Uneven Dice
Say I have a six-sided die that lands on an even number with probability \(\frac{1}{4}\). If it lands on an even number, then it lands on each one with equal probability. The same is true for odd numbers. Find the expected value of this die. Please answer as a decimal with three significant figures.
Let \(R\) be a random variable for the value of the die roll.
From the description we know that \(\Pr[R=i]=\frac{1}{4}\cdot\frac{1}{3}=\frac{1}{12}\) for \(i\in\{2, 4, 6\}\) and \(\Pr[R=i]=\frac{3}{4}\cdot\frac{1}{3}=\frac{1}{4}\) for \(i\in\{1, 3, 5\}\). Then, \(E[R]=\sum_{i=1}^6 \Pr[R=i] = 1\cdot\frac{1}{4}+2\cdot\frac{1}{12}+3\cdot\frac{1}{4}+4\cdot\frac{1}{12}+5\cdot\frac{1}{4}+6\cdot\frac{1}{12}=\frac{13}{4}\)
From the description we know that \(\Pr[R=i]=\frac{1}{4}\cdot\frac{1}{3}=\frac{1}{12}\) for \(i\in\{2, 4, 6\}\) and \(\Pr[R=i]=\frac{3}{4}\cdot\frac{1}{3}=\frac{1}{4}\) for \(i\in\{1, 3, 5\}\). Then, \(E[R]=\sum_{i=1}^6 \Pr[R=i] = 1\cdot\frac{1}{4}+2\cdot\frac{1}{12}+3\cdot\frac{1}{4}+4\cdot\frac{1}{12}+5\cdot\frac{1}{4}+6\cdot\frac{1}{12}=\frac{13}{4}\)