The original RBM had binary visible and hidden nodes. They were showne to be universal approximators of discrete distributions. It was also shown that adding hidden units yields strictly improved modelling power. The common choice of binary values are 0 and 1. However, in some physics applications, -1 and 1 might be a more natural choice. We will here use 0 and 1.
Here we look at some of the relevant equations for a binary-binary RBM. $$ \begin{align} E_{BB}(\boldsymbol{x}, \mathbf{h}) = - \sum_i^M x_i a_i- \sum_j^N b_j h_j - \sum_{i,j}^{M,N} x_i w_{ij} h_j. \tag{17} \end{align} $$ $$ \begin{align} p_{BB}(\boldsymbol{x}, \boldsymbol{h}) =& \frac{1}{Z_{BB}} e^{\sum_i^M a_i x_i + \sum_j^N b_j h_j + \sum_{ij}^{M,N} x_i w_{ij} h_j} \tag{18}\\ =& \frac{1}{Z_{BB}} e^{\boldsymbol{x}^T \boldsymbol{a} + \boldsymbol{b}^T \boldsymbol{h} + \boldsymbol{x}^T \boldsymbol{W} \boldsymbol{h}} \tag{19} \end{align} $$
with the partition function $$ \begin{align} Z_{BB} = \sum_{\boldsymbol{x}, \boldsymbol{h}} e^{\boldsymbol{x}^T \boldsymbol{a} + \boldsymbol{b}^T \boldsymbol{h} + \boldsymbol{x}^T \boldsymbol{W} \boldsymbol{h}} . \tag{20} \end{align} $$
In order to find the probability of any configuration of the visible units we derive the marginal probability density function. $$ \begin{align} p_{BB} (\boldsymbol{x}) =& \sum_{\boldsymbol{h}} p_{BB} (\boldsymbol{x}, \boldsymbol{h}) \tag{21}\\ =& \frac{1}{Z_{BB}} \sum_{\boldsymbol{h}} e^{\boldsymbol{x}^T \boldsymbol{a} + \boldsymbol{b}^T \boldsymbol{h} + \boldsymbol{x}^T \boldsymbol{W} \boldsymbol{h}} \nonumber \\ =& \frac{1}{Z_{BB}} e^{\boldsymbol{x}^T \boldsymbol{a}} \sum_{\boldsymbol{h}} e^{\sum_j^N (b_j + \boldsymbol{x}^T \boldsymbol{w}_{\ast j})h_j} \nonumber \\ =& \frac{1}{Z_{BB}} e^{\boldsymbol{x}^T \boldsymbol{a}} \sum_{\boldsymbol{h}} \prod_j^N e^{ (b_j + \boldsymbol{x}^T \boldsymbol{w}_{\ast j})h_j} \nonumber \\ =& \frac{1}{Z_{BB}} e^{\boldsymbol{x}^T \boldsymbol{a}} \bigg ( \sum_{h_1} e^{(b_1 + \boldsymbol{x}^T \boldsymbol{w}_{\ast 1})h_1} \times \sum_{h_2} e^{(b_2 + \boldsymbol{x}^T \boldsymbol{w}_{\ast 2})h_2} \times \nonumber \\ & ... \times \sum_{h_2} e^{(b_N + \boldsymbol{x}^T \boldsymbol{w}_{\ast N})h_N} \bigg ) \nonumber \\ =& \frac{1}{Z_{BB}} e^{\boldsymbol{x}^T \boldsymbol{a}} \prod_j^N \sum_{h_j} e^{(b_j + \boldsymbol{x}^T \boldsymbol{w}_{\ast j}) h_j} \nonumber \\ =& \frac{1}{Z_{BB}} e^{\boldsymbol{x}^T \boldsymbol{a}} \prod_j^N (1 + e^{b_j + \boldsymbol{x}^T \boldsymbol{w}_{\ast j}}) . \tag{22} \end{align} $$
