As Robert Israel has pointed out, the sum of squares of $n$ independent random variables with a standard normal distribution has a chi-square distribution with $n$ degrees of freedom.
Take them from a normal distribution whose expectation is $\mu$ and whose standard deviation is $\sigma$, you have have
$$
\left(\frac{X_1-\mu}{\sigma}\right)^2 + \cdots + \left(\frac{X_n-\mu}{\sigma}\right)^2
$$
has chi-square distribution with $n$ degrees of freedom.
So why might it appear that one of them is not counted? The answer to that comes from such results as this: Suppose instead of the population mean $\mu$, you subtract the sample mean $\overline X$. Then you have
$$
\left(\frac{X_1-\overline X}{\sigma}\right)^2 + \cdots + \left(\frac{X_n-\overline X}{\sigma}\right)^2,\tag{1}
$$
and this has a chi-square distribution with $n-1$ degrees of freedom. In particular, if $n=1$, then the sample mean is just the same as $X_1$, so the numerator in the first term is $X_1-X_1$, and the sum is necessarily $0$, so you have a chi-square distribution with $0$ degrees of freedom.
Notice that in $(1)$, you have $n$ terms in the sum, not $n-1$, and they're not independent (since if you take away the exponents, you get $n$ terms that necessarily always add up to $0$) and the standard deviation of the fraction that gets squared is not actually $1$, but less than $1$. So why does it have the same probability distribution as if there were $n-1$ of them, and they were indepedent, and those standard deviations were each $1$? The simplest way to answer that may be this:
$$
\begin{bmatrix} X_1 \\ \vdots \\ X_n \end{bmatrix} = \begin{bmatrix} \overline X \\ \vdots \\ \overline X \end{bmatrix} + \begin{bmatrix} X_1 - \overline X \\ \vdots \\ X_n - \overline X \end{bmatrix}
$$
This is the decomposition of a vector into two components orthogonal to each other: one in a $1$-dimensional space and the other in an $n-1$ dimensional space. Now think about the spherical symmetry of the joint probability distribution, and about the fact that the second projection maps the expected value of the random vector to $0$.
Later edit:
Sometimes it might seem as if two of them are not counted. Suppose $X_i$ is a normally distributed random variable with expected value $\alpha+\beta w_i$ and variance $\sigma^2$, and they're independent, for $i=1,\ldots,n$. When $w_i$ is observable and $\alpha$, $\beta$, are not, one may use least-squares estimates $\hat\alpha$, $\hat\beta$. Then
$$
\left(\frac{X_1-(\alpha+\beta w_1)}{\sigma}\right)^2 + \cdots + \left(\frac{X_n-(\alpha+\beta w_n)}{\sigma}\right)^2 \sim \chi^2_n
$$
but
$$
\left(\frac{X_1-(\hat\alpha+\hat\beta w_1)}{\sigma}\right)^2 + \cdots + \left(\frac{X_n-(\hat\alpha+\hat\beta w_n)}{\sigma}\right)^2 \sim \chi^2_{n-2}.
$$
A similar sort of argument involving orthogonal projections explains this.
One needs these results in order to derive things like confidence intervals for $\mu$, $\alpha$, and $\beta$.
Since you're probably performing a chi-squared test, then test-statistic
$$
X^2=\sum_{i=1}^n \frac{(O_i-E_i)^2}{E_i}
$$
follows a $\chi^2(p)$ distribution with $p=n-1$ degrees of freedom. Since large values are critical, the corresponding $p$-value is given by
$$
P(\chi^2(p)\geq X^2)=1-F_{\chi^2(p)}(X^2),
$$
i.e. the probability of a $\chi^2(p)$-variable being larger than what we have observed. Here $F_{\chi^2(p)}$ is the distribution function of a $\chi^2(p)$ distribution.
This is easily calculated in R with the command 1-pchisq(x,p), where x denotes the test-statistic $X^2$ and p is the number of degrees of freedom.
Best Answer
The chi-squared distribution with $2$ degrees of freedom is just the exponential with density function $\frac{1}{2}e^{-x/2}$ for $x\gt 0$, and $0$ elsewhere. In particular, $\Pr(X\ge x)=e^{-x/2}$ for $x\gt 0$.
We first find the cdf $F_Y(y)$ of $Y$. Clearly $\Pr(Y\le y)=0$ if $y\le 0$, and $\Pr(Y\le y)=1$ if $y\ge 1$. For $0\lt y\lt 1$ we have $$\small F_Y(y)=\Pr(Y\le y)=\Pr(e^{-X/2}\le y)=\Pr\left(\frac{-X}{2}\le \ln y\right)=\Pr(X\ge -2\ln y)=e^{(2\ln y)/2}=y.$$ To find the density function $f_Y(y)$ of $Y$, differentiate $F_Y$. We get $f_Y(y)=1$ for $0\lt y\lt 1$.