probability of dice throws, uneven dice

Question

We have a story telling dice game, and 3 of the 8 dice have a certain symbol on them (these are 6-sided dice). We roll every dice each time.

I told my son the probability of getting that symbol was 3 in 48 (1 in 16). My husband said it was 1 in 6, as in every throw one of the 3 dice could come up with the symbol.

So basically for the calculation we can actually ignore the 5 dice without the symbol totally, as they have no effect.

I think the 1 in 6 is wrong so I found this page, which says the answer is $\frac{91}{216}$ or 42.13%.

Am I correct in my assumption that we can ignore all the dice which don't have the symbol from the calculation, so it effectively just becomes a rolling a six on 1 of 3 dice question?

Answer 1

Yes, you are correct in your assumption that you can discard all other dice. The correct answer is indeed $42.13\%$. To see this, let $X$ denote the number of times that you roll the symbol. Then, $X$ can take values in $\{0,1,2,3\}$. You want to find the probability $P(X\ge1)$:

\begin{align}P(X\ge 1)=1-P(X=0)=1-\prod_{i=1}^3P(X_i\neq \text{symbol})=1-\left(\frac{5}{6}\right)^3=1-\frac{125}{216}=0.4213\end{align}

Answer 2

You can discard the dice which don't have the symbols on them.

We first calculate the probability of the symbols not appearing on any die, i.e., neither of the die shows a symbol. That will be $\frac{5}{6}$ for one die not showing the symbol. And it will be $\left(\frac{5}{6}\right)^3$ for all of the three dice not showing a symbol.

Since we want at least one symbol, we subtract the probability of no symbols with the total probability, i.e, $1$:

$$ 1-\left(\frac{5}{6}\right)^3 = 0.4213$$

Hence the probability is $42.13\%$