lottery probability - Page 2 - Gambling and Probability

That script I posted wasn't right since you are choosing 10 numbers. It shifts the numbers by 2, subtracts, and looks for a difference of 2, but then it needs to check positions 3:10 not 3:6. I made a new one below which should also be able to do more than 3 consecutive. It can do r consecutive for c(n,k) by shifting by r-1, subtracting, and checking for a difference of r-1 from r to k. The answer for c(20,10) with 3 consecutive is coming out to 87% now. That is 160688 total. So

C(17,7) + 17*C(16,7) - 53240 = 160688

to answer your question.

I thought about the recursion, but I didn't see an easy way to do it. The placement of the first 3 affects the calculation for the others. You have to consider that some can come before or after. With C(49,6) it was easy because once you place the first 3, there are only 3 more that can come before with no extras. I'll let you know if I think of a way. Can you just use the R script to compute the ones you need? You could put them in a table. Why do you need a formula?

Code:

n = 20
k = 10
r = 3

consec = function(x) {
  any((c(x,rep(0,r-1)) - c(rep(0,r-1),x))[r:k] == (r-1))
}

combs = combn(n,k)
sum(apply(combs, 2, consec))/choose(n,k)

Output:

[1] 0.8697309

Last edited by BruceZ; 12-22-2012 at 03:08 AM. Reason: combn(20,10) -> combn(n,k)

Quote

12-22-2012 , 03:33 AM

#27

BruceZ

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I made this into a function that you can call. For example, just enter consec(20,10,3).

> consec(20,10,3)
[1] 0.8697309

Code:

consec = function(n,k,r) {
  combs = combn(n,k)
  sum(apply(combs, 2, shiftsub, k = k, r = r)) / choose(n,k)
}

shiftsub = function(x,k,r) {
  any((c(x,rep(0,r-1)) - c(rep(0,r-1),x))[r:k] == (r-1))
}

Quote

12-22-2012 , 04:36 AM

#28

Mirapep

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Join Date: Dec 2012 Posts: 53

Although I have difficulty with the language I will try to explain. I plan beginner in visual basic 6 and I need to place money management in my own software. Now I have no need to develop columnar, which slows down the processor, but only a number. In fact, I gave an example of C (20,10), but C (k, n) may be high.
You need to explain how does the money management and it is not easy. I'll try with a simple example. Suppose that:
- Total events k = 20,
- At least n = 10 winning events,
- Equal share for each event p = 2, for simplicity, but it could be different for each event,
- Cash available s = $ 100

Now we have to calculate the overall performance of our system x and t determine the amount of the first bet:

for a = n to k
y = y + C (k, a) * 1 / p ^ a * (1-1 / p) ^ (k-a)
next

x = 1 / y

hours run this operation for determining t

k = k - 1

for a = n to k
y1 = y1 + C (k, a) * 1 / p ^ a * (1-1 / p) ^ (k-a)
next

x1 = 1/y1

t = (1 - x/x1) * s

Then, in the outcome of the first bet are calculated after t

I hope I have made mistakes, work is in progress

Now, I would like to increase the yield x inserting some constraints such as by removing the z combinations with at least r consecutive

Therefore .......

for a = n to k
y = y + (C (k, a) - z) * 1 / p ^ a * (1-1 / p) ^ (k-a)
next

...............

Quote

12-22-2012 , 07:22 AM

#29

Mirapep

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Join Date: Dec 2012 Posts: 53

Quote:

Originally Posted by BruceZ

I made this into a function that you can call. For example, just enter consec(20,10,3).

> consec(20,10,3)
[1] 0.8697309

Code:

consec = function(n,k,r) {
  combs = combn(n,k)
  sum(apply(combs, 2, shiftsub, k = k, r = r)) / choose(n,k)
}

shiftsub = function(x,k,r) {
  any((c(x,rep(0,r-1)) - c(rep(0,r-1),x))[r:k] == (r-1))
}

I applied this function, it is perfect and wonderful. I'm sorry to bother you, but I do not know R, can you give me instructions to play it in vb6?
For the function combn I have no problems

Quote

12-23-2012 , 12:09 AM

#30

BruceZ

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Quote:

