
Probability and Statistics Basic Probability and Statistics Math Forum 
 LinkBack  Thread Tools  Display Modes 
March 4th, 2017, 09:59 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 10:00 AM. Reason: typo 
March 4th, 2017, 12:16 PM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 461 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: $\displaystyle \newcommand\T{\Rule{0pt}{1em}{1em}} \begin{array}{cc} \hline S & P(S) \T \\\hline 3 \T & \dfrac{1}{216} \\\hline 4 \T & \dfrac{1}{36} \\\hline 5 \T & \dfrac{1}{18} \\\hline 6 \T & \dfrac{5}{108} \\\hline 7 \T & \dfrac{5}{72} \\\hline 8 \T & \dfrac{7}{72} \\\hline 9 \T & \dfrac{25}{216} \\\hline 10 \T & \dfrac{1}{8} \\\hline 11 \T & \dfrac{1}{8} \\\hline 12 \T & \dfrac{25}{216} \\\hline 13 \T & \dfrac{7}{72} \\\hline 14 \T & \dfrac{5}{72} \\\hline 15 \T & \dfrac{5}{108} \\\hline 16 \T & \dfrac{1}{18} \\\hline 17 \T & \dfrac{1}{36} \\\hline 18 \T & \dfrac{1}{216} \\\hline \end{array}$ 
March 4th, 2017, 04:52 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 112 using three dice but locking two of them (212) and using the third one by itself (16). 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 112 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 05:04 PM. 
March 4th, 2017, 05:27 PM  #4 
Senior Member Joined: May 2016 From: USA Posts: 595 Thanks: 251 
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.

March 4th, 2017, 06:02 PM  #5 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 461 Math Focus: Calculus/ODEs  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, 03:43 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 $\displaystyle (x^1+x^2+x^3+x^4+x^5+x^6)^n=\frac{x^n(1x^6)^n}{(1x)^n}=x^n(\sum_{r=0}^n C_n^r (1)^r x^{6r})(\sum_{k=0}^\infty C_{n1+k}^k x^k)$ 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, 07:20 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, 05:07 PM  #8 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 507 Thanks: 74 
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, 06:23 PM  #9  
Newbie Joined: Mar 2017 From: Florida Posts: 4 Thanks: 0  Quote:
"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%  

Tags 
dice, probabilities, probablities 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Dice, is this right?  johnmushroom  Probability and Statistics  30  October 6th, 2014 02:11 PM 
If you roll 10sided dice, what is the prob that two dice add up to either 19 or 3?  mepoom  Probability and Statistics  2  September 12th, 2014 05:05 PM 
I really need help in my probablities home work  KhalDrogo  Algebra  3  October 14th, 2012 11:00 AM 
Calculating Probablities with Poker hands  Nicolaos  Advanced Statistics  1  June 12th, 2009 02:52 PM 
Dice probability with multicolored dice pips  Brimstone  Algebra  10  May 30th, 2008 07:36 PM 