# Differential privacy RAPPOR article proof doubts

Recently I've been trying to understand the RAPPOR proof:

$$\begin{eqnarray*} P(B' = b' | B = b^*) & = & \left(\frac{1}{2}f\right)^{b'_1}\left(1 - \frac{1}{2}f\right)^{1 - b'_1} \times \ldots \\ & & \times \left(\frac{1}{2}f\right)^{b'_h}\left(1 - \frac{1}{2}f\right)^{1 - b'_h} \times \ldots \\ & & \times \left(1 - \frac{1}{2}f\right)^{b'_{h + 1}}\left(\frac{1}{2}f\right)^{1 - b'_{h + 1}} \times \ldots \\ & & \times \left(1 - \frac{1}{2}f\right)^{b'_k}\left(\frac{1}{2}f\right)^{1 - b'_k}. \end{eqnarray*}$$

but the authors have defined:

$$\begin{eqnarray*} P(b'_i = 1 | b_i = 1) & = & \frac{1}{2}f + 1 - f = 1 - \frac{1}{2}f \;\;\;\text{~and~}\\ P(b'_i = 1 | b_i = 0) & = & \frac{1}{2}f. \end{eqnarray*}$$

and we know that $$b^* = \{b_1 = 1, \ldots, b_h = 1, b_{h+1} = 0, \ldots, b_{k} = 0\}$$ (defined by the article).

Have the exponents been swapped? I have tried to recreate the proof, but the result shows inverse exponents.