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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20 Responses to “Simulating Dice throws – the correct way to do it in excel”
You have an interesting point, but the bell curve theory is nonsense. Certainly it is not what you would want, even if it were true.
Alpha Bravo - Although not a distribution curve in the strict sense, is does reflect the actual results of throwing two physical dice.
And reflects the following . .
There is 1 way of throwing a total of 2
There are 2 ways of throwing a total of 3
There are 3 ways of throwing a total of 4
There are 4 ways of throwing a total of 5
There are 5 ways of throwing a total of 6
There are 6 ways of throwing a total of 7
There are 5 ways of throwing a total of 8
There are 4 ways of throwing a total of 9
There are 3 ways of throwing a total of 10
There are 2 ways of throwing a total of 11
There is 1 way of throwing a total of 12
@alpha bravo ... welcome... 🙂
either your comment or your dice is loaded 😉
I am afraid the distribution shown in the right graph is what you get when you throw a pair of dice in real world. As Karl already explained, it is not random behavior you see when you try to combine 2 random events (individual dice throws), but more of order due to how things work.
@Karl, thanks 🙂
When simulating a coin toss, the ROUND function you used is appropriate. However, your die simulation formula should use INT instead of ROUND:
=INT(RAND()*6)+1
Otherwise, the rounding causes half of each number's predictions to be applied to the next higher number. Also, you'd get a count for 7, which isn't possible in a die.
To illustrate, I set up 1200 trials of each formula in a worksheet and counted the results. The image here shows the table and a histogram of results:
http://peltiertech.com/WordPress/wp-content/img200808/RandonDieTrials.png
@Jon: thanks for pointing this out. You are absolutely right. INT() is what I should I have used instead of ROUND() as it reduces the possibility of having either 1 or 6 by almost half that of having other numbers.
this is such a good thing to learn, helps me a lot in my future simulations.
Btw, the actual graphs I have shown were plotted based on randbetween() and not from rand()*6, so they still hold good.
Updating the post to include your comments as it helps everyone to know this.
By the way, the distribution is not a Gaussian distribution, as Karl points out. However, when you add the simulations of many dice together (i.e., ten throws), the overall results will approximate a Gaussian distribution. If my feeble memory serves me, this is the Central Limit Theorem.
@Jon, that is right, you have to nearly throw infinite number of dice and add their face counts to get a perfect bell curve or Gaussian distribution, but as the central limit theorem suggests, our curve should roughly look like a bell curve... 🙂
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I'm afraid to say that this is a badly stated and ambiguous post, which is likely to cause errors and misunderstanding.
Aside from the initial use of round() instead of int(),.. (you've since corrected), you made several crucial mistakes by not accurately and unambiguously stating the details.
Firstly, you said:
"this little function generates a random fraction between 0 and 1"
Correctly stated this should be:
"this little function generates a random fraction F where 0 <= F < 1".
Secondly, I guess because you were a little fuzzy about the exact range of values returned by rand(), you have then been just as ambiguous in stating:
"I usually write int(rand()*12)+1 if I need a random number between 0 to 12".
(that implies 13 integers, not 12)
Your formula, does not return 13 integers between 0 to 12.
It returns 12 integers between 1 and 12 (inclusive).
-- As rand() returns a random fraction F where 0 <= F < 1, you can obviously can only get integers between 1 and 12 (inclusive) from your formula as stated above, but clearly not zero.
If you had said either:
"I usually write int(rand()*12) if I need a random number between 0 to 11 (inclusive)",
or:
"I usually write int(rand()*12)+1 if I need a random number between 1 to 12 (inclusive)"
then you would have been correct.
Unfortunately, you FAIL! -- repeat 5th grade please!
