What is the probability of a rolling 8 biased dice to get a sum of 28?

Question

You roll 8 6-sided dice. What is the probability that the sum is 28?

Given that the P(1)=8/117, P(2)=14/117, P(3)=34/117, P(4)=39/117, P(5)=14/117, and P(6)=8/117

At first I've been trying to find the permutations of rolling 8 dice to get a sum of 28 but I don't know if the order matters so I tried to find combinations as well. Having found the permutations and combinations, because the probabilities are different for each number, I don't know how to find the probability for each combination/permutation.

Please explain it in detail if you can. Thank you

Answer 1

We have that $$ \eqalign{ & 1^{\,8} = \left( {p_{\,1} + \,p_{\,2} + \, \cdots \, + p_{\,6} } \right)^{\,8} = \cr & = \sum\limits_{\left\{ {\matrix{ {0\, \le \,k_{\,j} } \hfill \cr {k_{\,1} + \,k_{\,2} + \, \cdots \, + k_{\,6} = \,8} \hfill \cr } } \right.} {\left( \matrix{ 8 \cr k_{\,1} ,\,k_{\,2} ,\, \cdots \,,k_{\,6} \cr} \right) p_{\,1} ^{\,k_{\,1} } \,p_{\,2} ^{\,k_{\,2} } \, \cdots \,p_{\,6} ^{\,k_{\,6} } } \cr} $$

Then consider the polynomial $$ \eqalign{ & \left( {p_{\,1} x^{\,1} + \,p_{\,2} x^{\,2} + \, \cdots \, + p_{\,6} x^{\,6} } \right)^{\,8} = \cr & = \sum\limits_{\left\{ {\matrix{ {0\, \le \,k_{\,j} } \hfill \cr {k_{\,1} + \,k_{\,2} + \, \cdots \, + k_{\,6} = \,8} \hfill \cr } } \right.} {\left( \matrix{ 8 \cr k_{\,1} ,\,k_{\,2} ,\, \cdots \,,k_{\,6} \cr} \right)\left( {p_{\,1} x^{\,1} } \right)^{\,k_{\,1} } \,\left( {p_{\,2} x^{\,2} } \right)^{\,k_{\,2} } \, \cdots \,\left( {p_{\,6} x^{\,6} } \right)^{\,k_{\,6} } } = \cr & = \sum\limits_{\left\{ {\matrix{ {0\, \le \,k_{\,j} } \hfill \cr {k_{\,1} + \,k_{\,2} + \, \cdots \, + k_{\,6} = \,8} \hfill \cr } } \right.} {\left( \matrix{ 8 \cr k_{\,1} ,\,k_{\,2} ,\, \cdots \,,k_{\,6} \cr} \right)\left( {p_{\,1} ^{\,k_{\,1} } \,p_{\,2} ^{\,k_{\,2} } \, \cdots \,p_{\,6} ^{\,k_{\,6} } } \right)x^{\,1 \cdot k_{\,1} + 2 \cdot k_{\,2} + \cdots + 6 \cdot k_{\,6} } } = \cr & = \sum\limits_{8\, \le \,m\, \le \,48} {\left( {\;\sum\limits_{\left\{ {\matrix{ {0\, \le \,k_{\,j} } \cr {k_{\,1} + \,k_{\,2} + \, \cdots \, + k_{\,6} = \,8} \cr {\,1 \cdot k_{\,1} + 2 \cdot k_{\,2} + \cdots + 6 \cdot k_{\,6} = m} \cr } } \right.} {\left( \matrix{ 8 \cr k_{\,1} ,\,k_{\,2} ,\, \cdots \,,k_{\,6} \cr} \right)\left( {p_{\,1} ^{\,k_{\,1} } \,p_{\,2} ^{\,k_{\,2} } \, \cdots \,p_{\,6} ^{\,k_{\,6} } } \right)} } \right)x^{\,m} } \cr} $$

For $x=1$ the polynomial has value $1$.
So clearly the probability you are looking for is the coefficient of $x^{28}$ in the above identity which is $$ \left[ {x^{\,28} } \right] = 0.11278 \ldots $$ The plot of all the coefficients, compared with case of $p_k = const. = 1/6$, is as given below.

You are further asking for the case without replacement.
It is not clear what kind of process you want actually to simulate.

If you mean that you can extract only once any of six tickets labelled $[1,6]$, then clearly you cannot do more than six extractions.

However I will expose briefly the principles underlying the construction of a Ordinary Generating Function

Consider to expand the following polynomial $$ \eqalign{ & F(x) = \left( {1 + x} \right)\left( {1 + x^{\,2} } \right) \cdots \left( {1 + x^{\,q} } \right) = \cr & = \cdots + 1 \cdot x^{\,k_{\,1} } \cdot \ldots \cdot x^{\,k_{\,2} } \cdot 1 \cdot \ldots \cdot x^{\,k_{\,m} } + \cdots \cr} $$ it clearly represents all the cases in which, out of $q$ tickets, some are extracted only once ($k_1 \ne k_2$ etc.), and the remaining are not extracted at all (the $1=x^0$).

If the probability of being extracted are different, then consider $$ F(x) = \left( {\left( {1 - p_{\,1} } \right) + p_{\,1} x} \right)\left( {\left( {1 - p_{\,2} } \right) + p_{\,2} x^{\,2} } \right) \cdots \left( {\left( {1 - p_{\,q} } \right) + p_{\,q} x^{\,q} } \right) $$ where $F(1) = 1$.

Finally pass to $$ G(x,y) = \left( {\left( {1 - p_{\,1} } \right) + p_{\,1} yx} \right)\left( {\left( {1 - p_{\,2} } \right) + p_{\,2} yx^{\,2} } \right) \cdots \left( {\left( {1 - p_{\,q} } \right) + p_{\,q} yx^{\,q} } \right) $$ it should be clear what the coefficient of $y^n$ represents, and what is the coefficient of $y^n x^m$.

Answer 2

In answer on your comment concerning "without replacement".

Let $(a_1,a_2,a_3,a_4,a_5,a_6)\in S$ if and only if the $a_i$ are nonnegative integers that satisfy:

$$\sum_{i=1}^6a_i=8\text{ and }\sum_{i=1}^6ia_i=28$$

Then to be found is:$$\sum_{(a_1,a_2,a_3,a_4,a_5,a_6)\in S}\frac{\binom{8}{a_1}\binom{14}{a_2}\binom{34}{a_3}\binom{39}{a_4}\binom{14}{a_5}\binom{8}{a_6}}{\binom{117}{8}}$$

It is like taking $8$ notes from an urn that contains $8$ notes with number $1$ on it, $14$ with number $2$ on it, et cetera.

Favourable outcomes give $28$ as summation of all drawn numbers.

So essential here is finding the set $S$.

Again that could be quite a job on its own.