<s>[INST] Pairing 8 numbers, counting probability with or without permutation This is a problem posted in this website yesterday, although it has been closed, this is quite an interesting problem.
Numbers $1,2,...,8$ are divided into pairs. Then the lower number in each pair is eliminated, leaving $4$ numbers. This is repeated, dividing the $4$ remaining numbers into pairs and eliminating the lower number in each pair, leaving $2$ numbers. What is the probability that $4$ is one of the two remaining numbers?
my attempt:
Alternative 1:
$4$ can survive twice if none of $1,2,3,4$ are paired with $5,6,7,8$ on both repetitions. Number of ways to divide $1,2,3,4$ into pairs is $3$, so is the number of ways to divide $5,6,7,8$ into pairs. Number of ways to divide $1,2,...,8$ into pairs is $\binom{8}{2\ 2\ 2\ 2}\frac{1}{4!}=105$. Thus the probability is $\frac{3\times 3}{105}=\frac{3}{35}$
Then $4$ needs to be paired with the other survivor from $1,2,3,4$. Probability is $\frac{1}{3}$. Therefore the probability of $4$ surviving is $\frac{3}{35}\times\frac{1}{3}=\frac{1}{35}$
Alternative 2:
We do tournament with $8$ slots. In order to have $4$ surviving to semifinal, we need to have $1,2,3,4$ in the first $4$ slots or the last $4$ slots ($2$ ways). Total number of choosing $4$ numbers from $1,2,...,8$ for the first $4$ slots is $\binom{8}{4}$. So the probability is $\frac{2}{\binom{8}{4}}=\frac{1}{35}$
Could You guys check my solutions and is there any other solutions? Thanks.
 [/INST] Your solutions look good.
A perhaps somewhat simpler solution: In the knockout tournament, $4$ survives the first two rounds exactly if the other three numbers in its bracket are $1,2,3$. That’s one out of $\binom73=35$ possible choices for these three numbers.
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