I am fascinated by board games. They provide immense fun, anyone can enjoy them, they are unpredictable and best of all they are great value for money. That is why whenever I get sometime I experiment with simulating games to know them better [read Why Monopoly board game is not as random as it appears]. So, out of curiosity I have created an excel sheet that can generate bingo / housie (housey) tickets – 24 of them at a time. To get new set of tickets you would hit F9 (recalculate).
Click here to download the bingo / housie ticket generator.
Note that these are Bingo UK / India / Australia variant I am talking about, not the US 5*5 type of bingo tickets.
Read on if you want to know how this is done:
According to Wikipedia:
A typical housie/bingo ticket .., contains fifteen numbers, arranged in nine columns by three rows. Each row contains five numbers and four blank spaces. Each column contains either one, two, or very rarely three, numbers:
* The first column contains numbers from 1 to 9,
* The second column numbers from 10 to 19,
* The third 20 to 29 and so on up until the last column, which contains numbers from 80 to 90.
I have removed number “90” from the list in order to reduce some complexity in generating the tickets.
The problem is now to “generate 15 random number between 1 to 89 and fill them in 15 random spots in a grid of 3 rows by 9 columns such that each row has exactly 5 numbers”
Now I could write a function in VBA to do this, but I wanted to do this only using formulas. So I started breaking the problem.
The first challenge is to select any random 5 cells in a 9 cell row
Once we select any random 5 cells in a 9 cell row, we will fill them with bingo numbers. Now, excel has a function to generate random numbers between 1 to 9 (=round(rand()*9,0)), but this is not good for us since each time we call this function we will get a random number between 1 to 9, where as we need a 5 random numbers without repetition between 1 to 9. The function is memoryless and could repeat numbers when called 5 times.
Instead we can list all the possible “5 cells with numbers and others are empty” combinations of a 9 cells region and select a random combination every time. There are essentially 9C5 i.e. 126 ways in which you can select any 5 cells out of 9 cells (without repetition of course).
So I listed all these combinations in a table and then randomly selected one of the combinations. You can see the first five such combinations in the image below:
Selecting any five cells out of nine cells
Now I created a 3*9 region and filled the cells with 1s or 0s, “1” when the cell in bingo ticket is supposed to have a value and “0” if the cell is empty as shown below:
Next challenge is to show random values in each cell
The trick here is that first column in our 3*9 bingo ticket has any number(s) from 1 to 9, second column has any number(s) from 10 to 19 …
Again, the challenge is the numbers should not repeat, otherwise we could simply use rand()*10, rand()*10+10, rand()*10+20 ... to generate the numbers.
This time it gets even more trickier because each column can have either no values, or 1 value or 2 values or 3 values.
The ticket generation logic now looks like:
- If the column has no values in it, then we will leave all the cells in that column of bingo ticket empty
- If the column has 1 value, we will generate any random number from that column’s range of possible values (1-9, 10-19,20-29,…80-89) and place it in the cell that is supposed to have a value and leave other cells empty.
- If the column has 2 values, we will generate 2 random numbers without repetition from that column’s range of possible values and place them in cells that are supposed to have them
- If the column has 3 values, we will generate 3 random numbers without repetition from that column’s range of possible values and place them in cells that are supposed to have them
As you can see, it is easy when the column has no or 1 value in it. But when the column has 2 or 3 I used the combinations trick described earlier.
First I created all 2 number combinations and 3 number combinations. Since the numbers on Bingo ticket are always sorted from top to bottom in a column, I just had to list down 45 combinations (10C2) for 2 numbers and 120 combinations (10C3) for 3 numbers.
The rest of the details are small enough that I can leave them to your imagination. So when the ticket is generated, it looks like this:
Remember to download housie / bingo ticket generator excel sheet and print your tickets at home. Just F9 to generate new set of tickets. Un-hide the rows from 43 if you want to see how this is done.
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13 Responses to “Using pivot tables to find out non performing customers”
To avoid the helper column and the macro, I would transpose the data into the format shown above (Name, Year, Sales). Now I can show more than one year, I can summarize - I can do many more things with it. ASAP Utilities (http://www.asap-utilities.com) has a new experimental feature that can easily transpose the table into the correct format. Much easier in my opinion.
David
Of course with alternative data structure, we can easily setup a slicer based solution so that everything works like clockwork with even less work.
David, I was just about to post the same!
In Contextures site, I remember there's a post on how to do that. Clearly, the way data is layed out on the very beginning is critical to get the best results, and even you may thinkg the original layout is the best way, it is clearly not. And that kind of mistakes are the ones I love ! because it teaches and trains you to avoid them, and how to think on the data structure the next time.
Eventually, you get to that place when you "see" the structure on the moment the client tells you the request, and then, you realized you had an ephiphany, that glorious moment when data is no longer a mistery to you!!!
Rgds,
Chandoo,
If the goal is to see the list of customers who have not business from yearX, I would change the helper column formula to :
=IF(selYear="all",sum(C4:M4),sum(offset(C4:M4,,selyear-2002,1,columns(C4:M4)-selyear+2002)))This formula will sum the sales from Selected Year to 2012.
JMarc
If you are already using a helper column and the combox box runs a macro after it changes, why not just adjust the macro and filter the source data?
Regards
I gotta say, it seems like you are giving 10 answers to 10 questions when your client REALLY wants to know is: "What is the last year "this" customer row had a non-zero Sales QTY?... You're missing the forest for the trees...
Change the helper column to:
=IFERROR(INDEX(tblSales[[#Headers],[Customer name]:[Sales 2012]],0,MATCH(9.99999999999999E+307,tblSales[[#This Row],[Customer name]:[Sales 2012]],1)),"NO SALES")
And yes, since I'm matching off of them for value, I would change the headers to straight "2002" instead of "Sales 2002" but you sort the table on the helper column and then and there you can answer all of your questions.
Hi thanks for this. Just can't figure out how you get the combo box to control the pivot table. Can you please advise?
Cheers
@Kevin.. You are welcome. To insert a combo box, go to Developer ribbon > Insert > form controls > combo box.
For more on various form controls and how to use them, please read this: http://chandoo.org/wp/2011/03/30/form-controls/
Thanks Chandoo. But I know how to insert a combobox, I was more referring to how does in control the year in the pivot table? Or is this obvious? I note that if I select the Selected Year from the PivotTable Field List it says "the field has no itens" whereas this would normally allow you to change the year??
Thanks again
worked it out thanks...
when =data!Q2 changes it changes the value in column N:N and then when you do a refreshall the pivottable vlaues get updated
Still not sure why PivotTable Field List says “the field has no itens"?? I created my own pivot table and could not repeat that.
Hi, I put the sales data in range(F5:P19) and added a column D with the title 'Last sales in year'. After that, in column D for each customer, the simple formula
=2000+MATCH(1000000,E5:P5)
will provide the last year in which that particular customer had any sales, which can than easily be managed by autofilter.
Somewhat longer but perhaps a bit more solid (with the column titles in row 4):
=RIGHT(INDEX($F$4:$P$19,1,MATCH(1000000,F5:P5)),4)
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