Chance of rolling two dices and throwing 1 and 2 7 times in a row

Question

I've just seen an episode of Red Dwarf and there was something that doesn't stop bugging me.

They played a game with $2$ dice and one of them didn't want to roll a $1$ on one dice and $$2 on the other one.
He rolled a combination of $1$ and $2$, $7$ times in a row.
The calculation in that show was around $\frac{1}{62000000}$. How would you calculate that?
I know just some basic stuff from school so I believe there is a $1$ in $36$ chance to get $1$ and $2$ when rolling two dices once ($36$ possible outcomes and only one correct), but I can't wrap my head around how to calculate multiple throws. Thank you. :)

Answer 1

There are 36 possible outcomes when throwing two dices, as there are 6 outcomes throwing only one dice. But, naming a dice $A$ and the other $B$, there are two outcomes you are looking for; $A=1$, $B=2$ and $A=2$, $B=1$.

So the probability of getting 1 and 2 by throwing two dices is $$ \frac{2}{36}=\frac{1}{18}. $$ Then, to get 1 and 2 seven times in a row, $$ \left(\frac{1}{18}\right)^7=\frac{1}{612220032}\cong1.63\cdot10^{-9}. $$ So the rate would be $612220032:1$. Still, this is 10 times bigger than the one you said of 62 million. I searched on the internet about this episode and I found a quote in IMDb, saying exactly what you said. I really believe that the show is wrong.

Answer 2

The chance of rolling a dice with a specific sequence (ANY sequence really) with length $n$ (in this case, $14$) is $\left(\dfrac{1}{6}\right)^n$

So calculating $\left(\dfrac{1}{6}\right)^{14}$, we get $\dfrac{1}{78364164096}$

I have no clue why they came up with $1$ in $62$ million chance. The show might've made a mistake.