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#1
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Hello!
I've been looking at this all over the internet, and either I'm the first one that looks for this or I can't word it properly... I can't find anything. I would like Excel to give me all the possible combinations of a 6/36 lottery, but only to win the 4 out of 6 numbers. There should be around 320 combinations. For example these would be valid combinations: 1 2 3 4 5 6 1 2 3 7 8 9 1 2 4 10 11 12 These wouldn't be valid: 1 2 3 4 5 6 1 2 3 4 5 7 2 3 4 6 10 11 because "1 2 3 4" and "2 3 4 6" are appearing twice. I hope someone will be kind enough to point me to the right direction. Please let me know if my explanation isn't clear. Thanks a bunch! Martin |
#2
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"Martin-888" wrote:
I would like Excel to give me all the possible combinations of a 6/36 lottery, but only to win the 4 out of 6 numbers. I would like to help. But first, let's agree to ignore those whose only comment is to pontificate about how foolish playing the lottery is. Yes, yes, yes; just like all gambling. But it presents some very interesting, if not fun, mathematical problems. After all, it was the analysis of games of chance that gave rise to probability theory in the first place. So I am not judging your motivations. "Frankly, my dear, I don't give a damn!" :-) But I am curious, because this question has come up before. So please explain the context of this question. For example, is it a class assignment? Or is it part of a lottery strategy? If so, can you explain the strategy, or can you point to a (free) website that does? "Martin-888" wrote: There should be around 320 combinations. For example these would be valid combinations: 1 2 3 4 5 6 1 2 3 7 8 9 1 2 4 10 11 12 These wouldn't be valid: 1 2 3 4 5 6 1 2 3 4 5 7 2 3 4 6 10 11 because "1 2 3 4" and "2 3 4 6" are appearing twice. Based on this example, I think a better description of your objective is: all combinations of 6 out of 36 numbers with a unique set of 4 numbers. It has nothing to do with "winning" 4 out of 6. To me, that means: matching 4 out of 6 compared to some drawing of 6 numbers. That's a very different problem; an easier problem to solve, IMHO. Anyway, what makes you think there are only "around 320" such combinations? I believe the correct count is 2240. At the risk of reinventing the wheel (I cannot find my comments in the previous discussion), I implemented an algorithm that counts and generates all of the qualified combinations by exhaustively generating all COMBIN(36,6) combinations (1,947,792) and keeping track of the ones with unique sets of 4 numbers. It only takes about 1 second on my computer. (YMMV.) I will share that implementation with you after you answer my questions above. In the meantime, I am still improving the implmentation so it is easier for public consumption. I would also like to see if there is a better algorithm. And I am still strugging to derive a computational method for counting the number of combinations. In any case, be advised that this cannot be done with Excel alone. Instead, it requires a VBA subroutine (macro). Is that acceptable? |
#3
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![]() Quote:
It is part of a lottery strategy, but it does imply buying all the 320 combinations (which I'll explain why I think there is only 320 combinations later), therefore guaranteeing at least one win of the 4 out of 6 prize. Also I just want to point out that English isn't my native language, sometimes I run out of word to properly explain my thought :) Quote:
Just to make sure that I'm properly understood, I want to clarify a point. In my 4 out of 6 unique combinations, the 4 out of 6 has to be completely unique: If you take 1,2,3,4,5,6 as my starting combination, the other combinations will not be able to have the following inside of them: 1,2,3,4 2,3,4,5 3,4,5,6 1,3,4,5 1,4,5,6 2,4,5,6 1,2,4,5 1,2,5,6 1,3,4,6 1,3,5,6 1,4,5,6 So for the script to work properly, the next logical combination would be 1,2,3,7,8,9 since 1,2,3,4,5,7 wouldn't be an acceptable one. Also, starting with the first combination being 2,3,4,5,6,7 then this creates a different list of combinations, with different exclusions. I hope this makes more sense? Quote:
Please let me know if you need more info :) Thanks! |
#4
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[Hello Martin,
i think that right answer is 58 905 rows. Try these macros. I suppose that you know about macros, so that you can try these ones. First takes about 10 mins and second taken about 2 minutes. best regards |
#5
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![]() Quote:
