This week in Formula Forensics we’ll look at, Zebra Stripes and Checker-board Conditional Formatting.
This idea is inspired by a number of posts over the past few years asking about zebra stripes but specifically BobR who in in June 2011, also asked about Checkerboards in the post: Want to be an excel conditional-formatting Rock Star, Comment No. 154.
I got the conditional format for alternating row and column colors,
Is there a conditional format to make it a checkerboard whereas the cell A2 will remove either the conditional for the row or column and then alternately to A4, B1, B3 etc?
Chandoo responded fairly quickly with this Conditional Formatting formula:
=IF(MOD(ROW(),2)=1,MOD((ROW()-1)*8+COLUMN(),2)=0,MOD((ROW()-1)*8+COLUMN(),2)=1)
Unbeknownst to Chandoo I posted this about a minute later:
=ISODD(ROW()+COLUMN())
Both formula correctly answer BobR’s question.
So today we’re going to pull apart Zebra Stripes and Checker Boards and see what makes them tick.
As always you can follow along in a download file here: Download File.
ZEBRA STRIPES
Zebra Stripes as Conditional Formatting is simply applied using a simple formula within Conditional Formatting.
=MOD(ROW(),2)=0
Conditional Formatting requires a formula that returns a boolean “True” to apply a format or a Boolean “False” to not Apply a format.
So the formula is better read as: If MOD(ROW(),2)=0
And If MOD(ROW(),2)=0, the formula will evaluate as True
This is best evaluated as 3 columns on a worksheet.
In cells
B5:B10 The formula =Row() returns the Row Number
C5:C10 The formula =Mod(Row() ,2) returns the Mod of Row Number, divided by 2
The Mod function returns the remainder of the division of the Row Number divided by 2,
So in Row 5, Mod(Row(),2) = Mod(5, 2) = 5/2 = 2 Remainder 1 = 1
and in Row 6, Mod(Row(),2) = Mod(6, 2) = 6/2 = 3 Remainder 0 = 0
D5:D10 The formula =Mod(Row() ,2)=0 checks the remainder against the value 0
This is what evaluates to either True or False depending on the Row number.
Where the Values are True the Format will be applied (Even Rows)
The Conditional Formatting can be applied to Odd Rows If the Formula is slightly altered
=Mod(Row() ,2)=1
Similarly the formatting can be applied to Columns using
=MOD(COLUMN(),2)=0/1
CHECKER BOARDS
RobR received two responses to his Checker-Board Conditional Formatting request.
=IF(MOD(ROW(),2)=1,MOD((ROW()-1)*8+COLUMN(),2)=0,MOD((ROW()-1)*8+COLUMN(),2)=1)
and
=ISODD(ROW()+COLUMN())
Lest see what’s inside these two formula.
=IF(MOD(ROW(),2)=1,MOD((ROW()-1)*8+COLUMN(),2) =0, MOD( (ROW() -1)*8+COLUMN(),2)=1)
This is a simple If Formula with 3 components
=IF(MOD(ROW(),2)=1,MOD((ROW()-1)*8+COLUMN(),2)=0,MOD((ROW()-1)*8+COLUMN(),2)=1)
If Condition MOD(ROW(),2)=1
Value if True: MOD((ROW()-1)*8+COLUMN(),2)=0
Value if False: MOD((ROW()-1)*8+COLUMN(),2)=1
The If Condition is already known to us, as it’s the same formula used in the Zebra Stripes above.
It evaluates to True when it is on an Odd Row.
So when it is an Odd numbered Row Excel will look at MOD((ROW()-1)*8+COLUMN(),2)=0
And when it is an Even numbered Row Excel will look at MOD((ROW()-1)*8+COLUMN(),2)=1
We can notice that these are the same formulas which have a different ending of =0 and =1
MOD((ROW()-1)*8+COLUMN(),2)=0
This section Takes each Row subtracts 1 and then multiplies this number by 8. This can be expressed as simply as saying multiply the Row * 8.
This will always return an Even Number and could have been simplified to Row()*2
MOD((ROW()-1)*8+COLUMN(),2)=0
The next bit adds the column number to the previous Even Number.
So now this part will be Odd when the column is Odd and Even when the column is Even.
MOD((ROW()-1)*8+COLUMN(),2)=0
The remainder of the formula is the same as the Zebra Stripes formula.
An Odd Number (Odd Columns) in the section above will return a 1 as the result of =Mod(Odd,2)
An Even Number (Even Columns) in the section above will return a 0 as the result of =Mod(Odd,2)
When evaluated against 0 will return True for Even Columns and False for Odd Columns.
Now the exact same happens in the False section of the If formula except that it is evaluated against 1.
=ISODD(ROW()+COLUMN())
I tackled this problem from a different direction to Chandoo.
Knowing that Even + Even = Even and Even + Odd = Odd and that the row and Column Numbers increase in each direction by 1 each Row/Column, it was simply a matter of adding the Row and Column numbers together and checking if it was Odd or Even
The Excel function IsOdd() and IsEven() both return a Boolean “True” if the contents are Odd or “Even” respectively. This negates an external truth check as described above.
This is easily shown by adding a formula to the Checker area
=Row()+Column()
Excel 2003: The above formula won’t work in Excel 2003.
Try this instead =Mod(Row()+Column(),2)=1
If the alternate shading is required a switch to
=ISEVEN(ROW()+COLUMN())
Does the trick.
Excel 2003: The above formula won’t work in Excel 2003.
Try this instead =Mod(Row()+Column(),2)=0
Learn More About Conditional Formatting Here:
http://chandoo.org/wp/2009/03/13/excel-conditional-formatting-basics/
and
http://chandoo.org/wp/2008/03/13/want-to-be-an-excel-conditional-formatting-rock-star-read-this/
and
http://chandoo.org/wp/2008/10/14/more-than-3-conditional-formats-in-excel/
DOWNLOAD
You can download a copy of the above file and follow along, Download Here.
OTHER POSTS IN THIS SERIES
You can learn more about how to pull Excel Formulas apart in the following posts
Formula Forensics 001 – Tarun’s Problem
Formula Forensics 002 – Joyce’s Question
Formula Forensics 003 – Lukes Reward
Formula Forensics 004 – Freds Problem
We Need Your Help !
If you have a neat formula that you would like to share and explain, try putting pen to paper and draft up a Post as Luke did in Formula Forensics 003. or this post.
If you have a formula that you don’t understand and would like explained but don’t want to write a post also send it in to Chandoo or Hui.
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