A game on the smaller face

Topic: Random variables and distributions · more questions on this topic
‹ Two components with the same mean life

Two fair dice are thrown and Z is the smaller face.

  1. Give the probability mass function of Z.

  2. Calculate \mathbb{E}(1/Z).

  3. A player is paid Z and pays 3 to play. Find the expected profit from 100 independent plays.

  1. \mathbb{P}(Z\ge z) is the probability that both faces are at least z, which is ((7-z)/6)^2. So

\mathbb{P}(Z=z)=\frac{(7-z)^2-(6-z)^2}{36}=\frac{13-2z}{36},\qquad z=1,\ldots,6,

that is 11/36, 9/36, 7/36, 5/36, 3/36 and 1/36.

\mathbb{E}(1/Z)=\sum_{z=1}^6\frac{13-2z}{36z}=\frac{11}{36}+\frac9{72}+\frac7{108}+\frac5{144}+\frac3{180}+\frac1{216}=\frac{397}{720}=0.551.

  1. \mathbb{E} Z=\sum z(13-2z)/36=91/36=2.53. The expected profit per play is 91/36-3=-17/36, so over 100 plays the player expects to lose about 47.2.