My Math Forum 3 dice probabilities

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 March 4th, 2017, 08:59 AM #1 Newbie   Joined: Mar 2017 From: Florida Posts: 4 Thanks: 0 3 dice probabilities Here are the well known probability results for rolling two dice. 2 -2.8% 3- 5.6 4 - 8.3 5- 11.1 6- 13.9 7 - 16.7 8 - 13.9 9 - 11.1 10 - 8.3 11 - 5.6 12 - 2.8 Can someone provide similar results for rolling 3 dice that includes summing 2 of the dice together? Example: rolling a 3, 4 and 5 could be 7 and 5, 9 and 3, or 8 and 4. Thanks in advance. Last edited by newbieDes; March 4th, 2017 at 09:00 AM. Reason: typo
 March 4th, 2017, 11:16 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs What I would do is first observe that the sums $S$ will range from 3 to 18, and the total number of outcomes is $N=6^3=216$. We should also expect the probabilities to be symmetric about $S=10.5$. And so we begin with $S=3$. There's only 1 way (the permutations of (1,1,1)) to get 3, so the probability of rolling a sum of 3 is: $\displaystyle P(3)=\frac{1}{216}$ For $S=4$, we find there are 3 permutations of (1,1,2), hence: $\displaystyle P(4)=\frac{3}{216}=\frac{1}{36}$ For $S=5$, we find there are 3 permutations of (1,2,2) and 3 permutations of (1,1,3), hence: $\displaystyle P(5)=\frac{3+3}{216}=\frac{1}{18}$ For $S=6$, we find 3 permutations of (1,1,4), 6 permutations of (1,2,3), and 1 permutation of (2,2,2), hence: $\displaystyle P(6)=\frac{3+6+1}{216}=\frac{5}{108}$ For $S=7$, we find 3 permutations of (1,1,5), 6 permutations of (1,2,4), 3 permutations of (1,3,3), and 3 permutations of (2,2,3), hence: $\displaystyle P(7)=\frac{3+6+3+3}{216}=\frac{5}{72}$ For $S=8$, we find 3 permutations of (1,1,6), 6 permutations of (1,2,5), 6 permutations of (1,3,4), 3 permutations of (2,2,4), 3 permutations of (2,3,3), hence: $\displaystyle P(8 )=\frac{3+6+6+3+3}{216}=\frac{7}{72}$ For $S=9$, we find 6 permutations of (1,2,6), 6 permutations of (1,3,5), 3 permutations (1,4,4), 3 permutations of (2,2,5), 6 permutations of (2,3,4) and 1 permutation of (3,3,3), hence: $\displaystyle P(9)=\frac{6+6+3+3+6+1}{216}=\frac{25}{216}$ For $S=10$, we find 6 permutations of (1,3,6), 6 permutations of (1,4,5), 3 permutations of (2,2,6), 6 permutations of (2,3,5), 3 permutations of (2,4,4), and 3 permutations of (3,3,4), hence: $\displaystyle P(10)=\frac{6+6+3+6+3+3}{216}=\frac{1}{8}$ Now we can use symmetry to fill in the table: $\displaystyle \newcommand\T{\Rule{0pt}{1em}{1em}} \begin{array}{|c|c|} \hline S & P(S) \T \\\hline 3 \T & \dfrac{1}{216} \\\hline 4 \T & \dfrac{1}{36} \\\hline 5 \T & \dfrac{1}{18} \\\hline 6 \T & \dfrac{5}{108} \\\hline 7 \T & \dfrac{5}{72} \\\hline 8 \T & \dfrac{7}{72} \\\hline 9 \T & \dfrac{25}{216} \\\hline 10 \T & \dfrac{1}{8} \\\hline 11 \T & \dfrac{1}{8} \\\hline 12 \T & \dfrac{25}{216} \\\hline 13 \T & \dfrac{7}{72} \\\hline 14 \T & \dfrac{5}{72} \\\hline 15 \T & \dfrac{5}{108} \\\hline 16 \T & \dfrac{1}{18} \\\hline 17 \T & \dfrac{1}{36} \\\hline 18 \T & \dfrac{1}{216} \\\hline \end{array}$ Thanks from topsquark and JeffM1
 March 4th, 2017, 03:52 PM #3 Newbie   Joined: Mar 2017 From: Florida Posts: 4 Thanks: 0 Mark, thanks for the reply. What I wanted to calculate is the odds of rolling 1-12 using three dice but locking two of them (2-12) and using the third one by itself (1-6). Since I mentioned every roll would include locking two dice, the value of the locked pair would never exceed 12 and the value of the single dice would never exceed 6. I don't want to calculate the sum of these two values but want to know the odds of how many times 1-12 will be rolled when each roll involves locking two dice and using one by itself. Hope that is more clear than my original post. Last edited by newbieDes; March 4th, 2017 at 04:04 PM.
 March 4th, 2017, 04:27 PM #4 Senior Member   Joined: May 2016 From: USA Posts: 1,310 Thanks: 550 I am confident that there is a way to calculate whatever probabilities you want provided that we understand what you want. I must admit that I do not want understand. Thanks from MarkFL
March 4th, 2017, 05:02 PM   #5
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From: St. Augustine, FL., U.S.A.'s oldest city

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Quote:
 Originally Posted by JeffM1 I am confident that there is a way to calculate whatever probabilities you want provided that we understand what you want. I must admit that I do not want understand.
Yes, my initial post we made based on my best guess as to what was actually being asked...and I have to admit I still don't understand either.

