In early February Sujit asked a question at Chandoo.org, original post.
I require a formula stating criteria [0%-25% output will be 0, 26%-50% output will be 0.1, 51%-75% output will be 0.2, 76%-100% output will be 0.3 & 100% + output will be 0.4]
Kyle, responded with a neat Sumproduct formula
=SUMPRODUCT((B3>{0.25,0.5,0.75,1})*0.1)
I think it is so neat that it is worthy of sharing and detailing here at Formula Forensics:
So today we will pull Kyle’s answer apart to see what’s inside.
Kyle’s Formula
As usual we will work through this formula using a sample file for you to follow along. Download Here.
Kyle’s formula is a Sumproduct based formula
=SUMPRODUCT((B3>{0.25,0.5,0.75,1})*0.1)
Lets look at cell C3 as our example.
;
In C3 we see the formula: =SUMPRODUCT((B3>{0.25,0.5,0.75,1})*0.1)
Which consists of a Sumproduct function and a formula inside the sumproduct.
We know from Formula Forensics 007 that Sumproduct, Sums the Product of the Arrays, and that when there is only 1 array it simply sums the array elements.
In this case the Sumproduct only has a single array as an element
=SUMPRODUCT((B3>{0.25,0.5,0.75,1})*0.1)
and so the (B3>{0.25,0.5,0.75,1})*0.1 component must return an Array of elements for the Sumproduct to sum.
If we now look at the (B3>{0.25,0.5,0.75,1})*0.1 component.
We can see that it consists of a comparison B3>{0.25,0.5,0.75,1}
The result of the comparison is Multiplied by 0.1.
Sujit’s orginal question asked: 0%-25% output will be 0, 26%-50% output will be 0.1, 51%-75% output will be 0.2, 76%-100% output will be 0.3 & 100% + output will be 0.4
And Kyles formula is using B3>{0.25,0.5,0.75,1} to work out which category the value in B3 belongs to.
We can see this if in a blank cell say C5: we enter the following:
= B3>{0.25,0.5,0.75,1} press F9 not Enter.
Excel will respond with ={TRUE,TRUE,TRUE,FALSE}
This is showing us that the 1st, 2nd and 3rd elements in the formula: B3>{0.25,0.5,0.75,1}, are True
In our example the value in B3 is 80% which is 0.8 which is Greater than 0.25 and Greater than 0.5 and Greater than 0.75, but Not Greater than 1.0.
The next part of Kyle’s formula is (B3>{0.25,0.5,0.75,1})*0.1
In a blank cell say C7: enter the following:
= B3>{0.25,0.5,0.75,1}*0.1 press F9 not Enter.
Excel will respond with ={0.1,0.1,0.1,0}
This is showing us the result of
=(B3>{0.25,0.5,0.75,1})*0.1
={TRUE,TRUE,TRUE,FALSE} *0.1
={0.1,0.1,0.1,0}
Sumproduct now only has to add up the Array
=Sumproduct({0.1,0.1,0.1,0})
Which it does returning 0.3.
The Neat Part
The neat part of this is that Kyle has used the 0.1 Multiplier to Force the array to an array of Numbers for Sumproduct to sum.
Had Kyle used: =SUMPRODUCT((B3>{0.25,0.5,0.75,1}))*0.1
Excel would have returned an answer of 0
This is because as we saw in Formula Forensics 007, Sumproduct doesn’t know what to do with the array of True/False, they need to be converted to numerical equivalents for Sumproduct to operate on.
In a spare cell, say C9, enter: =SUMPRODUCT((B9>{0.25,0.5,0.75,1}))*0.1
Excel will respond with 0
Of course that can be fixed by using a double degative of a 1* inside the formula
In a spare cell, say C10, enter either:
=SUMPRODUCT(1*(B9>{0.25,0.5,0.75,1}))*0.1
or
=SUMPRODUCT(- -(B9>{0.25,0.5,0.75,1}))*0.1
Excel will respond with 0.3 as it should
Except that the formula is longer and now has to do 1 more multiplication.
Download
You can download a copy of the above file and follow along, Download Here.
Formula Forensics “The Series”
You can learn more about how to pull Excel Formulas apart in the following posts
We Need Your Help
I have received a few more ideas since last week and these will feature in coming weeks.
I do need more ideas though and so I 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 like above or;
If you have a formula that you would like explained but don’t want to write a post also send it 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