Suppose $A$ and $B$ are independent random variables with continuous Bernoulli distribution: \begin{align} f_A(x)&=\lambda_a^x(1-\lambda_a)^{1-x}\frac{2 \tanh ^{-1}\left(1-2 \lambda_a\right)}{1-2 \lambda_a}\cdot\mathsf 1_{(0,1)}(x) \\ f_B(y)&=\lambda_b^y(1-\lambda_b)^{1-y}\frac{2 \tanh ^{-1}\left(1-2 \lambda_b\right)}{1-2 \lambda_b}\cdot\mathsf 1_{(0,1)}(y) \end{align} What's the probability $\mathbb P(A>B)$?
The question is simple in discrete case: $\mathbb P(A >B)=P(A=1)P(B=0)$, but it seems harder in continuous case. Should I first compute the joint distribution \mathbb $P(A=x,B=y)$?
The computation is a routine integral of the joint density over the appropriate region: \begin{align} \mathbb P(A>B) &= \small\int _0^1\int _y^1\frac{\lambda _a^x \left(1-\lambda _a\right){}^{1-x} \lambda _b^y \left(1-\lambda _b\right){}^{1-y} \left(4 \tanh ^{-1}\left(1-2 \lambda _a\right) \tanh ^{-1}\left(1-2 \lambda _b\right)\right)}{\left(2 \lambda _a-1\right) \left(2 \lambda _b-1\right)}\ \mathsf dx\ \mathsf dy\\ &= \small \frac{\lambda _a \left(2 \lambda _b \left(\tanh ^{-1}\left(1-2 \lambda _a\right)-\tanh ^{-1}\left(1-2 \lambda _b\right)\right)-\tanh ^{-1}\left(1-2 \lambda _a\right)-2 \tanh ^{-1}\left(1-2 \lambda _b\right)\right)+\left(1-\lambda _b\right) \tanh ^{-1}\left(1-2 \lambda _b\right)}{\left(2 \lambda _a-1\right) \left(2 \lambda _b-1\right) \left(\tanh ^{-1}\left(1-2 \lambda _a\right)+\tanh ^{-1}\left(1-2 \lambda _b\right)\right)}. \end{align}