Originally Posted by Mirapep

I applied this function, it is perfect and wonderful. I'm sorry to bother you, but I do not know R, can you give me instructions to play it in vb6?
For the function combn I have no problems

I've never called an R function from VB, but I found some info on how to do it. There are several options.

http://www.talkstats.com/showthread....unctions-in-VB

http://quantum.uncronopio.org/2005/02/04/114/

It may be faster if you code it in VB. The combn function returns a 2-D array with every combination of numbers in the columns. If you don't have combn, you may be able to find one online. There are a some here. You might have to translate them to VB. Then you have to examine every column, shift, and subtract like this for consec(20,10,3):

Code:

1  3  5  7  8  9 11 13 15 17  0  0
0  0  1  3  5  7  8  9 11 13 15 17
--------------------------------------------------subtract
1  3  4  4  3  2  3  4  4  4 -15 -17

The 2 means that there were 3 consecutive. For r consecutive, shift by r-1 with r-1 zeros, and look for r-1 from index r:k. This is what shiftsub does in 1 line!

I am still working on a formula. I think I'm getting closer. That would be easier, use less memory, and may be faster.

Last edited by BruceZ; 12-23-2012 at 02:38 AM.

Quote

12-23-2012 , 05:36 AM

#31

Mirapep

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Join Date: Dec 2012 Posts: 53

You are an amazing person, I am sincerely grateful for your kindness

Quote

12-23-2012 , 07:05 AM

#32

Mirapep

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Join Date: Dec 2012 Posts: 53

1 3 5 7 8 9 11 13 15 17 0 0
0 0 1 3 5 7 8 9 11 13 15 17

How do you get this?

I have to understand what performs this function:

c (x, rep (0, r-1))

and what values

function (x, k, r)

Quote

12-23-2012 , 02:22 PM

#33

BruceZ

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Quote:

Originally Posted by Mirapep

1 3 5 7 8 9 11 13 15 17 0 0
0 0 1 3 5 7 8 9 11 13 15 17

How do you get this?

I have to understand what performs this function:

c (x, rep (0, r-1))

and what values

function (x, k, r)

x is a single combination of lottery numbers in ascending order that got created by combn and passed to function shiftsub. The function apply applies the function shiftsub to every combination. rep(0,r-1) inserts r-1 zeroes. c(x,rep(0,r-1)) is an array with r-1 zeroes inserted at the end of x. Then c(rep(0,r-1),x) is an array with r-1 zeros and inserted at the beginning of x That shifts x and makes both the same length for subtract. k and r are the values you passed to consec. So for consec(20,10,3), n = 20, k = 10, and r = 3.

Code:

> x = c(1,3,5,7,8,9,11,13,15,17)
> x
 [1]  1  3  5  7  8  9 11 13 15 17
>
> r = 3
> c(rep(0,r-1),x)
 [1]  0  0  1  3  5  7  8  9 11 13 15 17
>
> c(x,rep(0,r-1))
 [1]  1  3  5  7  8  9 11 13 15 17  0  0
>
> c(x,rep(0,r-1))- c(rep(0,r-1),x)
 [1]   1   3   4   4   3   2   3   4   4   4 -15 -17
>
> k = 10
>  (c(x,rep(0,r-1))- c(rep(0,r-1),x))[r:k]
[1] 4 4 3 2 3 4 4 4
>
> any((c(x,rep(0,r-1))- c(rep(0,r-1),x))[r:k] == 2)
[1] TRUE

You get help on any R function like rep or any by typing ?rep or ?any.

Last edited by BruceZ; 12-23-2012 at 03:13 PM.

Quote

12-24-2012 , 11:38 AM

#34

Mirapep

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Join Date: Dec 2012 Posts: 53

Ok, I realized that examines each combination. If it does this is still a fast algorithm. I was thinking of a recursive formula only on the combinations to subtract.
Merry Christmas

Quote

12-24-2012 , 11:40 AM

#35

Mirapep

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Join Date: Dec 2012 Posts: 53

Applying triangle Tartaglia

Quote

12-26-2012 , 03:04 PM

#36

BruceZ

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I've done it! I wrote a recursive program to quickly compute the probability of r consecutive numbers in a row when choosing k numbers from n numbers. I verified that it gives exactly the same numbers as the program that searched every combination for these test cases:

(n,k,r) =

(20,10,3)
(25,10,3)
(49,6,3)
(20,11,3)
(20,12,3)
(20,10,4)
(20,11,4)
(20,12,4)
(20,16,5)

You really need this because the 25 takes several minutes with the old program. The new one does it instantly. (49,6,3) used to take over 20 minutes, and now that is instantaneous.

