For iid random variables $Y_1, \ldots, Y_n$ with cdf $F$, assuming continuous rv's with $F$ strictly increasing on its support, what is $P \left(Y_n > \max(Y_1, \ldots, Y_{n-1}\right)$?
My method is:
$$ P\left(Y_n > \max(Y_1, \ldots, Y_{n-1}\right) = P(Y_n > Y_{n-1})\cdots P(Y_2>Y_1). $$
Now, intuitively it seems each of these is a half, but I'm not sure if thats correct?
Your equality is not correct. The condition $Y_n > \max\{Y_1, \cdots, Y_{n-1}\}$ is not equivalent to $Y_n > Y_{n-1} > \cdots > Y_1$, and even if this is true, you cannot split the probability in your way because they are not independent.
The idea is to use the symmetry. By the assumption, the probability that any two of $Y_1, \cdots, Y_n$ is equal is zero. Thus we have
$$ \Bbb{P}(Y_n > \max\{Y_1, \cdots, Y_{n-1}\}) = \Bbb{P}(Y_n = \max\{Y_1, \cdots, Y_n\})$$
Moreover, since $Y_1, \cdots, Y_n$ are i.i.d, the joint distribution does not change if we relabel them. Thus
$$ \Bbb{P}(Y_n = \max\{Y_1, \cdots, Y_n\}) = \Bbb{P}(Y_k = \max\{Y_1, \cdots, Y_n\}), \qquad k = 1, \cdots, n.$$
Therefore
$$ \Bbb{P}(Y_n > \max\{Y_1, \cdots, Y_{n-1}\}) = \frac{1}{n}\underbrace{\sum_{k=1}^{n} \Bbb{P}(Y_k = \max\{Y_1, \cdots, Y_n\})}_{=1} = \frac{1}{n}. $$
$\bullet\, \mathbb{P}(\max\{Y_1,...,Y_n\}=Y_n)=\mathbb{P}(\max\{Y_1,...,Y_n\}=Y_k),\,\forall k
If X and Y are independent random variables with someone of distribuition funtion continuous, then $$\mathbb{P}(X=Y)=0.\hspace{3cm}(1)$$
Indeed in the same sense of above, if $\mu_X$ is the distribuition of X, and $F_X(y)=\mu_X\left((-\infty,y]\right)$, then
$$\displaystyle \mathbb{P}(X-Y\le z)=\int_{\mathbb{R}}F_X(z+y) d\mu_Y(y).$$
Choose $z=0$ and using $(1)$
$$\displaystyle \mathbb{P}(X-Y<0)=\int_{\mathbb{R}}F_X(y) d\mu_X(y),$$
and finally using that $Y_i$ is i.i.d, you have $Y_n$ , $\max\{Y_1,...,Y_{n-1}\}$ are independent and for $k
$$\mathbb{P}(\max\{Y_1,...,Y_n\}=Y_n)=\mathbb{P}(\max\{Y_1,...,Y_{n-1}\}-Y_n<0)=\int_{\mathbb{R}}F_{\max\limits_{i