A similar derivation yields the marginal probability of the hidden units $$ \begin{align} p_{BB} (\boldsymbol{h}) = \frac{1}{Z_{BB}} e^{\boldsymbol{b}^T \boldsymbol{h}} \prod_i^M (1 + e^{a_i + \boldsymbol{w}_{i\ast}^T \boldsymbol{h}}) . \tag{23} \end{align} $$
We derive the probability of the hidden units given the visible units using Bayes' rule $$ \begin{align} p_{BB} (\boldsymbol{h}|\boldsymbol{x}) =& \frac{p_{BB} (\boldsymbol{x}, \boldsymbol{h})}{p_{BB} (\boldsymbol{x})} \nonumber \\ =& \frac{ \frac{1}{Z_{BB}} e^{\boldsymbol{x}^T \boldsymbol{a} + \boldsymbol{b}^T \boldsymbol{h} + \boldsymbol{x}^T \boldsymbol{W} \boldsymbol{h}} } {\frac{1}{Z_{BB}} e^{\boldsymbol{x}^T \boldsymbol{a}} \prod_j^N (1 + e^{b_j + \boldsymbol{x}^T \boldsymbol{w}_{\ast j}})} \nonumber \\ =& \frac{ e^{\boldsymbol{x}^T \boldsymbol{a}} e^{ \sum_j^N (b_j + \boldsymbol{x}^T \boldsymbol{w}_{\ast j} ) h_j} } { e^{\boldsymbol{x}^T \boldsymbol{a}} \prod_j^N (1 + e^{b_j + \boldsymbol{x}^T \boldsymbol{w}_{\ast j}})} \nonumber \\ =& \prod_j^N \frac{ e^{(b_j + \boldsymbol{x}^T \boldsymbol{w}_{\ast j} ) h_j} } {1 + e^{b_j + \boldsymbol{x}^T \boldsymbol{w}_{\ast j}}} \nonumber \\ =& \prod_j^N p_{BB} (h_j| \boldsymbol{x}) . \tag{24} \end{align} $$
From this we find the probability of a hidden unit being "on" or "off": $$ \begin{align} p_{BB} (h_j=1 | \boldsymbol{x}) =& \frac{ e^{(b_j + \boldsymbol{x}^T \boldsymbol{w}_{\ast j} ) h_j} } {1 + e^{b_j + \boldsymbol{x}^T \boldsymbol{w}_{\ast j}}} \tag{25}\\ =& \frac{ e^{(b_j + \boldsymbol{x}^T \boldsymbol{w}_{\ast j} )} } {1 + e^{b_j + \boldsymbol{x}^T \boldsymbol{w}_{\ast j}}} \tag{26}\\ =& \frac{ 1 }{1 + e^{-(b_j + \boldsymbol{x}^T \boldsymbol{w}_{\ast j})} } , \tag{27} \end{align} $$ and $$ \begin{align} p_{BB} (h_j=0 | \boldsymbol{x}) =\frac{ 1 }{1 + e^{b_j + \boldsymbol{x}^T \boldsymbol{w}_{\ast j}} } . \tag{28} \end{align} $$
Similarly we have that the conditional probability of the visible units given the hidden ones are $$ \begin{align} p_{BB} (\boldsymbol{x}|\boldsymbol{h}) =& \prod_i^M \frac{ e^{ (a_i + \boldsymbol{w}_{i\ast}^T \boldsymbol{h}) x_i} }{ 1 + e^{a_i + \boldsymbol{w}_{i\ast}^T \boldsymbol{h}} } \tag{29}\\ &= \prod_i^M p_{BB} (x_i | \boldsymbol{h}) . \tag{30} \end{align} $$ $$ \begin{align} p_{BB} (x_i=1 | \boldsymbol{h}) =& \frac{1}{1 + e^{-(a_i + \boldsymbol{w}_{i\ast}^T \boldsymbol{h} )}} \tag{31}\\ p_{BB} (x_i=0 | \boldsymbol{h}) =& \frac{1}{1 + e^{a_i + \boldsymbol{w}_{i\ast}^T \boldsymbol{h} }} . \tag{32} \end{align} $$