Loto 49/6
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Loto 49/6

how to make all permutations of 5 garanteded in permutations of 6 numbers to
49 numbers?

ex permutations of 6 numbers:

1 2 3 4 5 6
1 2 3 4 5 7
1 2 3 4 5 8
1 2 3 4 5 9
etc..

5 garanteded in permutations of 6 numbers:

?????????????????????

Sun, 24 Jul 2005 04:20:55 GMT
Loto 49/6
You do realise don't you that this will yield millions of combinations? Why
do you need to do it? Whatever your reasons, there's probably a much easier
approach.

Rob

Quote:
> how to make all permutations of 5 garanteded in permutations of 6 numbers
to
> 49 numbers?

> ex permutations of 6 numbers:

> 1 2 3 4 5 6
> 1 2 3 4 5 7
> 1 2 3 4 5 8
> 1 2 3 4 5 9
>  etc..

> 5 garanteded in permutations of 6 numbers:

> ?????????????????????

Sun, 24 Jul 2005 07:59:06 GMT
Loto 49/6
i know that there are millions of permutations
but you know the code to do that?

Tue, 26 Jul 2005 01:48:30 GMT
Loto 49/6
Actually, the method you specify for listing the number of possible "sets of
six" from a "set of of 49" is not correct, because if you carry on your list
in the way that you have started it you will end up with all possible
arrangements of six objects from a set of 49 objects, which will include
many, many cases where the same set of six objects is presented in all the
possible ways it can be. In other words, if you start with a list of 49
numbered lottery balls and draw six balls at random then your entire list
will contain that specific set of six in the exact order in which they were
drawn. The total number of these combinations is 49 * 48 * 47 * 46 * 45 *
44, which is 10068347520. However, in a lottery you just need the six drawn
numbers, and you do not need them in a specified order. Since any set of six
objects can be arranged in 720 different ways (6 * 5 * 4 * 3 * 2 * 1) then
the total number of possible arrangements (10068347520 as already stated)
will contain *all 720 arrangements* of each set of six numbers. This means
that in order to guarantee that you will get the six drawn numbers you would
have to buy 10068347520 divided by 720 tickets. That is 13983816 tickets
(nearly four{*filter*} million tickets). Now we come to examine what happens if
you only need five of these six numbers to be correct . . . Oh {*filter*} . . .
I've lost interest now. I've just realised how much chance I've really got
of winning the lottery this week!

Mike

Quote:
> how to make all permutations of 5 garanteded in permutations of 6 numbers
to
> 49 numbers?

> ex permutations of 6 numbers:

> 1 2 3 4 5 6
> 1 2 3 4 5 7
> 1 2 3 4 5 8
> 1 2 3 4 5 9
>  etc..

> 5 garanteded in permutations of 6 numbers:

> ?????????????????????

Tue, 26 Jul 2005 21:37:48 GMT
Loto 49/6

rdlemSPAMOK.freeserve.co.uk> writes

Quote:
>Actually, the method you specify for listing the number of possible "sets of
>six" from a "set of of 49" is not correct, because if you carry on your list
>in the way that you have started it you will end up with all possible
>arrangements of six objects from a set of 49 objects, which will include

Just my one pennyworth added to Mike's lucid explanation. Think
'combinations' not 'permutations'. They're not the same.

Regards.

--
Martin Trump

Wed, 27 Jul 2005 01:35:33 GMT
Loto 49/6

Quote:

> i know that there are millions of permutations
> but you know the code to do that?

Hi there...
Try this code..... I am sure you can improve on it
But its a start

Good luck
George

----------------------------------------------------------
Option Explicit

Private Sub cmdGo_Click()

'This program displays the combinations
'of n items taken k at a time

'   On a form, place
'   2 text boxes    Text1, Text2
'   1 button        cmdGo
'   2 list boxes    List1(0), List1(1)
'   2 Labels        Label1, Label2
'
'   Text1 is n, Text2 is k;  start with Text1.Text=5 and Text2.Text=3
'   Program Limitation: Memory

Dim n%, k%
Dim D\$, d1\$, lst\$
Dim i%, j%, l%, ii%
Dim Sum1#, Sum2#, Num&
Dim Times#
Dim ret&

List1(1).Clear

n = Val(Text1.Text)
k = Val(Text2.Text)
If k > n Then Exit Sub
If k = 0 Then k = n

Sum1 = 1
Sum2 = 1

'   n!/(n-k)!
For i = n To n - k + 1 Step -1
Sum1 = Sum1 * i
Next

'   k!
For i = k To 1 Step -1
Sum2 = Sum2 * i
Next

'   Number of combinations
Num = Sum1 / Sum2
Label1.Caption = Num & " combinations"

If Num >= 6430 Then
ret = MsgBox("There are " & Num & " combinations" & vbCrLf &
"Do you want to go on ?", vbYesNo)
If ret = 7 Then Exit Sub
End If

Times = Timer

For i = 1 To n
Next

ii = 1
For j = 2 To k
lst = j Mod 2
List1(lst).Clear

For i = 1 To n
For l = 0 To List1((lst + 1) Mod 2).ListCount - 2
D = List1((lst + 1) Mod 2).List(l + ii)
d1 = Format(i, "00 ")
If i <= Val(Left\$(D, 3)) And InStr(D, d1) = 0 Then
List1(lst).AddItem Format(i, "00 ") & D
End If
Next
Next
ii = ii + 1
Next

Times = Timer - Times
Label2.Caption = Format(Times, "0.0000") & " sec"

lst = j Mod 2
List1(lst).Clear
lst = (j + 1) Mod 2
List1(lst).ZOrder
List1(lst).Refresh
End Sub

Fri, 29 Jul 2005 06:25:58 GMT
Loto 49/6

Quote:
> i know that there are millions of permutations
> but you know the code to do that?