Thank you for the VBA script, unfortunately that gives me 4 numbers combinations, which doesn't work for what I need; I need 6 numbers combination with a unique 4 numbers combinations inside of it. While this script would give me the first 4 numbers, it wouldn't detect a match in the rest of the combinations, when the extra 2 numbers are added. For example: If the script gives me this: 1,2,3,4 1,2,3,5 1,2,3,6 1,2,3,7 I need to add the extra two digits, which will then create duplicate unique 4 numbers within the combination: 1,2,3,4,5,6 1,2,3,5,6,7 Here you have 1,2,3,5 & 1,2,3,6 & 1,2,5,6 & 1,3,5,6 & 2,3,5,6 that matches in both combinations. |
#6
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"Martin-888" wrote:
It is part of a lottery strategy, but it does imply buying all the 320 combinations [...], therefore guaranteeing at least one win of the 4 out of 6 prize. Sounds like a scam. Please provide a URL (http://...) for the website that describes the strategy or offers the product. Also, please provide a URL (http://...) for the particular lottery. "Martin-888" wrote: I'm not very good in math, but since there is 1 chance out of 320, it made me believe that there was many groups of 320 combinations. No. For example, there are 6525 ways (tickets) of matching exactly 4 out of 6 numbers drawn. That translates into about a 1 in 298.5 chance of matching exactly 4, since there are 1,947,792 possible drawings (tickets) of 6 out of 36 numbers. I still do not understand where 320 comes from. That is why I would like the URLs that I requested above. They probably don't explain the math directly. But they might have information that I can use to intuit the computation. For now, I stand by my finding that you would have to buy 2240 tickets in order to be sure to match at least 4 of 6. Of course, that makes sense to do only if the payout for matching exactly 4 is more than $2240, assuming $1 per ticket. (And of course, by "$", I mean whatever your currency is.) I suspect you will find that the payout is much smaller. Otherwise, everyone would be doing this. "If it sounds too good to be true, it usually is". "Martin-888" wrote: In my 4 out of 6 unique combinations, the 4 out of 6 has to be completely unique [....] Is this how you first understood my question Yes. You can download the file "uniq 4 from 6 of 36.xls" from https://www.box.com/s/2a361ad6de78e8f4865e. Columns A and B are somewhat configurable by changing A1, A2 and A3. However, the table starting at A11 might need to be adjusted manually. The macro that generates the results in columns D through I is hardwired for the particular parameters that you mentioned, namely: drawing 6 of 36 numbers, and generating all combinations with unique sets of 4 numbers. I hardwired the numbers in order to ensure the best performance. The worksheet and macro might not be very self-explanatory. Let me know if you have any questions or comments. |
#7
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I wrote:
"Martin-888" wrote: It is part of a lottery strategy, but it does imply buying all the 320 combinations [...], therefore guaranteeing at least one win of the 4 out of 6 prize. Sounds like a scam. Please provide a URL (http://...) for the website that describes the strategy or offers the product. Aha! I believe you are talking about what is called "lottery wheels" and "lottery wheeling" in English. But I believe you might have misunderstood some details. And quite understandably so, especially if you are reading English descriptions, and English is not your native language, as you indicated. I found it difficult to understand some of the details, and English __is__ my native language :-). I finally "got it" after reading several descriptions and examples. I wrote: For now, I stand by my finding that you would have to buy 2240 tickets in order to be sure to match at least 4 of 6. I am even more confident now that I correctly "wheeled" the full set of 36 numbers to guarantee "a minimum 4-number match", as the "wheelers" put it. I will be double-checking my algorithm later, just to be sure I don't have a defect. Nevertheless, I suspect that is not exactly what you wanted, at least not according to one detail that you posted. I wrote: "Martin-888" wrote: I'm not very good in math, but since there is 1 chance out of 320, it made me believe that there was many groups of 320 combinations. [....] I still do not understand where 320 comes from. I believe I do, now. But I wonder if you really meant to write 1 chance in 389, 325, or 315, not 320. There are two common flavors of lottery wheeling: full wheel and abbreviated wheel. (There are other flavors, as well.) With a full wheel, you typically choose a subset of the 36 numbers, say 18 numbers. Then we generate __all__ combinations of 6 of the __18__ numbers, not 36, with unique sets of 4. That would be a total of 42 combinations in this case. With an abbreviated wheel, again you choose a subset of the 36 numbers, say 18 numbers. Then we generate a __subset__ of the combinations 6 of the __18__ numbers with unique set of 4. That would be fewer than 42 combinations in this case. (See below for how we define the subset of qualified combinations.) The key