There's just one problem. I got an error in R that I was nesting too deeply when I tried to use a for loop to continue the pattern for all k. I increased the limit on nesting, but then I got a protection stack size error. So I increased the size of the protection stack when R starts, but then I got an error that the C stack is too close to the limit! I have to try to change that through the OS, but I don't even know if I can. It may just be that I'm low on disk space, so I need to clean up. The first program below works, but you have to add lines for each k. Right now it will do up to k=2r+6, so 10 for r=3, 14 for r=4, 16 for r=5, etc. You should be able to do any n, and any r and k as long as you add enough lines for the k. You could do this automatically using macro expansion. The second program does it automatically with for loops, but I haven't been able to test it yet because of the problems. You should be able to easily translate that into VB and try it. The function choose(n,k) is combin(n,k) in VB. I will also try to get it to work, perhaps in another language besides R which really isn't designed for recursion.

Enter pconsec(20,10,3) for example. If you just want the number instead of the probability, use nconsec(20,10,3).

This one works, but you have to expand it manually for k > 2r+6:

Code:

pconsec = function(n,k,r) { nconsec(n,k,r) / choose(n,k)}

nconsec = function(n,k,r) {
  if (n == k) {s = 1} else {

    s = choose(n-r,k-r) + (n-r)*choose(n-r-1,k-r)

    if (k >= 2*r) {
      for (m in r:(n-k+r-1)) {s = s - nconsec(m,r,r)*choose(n-r-1-m,k-2*r)} }

    if (k >= 2*r+1) {
       for (m in (r+1):(n-k+r)) {s = s - nconsec(m,r+1,r)*choose(n-r-1-m,k-2*r-1)} }

    if (k >= 2*r+2) {
      for (m in (r+2):(n-k+r+1)) {s = s - nconsec(m,r+2,r)*choose(n-r-1-m,k-2*r-2)} }

    if (k >= 2*r+3) {
      for (m in (r+3):(n-k+r+2)) {s = s - nconsec(m,r+3,r)*choose(n-r-1-m,k-2*r-3)} }

    if (k >= 2*r+4) {
      for (m in (r+4):(n-k+r+3)) {s = s - nconsec(m,r+4,r)*choose(n-r-1-m,k-2*r-4)} }

    if (k >= 2*r+5) {
      for (m in (r+5):(n-k+r+4)) {s = s - nconsec(m,r+5,r)*choose(n-r-1-m,k-2*r-5)} }

    if (k >= 2*r+6) {
      for (m in (r+6):(n-k+r+5)) {s = s - nconsec(m,r+6,r)*choose(n-r-1-m,k-2*r-6)} }

### Continue these for higher values of k

  }
  s
}

> pconsec(20,10,3)
[1] 0.8697309
> nconsec(20,10,3)
[1] 160688

This one should do the same thing, but it gives stack errors on my machine:

Code:

pconsec = function(n,k,r) { nconsec(n,k,r) / choose(n,k)}

nconsec = function(n,k,r) {
  if (n == k) {s = 1} else {
    s = choose(n-r,k-r) + (n-r)*choose(n-r-1,k-r)
    for (j in (2*r):k) {
      if (k >= j) {
        i = j-2*r
        for (m in (r+i):(n-k+r+j-1)) {s = s - nconsec(m,r+j,r)*choose(n-r-1-m,k-2*r)}
      }
    }
  }
  s
}

If you want to try this one in R on your machine, first do

>options(expressions = 30000)

The default is 5000 which doesn't work. You can go as high as 500000. Then if that doesn't work, try invoking R with

R --max--ppsize=100000

or even 500000. The default is 10000.

Last edited by BruceZ; 12-26-2012 at 11:37 PM.

Quote

12-26-2012 , 07:53 PM

#37

BruceZ

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I got the short recursive code working now in both VB and R. The stack error was caused by my own bug.

