Probability an 6 sided die will be higher than a 8 sided die?

Question

Say one person rolls an 8 sided die and the other rolls a six, what is the probability that the six sided die is higher than the 8?

I know that the expected value of the eight is 4.5 and the six is 3.5 but am having trouble figuring out how to find the probability.

EDIT: Answer is 15/48 but still curious if there's a way of doing this without creating a grid.

Answer 1

Probability That The $\boldsymbol{6}$-Sided Die Will Be Higher (from the body of the question) $$ \sum_{k=1}^6\overbrace{\frac16}^{k\text{ on d}6}\cdot\overbrace{\frac{k-1}8}^{\lt k\text{ on d}8} =\frac{5}{16} $$

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Probability That The $\boldsymbol{8}$-Sided Die Will Be Higher (from the original title to the question) $$ \sum_{k=1}^7\overbrace{\frac18}^{k\text{ on d}8}\cdot\overbrace{\frac{k-1}6}^{\lt k\text{ on d}6} +\overbrace{\frac18}^{8\text{ on d}8}\cdot\overbrace{1\vphantom{\frac16}}^{\lt8\text{ on d}6} =\frac{9}{16} $$

Answer 2

This is rather easy. You don't even need a grid. We just have to count the number of tuples $(x,y) ; x \subset A$ and $y\subset B$ [where A is the est of all possible numbers of the six sided dice(1-6) and B is the set of all possible numbers of the eight sided dice(1-8)] such that $x>y$(by the problem). There are 15 such tuples: 5 for x = 6, 4 for x = 5 so on till 1 for x = 2. Thus the probability is P= $\frac{15}{48}$. PS: I don't know how your edit makes sense with the probability being $\frac{120}{48}$,which i believe is greater than 1.

Answer 3

If there is a die with $m$ sides and a die with $n$ sides, assuming that $m<n$, the probability that $m$ will be greater than $n$ assuming that those are the names given to the outcomes when both of them are rolled is

$$\frac{m-1}{2n}$$

Full disclaimer: I made this up with help from the grid. This is a formula very specific to your question.

Answer 4

Let $d_8$ be the result of the 8-sided die, and $d_6$ the result of the the 6-sided die.

$$P(d_8<d_6) = \sum_{i=1}^6 P((d_8 < d_6) \cap(d_6=i)) = \sum_{i=1}^6 P(d_8 < d_6|d_6=i)\cdot P(d_6=i)$$

$$=\sum_{i=1}^6 \frac{i-1}{8} \cdot \frac{1}{6} = \frac{1}{48}\sum_{i=1}^6 (i-1) = \frac{\frac{6\cdot 7}{2}-6}{48} = \frac{15}{48} = \frac{5}{16}$$

You can generalise for any $n,m$ sided dice. Assume $n>m$ we have:

$$P(d_n<d_m) = \sum_{i=1}^m P((d_n < d_m) \cap(d_m=i)) = \sum_{i=1}^m P(d_n < d_m|d_m=i)\cdot P(d_m=i)$$

$$=\sum_{i=1}^m \frac{i-1}{k} \cdot \frac{1}{m} = \frac{1}{nm}\sum_{i=1}^m (i-1) = \frac{\frac{m\cdot (m+1)}{2}-m}{nm} = \frac{\frac{m+1}{2}-1}{n} = \frac{\frac{m-1}{2}}{n} = \frac{m-1}{2n}$$

Answer 5

$$\begin{align}\mathsf P(X_6>X_8) ~=~& \mathsf P(X_8<7)~\mathsf P(X_6>X_8\mid X_8<7)+\mathsf P(X_8>6)~\mathsf P(X_6>X_8\mid X_8>6) \\ ~=~& \frac 68\cdot\frac {15}{36}+\frac 28\cdot 0 \\ ~=~& \frac {5}{16} \end{align}$$

Answer 6

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From the OP:

1. " Say one person rolls an $8$ sided die and the other rolls a six ".
2. " What is the probability that the six sided die $\underline{is\ higher\ than}$ the $8$ ? ".

$$\bbx{\ds{\large% \mbox{Hereafter,}\ \bracks{\cdots}\ \mbox{is the}\ Iverson\ Bracket}} $$

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The answer to $2.$ is given by: \begin{align} {1 \over 8}\sum_{d_{8} = 1}^{8}{1 \over 6}\sum_{d_{6} = 1}^{6} \bracks{d_{6} > d_{8}} & = {1 \over 48}\sum_{d_{8} = 1}^{8}\braces{\bracks{d_{8} \leq 5} \sum_{d_{6} = d_{8} + 1}^{6}} = {1 \over 48}\sum_{d_{8} = 1}^{8}\bracks{d_{8} \leq 5}\pars{6 - d_{8}} \\[5mm] & = {1 \over 48}\sum_{d_{8} = 1}^{5}\pars{6 - d_{8}} = {1 \over 48}\bracks{5 \times 6 - {5\pars{5 + 1} \over 2}} = \bbx{\ds{5 \over 16}} = 0.3125 \end{align}