My Math Forum Help with Dice Probability

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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 6-sided 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 1-6. 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: CA Posts: 1,112 Thanks: 580 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
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 Originally Posted by romsek 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
Why would rolling at least a 2 not be a probability of 1, when every value is greater than or equal to 2? You should always get at least 2 when rolling one die.

April 7th, 2017, 06:13 AM   #4
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 Originally Posted by Kvan79 Why would rolling at least a 2 not be a probability of 1, when every value is greater than or equal to 2? You should always get at least 2 when rolling one die.
that's not a CDF. A CDF is the function that the roll is less than or equal to k.

What you are describing is 1 - CDF.

April 7th, 2017, 06:38 AM   #5
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Quote:
 Originally Posted by romsek that's not a CDF. A CDF is the function that the roll is less than or equal to k. What you are describing is 1 - CDF.
I'm trying to figure out a general formula for a given number of dice and their total, in the format of "at least n". Any ideas on that?

April 7th, 2017, 06:58 AM   #6
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Quote:
 Originally Posted by Kvan79 I'm trying to figure out a general formula for a given number of dice and their total, in the format of "at least n". Any ideas on that?
$\left( \begin{array}{cc} 2 & 1 \\ 3 & \frac{2}{3} \\ 4 & \frac{1}{3} \\ 5 & \frac{1}{6} \\ 6 & 0 \\ \end{array} \right)$

$\left( \begin{array}{cc} 3 & 1 \\ 4 & 1 \\ 5 & \frac{8}{9} \\ 6 & \frac{2}{3} \\ 7 & \frac{4}{9} \\ 8 & \frac{2}{9} \\ 9 & \frac{1}{12} \\ 10 & \frac{1}{36} \\ 11 & 0 \\ \end{array} \right)$

Last edited by romsek; April 7th, 2017 at 07:11 AM.

April 7th, 2017, 07:04 AM   #7
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Quote:
 Originally Posted by romsek take 1 - the tables I posted.
That works for 1 or 2 dice, but it isn't a general formula for any given number of dice.

April 7th, 2017, 07:56 AM   #8
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 Originally Posted by Kvan79 That works for 1 or 2 dice, but it isn't a general formula for any given number of dice.
ok.. I have a formula for you.

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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