This VB version runs in Excel, so replace Application.Combin with Combin if you use it standalone. These often run very fast, but they will slow down as you increase the ratio of k to r, and when n becomes large.

Code:

Function pconsec(n As Long, k As Long, r As Long)
    pconsec = nconsec(n, k, r) / Application.Combin(n, k)
End Function
    
Function nconsec(n As Long, k As Long, r As Long)
    
    Dim s   As Long
    Dim j   As Long
    Dim i   As Long
    Dim m   As Long

    If (n = k) Then
        nconsec = 1
    Else
        nconsec = Application.Combin(n - r, k - r) + (n - r) * Application.Combin(n - r - 1, k - r)
        For i = 0 To k - 2 * r
            For m = r + i To n - k + r + i - 1
                nconsec = nconsec - nconsec(m, r + i, r) * Application.Combin(n - r - 1 - m, k - 2 * r - i)
            Next m
        Next i
    End If
    
End Function

R version:

Code:

pconsec = function(n,k,r) { nconsec(n,k,r) / choose(n,k)}

nconsec = function(n,k,r) {
  if (n == k) {s = 1} else {
    s = choose(n-r,k-r) + (n-r)*choose(n-r-1,k-r)
    if (k >= 2*r) {
      for (i in 0:(k-2*r)) {
          for (m in (r+i):(n-k+r+i-1)) {s = s - nconsec(m,r+i,r)*choose(n-r-1-m,k-2*r-i)}
      }
    }
  }
  s
}

> pconsec(20,10,3)
[1] 0.8697309
> nconsec(20,10,3)
[1] 160688

Last edited by BruceZ; 12-27-2012 at 02:12 AM.

Quote

12-27-2012 , 07:45 AM

#38

Mirapep

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Join Date: Dec 2012 Posts: 53

Thank you very much for the work you have done for me.
This evening, after returning home, I'll try the functions, you'll know the results

Quote

12-27-2012 , 10:00 AM

#39

BruceZ

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Quote:

Originally Posted by Mirapep

Thank you very much for the work you have done for me.
This evening, after returning home, I'll try the functions, you'll know the results

This will run a lot faster for you if you are using compiled VB. I'm running it in VBA with Excel which isn't compiled. If you do use it in VBA with Excel, I just found a way to speed it up by a factor of 4 by using the With WorksheetFunction statement instead of calling .Application Combin every time. Now it's 4 times faster than R even uncompiled. Speed may not even be an issue for you at all depending on your inputs.

I'm working on a way to get rid of the for loops in R by vectorizing, That should make the R program much faster too.

Code:

Function pconsec(n As Long, k As Long, r As Long)
    pconsec = nconsec(n, k, r) / Application.Combin(n, k)
End Function

Function nconsec(n As Long, k As Long, r As Long)
    
    Dim s   As Long
    Dim j   As Long
    Dim i   As Long
    Dim m   As Long
    
    With WorksheetFunction

        nconsec = .Combin(n - r, k - r) + (n - r) * .Combin(n - r - 1, k - r)
        For i = 0 To k - 2 * r
            nconsec = nconsec - .Combin(n - 2 * r - i - 1, k - 2 * r - i)
            For m = r + i + 1 To n - k + r + i - 1
                nconsec = nconsec - nconsec(m, r + i, r) * .Combin(n - r - 1 - m, k - 2 * r - i)
            Next m
        Next i
       
    End With
    
End Function

Last edited by BruceZ; 12-28-2012 at 02:37 AM. Reason: Removed if

Quote

12-28-2012 , 06:01 AM

#40

Mirapep

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Join Date: Dec 2012 Posts: 53

Hello Bruce,
I tried the function in vb6, it is fast and perfect. However, there seems to be a problem for high values of n and k. I did pc work all night and in the morning he had not yet finished. I also tried in R:
- Nconsec (100,15,5) has provided quick response;
- Nconsec (100,50,5) still work for several minutes.
Is a possible conflict in the routine?

