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April 6th, 2017, 02:42 PM   #1
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Thumbs up 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
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April 6th, 2017, 10:13 PM   #2
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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
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April 7th, 2017, 06:51 AM   #3
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Quote:
Originally Posted by romsek View Post
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.
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April 7th, 2017, 07:13 AM   #4
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Quote:
Originally Posted by Kvan79 View Post
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.
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April 7th, 2017, 07:38 AM   #5
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Quote:
Originally Posted by romsek View Post
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?
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April 7th, 2017, 07:58 AM   #6
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Quote:
Originally Posted by Kvan79 View Post
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 08:11 AM.
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April 7th, 2017, 08:04 AM   #7
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Quote:
Originally Posted by romsek View Post
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.
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April 7th, 2017, 08:56 AM   #8
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Quote:
Originally Posted by Kvan79 View Post
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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