If 5 dice are thrown, what's the probability of getting a 1 or a 5

Question

You can have more than one 1 or 5.

The dice an be all 1s, all fives, or have one 1... or just one 5, and everything in between.

This confuses me because, in the basic sense, adding probabilities (such as 2/6) five times brings a number that exceeds 1, which obviously doesn't make sense. Obviously, the five dice being thrown have a very real chance of turning up nothing but 2s, 3s, 4s, and 6s.

What is the appropriate process?

Answer 1

Let $ A_i $ be the event " dice $i$ is not 1 nor 5" $$ \forall i \in \{1, \cdots ,5\}, \ P(A_i)=\frac 2 3$$ The events are independent therefore $P(A_1\wedge A_2\wedge A_3 \wedge A_4 \wedge A_5)=\left(\frac 2 3\right) ^5$

In reality

$P(A_1\wedge A_2\wedge A_3 \wedge A_4 \wedge A_5)=P(A_1\mid A_2\wedge A_3 \wedge A_4 \wedge A_5)P(A_2\mid A_3 \wedge A_4 \wedge A_5)P(A_3\mid A_4\wedge A_5)P(A_4\mid A_5)P(A_5)$

Let $B$ the event "obtaining a 1 or 5" $$P(B)=1-P(A_1\wedge A_2\wedge A_3 \wedge A_4 \wedge A_5)=1-\left(\frac 2 3\right) ^5$$

Answer 2

This situation would have a Binomial Distribution with n=5 and p=$\frac{2}{6}$.

Each individual die has a $\frac{2}{6}$ chance of "success". When rolling 5 at a time, you can't add these probabilities together, as a set of 'or's, because if you add up the $\frac{4}{6}$ of "fails", there is another probability that's > 1. In short, that method only works for single events.

The Binomial Distribution takes into account all possible probabilities that can happen across multiple dice-rolls.

$P(k) = \binom{n}{k}p^{k}(1-p)^{n-k}$

Where n=the number of trials (5), and p is the probability of success in any given trial $(\frac{2}{6}=\frac{1}{3})$, and $\binom{n}{k}$ is the binomial coefficient $\binom{n}{k}=\frac{n!}{k!(n-k)!}$. We'll set $k=0$ since if we can find the probability of $zero$ dice being "successes" then we can deduce the probability of $any$ dice being a success as $1-P(0)$

In this instance,

$P(Any)=1-\binom{5}{0}\frac{1}{3}^{0}(1-\frac{1}{3})^{5-0}$

$\quad = 1-(\frac{32}{243})$

$P(Any) = \frac{211}{243}$

This should take into consideration each and every possible combination of dice, and those that result in a "success" defined by having any of the 5 dice displaying a 1 or a 5 (which for each dice has a chance of $\frac{1}{3}$)