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April 6th, 2017, 01:42 PM  #1 
Newbie Joined: Apr 2017 From: Lansing, MI Posts: 4 Thanks: 0 Math Focus: Multivariable calculus  Help with Dice Probability
Hello, I'm trying to figure this probability problem out. There are 6sided dice, and I'm trying to figure out a general formula of odds for a given number of dice and their total, in the format of "at least n". The sides are not a straight 16. Two sides have 2, two sides have 3, one side has 4, and one has 5. I made a table of odds for 1 and 2 dice that I think is right, but I'm having trouble with a general rule besides the denominator is 6^n, the minimum is 2*n, and maximum is 5*n for n= number of dice. Here is what I have so far: 1 die 2 dice At least... At least... 2  1  4  1 3  .667  5  .889 4  .333  6  .667 5  .167  7  .444  8  .222  9  .083  10  .028 
April 6th, 2017, 09:13 PM  #2 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,493 Thanks: 752 
i get cdfs of 1 die roll $\left( \begin{array}{cc} 2. & 0.333333 \\ 3. & 0.666667 \\ 4. & 0.833333 \\ 5. & 1. \\ \end{array} \right)$ 2 die rolls $\left( \begin{array}{cc} 4. & 0.111111 \\ 5. & 0.333333 \\ 6. & 0.555556 \\ 7. & 0.777778 \\ 8. & 0.916667 \\ 9. & 0.972222 \\ 10. & 1. \\ \end{array} \right)$ values below those listed in the tables have probability 0 values above those listed in the tables have probability 1 
April 7th, 2017, 05:51 AM  #3  
Newbie Joined: Apr 2017 From: Lansing, MI Posts: 4 Thanks: 0 Math Focus: Multivariable calculus  Quote:
 
April 7th, 2017, 06:13 AM  #4  
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,493 Thanks: 752  Quote:
What you are describing is 1  CDF.  
April 7th, 2017, 06:38 AM  #5 
Newbie Joined: Apr 2017 From: Lansing, MI Posts: 4 Thanks: 0 Math Focus: Multivariable calculus  
April 7th, 2017, 06:58 AM  #6  
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,493 Thanks: 752  Quote:
\begin{array}{cc} 2 & 1 \\ 3 & \frac{2}{3} \\ 4 & \frac{1}{3} \\ 5 & \frac{1}{6} \\ 6 & 0 \\ \end{array} \right)$ $\left( \begin{array}{cc} 3 & 1 \\ 4 & 1 \\ 5 & \frac{8}{9} \\ 6 & \frac{2}{3} \\ 7 & \frac{4}{9} \\ 8 & \frac{2}{9} \\ 9 & \frac{1}{12} \\ 10 & \frac{1}{36} \\ 11 & 0 \\ \end{array} \right)$ Last edited by romsek; April 7th, 2017 at 07:11 AM.  
April 7th, 2017, 07:04 AM  #7 
Newbie Joined: Apr 2017 From: Lansing, MI Posts: 4 Thanks: 0 Math Focus: Multivariable calculus  
April 7th, 2017, 07:56 AM  #8  
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,493 Thanks: 752  Quote:
suppose you have $n$ dice let $p(z) =\displaystyle \frac{1}{6 z^3}+\frac{1}{6 z^2}+\frac{1}{3 z}+\frac{1}{3}$ expand out $\left(p(z)\right)^n = \displaystyle \sum_{k=0}^{3n} c_k z^{k}$ $P[\text{rolling a sum of 2n+k}]=c_k$ so for example with 2 dice $(p(z))^2 =\displaystyle \frac{1}{36 z^6}+\frac{1}{18 z^5}+\frac{5}{36 z^4}+\frac{2}{9 z^3}+\frac{2}{9 z^2}+\frac{2}{9 z}+\frac{1}{9}$ we can read off the probabilities as $P[4]= \dfrac 1 9$ $P[5]=\dfrac 2 9$ $P[6]=\dfrac 2 9$ $P[7]=\dfrac 2 9$ $P[8]=\dfrac {5}{36}$ $P[9]=\dfrac {1}{18}$ $P[10]=\dfrac {1}{36}$ There's not going to be a closed form formula for the CDF of this. For a given number of dice you're just going to have to compute the probability table and then make the appropriate sums to obtain the probability of being at least a given value. NOTE: This works for this die you described only. A different die will have a different polynomial.  

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