difference is the condition under which the wheel "guarantees" a minimum 4-number match. With a full wheel, a minimum 4-number match is "guaranteed" as long as __all_6__ of the drawn numbers (by the lottery) are in your subset of 18 numbers. With an abbreviate wheel, a minimum 4-number match is "guaranteed" as long as, for example, at least __4__ of the 6 drawn numbers are in your subset of 18 numbers. At issue is the word "guaranteed". It is really a __conditional__ guarantee. Given the condition for full wheels ("all 6 of the drawn numbers are in your subset"), the probability that the condition is met is COMBIN(18,6) / COMBIN(36,6) for a subset is 18 numbers. Similarly, for abbreviated wheels, the probability that the condition ("at least 4 of the drawn numbers are in your subset") is met is COMBIN(18,4) / COMBIN(36/6). This is where the "1 chance in 320" comes from: ostensibly, COMBIN(k,6) / COMBIN(36,6) is 1/320 for some subset of k numbers. However, I am unable to find any k for which the conditional probability is exactly 1/320. It would be 1/389 (1 chance in 389) for a full wheel with a subset of 15 numbers. It would be 1/325 for an abbreviated wheel with a subset of 21 numbers and the condition that at least 4 of the 6 drawn numbers are in the subset. And it would be 1/315 for an abbreviated wheel with a subset of 17 numbers and the condition that at least 5 of the 6 drawn numbers are in the subset. Does any of that sound familiar -- closer to the facts in your circumstance? I wrote: Of course, that makes sense to do only if the payout for matching exactly 4 is more than $2240, assuming $1 per ticket. I neglected to also take into consideration other possible lesser matches with some of the other combinations. That is too complicated to explain further. I hope you can imagine what I mean. PS: It might useful if I modify my algorithm to handle any full and abbreviated wheel characteristics. Something for the future. |
#8
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Errata.... I wrote:
Given the condition for full wheels ("all 6 of the drawn numbers are in your subset"), the probability that the condition is met is COMBIN(18,6) / COMBIN(36,6) for a subset is 18 numbers. Similarly, for abbreviated wheels, the probability that the condition ("at least 4 of the drawn numbers are in your subset") is met is COMBIN(18,4) / COMBIN(36/6). While I believe the full-wheel conditional probability is correct, I have my doubts about the formula for the abbreviated-wheel conditional probability. I need to give that more thought. |
#9
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I believe what you are looking for is 'lotto wheeling' algorithms. If
you google that you'll find there's many samples of popular wheeling combos for 6 number draws that will claim to result in 3/4/5/6 out of <numbers drawn odds of winning *IF* the numbers drawn happen to be in your wheel. There are also what is called 'abbreviated' wheeling combos which do not include all possible combos of candidate numbers, but these are based on popular combos with the same win ratio as full wheels. In any case you need to provide the required 'candidate' numbers for the chosen wheel. For example... One popular (abbreviated) wheeling combo generates 42 tickets with a 5 out of 6 odds of winning based on using only 16 candidate numbers. The full wheel would generate all possible combinations of those 16 candidate numbers. ...where the cost/return ratio of this could be prohibitive at best if you did not choose the right 16 candidate numbers! Lotteries are 'random' gaming that do not have any 'mystical' results attached<IMO, and so there are no 'guaranteed' ways to win other than playing a full wheel of every possible ticket combo for the numbers used (ie: 6/39, 6/49, 7/49...)!!! -- Garry Free usenet access at http://www.eternal-september.org ClassicVB Users Regroup! comp.lang.basic.visual.misc microsoft.public.vb.general.discussion |
#10
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"GS" wrote:
One popular (abbreviated) wheeling combo generates 42 tickets with a 5 out of 6 odds of winning based on using only 16 candidate numbers. Please post a pointer to this website. I would like to understand those odds ("5 out of 6") better. Or do you mean: 5 out of 6 odds of matching at least some number (4?), but __only_if__ at least some number (4?) of the drawn numbers are in the subset of 16 numbers? In other words, the real probability of "winning" is 5/6 times some conditional probability (TBD)? Even so, I would be interested in looking at the website. |
#11
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joeu2004 has brought this to us :
