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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12 Responses to “29 Excel Formula Tips for all Occasions [and proof that PHD readers truly rock]”
Some great contributions here.
Gotta love the Friday 13th formula 😀
Great tips from you all! Thanks a lot for sharing! bsamson, particularly you helped me on a terribly annoying task. 🙂
(BTW, Chandoo, it's not exactly "Find if a range is normally distributed" what my suggestion does. It checks if two proportions are statistically different. I probably gave you a bad explanation on twitter, but it'd be probably better if you fix it here... 🙂 )
Great compilation Chandoo
For the "Clean your text before you lookup"
=VLOOKUP(CLEAN(TRIM(E20)),F5:G18,2,0)
I would like to share a method to convert a number-stored-as-text before you lookup:
=VLOOKUP(E20+0,F5:G18,2,0)
@Peder, yeah, I loved that formula
@Aires: Sorry, I misunderstood your formula. Corrected the heading now.
@John.. that is a cool tip.
Hey Chandoo,
That p-value formula is really great for a statistics person like me.
What a p-value essentially is, is the probability that the results obtained from a statistical test aren't valid. So for example, if my p value is .05, there's a 5% probability that my results are wrong.
You can play with this if you install the Data Analysis Toolpak (which will perform some statistical tests for you AND provide the P Value.)
Let's say for example I've got two weeks of data (separated into columns) with the number of hours worked per day. I want to find out if the total number of hours I worked in week two were really all the different than week one.
Week1 Week2
10 11
12 9
9 10
7 8
5 8
Go to Data > Data Analysis > T-Test Assuming Unequal Variances > OK
In the Variable 1 Box, select the range of data for week 1.
In the Variable 2 Box, select the range of data for week 2.
Check "Labels"
In the Alpha box, select a value (in percentage terms) for how tolerant you are of error.
.05 is the general standard; that is to say I am willing to accept a 95% level of confidence that my result is accuarate.
Select a range output.
Excel calculates a number of results: Average (mean) for each week's data, etc.
You'll notice however that there are two P Values; one-tail and two-tail. (one tail tests are for > or .05), the number of hours I worked in week two is statistically equivalent to the number of hours I worked in week one.
So here’s a way you might want to use this. You put up a new entry on your blog. You think it’s the best entry ever! So you pull your webstats for this week and compare it to last week. You gather data for each week on the length of time a visitor spends on your website. The question you’re trying to prove statistically is whether there’s an average increase in the amount of time spent on your website this week as compared to last week (as a result of your fancy new blog post). You can run the same statistical test I illustrated above to find out. Incidentally, it matters very little to the stat test whether the quantity of visitors differs or not.
Anyhow, the Data Analysis toolpack doesn't perform a lot of stat tests that folks like me would like to have access to. In those cases I have to either use different software, or write some very complicated mathematical formulas. Having this p-value formula makes my life a LOT easier!
Thanks!
Eric~
Fantastic stuf..One line explanation is cool.
Thanks to all the contributors
OS
Take FirstName, MI, LastName in access (you can fix it to work in excel) capitalize first letter of each and lowercase the rest and add ". " if MI exists then same for last name:
Full Name: Format(Left([FirstName],1),">") & Format(Right([FirstName]),Len([FirstName])-1),"") & ". ","") & Format(Left([LastName],1),">") & Format(Right([LastName],Len([LastName])-1),"<")
I teach excel, access, etc etc for a living and i have my access students build this formula one step at a time from the inside out to show how formulas can be made even if it looks complicated. Yes I know I could just do IsNull([MI]) and reverse the order in the Iif() function but the point here is to nest as many functions as possible one by one (also I illustrate how it will fail without the Not() as it is)
Extract the month from a date
The easiest formula for this is =MONTH(a1)
It will return a 1 for January, 2 for February etc.
if in a column we write the value of total person for eg. 10 if we spent 1.33 paise each person then how we get total amount in next column and the result will in round form plzzzzz solve my problem sir................... thank u
@Anjali
If the value 10 is in B2 and 1.33 paise is in C2 the formula in D2 could be =B2*C2
If the values are a column of values you can copy the formula down by copy/paste or drag the small black handle at the bottom right corner of cell D2
kindly share with me new forumulas.
How to convert a figure like 870.70 into 870 but 871.70 into 880 using excel formula ? Please help.