Two dice and a seven

My brother and I play with two dice. Each throw, we add the dots. He always picks twelve and I pick seven, and whichever total comes up first wins.

Two dice; each throw I add the dots and call the total $S$. He picks $S=12$, I pick $S=7$. First to appear wins.

$S$ the sum, $A=\{S=12\},B=\{S=7\}$.

Each die has six faces, so two dice land in six times six, thirty-six ways. Twelve needs two sixes, one way only. Seven can be made as one and six, two and five, three and four, and each of those the other way round: six ways.

Each die has $6$ faces, so $|\Omega |=6\times 6=36$ outcomes. $S=12$ needs $(6,6)$ only. $S=7$ comes from $(1,6),(2,5),(3,4)$ and the reverses: $6$ ways.

$|\Omega |=6^2=36,|A|=1,|B|=6$.

So twelve has a chance of one in thirty-six, and seven has six in thirty-six, which is one in six. Seven is six times as likely as twelve. No other total has as many as six ways; seven is the king.

$P(S=12)=\frac{1}{36}$ and $P(S=7)=\frac{6}{36}=\frac{1}{6}$. Seven is $6$ times as likely, and for every other total $s\ne 7$ there are fewer than $6$ ways.

$P(A)=\frac{1}{36},P(B)=\frac{1}{6}=6P(A)$; $\forall s\ne 7$ we have $|\{S=s\}|<6$.

Now my brother has learned. Today he picks seven, and I pick six, which has five ways. We play until one of the two appears. His chance is six out of eleven, mine five out of eleven. I still enjoy it.

Today he picks $S=7$ and I pick $S=6$, with $5$ ways. First of the two to appear wins: he has $\frac{6}{6+5}=\frac{6}{11}$, I have $\frac{5}{11}$.

$C=\{S=6\}$, $|C|=5$: $\frac{|B|}{|B|+|C|}=\frac{6}{11},\frac{|C|}{|B|+|C|}=\frac{5}{11}$.