A few weeks back I was asked “Is it possible to setup a control and then drag it down a range, so that it links to all the cells below it?”
The answer is, of course, No.
But it got me thinking about why not allow one control to control a number of cells.
This post describes the solution, One Control Three Cells.
But it could just as easily be applied to a larger group of controls in a much larger system.
I have attached a sample file demonstrating the technique: Download sample file
The Old
In the sample file select the Old worksheet.
Typically if you had 3 cells and wanted to add automation, you would add a control to each cell.
Here I have added 3 controls. Each Control in Column E controls the Cells value to the left of it.
Each Control is independent and has no relationship to other cells or other controls.
Each control is setup and linked as shown below to a single cell.
This whole setup has to be applied individually to each control and associated cell.
The Cell link: dialog above cannot have a range
Well it can hold a range, but it only links the control to the upper left cell of the range, C3 in the example above.
But this got me thinking, why not link the control’s Cell Link to a Named Formula, which would return the range based on say where the active cell was.
The New
Change to the New Worksheet.
Notice how we now have a single control next to the 3 cells we wish to control.
You can see that in action here
Lets first examine what has been setup, then we will work through how it works.
First, Goto the Name Manager in the Formulas, Name Manager tab.
There are 3 Named Formula setup
SelectedRow : is a direct Link to cell A1
ControlRange: is a direct Link to cells C3:C5
ControlLink : is a named Formula containing a formula =OFFSET(New!$C$1,SelectedRow-1,0)
Next Right click on the Control and notice that it is linked to the ControlLink Named Formula.
There is more, but lets follow this through first.
Cell A1 “SelectedRow” contains the value 4.
The Named Formula ControlLink has a formula =OFFSET(New!$C$1, SelectedRow-1, 0)
which evaluates to =OFFSET(New!$C$1, 4-1, 0)
which simplifies to =OFFSET(New!$C$1, 3, 0)
The offset of C1 by 3 rows and 0 columns is C4
so the Named Formula ControlLink =OFFSET(New!$C$1, SelectedRow-1, 0)
returns the address of C4
So the Control uses an Address of C4 when the value of A1 is 4
But we didn’t change cell A1 ?
I did say there was more, and the more is a small piece of VBA code, which does some checking for us and places an appropriate value in A1
Goto VBA by pressing Alt+F11
Double click on the Sheet1(New) object and you should now see the code in the Code Pane
This tiny piece of code is the secret behind what makes this technique work.
Lets look at what it does
Private Sub Worksheet_SelectionChange(ByVal Target As Range)
If Intersect(Target, Range(“ControlRange”)) Is Nothing Then
Range(“SelectedRow”).Value = 0
Exit Sub
End If
Range(“SelectedRow”).Value = Target.Row
Application.CalculateFull
End Sub
The code is encapsulated in what is known as a Worksheet event.
Worksheet events, as the name implies, are events that occur on the worksheet.
In this case it is the SelectionChange event. That is every time you change the cell by clicking on it or using the keyboard arrows etc to change the active cell, this event is triggered and the enclosed code executed.
When the event is triggered the code starts and a variable Target is assigned to the new active cell. It is the Target of the events occurrence, ie: Your click on another cell.
The next piece of code handles what happens next
If Intersect(Target, Range(“ControlRange”)) Is Nothing Then
Range(“SelectedRow”).Value = 0
Exit Sub
End If
It basically says If the Target and the ControlRange Don’t Intersect then do the enclosed code
That is if the Target doesn’t intersect with the ControlRange, then set the SelectedRange cell A1 to 0
Then exit the subroutine
This is done so that cells that are selected whilst using the worksheet don’t interfere with the control.
But the important thing is what happens if the Target and ControlRange do intersect
The code says If there is not an intersection do what is inside the If / End If statements
If Intersect(Target, Range(“ControlRange”)) Is Nothing Then
Range(“SelectedRow”).Value = 0
Exit Sub
End If
But if the two ranges Do Intersect, the code simply passes over the included code and continues past to the next code.
The next code is
Range(“SelectedRow”).Value = Target.Row
Application.CalculateFull
This is where the SelectedRow cell A1 is assigned the value which is the Row number of the Target cell.
That is if we click in a cell in the ControlRange, the SelectedRow is assigned the value of the Target cells Row.
The worksheet is then calculated. This simply forces the named Formula to update.
Then the VBA finishes executing.
When the Worksheet was recalculated just above, the LinkedCell was updated.
Now when a user presses the Spin Button Control, it will use the new value in the LinkedCell named range as the Link cell and update the value of the cell according to whether you pressed the Up or Down arrow.
Final
This code can be applied to any number of controls as well as to complex ranges
If you wanted to control the values in the 9, dashed green, cells shown below highlighted
You would change the formula for ControlRange to
ControlRange : =New!$B$8:$B$10,New!$C$11:$C$13,New!$B$14:$B$16
Comments:
What do you think about this technique?
Let me know in the comments below:
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