Question

How to do a simulation of three fair dice 1000 times? Could someone give me some help on how to start this?

Answer 1

You can use any programming language or even Excel. In Excel use the formula =randbetween(1,6). Place it in three columns then copy down 1000 rows. A bit tedious but works.

Answer 2

Fill series in Excel = home ribbon, fill on right side of screen, select series, then 1, 1000, and columns

Answer 3

okay thank you, I now have to calculate the sum of each column, I think. It says: " Create a relative frequency graph of the sums obtained" ...my question is...how do i obtain the sums? Would it be the 1001th cell in the first second and third columns? or a new cell block

Answer 4

Guessing about your problem, I believe that you want to create a sum of each set of 3 digits. IN the fourth column, type =sum(A1:B1) and fill down. From there you need to calculate the lookup.

Answer 5

Here is a Java solution


public static void main(String[] args) {
// TODO code application logic here
int die1=0;
int die2=0;
int die3 = 0;
int roll = 0;
int totals[] = {0,0,0,0,0,0,0,0,0,0,00,0,0,0,0,0,0,0,0};
for(int x=0; x < 1000; x++){
die1 = (int) (Math.random()*6) + 1;
die2 = (int) (Math.random()*6) + 1;
die3 = (int) (Math.random()*6) + 1;
roll = die1 + die2 + die3;
totals[roll]++;
System.out.println(die1+ " "+ die2+ " "+ die3);
}

for(int x=3; x < 18; x++){
System.out.println(x + " = " + totals[x]);

}
}
}

Answer 6

to complete the counting you need to use countif, see my file I attached.

Answer 7

i am so confused, ton the attatched file, there are the first 10 trials out of the 1000 i have, how do i create the relative frequency?

Answer 8

in the next blank column, enter the numbers 3, 4,5,... 18 each in their own cell,down

then next to the 3 enter the expression =countif(your range of sums here, 3)

repeat for each number. If you are good, you can use the cell nuberholding 3 instead of 3 and simply drag down.

Answer 9

what would i enter as my range?

Answer 10

or how would i enter it, because i know my range of sums is 3-18

Answer 11

based on your picture D2:d10

Answer 12

so it would be D2:D1000

Answer 13

ex: =countif(d2:d1000,3)
yes

Answer 14

once you get the values, highlite from the 3 to the formula next to countif and create a graph
It should represent a curve with the middle values the highest

Answer 15

please assume that the 25 rows are really all of my 1000. The D block is: A+B+C blocks, for the E block, i would type in the function: = ? inorder to get the relatvie frequency

Answer 16

i type into E2: =Countif(D2:D1000,3) and it is not excepting it

Answer 17

=countif(d2:d25,3)
=countif(d2:d25,4)
=countif(d2:d25,5)
=countif(d2:d25,6)
=countif(d2:d25,7)
=countif(d2:d25,8)
=countif(d2:d25,9)
=countif(d2:d25,10)
=countif(d2:d25,11)
=countif(d2:d25,12)

etc

Answer 18

it worked! and if went from 666666...55555555...4444...3333333.....222222.....11111 all the way down to 1000?

Answer 19

numbers are all too high, I think that your fill series filled numbers instead of random functions, check out my file - I assure you it is clean and malware free.

Answer 20

i opened up the attatchment, and i sort of understand what is going on, however, what is the G and H blocks? and why do they start at the 11th row?

Answer 21

I am not sure why i started them there, blame it on my really slow laptop. The numberss in G represents the possible sums. The numbers in H are the actual counts of each possible sum

Answer 22

okay, i understand, i am going to try to enter what you did in the G and H blocks and see what i come up with

Answer 23

when i enter =Countif(D2:D1000,3) it just gives me a whole bunch of sixes, looks like this: 6666666666.....5555555.........444444444........3333333.......22222222.....1111111......00000000 ):

Answer 24

whatever, i will continue tomorrow, thankk you for your help!

Answer 25

good luck

Answer 26

thanks