What is the average result of rolling two dice, and only taking the value of the higher dice roll?
To make sure the situation I am asking about is clear, here is an example:
I roll two dice and one comes up as a four and the other a six, the result would just be six.
Would the average dice roll be the same or higher than just rolling one dice?
Best Answer
For $k=1,\dots,6$ there are $k^2$ ways to get two numbers less than or equal to $k$. To get two numbers whose maximum is $k$ I must get two numbers that are less than or equal to $k$, but not two numbers that are less than or equal to $k-1$, so there are $k^2-(k-1)^2=k^2-(k^2-2k+1)=2k-1$ ways to get two numbers whose maximum is $k$. Thus, the probability of getting a maximum of $k$ is
$$\frac{2k-1}{36}\;,$$
and the expected value of the maximum is
$$\begin{align*} \sum_{k=1}^6k\cdot\frac{2k-1}{36}&=\frac1{36}\sum_{k=1}^6\left(2k^2-k\right)\\ &=\frac1{18}\sum_{k=1}^6k^2-\frac1{36}\sum_{k=1}^6k\\ &=\frac{6\cdot7\cdot13}{18\cdot6}-\frac{6\cdot7}{36\cdot2}\\ &=\frac{91}{18}-\frac{21}{36}\\ &=\frac{161}{36}\\ &=4.47\overline{2}\;. \end{align*}$$
Of course this is larger than the expected value of $\frac72=3.5$ for a single roll of a die: picking the maximum of the two numbers can be expected to bias the result upwards.