 March 5th, 2017, 02:43 AM #6 Newbie     Joined: Jun 2016 From: Hong Kong Posts: 20 Thanks: 2 We have a generating function for rolling n dice $\displaystyle (x^1+x^2+x^3+x^4+x^5+x^6)^n$ $\displaystyle (x^1+x^2+x^3+x^4+x^5+x^6)^2=x^2+2x^3+3x^4+4x^5+5x^ 6+6x^7+5x^8+4x^9+3x^{10}+2x^{11}+x^{12}$ observe the coefficient of each $\displaystyle x^m$,you will get the number of ways to get the sum of m for rolling n dice $\displaystyle (x^1+x^2+x^3+x^4+x^5+x^6)^n=\frac{x^n(1-x^6)^n}{(1-x)^n}=x^n(\sum_{r=0}^n C_n^r (-1)^r x^{6r})(\sum_{k=0}^\infty C_{n-1+k}^k x^k)$ when n=3,m=4 $\displaystyle x^3(\sum_{r=0}^3 C_3^r (-1)^r x^{6r})(\sum_{k=0}^\infty C_{2+k}^k x^k)$ look for r=0,k=1 $\displaystyle C_{2+1}^1=3$ there are 3 ways to get 4,which are "1+1+2","1+2+1","2+1+1" when n=3,m=18 $\displaystyle x^3(\sum_{r=0}^3 C_3^r (-1)^r x^{6r})(\sum_{k=0}^\infty C_{2+k}^k x^k)$ look for r=0,1,2 and k=15,9,3 $\displaystyle C_{2+15}^{15}-3C_{2+9}^9+3C_{2+3}^3=1$
 March 5th, 2017, 06:20 AM #7 Newbie   Joined: Mar 2017 From: Florida Posts: 4 Thanks: 0 Let’s try this. Bob goes to the local 12 theatre Cineplex every day and chooses two movies to see based on the results of rolling three dice. One dice is chosen to determine his first theatre choice, and the other two dice are added together to determine his 2nd theatre choice. Example: Bob rolls three dice and can choose to see two movies in these theatres: 6,6,6= Theatre 12 & 6 1,1,1= Theatre 1 and 2 2,4,5= Theatre 2,4,5,6,7,9 3,4,5= Theatre 3,4,5,7,8,9 4,4,6= Theatre 4,6,8,10 Question- How often will Bob be able to see a movie in theatre 1? Theatre2? Etc.
 March 7th, 2017, 04:07 PM #8 Senior Member   Joined: Oct 2013 From: New York, USA Posts: 637 Thanks: 85 Here are the probabilities based on a spreadsheet I made (hopefully without any mistakes): 1: 91/216 2: 104/216 3: 115/216 4: 128:216 5: 139/216 6: 152/216 7: 92/216 8: 75/216 9: 61/216 10: 46/216 11: 30/216 12: 16/216 Sum: 1,049/216
March 7th, 2017, 05:23 PM   #9
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Quote:
 Originally Posted by newbieDes Let’s try this. Bob goes to the local 12 theatre Cineplex every day and chooses two movies to see based on the results of rolling three dice. One dice is chosen to determine his first theatre choice, and the other two dice are added together to determine his 2nd theatre choice. Example: Bob rolls three dice and can choose to see two movies in these theatres: 6,6,6= Theatre 12 & 6 1,1,1= Theatre 1 and 2 2,4,5= Theatre 2,4,5,6,7,9 3,4,5= Theatre 3,4,5,7,8,9 4,4,6= Theatre 4,6,8,10 Question- How often will Bob be able to see a movie in theatre 1? Theatre2? Etc.
Someone posted this on another forum:

"There are four variables here: A, B and C are dice (where Nature chooses a value of 1 to 6 for each) and D (the player's decision of which die to stand alone). This gives you 6x6x6x3 = 648 total states."

1 36 00 36 / 216 = 16.67%
2 36 06 42 / 216 = 19.44%
3 36 12 48 / 216 = 22.22%
4 36 18 54 / 216 = 25.00%
5 36 24 60 / 216 = 27.78%
6 36 30 66 / 216 = 30.56%
7 00 36 36 / 216 = 16.67%
8 00 30 30 / 216 = 13.89%
9 00 24 24 / 216 = 11.11%
10 00 18 18 / 216 = 08.33%
11 00 12 12 / 216 = 05.56%
12 00 06 06 / 216 = 02.78%

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