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

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.

• Draw a $6 \times 8$ array and fill it with all the elementary events that can occur (each one with probability $1/48$, assuming equidistribution) then count... – Jean Marie Oct 17 '16 at 14:32
• Ok that'll work, I am curious if there's a simpler way though. What if I'm dealing with numbers too large to draw out? I suppose I could always whip up a script to do it but curious if there's a formula. – i_am_so_stupid Oct 17 '16 at 14:42
• There are cases like this one where no "witty" way exist. One thing is sure, the fact that you know the expected values of the "6" and the "8" dice is of no help for your problem. – Jean Marie Oct 17 '16 at 14:48
• Fair enough, thank you for your help – i_am_so_stupid Oct 17 '16 at 14:57
• @Skeleton Bow You are perfectly right: $n(n+1)/2=15$ for $n=5$. – Jean Marie Oct 17 '16 at 15:14

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

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

• Can you explain what you did there? Your answer does not agree with what I (and others) have found. – Thanassis Oct 17 '16 at 15:20
• It is the six sided die that is to score higher. – true blue anil Oct 17 '16 at 15:36
• Sorry. I read the title. I have amended my answer. – robjohn Oct 17 '16 at 15:36
• I made the same "mistake" and deleted, I now see that OP confused us with conflicting title and body ! – true blue anil Oct 17 '16 at 15:49
• Thanks. Sorry for the conflicting title, it's been fixed. – i_am_so_stupid Oct 17 '16 at 16:10

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 is 5!/48? So greater than one probability?

Of course you mean the the 5th triangle number / 48.

Use theorem of total probability to reach this answer, conditioning on the value the 6 sided die takes.

That is to say if you roll x with the 6 sided die. You have an x-1/8 chance of it being higher than the 8 sided die.

The generalisation is pretty clear

\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}

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:

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

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

• It could perhaps be more simply derived as $\frac12$ (total squares minus diagonal squares in smaller die)/(total squares in grid) $= \frac{(m^2-m)/2}{mn} = \frac{m-1}{2n}$ – true blue anil Oct 17 '16 at 15:44


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 $8$ ? ".

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

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}

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.

• I don't think this is a general formula. – true blue anil Oct 17 '16 at 15:28
• @trueblueanil it is the general formula (look at my answer), I just did not see any explanation on how SkeletonBow arrived there. – Thanassis Oct 17 '16 at 15:31
• Oh, I misread the question as the $8$ sided die scoring higher. Drat, can't up vote now, so sorry ! – true blue anil Oct 17 '16 at 15:33
• @trueblueanil no problem. I saw your answer (which you deleted), and figured that you thought it was for numbers higher than or equal to the ones on the 8-die. – Skeleton Bow Oct 17 '16 at 15:37
• @Thanassis to be honest, it was quite simple: I drew it out on a grid and realized that every time you add a number to the smaller $m$, the number of possibilities grew by $m-1$. This meant that I needed to find the total sum of the first $m-1$ integers and then divide that by the total number of squares on the grid, which would be $m\cdot n$ and then got the simplified result. – Skeleton Bow Oct 17 '16 at 15:40