values of individual elements to count only those matching your question. I also get 39,389 for that.

If you have access to a VB programming environ... [ More ]

Reply #8

lakerben — Mon, 28 Dec 2015 16:29:53 GMT

Thanks

Reply #7

JADELottery — Mon, 28 Dec 2015 04:05:12 GMT

There's only 73,815 combinations.

We ran through all of them using our =Index2Combin(N, R, Index) function in Excel.

The files can be found here:

Excel 2007 - ftp://www.jadexcode.com/Excel/CombinatorialIndex.xlsm

Excel 2003 - ftp://www.jadexcode.com/Excel/CombinatorialIndex.xls

Here are all the combinations that meet your criteria.

01 05 12 19 02 05 12 19 03 05 12 19 04 05 12 19 05 06 12 19 06 07 12 19 07 08 12 19 08 09 12 19 09 10 12 19 10 11 12 19 11 12 13 19

01 05 12 2... [ More ]

Reply #6

RJOh — Mon, 28 Dec 2015 02:19:36 GMT

The lowest ball can always be in a range of 1 to 34 and the second highest in a range of 2 to 35 but neither can be in both.

Reply #5

RJOh — Mon, 28 Dec 2015 02:11:45 GMT

1 to 11 for first position, 5 to 24 for second position, 12 to 34 for third position, 19 to 37 for fourth position don't include all the possible combinations or it would.

Reply #4

tdottlotto — Sat, 26 Dec 2015 23:45:47 GMT

your first pick 4 question (googled 38 choose 4 gave 73,815) is correct for when 4 balls are coming from the same pool.

But your second question is unclear. Is each ball coming from a separate pool? If each ball is coming from a separate pool then you just multiple the size of pools together (11*19*23*19) there's no need to divided by 4/3/2/1.

If they're coming from the same pool..then i don't see how the first ball can be from a different range than the second ball and so on.... [ More ]

Reply #3

ithastobesaid — Wed, 23 Dec 2015 21:10:24 GMT

You are both wrong. Determining the # of unique combinations from picking 4 balls from 38 is a factorial problem.

In excel use =combin(38,4)

Or: (38 * 37 * 36 * 35) / (4 * 3 * 2 * 1) = 73,815

Reply #2

A649KING2B — Wed, 23 Dec 2015 20:14:03 GMT

Wow, I thought formula would be similar to first question, 11 numbers x 20 numbers x 23 numbers x 19 numbers divided by 4x3x2x1 = 4006.

May I ask formula to use? Thanks

Reply #1

JADELottery — Wed, 23 Dec 2015 18:24:05 GMT

Pick 4, 1 to 38, ? - ? - ? - ?, = 73,815 combinations, is this correct ?

Yes.

Pick 4, ?-?-?-?, 1 to 11 for first position, 5 to 24 for second position, 12 to 34 for third position, 19 to 37 for fourth position, = ? combinations.

is it 4006?

No.

it is 39,389

I have a math question

A649KING2B — Wed, 23 Dec 2015 05:22:41 GMT

Some help would be greatly appreciated, Thanks in advance.

Pick 4, 1 to 38, ? - ? - ? - ?, = 73,815 combinations, is this correct ?

Pick 4, ?-?-?-?, 1 to 11 for first position, 5 to 24 for second position, 12 to 34 for third position, 19 to 37 for fourth position, = ? combinations.

is it 4006