My question regards a simple dice game. You roll one six sided die and win, say, a dollar for each dot on the roll. e.g. five dollars for rolling a five, two dollars for rolling a two, etc.
You are given the option to reroll once.
Letting $X$ be the outcome of the first roll, and $Y$ be the outcome of the (potential) second roll, consider the strategy $W$
\begin{equation}
W=\begin{cases}
X&\text{if $X\geq 4$}\\
Y&\text{if $X< 4$}
\end{cases}
\end{equation}
I.e. if you roll 4 or higher on first try you keep it, if you roll lower you try again.
What is the average winnings using this strategy?
Solution
My issue here is, really, that I don't have more mathematically stringent way to prove this, and I got the answer rather intuitively.
I consider two "macroscopic" outcomes;
- a) We roll a 4, 5, or 6 in the first attempt and stop.
- b) We take a second attempt, and roll either 1, 2, 3, 4, 5, or 6.
The average winnings is the average of these two options, which in itself is the average of a roll of 4/5/6 and the average of a roll of 1/2/3/4/5/6. That is
\begin{equation}
E[W] = (4+5+6)\frac{1}{6}+(1+2+3+4+5+6)\frac{1}{2} = \frac{15}{6} + \frac{3.5}{2} = 4.25
\end{equation}
Which indeed is a higher expectation than winning than rolling the die just once ($\mu=3.5$)
I feel like my argumentation is mostly just intuition, and could benefit from a more mathematically sound description.
Best Answer
First note that $$E[Y]=\sum_{k=1}^6 k P[Y=k] = \frac{1}{6}\sum_{k=1}^6 k = \frac{7}{2}.$$ Now apply the law of total expectation: \begin{align} E[W] &= E[E[W|X]] \\ &= \sum_{k=1}^6 E[W|X=k] P[X=k] \\ &= \frac{1}{6} \sum_{k=1}^6 E[W|X=k] \\ &= \frac{1}{6} \sum_{k=4}^6 E[W|X=k] + \frac{1}{6} \sum_{k=1}^3 E[W|X=k] \\ &= \frac{1}{6} \sum_{k=4}^6 E[X|X=k] + \frac{1}{6} \sum_{k=1}^3 E[Y|X=k] \\ &= \frac{1}{6} \sum_{k=4}^6 k + \frac{1}{6} \sum_{k=1}^3 E[Y] \\ &= \frac{15}{6} + \frac{3}{6}\cdot \frac{7}{2} \\ &= \frac{17}{4} \end{align}