Probability of having $5$ and $6$ in $12$ dice throws

Question

Let an experiment consists in throwing $12$ dice and let us count $5$ and $6$ as "success." With perfect dice the probability of success is $p = 1/3$ and the number of successes should follow the binomial distribution $b(k; 12, l)$.

I was reading Feller's Intro. to Probability theory and its applications and was wondering how he got the $p=1/3$?

As I understand, he calculated the probability, that in $12$ throws of one dice he get at least one 5 or 6. Denote it with $P(A_5 \text{ or }A_6)$.

$P(A_5 \text{ or } A_6)=P(A_5)+P(A_6)-P(A_5 A_6)$;

$P(A_5)=P(A_6)=1-(\frac {5} {6})^{12}$;

$P(A_5 A_6)=P(A_5)\cdot P(A_6)$; since the event outcomes are independent,

but in this case I my result is $0.987420884$

Answer 1

Denoting the number that appears from the $i$-th die by $D_i$ there is a probability on success at the $i$-th experiment that equals: $$P(D_i=5\vee D_i=6)=P(D_i=5)+P(D_i=6)=\frac16+\frac16=\frac13$$

Answer 2

It seems like he meant that the probability of success in one Bernoulli trial (i.e., one dice roll) is $\frac13$, then he's going to use this as a parameter in the binomial distribution for $12$ dice rolls.

I'll also note that if $A_5$ means rolling at least one $5$ among $12$ dice, and $A_6$ means rolling at least one $6$, they are not independent -- this can be seen intuitively by noting that if we know you rolled at least one $5$, it's lightly less likely that you rolled at least one $6$ (we can discount the possibility that you rolled all twelve $6$s).