Quote

12-28-2012 , 06:33 AM

#41

Mirapep

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Still working

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12-28-2012 , 11:03 AM

#42

BruceZ

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Quote:

Originally Posted by Mirapep

There is no conflict, it just takes a very long time to compute all the terms when k is a large multiple of r. With the current algorithm, the time increases exponentially with k/r. It goes up by about a factor of 30 for every increase of 5. So (100,20,5) takes 30 times longer than (100,15,5), and (100,25,5) takes 30^2 = 900 times longer than (100,15,5), etc. So (100,50,5) takes about 30^7 = 20 billion times longer than (100,15,5). So even if (20,15,5) took only 0.1 seconds, (100,50,5) would take over 69 years! But if you did (100,50,17), that would be very fast. In fact, (10000,50,17) is very fast. It's also a tiny probability.

I'm working on a much faster algorithm. The current one computes many of the same numbers over and over recursively. I think I can compute them once, save them, and then reuse them. So instead of calling nconsec recursively, those numbers will already be in a precomputed in a table. Then I think the time will only increase linearly, not exponentially. So instead of taking 20 billion times longer, it should only take 35 times longer. The current algorithm is stupid, but it was an important first step to prove that the formula is right.

Last edited by BruceZ; 12-28-2012 at 02:55 PM.

Quote

12-29-2012 , 04:05 AM

#43

BruceZ

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I got the fast routine working in R. It can do (100,15,5) in 40 seconds in R, and in compiled VB it will probably be just a few seconds. The time only increases linearly with k now, not exponentially.

> pconsec(100,50,5)
[1] 0.8428898
> nconsec(100,50,5)
[1] 8.504029e+28

It makes 2 tables of numbers that it reuses. You can still call it like that with just 3 parameters and it will make the tables for you, but if you want to do a lot of these with the same r, you can just make the main table 1 time like this:

> t = make_tables(20,10,3)
> pconsec(20,10,3,t)
[1] 0.8697309
> nconsec(20,10,3,t)
[1] 160688

Then every time you call it, it will use same table, and it won't have to recompute it every time. Just make it big enough for your largest n and k. You have to remake it if you change r though. You won't have to do this if it is already very fast for your inputs. Here is a sample table for (20,10,3):

Code:

      [,1] [,2] [,3] [,4] [,5] [,6] [,7]
 [1,]    0    0    0    0    0    0    0
 [2,]    0    0    0    0    0    0    0
 [3,]    0    0    1    0    0    0    0
 [4,]    0    0    2    1    0    0    0
 [5,]    0    0    3    4    1    0    0
 [6,]    0    0    4    9    6    1    0
 [7,]    0    0    5   16   18    7    1
 [8,]    0    0    6   25   40   27    8
 [9,]    0    0    7   36   75   74   36
[10,]    0    0    8   49  126  165  116
[11,]    0    0    9   64  196  321  300
[12,]    0    0   10   81  288  567  666
[13,]    0    0   11  100  405  932 1323
[14,]    0    0   12  121  550 1449 2416
[15,]    0    0   13  144  726 2155 4131
[16,]    0    0   14  169  936 3091 6700

Interesting number sequence. I don't recognize it. If we could identify it as something, we may be able to compute it from built in functions. It changes with every r.

Code:

pconsec = function(n,k,r,...) {nconsec(n,k,r,...) / choose(n,k)}

nconsec = function(n,k,r,table) {
  if (n == k) {
    s = 1 } 
  else if (k < 2*r) {
    s = choose(n-r,k-r) + (n-r)*choose(n-r-1,k-r)} 
  else {
    ctable = make_ctable(n,k,r)
    if (missing(table)) {table = make_table(n,k,r,ctable)}
    s = recurse(n,k,r,ctable,table)
  }
  s
}
  

make_ctable = function(n,k,r) {
  ctable = matrix(rep(0,(n-r+1)*(k-r+1)),nrow=(n-r+1))
  ctable = choose(row(ctable)-1,col(ctable)-1)
  ctable
}


make_table = function(n,k,r,ctable) {
  table = matrix(rep(0,(n-r-1)*(k-r)),nrow=(n-r-1))
  for (x in r:(n-r-1)) {
    for (y in r:min(x,(k-r))) {table[x,y] = ifelse(x==y,1,recurse(x,y,r,ctable,table))} }
  table
}


make_tables = function(n,k,r) {
  ctable = make_ctable(n,k,r)
  make_table(n,k,r,ctable)
}


recurse = function(n,k,r,ctable,table) {
  s = ctable[n-r+1,k-r+1] + (n-r)*ctable[n-r,k-r+1]
  if (k >= 2*r) {
    for (i in 0:(k-2*r)) {
      for (m in (r+i):(n-k+r+i-1)) {s = s - table[m,r+i]*ctable[n-r-m,k-2*r-i+1] } }
   }
  s
}

The ... is for the optional parameter. The function missing(table) determines if you didn't supply the table, so it will then make it for you.

