Say we have $X \sim \text{uniform}\{1 ,\dots, T\}$ and we denote $X_{(n)} = \max{\{X_1 , \dots , X_n\}}$. I would like to find the conditional probability of $X$ conditioned on $X_{(n)}$.
Here is an attempt to solve this.
$$P(X = x \mid X_{(n)} = t) = \frac{P(X = x , X_{(n)} = t)}{P(X_{(n)} = t)}$$
We will look at three cases, when $x > t$ , $x = t$ , $x < t$. In the first case we notice: $P(X = x \mid X_{(n)} = t) = 0$. In the second, take WLOG the observation of $X$ to be the first one so $X_1$:
$$\frac{P(X_1 = t , X_{(n)} = t)}{P(X_{(n)} = t)} = \frac{P(X_1 = t)}{P(X_{(n)} = t)} = \frac{1/T}{(1/T)^n(t^n-(t-1)^n)}$$
In the third we have:
$$P(X_1 = x , X_{(n)} \leq t) = P(X_2 \leq t , \dots , X_n \leq t)P(X_1 = x) = \left(\frac{t}{T}\right)^{n-1}\left(\frac{1}{T}\right)$$
Hence:
\begin{align}P(X_1 = x , X_{(n)} = t) &= P(X_1 = x , X_{(n)} \leq t) - P(X_1 = x , X_{(n)} \leq t-1)\\&= \left(\frac{1}{T}\right)^n\left(t^{n-1} - (t-1)^{n-1}\right)\end{align}
and so:
$$P(X_1 = x \mid X_{(n)} = t) = \frac{t^{n-1} - (t-1)^{n-1}}{t^{n} - (t-1)^{n}}$$
However I do not see how $\sum_{x=1}^tP(X_1 = x \mid X_{(n)} = t) = 1$.
I feel I made a mistake. Could I have some insight on the problem?
The second is not correct: $P(X_1=t, X_{(n)}=t)\neq P(X_1=t)$, since you also need to specify that $X_2,\dots, X_n$ are less or equal than $t$. Specifically, $$P(X_1=t, X_{(n)}=t)=P(X_1=t)\cdot P(X_2\le t, \dots, X_n\le t)=t^{n-1}/T^n$$ Hence, $P(X_1=t\mid X_{(n)}=t)=\dfrac{t^{n-1}}{t^n-(t-1)^n}$. So, \begin{align}\sum_{x=1}^t P(X_1=t\mid X_{(n)}=t)&=(t-1)\cdot \frac{t^{n-1}-(t-1)^{n-1}}{t^n-(t-1)^n}+\frac{t^{n-1}}{t^n-(t-1)^n}\\&=\frac{t^n-t(t-1)^{n-1}-t^{n-1}+(t-1)^{n-1}+t^{n-1}}{t^n-(t-1)^n}\\&=\frac{t^n-(t-1)\cdot(t-1)^{n-1}}{t^n-(t-1)^n}=1\end{align}