X~Unif(0, 1) ; X1 + X2 + ... X6 = 1 ; Y = sum(X1...X6) ; VAR(Y) =? Let $X_i$ ~ Unif(0, 1) s.t.
$X_1 + X_2 + ... + X_6 = 1$
Let $Y = X_1 + … X_6$
What is $Var(Y)$?
(Also the case when it's $X_n$)

Purpose for the curious:
I'm trying to rank confidence for softmax predictions using variance.
6 classes, 
1 | .01, .01, .01, .01, .01, .95 ; var=.1227, max=.95
2 | .01, .01, .01, .01, .50, .46 ; var=.0492, max=.50
3 | .01, .01, .01, .33, .33, .30 ; var=.0241, max=.33
4 | .01, .01, .25, .25, .25, .23 ; var=.0123, max=.25
5 | .01, .20, .20, .20, .20, .19 ; var=.0049, max=.20
6 | .16, .16, .16, .16, .16, .20 ; var=2.2e-4, max=.20
… etc
If we rank them by max or var, will both list be same?
Hypothesis: Similar but no.
In this case,
 - Var would rank obs#5  higher than #6; bc var says there’s no way the pred can class 1 in #5
 - Max, on the other hand would say there’s no difference in information between #5 and #6  
For cutoff idk if the variance of Unif(0, 1) would be a good cutoff or if I need to solve the problem I mentioned above for insight.
 A: jld has already shown in his comment that $Var(Y)=0$ because $Y=0$ and a constant must have variance zero.
It should also be noted that the question is only possible when $n=1$.
From the question, generalized to n:
Let $X_i$ ~ Unif(0, 1) s.t.
$X_1 + ... + X_n = 1$
That is, each $X_i$ is an uniform variable but the sum of all $X_i$ is 1. Since the expectation of sum is sum of expectations and all $X_i$ have the same distribution:
$E(X_1 + ... + X_n) = n·E(X_i)$
and 
$E(X_1 + ... + X_n) = E(1) = 1$
and
$E(X_i)=0.5$ because $X_i$ ~ Unif(0, 1) 
$n=\frac{1}{0.5}=2$
Furthermore, we can compute $Cov(X_1,X_2)$ and $Cor(X_1,X_2)$
$$0=Var(1)=Var(X_1+X_2)=var(X_1)+Var(X_2)+2\cdot Cov(X_1,X_2)=$$
$$=\frac{1}{12}+\frac{1}{12}+2\cdot Cov(X_1,X_2)$$
$$Cov(X_1,X_2)=-\frac{1}{12}$$
$$Cor(X_1,X_2)=\frac{Cov(X_1,X_2)}{\sqrt{(Var(X_1)\cdot Var(X_2))}} =\frac{-\frac{1}{12}}{\frac{1}{12}}=-1$$
In fact, that matches the only solution of two random variables with the same distribution and constant sum $X_2=1-X_1$
And returning to the question (modified in comments) after that digression: $Var(X_i)=\frac{1}{12}$ just as with any uniform variable.
And interestingly, the sample variances in the original question are different from 1/12 because those samples don't add to 1 or are uniform.