Your Fifth Grade Maths Teacher
Idk if I'm on the right forum for this or how soon one can reply, but I'm working on a test using Excel and I have a table set up to get all my answers from BUT I need to generate 10,000 answers from this one table. Every time, I try to do this I get 10,000 duplicate answers. I know there has to be some simple command I have left out or not used at all, any help would be extremely helpful! (And I already have the dice figured out lol)
Roll 4Dice with 20Sides (4D20) if the total < 20 add the sum of a rerolled 2D20. What is the average total over 10,000 turns? (Short and sweet)
Like I said when I try to simulate 10,000turns I just get "67" 10,000times -_- help please! 😀
@Justin
This is a good example to use for basic simulation
have a look at the file I have posted at:
https://rapidshare.com/files/1257689536/4_Dice.xlsx
It uses a variable size dice which you set
Has 4 Dice
Throws them 10,000 times
If Total per roll < 20 uses the sum of 2 extra dice Adds up the scores Averages the results You can read more about how it was constructed by reading this post: http://chandoo.org/wp/2010/05/06/data-tables-monte-carlo-simulations-in-excel-a-comprehensive-guide/
Oh derp, i fell for this trap too, thinking i was makeing a good dice roll simulation.. instead of just got an average of everything 😛
Noteably This dice trow simulate page is kinda important, as most roleplay dice games were hard.. i mean, a crit failure or crit hit (rolling double 1's or double 6's) in a a game for example dungeons and dragons, if you dont do the roll each induvidual dice, then theres a higher chance of scoreing a crit hit or a crit failure on attacking..
I've been working on this for awhile. So here's a few issues I've come across and solved.
#1. round() does work, but you add 0.5 as the constant, not 1.
trunc() and int() give you the same distributions as round() when you use the constant 1, so among the three functions they are all equally fair as long as you remember what you're doing when you use one rather than the other. I've proven it with a rough mathematical proof -- I say rough only because I'm not a proper mathematician.
In short, depending on the function (s is the number of sides, and R stands in for RAND() ):
round(f), where f = sR + 0.5
trunc(f), where f = sR + 1
int(f), where f = sR + 1
will all give you the same distribution, meaning that between the three functions they are fair and none favors something more than the others. However...
#2. None of the above gets you around the uneven distribution of possible outcomes of primes not found in the factorization of the base being used (base-10, since we're using decimal; and the prime factorization of 10 is 2 and 5).
With a 10-sided die, where your equation would be
=ROUND(6*RAND()+0.5)
Your distribution of possible values is even across all ten possibilities.
However, if you use the most basic die, a 6-sided die, the distributions favor some rolls over others. Let's assume your random number can only generate down to the thousandths (0.000 ? R ? 0.999). The distribution of possible outcomes of your function are:
1: 167
2: 167
3: 166
4: 167
5: 167
6: 166
So 4 and 6 are always under-represented in the distribution by 1 less than their compatriots. This is true no matter how many decimals you allow, though the distribution gets closer and closer to equal the further towards infinite decimal places you go.
This carries over to all die whose numbers of sides do not factor down to a prime factorization of some exponential values of 2 and 5.
So, then, how can we fix this one, tiny issue in a practical manner that doesn't make our heads hurt or put unnecessary strain on the computer?
Real quick addendum to the above:
Obviously when I put the equation after the example of the 10-sided die, I meant to put a 10*RAND() instead of a 6*RAND(). Oops!
Also, where I have 0.000 ? R ? 0.999, the ?'s are supposed to be less-than-or-equal-to signs but the comments didn't like that. Oh well.
How do you keep adding up the total? I would like to have a cell which keeps adding up the total sum of the two dices, even after a new number is generated in the cells when you refresh or generate new numbers.
So, how do you simulate rolling 12 dice? Do you write int(rand()*6) 12 times?
Is there a simpler way of simulating n dice in Excel?
I've run this code in VBA
Sub generate()
Application.ScreenUpdating = False
Application.Calculation = False
Dim app, i As Long
Set app = Application.WorksheetFunction
For i = 3 To 10002
Cells(i, 3).Value = i - 2
Cells(i, 4).Value = app.RandBetween(2, 12)
Cells(i, 5).Value = app.RandBetween(1, 6) + app.RandBetween(1, 6)
Next
Application.ScreenUpdating = True
Application.Calculation = True
End Sub
But I get the same distribution for both columns 4 and 5
Why ?
@Mohammed
I would expect to get the same distribution as you have effectively used the same function