Try the following...

Option Explicit
Private Sub Command1_Click()

Dim a As Long, b As Long, c As Long
Dim d As Long, e As Long, f As Long
Dim n As Long, count As Long

n = 49  '6 from 49 lotto
For a = 1 To n - 5
For b = a + 1 To n - 4
For c = b + 1 To n - 3
For d = c + 1 To n - 2
For e = d + 1 To n - 1
For f = e + 1 To n
count = count + 1
Print "No"; count; ":"; a; b; c; d; e; f
Print , "subsets of 5 From 6"
Print , a; b; c; d; e
Print , a; b; c; d; f
Print , a; b; c; e; f
Print , a; b; d; e; f
Print , a; c; d; e; f
Print , b; c; d; e; f
Print "------------------------"

If count > 4 Then Exit Sub 'demo only, change to suit
Next f, e, d, c, b, a
Print count
End Sub

Sat, 30 Jul 2005 06:58:15 GMT
Loto 49/6
Right. We're getting somewhere now. Your code (correctly, I think)
calculates that there are 83,0902,896 distinct combinations of "any five
from 6" drawn from a total set of 49, and it correctly lists them. That's
more of a result than I have achieved so far! Additionally, my own posting
tells the OP that there are a total of 13,983,816 combinations of "any six
from 49". That means that it is six times easier to get "five from six" than
it is to get all six. Sounds good to me. So, since each of these
combinations of "any six from 49" itself contains a total of six separate
"sets of five" (6*5*4*3*2*1 / 5*4*3*2*1) then it follows (or does it?) that
the number of tickets the OP needs to buy to guarantee that he has a set of
five from his six "ticked" ball numbers (drawn from a full set of 49 balls)
is 13,983,816 / 6 = 2,330,636 tickets. That's a lot of tickets, and I
suspect that the OP wants to know which set of six numbers to write on each
of them! {*filter*}! My head is really spinning now! Hope I've got this stuff
right! Over to you, Michael ;-)

But wait! Surely that can't be right, because it would mean that the chances
of getting all 6 of the six drawn numbers after ticking six of the 49 boxes
would actually be better than the chances of getting just five of the six
drawn numbers together with a seventh drawn "bonus" ball! Surely that can't
be right! They pay lots more cash for "all six" than they do for "all five
from six plus the bonus ball" {*filter*}! My head really hurts now!

Perhaps I should drink this Budweiser a little more slowly. Or maybe even go
back to my old "standby" of whisky and Coke. Gets you drunk even more
quickly, and you don't have to go to the toilet so often! Lovely stuff :-)

Mike

Quote:

> > i know that there are millions of permutations
> > but you know the code to do that?

> Try the following...

> Option Explicit
> Private Sub Command1_Click()

> Dim a As Long, b As Long, c As Long
> Dim d As Long, e As Long, f As Long
> Dim n As Long, count As Long

> n = 49  '6 from 49 lotto
> For a = 1 To n - 5
> For b = a + 1 To n - 4
> For c = b + 1 To n - 3
> For d = c + 1 To n - 2
> For e = d + 1 To n - 1
> For f = e + 1 To n
> count = count + 1
>     Print "No"; count; ":"; a; b; c; d; e; f
>     Print , "subsets of 5 From 6"
>     Print , a; b; c; d; e
>     Print , a; b; c; d; f
>     Print , a; b; c; e; f
>     Print , a; b; d; e; f
>     Print , a; c; d; e; f
>     Print , b; c; d; e; f
> Print "------------------------"

>     If count > 4 Then Exit Sub 'demo only, change to suit
> Next f, e, d, c, b, a
> Print count
> End Sub

Sat, 30 Jul 2005 02:17:48 GMT
Loto 49/6
Thanks for your help but its not like you say
Its hard to me explain this in englis but ill try

there are 13983816 combinations of 6 in 49 numbers
there are 1906884 combinations of 5 in 49 numbers

with 6 numbers we have 6 combinations of 5

exemple:
1,2,3,4,5,6:

1,2,3,4,5
1,2,3,4,6
1,2,3,5,6
1,2,4,5,6
1,3,4,5,6
2,3,4,5,6
So we have 1906884 / 6 = 317814 results for 5 combinations using 6 numbers
to 49 total numbers
i wrote a program that gives all combinations of 6 numbers in 49
What i want to do know is a program that gives me all combinations
of 5 with 6 numbers for 49 in total
it maybe looks like this:

1  2  3  4  5  6
1  2  3  4  7  8
1  2  3  4  9 10
1  2  3  4 11 12
1  2  3  4 13 14
1  2  3  4 15 16
1  2  3  4 17 18
1  2  3  4 19 20
1  2  3  4 21 22
1  2  3  4 23 24
1  2  3  4 25 26
1  2  3  4 27 28
and so one..

my program that gives all combinations of 6 numbers to 49 is:

Do While x1 <= 44
Do While x2 <= 45
Do While x3 <= 46
Do While x4 <= 47
Do While x5 <= 48
Do While x6 <= 49
x6 = x6 + 1
Loop
x5 = x5 + 1
x6 = x5 + 1
Loop
x4 = x4 + 1
x5 = x4 + 1
x6 = x5 + 1
Loop
x3 = x3 + 1
x4 = x3 + 1
x5 = x4 + 1
x6 = x5 + 1
Loop
x2 = x2 + 1
x3 = x2 + 1
x4 = x3 + 1
x5 = x4 + 1
x6 = x5 + 1
Loop
x1 = x1 + 1
x2 = x1 + 1
x3 = x2 + 1
x4 = x3 + 1
x5 = x4 + 1
x6 = x5 + 1
Loop

Sat, 30 Jul 2005 08:18:44 GMT
Loto 49/6

Quote:
> Right. We're getting somewhere now. Your code (correctly, I think)
> calculates that there are 83,0902,896 distinct combinations of "any five
> from 6" drawn from a total set of 49, and it correctly lists them. That's
> more of a result than I have achieved so far! Additionally, my own posting
> tells the OP that there are a total of 13,983,816 combinations of "any six
> from 49". That means that it is six times easier to get "five from six"
than
> it is to get all six. Sounds good to me. So, since each of these
> combinations of "any six from 49" itself contains a total of six separate
> "sets of five" (6*5*4*3*2*1 / 5*4*3*2*1) then it follows (or does it?)
that
> the number of tickets the OP needs to buy to guarantee that he has a set
of
> five from his six "ticked" ball numbers (drawn from a full set of 49
balls)
> is 13,983,816 / 6 = 2,330,636 tickets. That's a lot of tickets, and I
> suspect that the OP wants to know which set of six numbers to write on
each
> of them! {*filter*}! My head is really spinning now! Hope I've got this stuff
> right! Over to you, Michael ;-)

> But wait! Surely that can't be right, because it would mean that the
chances
> of getting all 6 of the six drawn numbers after ticking six of the 49
boxes
> would actually be better than the chances of getting just five of the six
> drawn numbers together with a seventh drawn "bonus" ball! Surely that
can't
> be right! They pay lots more cash for "all six" than they do for "all five
> from six plus the bonus ball" {*filter*}! My head really hurts now!

> Perhaps I should drink this Budweiser a little more slowly. Or maybe even
go
> back to my old "standby" of whisky and Coke. Gets you drunk even more
> quickly, and you don't have to go to the toilet so often! Lovely stuff :-)

> Mike

Mike, I spotted the error in your logic straight away! Although your move
from whiskey & coke to Bud was sound, you should have further upgraded
from Bud to Toohey's New (bottled, not canned). This makes for smoother
mellower coding sessions and gives searing insights into combinatronics!
Oh, and the lotto..
I use hypogemetric function (in Excel but easily whipped up in VB) for lotto
and keno computations. So odds of matching 5 of our selected six are..
6C5 *times(49-6)C(6-5) divided by 49C6 = (6*43)/13983816.
Now, 5 of our 6 numbers have been matched. What are the odds that our
6th number will be the supplementary? There are only 43 left in the barrel
so
odds of 5 plus supp are...(6*43)/(13983816*43) = 6/13983816.

The way I see it, if the OP wants to be certain of matching 5 from six he
must still blanket the lotto with all 13,983,816 tickets but will be
consoled by
getting the major prize once and 5 out of 6 some 43 times!
Cheers Mick.

Sat, 30 Jul 2005 20:25:24 GMT
Loto 49/6
Yes. The fog is lifting. Now I can see exactly what you mean. Everything
suddenly becomes clear.  This Budweiser really *is* too "watery". Toohey's
New (bottled) *is* much smoother than Bud. A delicious, smooth, creamy
flavour that tantalises the taste buds and certainly seems to reach the
parts that other beers just cannot reach. Visual Basic code flows much more
freely. Even some of the MSDN help files begin to make sense. I wonder why
they didn't include a mention of this wonderful Toohey's beverage in Service
Pack 5? Would have solved a lot of problems.

Mike

Quote:

> Mike, I spotted the error in your logic straight away! Although your move
> from whiskey & coke to Bud was sound, you should have further upgraded
> from Bud to Toohey's New (bottled, not canned). This makes for smoother
> mellower coding sessions and gives searing insights into combinatronics!
> Oh, and the lotto..
> snip <

Sat, 30 Jul 2005 14:33:25 GMT

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