I ported it to VB, but it is NOT working yet. For some reason the ctable is not being made. This should get you started though. Let me know if you find the problem. I may not be passing arrays correctly in VB.

Code:

Option Explicit

Function pconsec(n As Long, k As Long, r As Long)
    pconsec = nconsec(n, k, r) / Application.Combin(n, k)
End Function

Function nconsec(n As Long, k As Long, r As Long)

    Dim ctable() As Long
    Dim table() As Long
    
    With WorksheetFunction
    
    If (n = k) Then
        nconsec = 1
    ElseIf (k < 2 * r) Then
        nconsec = .Combin(n - r, k - r) + (n - r) * .Combin(n - r - 1, k - r)
    Else
        ctable = make_ctable(n, k, r)
        table = make_table(n, k, r, ctable)
        nconsec = recurse(n, k, r, ctable, table)
    End If
  
  End With

End Function

 
Function make_ctable(n As Long, k As Long, r As Long) As Long()
  
  ReDim ctable(0 To n - r, 0 To k - r) As Long
  Dim i As Long
  Dim j As Long
  
  With WorksheetFunction
  
  For i = 0 To n - r
    For j = 0 To k - r
     ctable(i, j) = .Combin(i, j)
    Next j
  Next i
  
  make_ctable = ctable()
  
  End With

End Function

Function make_table(n As Long, k As Long, r As Long, ctable() As Long) As Long()

    ReDim table(1 To n - r - 1, 1 To k - r) As Long
    Dim x As Long
    Dim y As Long
    
    For x = r To n - r - 1
        For y = r To Min(x, k - r)
            If x = y Then
                table(x, y) = 1
            Else
                table(x, y) = recurse(x, y, r, ctable, table)
            End If
        Next y
    Next x
    
    make_table = table()

End Function

Function recurse(n As Long, k As Long, r As Long, ctable() As Long, table() As Long)

    recurse = ctable(n - r, k - r) + (n - r) * ctable(n - r - 1, k - r)
    If k >= 2 * r Then
        For i = 0 To k - 2 * r
            For m = r + i To n - k + r + i - 1
                recurse = recurse - table(m, r + i) * ctable(n - r - m - 1, k - 2 * r - i)
            Next m
        Next i
    End If

End Function

Last edited by BruceZ; 12-29-2012 at 04:28 AM.

Quote

12-29-2012 , 05:47 AM

#44

Mirapep

enthusiast

Join Date: Dec 2012 Posts: 53

Bruce Ok, I'll try these functions. You have been exhaustive.
You've got to download the excel sheet Multimasaniello? I'm planning this in vb consecutive with the filter in order to increase the yield. This is to determine the progressive stakes in software for the lottery.
For this purpose I created a very interesting program that predicts the Italian lottery numbers to play. In an Italian lottery one winning number pays about 11 times the stake money. I use three numbers at a time so if they win I get about 3.5 times the amount bet. Now I have set the money management Multimasaniello total of three winds with height 3.5 that will be successful with one. The final yield is 1.5 times the initial cash. Next, I set up a second multimasaniello with 20 events total share for each 1.5, winning at least 5 out of 20. The final yeld is 0. From a statistical study of all winning lottery I found that very rarely will win 5 times in a row, so eliminating these combinations get a positive return final.
This explanation is to make known my motives. If you feel it interesting I'll try to be more specific

Quote

12-29-2012 , 11:37 AM

#45

Mirapep

enthusiast

Join Date: Dec 2012 Posts: 53

So Bruce,
I wrote your function in vb6 and there is error undefined function
Min (x, k - r)

Also I do not use combin excel. Using this function declared in a module

' returns: (n k) Binomial Coefficient = n! / (K! * (n-k)!)
' -1 If result overflows a long
' the number of ways to choose k elements from an n element set

