Beating a product of uniforms

Let X,Y,Z∼Unif(0,1)X, Y, Z \sim \text{Unif}(0,1) be independent. Find P[X>YZ]\mathbb{P}[X > YZ].

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  1. For X∼Unif(0,1)X\sim\text{Unif}(0,1) and a fixed pp, P[X>p]=1−p\mathbb{P}[X>p]=1-p. Condition on p=YZp=YZ.
  2. P[X>YZ]=1−E[YZ]\mathbb{P}[X>YZ]=1-\mathbb{E}[YZ], and E[YZ]=E[Y]E[Z]=14\mathbb{E}[YZ]=\mathbb{E}[Y]\mathbb{E}[Z]=\tfrac14 by independence.

Answer

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0.75 (± 0.001)

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