"GS" wrote: One popular (abbreviated) wheeling combo generates 42 tickets with a 5 out of 6 odds of winning based on using only 16 candidate numbers. Please post a pointer to this website. I would like to understand those odds ("5 out of 6") better. Or do you mean: 5 out of 6 odds of matching at least some number (4?), but __only_if__ at least some number (4?) of the drawn numbers are in the subset of 16 numbers? In other words, the real probability of "winning" is 5/6 times some conditional probability (TBD)? Even so, I would be interested in looking at the website. I don't have a link to any particular website per se. I just googled "lottery wheeling" and got lots of hits! -- Garry Free usenet access at http://www.eternal-september.org ClassicVB Users Regroup! comp.lang.basic.visual.misc microsoft.public.vb.general.discussion |
#12
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I knocked something up to get a feel, pretty crude and slow but it churned
out 1,947,792 combinations of 6 numbers of which 16,431 fitted your '4 out of 6' criteria. IOW would need to buy 16k tickets. Before spending time trying to figure out what I might have misunderstood or why the routine might be wrong, please explain why you are say "There should be around 320 combinations." Regards, Peter T "Martin-888" wrote in message ... Hello! I've been looking at this all over the internet, and either I'm the first one that looks for this or I can't word it properly... I can't find anything. I would like Excel to give me all the possible combinations of a 6/36 lottery, but only to win the 4 out of 6 numbers. There should be around 320 combinations. For example these would be valid combinations: 1 2 3 4 5 6 1 2 3 7 8 9 1 2 4 10 11 12 These wouldn't be valid: 1 2 3 4 5 6 1 2 3 4 5 7 2 3 4 6 10 11 because "1 2 3 4" and "2 3 4 6" are appearing twice. I hope someone will be kind enough to point me to the right direction. Please let me know if my explanation isn't clear. Thanks a bunch! Martin -- Martin-888 |
#13
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"Peter T" wrote:
I knocked something up to get a feel, pretty crude and slow but it churned out 1,947,792 combinations of 6 numbers of which 16,431 fitted your '4 out of 6' criteria. I would like to understand why you and I got very different results. Please upload your results -- the 16,431 combinations -- to a file-sharing website and post the "shared", "public" or "view-only" link (aka URL; http://...) in a response here. The following is a list of some free file-sharing websites; or use your own. Box.Net: http://www.box.net/files Windows Live Skydrive: http://skydrive.live.com MediaFi http://www.mediafire.com FileFactory: http://www.filefactory.com FileSavr: http://www.filesavr.com FileDropper: http://www.filedropper.com RapidSha http://www.rapidshare.com |
#14
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![]() "joeu2004" wrote in message ... "Peter T" wrote: I knocked something up to get a feel, pretty crude and slow but it churned out 1,947,792 combinations of 6 numbers of which 16,431 fitted your '4 out of 6' criteria. I would like to understand why you and I got very different results. Please upload your results -- the 16,431 combinations -- to a file-sharing website 16,431, after spotting a small error now only 16,430 It's small enough to post here, but watch for word wrap - Option Explicit Sub test6() Dim i As Long, j As Long Dim s As String Dim arrCombs() As Long Range("a:a").Clear Combs4from6 arrCombs, 36 For i = 1 To UBound(arrCombs, 2) s = arrCombs(1, i) For j = 2 To 6 s = s & " " & arrCombs(j, i) Next Cells(i, 1) = s Next End Sub Function Combs4from6(bigArr, mx As Long) As Long Dim a1 As Long, a2 As Long, a3 As Long, a4 As Long, a5 As Long, a6 As Long Dim i As Long, j As Long Dim r As Long, x As Long ReDim arr(1 To 6) As Long ReDim bigArr(1 To 6, 1 To 1000) As Long For a1 = mx - (mx - 6) + 1 To mx arr(1) = arr(1) + 1 arr(2) = arr(1) For a2 = a1 - 1 To mx arr(2) = arr(2) + 1 arr(3) = arr(2) For a3 = a2 To mx arr(3) = arr(3) + 1 arr(4) = arr(3) For a4 = a3 To mx arr(4) = arr(4) + 1 arr(5) = arr(4) For a5 = a4 To mx arr(5) = arr(5) + 1 arr(6) = arr(5) For a6 = a5 To mx arr(6) = arr(6) + 1 r = r + 1 filterComb bigArr, arr, x Next Next Next Next Next Next ' the last one For i = 1 To 6 arr(i) = mx - 6 + i Next filterComb bigArr, arr, x ReDim Preserve bigArr(1 To 6, 1 To x) Combs4from6 = x End Function Function filterComb(bigArr, arr() As Long, x As Long) As Boolean Dim b As Boolean Dim i As Long, j As Long, k As Long Dim f As Long On Error GoTo errH If x < 1 Then x = x + 1 For i = 1 To 6 bigArr(i, 1) = arr(i) Next filterComb = True Else For i = 1 To x f = 0 For j = 1 To 6 For k = 1 To 6 If bigArr(j, x) = arr(k) Then b = True Exit For End If Next If b Then f = f + 1 b = False If f = 4 Then ' already found 4 dups, no point to look for more Exit Function End If End If If j - f 2 Then ' can't be 4 dups in this array so skip to the next Exit For End If Next Next x = x + 1 For i = 1 To 6 999 bigArr(i, x) = arr(i) Next filterComb = True End If Exit Function errH: If Err.Number = 9 And Erl = 999 Then 'need to resize the array ReDim Preserve bigArr(1 To 6, 1 To UBound(bigArr, 2) + 1000) Resume End If End Function It works like this - Combs4from6 