Roll Initiative

While it is true that the second die (which has the numbers $1$ , $1$ , $1$ , $1$ , $1$ , and $19$ ) has the expected value of $4$ , and the first die (which has the numbers $1$ , $2$ , $3$ , $4$ , $5$ , and $6$ ) has the expected value of 3.5, the expected value doesn’t actually matter in this problem.

We can draw a probability tree to work out the odds that the second die will win.

As you can see, there’s only a $1/6$ chance that the second die wins, and a $25/36$ chance that the first die wins.

Therefore, for the best odds in beating Gerald is by using the first die.

We note that the probability of a person winning is equal to the probability of the other person losing. The probability of draws is the same for both persons. So we need only consider choosing one die. Let us consider you choose the regular 1-2-3-4-5-6 die.

Then the probabilities of you wining, losing and there is a draw are as follows:

$\begin{array} {cccc} \text{When} & \color{#3D99F6} \text{your throws} & \color{#D61F06} \text{Gerald's throws} & \text{Probability} \\ \hline \text{you win} & \color{#3D99F6} 2, 3, 4, 5, 6 & \color{#D61F06} 1, 1, 1, 1, 1 & {\color{#3D99F6} \dfrac 56} \times {\color{#D61F06}\dfrac 56} = \dfrac {25}{36} \\ \text{you lose} & \color{#3D99F6} 1, 2, 3, 4, 5, 6 & \color{#D61F06} 19 & {\color{#3D99F6} \dfrac 66} \times {\color{#D61F06}\dfrac 16} = \dfrac 16 \\ \text{draw} & \color{#3D99F6} 1 & \color{#D61F06} 1, 1, 1, 1, 1 & {\color{#3D99F6} \dfrac 16} \times {\color{#D61F06}\dfrac 56} = \dfrac 5{36} \\ & & \small \text{Note that} & \small \frac {25}{36} + \frac 16 + \frac 5{36} = 1 \end{array}$

Therefore choosing the 1-2-3-4-5-6 die gives a higher probability of winning.