Public Function Binomial (n As Long, k As Long) As Double
Dim i As Long, j As Long, u ()
On Error GoTo ErrOverflow

Select Case k
Case Is <0, Is> n 'valid range
Binomial = 0
Case 0, n 'trivial case 1
Binomial = 1
Case 1, n - 1 'trivial case n
Binomial = n
Case 2, n - 2 'simple case n * (n - 1) / 2
Binomial = (n \ (2 - (n And 1))) * ((n - 1) \ (1 + (n And 1)))
Case Else
ReDim u (n - k)
For j = 0 To (n - k)
u (j) = j + 1
Next j
For i = 2 To k 'recurrence: (nk) = (n-1 k-1) + (n-1 k)
For j = 1 To (n - k)
u (j) = u (j) + u (j - 1)
Next j
Next i
Binomial = u (n - k)
end Select

Exit Function
ErrOverflow:
Binomial = -1
end Function

Quote

12-29-2012 , 01:07 PM

#46

BruceZ

Carpal \'Tunnel

Join Date: Sep 2002 Posts: 11,877

Did you run the function in R? It is very fast now. 40 seconds for (100,50,5) instead of 69 years.

Min(x,k-r) returns the minimum of x and k-r (smaller one). Combin is just C(n,k) = n!/(n-k)!/k!. It is also the binomial coefficient. I'm sure you have equivalent functions in VB. I thought VB had a function called combin. This is not the same as combn which gives all the combinations, it is just a number. Did you get make_ctable to work?

I don't understand what you are doing with the lottery. There is no way to gain a +EV in the lottery by any money management or elimination of consecutive numbers if the numbers are random. Numbers with consecutive digits may not occur often, but that doesn't mean that such a sequence is any less likely to win unless there is a problem with the way the numbers are drawn. 123456 may rarely come up, but it should be as likely to win as any other combination. There will always be patterns in any sequence of truly random numbers. You cannot use that to gain an edge. Past numbers do not influence future numbers. Progressive betting systems do not work, and they are frauds.

Last edited by BruceZ; 12-29-2012 at 01:20 PM.

Quote

12-29-2012 , 03:35 PM

#47

Mirapep

enthusiast

Join Date: Dec 2012 Posts: 53

All correct your observations on the lottery. The probability is always the same. However, by continuing to apply the same method of prediction I observed his behavior by comparison with thousands of real extractions. The result is that the method is constantly excellent. But for the chance to wish you rightly given to implement a system for betting operations that provide additional added protection.
That said, the filter row should not be applied to the method of forecasting numbers to play. It should be applied to the system of money management.

Quote

12-29-2012 , 04:23 PM

#48

BruceZ

Carpal \'Tunnel

Join Date: Sep 2002 Posts: 11,877

Quote:

Originally Posted by Mirapep

So there is something wrong with the drawing so not all combinations are equally likely? What kind of analysis did you do to determine that your results weren't due to random chance?

Progressive betting schemes where your bets depend on patterns of numbers is a sign of a fraudulent system. Such schemes can appear to work well for awhile, but when it fails, you have a big loss, and your EV is always the edge for each bet times the total money you have invested. Typically such schemes are designed to fool people into thinking that the scheme works when it does not. A large number of people can observe positive results from such schemes due to random luck.

Quote

12-29-2012 , 05:59 PM

#49

Mirapep

enthusiast

Join Date: Dec 2012 Posts: 53

Bruce is difficult to explain to the English language. My method is not for sale. In Italy are extracted 5 numbers out of 90 and are paid various bets. Exactly one, two, three, four and five. In the history of mining has been slow for a number of its release also draws about 230. Play only one number has a probability of extraction of 90/5 and the dealer only pays 11 times your total bet. The draws are repeated three times a week for 10 wheels. I understand your skepticism because I sense that you are a mathematician. So, I said that a number has been slow for its release on a wheel also 230 extractions. Some scholars have applied a lottery process of competition among the ten wheels and when a number of the wheel becomes the first to delay more you begin to count an additional delay. In summary, this process has resulted that the delay thus counted has had a maximum of about 70 against the 230 first highlighted. Still too much to think about a play in progress without the risk of gambler's ruin