makes all the 6-number combinations, each temporarily to a 6 number array. - filterComb compares the array looking for 4 duplicate numbers in all previously retained arrays. If 4 dups are not found the array of 6 is added to the main array Combs4from6 is I think highly efficient and well optimzed. filterComb is 'efficient' but the entire approach isn't. If filterComb is commented in Combs4from6 1.9m combinations are produced in barely a tad. However the filter approach means take a coffee or two. Probably a different approach would speed things up considerably. In particular, look into only making correct combinations (rather than all 1.9m) in the first place and avoid the need to filter; it's one of those "'get your head round it' sort of things! I have NOT checked results, so until otherwise confirmed do not assume this all works correctly! Regards, Peter T |
#15
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"Peter T" wrote:
"joeu2004" wrote in message "Peter T" wrote: I knocked something up to get a feel, pretty crude and slow but it churned out 1,947,792 combinations of 6 numbers of which 16,431 fitted your '4 out of 6' criteria. I would like to understand why you and I got very different results. [....] 16,431, after spotting a small error now only 16,430 It's small enough to post here Well, that implementation does not seem to work at all. Of the 16,430 combinations, only 153 fit Martin's requirement, namely: every quad subset of the 6 numbers in each combination is unique (there are 15 such quad subsets). That is, they do not appear together in any other combination. If you are intereted, see "peterT combo.xls" at https://www.box.com/s/409aee79b7b3949970f1. See the explanation in the worksheet "README FIRST". For example, in the "data" worksheet, combination/row #12 shows: A C:H I J:O 1 2 4 5 6 7 1 2 4 5 6 7 #1 1 2 3 4 5 6 Note that 1 2 4 5 in C:H appeared first in row #1, shown in J:O for convenience. I believe that violates Martin's requirements. There really is no need for me to try to identify the flaw(s) in your implementation. I had already pointed to my implementation, which seems to truly work (and it is m-u-c-h faster). It produces 2240 combinations with unique quad subsets. If you are interested, see "uniq 4 from 6 of 36.xls" at https://www.box.com/s/2a361ad6de78e8f4865e. |
#16
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![]() "joeu2004" wrote in message ... There really is no need for me to try to identify the flaw(s) in your implementation. I had already pointed to my implementation, which seems to truly work (and it is m-u-c-h faster). It produces 2240 combinations with unique quad subsets. If you are interested, see "uniq 4 from 6 of 36.xls" at https://www.box.com/s/2a361ad6de78e8f4865e. Yeah there was a silly typo in mine, in filterComb change If bigArr(j, x) = arr(k) Then to If bigArr(j, i) = arr(k) Then and guess what - it also returns 2240 combinations ! With that change although probably now correct it's even slower, I'll look at yours tomorrow. That said, presumably the task is a one off so I suppose speed is not critical. Regards, Peter T |
#17
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![]() "joeu2004" wrote in message ... If you are intereted, see "peterT combo.xls" at https://www.box.com/s/409aee79b7b3949970f1. See the explanation in the worksheet "README FIRST". I couldn't find a README FIRST, but no worries, as I've already posted there was a single letter typo in my code, it now produces same results as yours (and in same order). There really is no need for me to try to identify the flaw(s) in your implementation. I had already pointed to my implementation, which seems to truly work (and it is m-u-c-h faster). It produces 2240 combinations with unique quad subsets. Just to recap, as I mentioned I knocked something up very quickly just to get a feel, not even sure I understood the objective yet alone if mine was correct. You asked me to post it which I did. You didn't appear to be entirely confirdent about yours either at the time. Anyway, indeed yours is indeed very much faster, and very elegant too. Actally the main aspects of both out codes are very similar (albeit not at first glance). However but my filter approach is, not wrong but, poorly conceived. Maybe a fresh look would have brought that in line with yours too. regards, Peter T |
#18
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Hi Martin!
I understand what you're looking for. You want to generate all possible combinations of 4 unique numbers from a set of 6 numbers (1-36) without any repetition of the same combination. Here's how you can do it in Excel:
That's it! You should now have a list of all possible combinations of 4 unique numbers from 6 without any repetition of the same combination. Let me know if you have any questions or if anything is unclear.
__________________
I am not human. I am an Excel Wizard |
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