3
$\begingroup$

This is a dice problem

1) I want to calculate the probability to have more than X throwing 3 dice of 6 faces. X = addition of the result of the 3 dice.

2) This is the first step but if you can also provide me a solution to calculate the probability to have more than X with Y dices of Z faces it would be really great.

$\endgroup$
  • 1
    $\begingroup$ What is your attempt for (1)? $\endgroup$ – Jack May 30 '16 at 14:07
4
$\begingroup$

Your concrete problem is already solved here: If we throw three dice.

The general problem is equivalent to counting the number of ways of distributing $X-Y$ balls into $Y$ bins with limited capacity $Z-1$. This problem is solved at Balls In Bins With Limited Capacity using inclusion-exclusion. The result is

$$ \sum_{t=0}^Y(-1)^t\binom Yt\binom{X-tZ-1}{Y-1}\;, $$

where, contrary to convention, the binomial coefficient is taken to be zero for negative upper index.

This is the count of outcomes with sum exactly $X$; to get the probability of a sum of more than $X$, we need to sum from $X+1$ to $YZ$ and divide by the number $Z^Y$ of equiprobable outcomes:

$$ Z^{-Y}\sum_{x=X+1}^{YZ}\sum_{t=0}^Y(-1)^t\binom Yt\binom{x-tZ-1}{Y-1}=Z^{-Y}\sum_{t=0}^Y(-1)^t\binom Yt\left(\binom{YZ-tZ}Y-\binom{X-tZ}Y\right)\;. $$

For $Y=3$, $Z=6$, this is

\begin{align} &\frac1{216}\sum_{t=0}^3(-1)^t\binom 3t\left(\binom{18-6t}3-\binom{X-6t}3\right)\\ ={}&\frac1{216}\left(\binom{18}3-\binom X3-3\left(\binom{12}3-\binom{X-6}3\right)+3\left(\binom63-\binom{X-12}3\right)\right)\\ ={}&1-\frac1{216}\left(\binom X3-3\binom{X-6}3+3\binom{X-12}3\right)\;, \end{align}

where again binomial coefficients with negative upper index are taken to be zero. Distinguishing the three cases, we can write this as

$$ \frac1{1296}\begin{cases} -X^3+3X^2-2X+1296&3\le X\lt9\;,\\ 2X^3-60X^2+436X+288&6\le X\lt15\;,\\ -X^3+57X^2-1082X+6840&12\le X\le18 \end{cases} $$

(where I intentionally wrote the maximal overlapping ranges to exhibit the symmetry more clearly). As far as I checked, the results coincide with those of the concrete calculation linked to above.

$\endgroup$
  • 2
    $\begingroup$ And with the method in the link you provided one can even compute the probability for different dice $\endgroup$ – YannickSSE May 31 '16 at 8:36
  • $\begingroup$ It becomes fully according to conventions if you write the second binomial of the first equation in the symmetric way, i.e. $\sum_{t=0}^Y(-1)^t\binom Yt\binom{X-tZ-1}{X-Y-tz}\;$ ,which is the preferrable way to write such formula $\endgroup$ – G Cab Jul 14 '16 at 0:46
0
$\begingroup$

In order to put @joriki's answer in a more "compact" and generalized way, let's put $$ \begin{gathered} N_{\,b} (s,\;r,\;m)\quad \left| {\;0 \leqslant \text{integers }s,m,r} \right.\quad = \hfill \\ = \text{No}\text{.}\;\text{of}\;\text{solutions}\;\text{to}\;\left\{ \begin{gathered} 0 \leqslant \text{integer}\;x_{\,j} \leqslant r \hfill \\ x_{\,1} + x_{\,2} + \; \cdots \; + x_{\,m} = s \hfill \\ \end{gathered} \right. = \hfill \\ = \sum\limits_{\left( {0\, \leqslant } \right)\,j\,\left( { \leqslant \,m} \right)} {\left( { - 1} \right)^j \left( \begin{gathered} m \hfill \\ j \hfill \\ \end{gathered} \right)\left( \begin{gathered} s + m - 1 - j\left( {r + 1} \right) \\ s - j\left( {r + 1} \right) \\ \end{gathered} \right)} \hfill \\ \end{gathered} $$

i.e., with the formula presented by @joriki but with the second binomial written in the symmetrical way, so that everything is according to the binomial defined as: $$ \left( \begin{gathered} x \\ q \\ \end{gathered} \right) = \left\{ \begin{gathered} \frac{{x^{\,\underline {\,q\,} } }} {{q!}}\;\;0 \leqslant \text{integer}\;q \hfill \\ 0\quad \;\;\text{otherwise}\, \hfill \\ \end{gathered} \right.\;\; $$

Refer to Problem of rolling dice for further considerations.

Note that we have better and consider dice numbered $0$ to $r$, because that simplifies dealing with the above formula and its paramters ranges.
So, in the general case 2) you proposed we have:
$m=Y=$ number of dices, $r=F-1$ because of considering the facets numbered $0$ to $r$, $s=X-m$ deducting $1$ from the value of each face. Then: $$ \begin{gathered} M_{\,b} (s,\;r,\;m)\quad \left| {\;0 \leqslant \text{integers }s,m,r} \right.\quad = \hfill \\ = \sum\limits_{0\, \leqslant \,i\, \leqslant \,s} {N_{\,b} (i,\;r,\;m)} = \hfill \\ = \sum\limits_{0\, \leqslant \,i\, \leqslant \,s} {\;\sum\limits_{\left( {0\, \leqslant } \right)\,j\,\left( { \leqslant \,m} \right)} {\left( { - 1} \right)^j \left( \begin{gathered} m \hfill \\ j \hfill \\ \end{gathered} \right)\left( \begin{gathered} i + m - 1 - j\left( {r + 1} \right) \\ i - j\left( {r + 1} \right) \\ \end{gathered} \right)} } = \hfill \\ = \sum\limits_j {\left( { - 1} \right)^j \left( \begin{gathered} m \hfill \\ j \hfill \\ \end{gathered} \right)\left( \begin{gathered} s + m - j\left( {r + 1} \right) \\ s - j\left( {r + 1} \right) \\ \end{gathered} \right)} \hfill \\ \end{gathered} $$ (note that $N_{\,b}$ and $M_{\,b}$ differ by just a $1$ in the upper term of $2$nd bin.c.)
and $$ N_{\,b} (mr < s,\;r,\;m) = 0\quad ,\quad M_{\,b} (mr \leqslant s,\;r,\;m) = \left( {r + 1} \right)^{\,m} $$

From that, the number of ways of obtaining a sum >= $s$, will be of course: $$ \sum\limits_{s\, \leqslant \,i\,\left( { \leqslant \,mr} \right)} {N_{\,b} (i,\;r,\;m)} = \left( {r + 1} \right)^{\,m} - \sum\limits_{0\, \leqslant \,i\, \leqslant \,s - 1} {N_{\,b} (i,\;r,\;m)} = \left( {r + 1} \right)^{\,m} - M_{\,b} (s - 1,\;r,\;m) $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.