Figure 1: Estimating expected value with finite sum |

This means that our guess as to the true expected values are below and the updated formula would be $\mathbb{E} N = f/p$.
\begin{align}
\mathbb{E} N_\text{infantry} & = 8 \\
\mathbb{E} N_\text{armor} & = 9 \\
\mathbb{E} N_\text{artillery} & = 12
\end{align}
This makes a lot more sense and perhaps I should have caught the previous error. Without the odd $-1$ hanging off the back end of the equation, this means that the expected value of proportional to the number of starting figures. This makes sense, especially in our one die at a time analysis. If it takes an average of $1/p$ dice to remove one figure, then it should take an average of $f/p$ dice to remove $f$ figures. First, we remove one figure, then the second, and then so on $f$ times. Each time takes an average of $1/p$ dice, so $f/p$ dice in total.
The expectation is also inversely proportional to $p$. While clearly in the correct direction, it'll take a little more analysis to establish this precise relationship. If we increase the probability of hitting with each die, it'll decrease the number of dice we need to roll before getting a hit. Let's analysis the expected number of dice rolled to get the first hit, $N_\text{hit}$, for a unit with a probability of getting hitting one die $p$. To get a hit on the $n$-th die, after missing previously, the probability is,
\begin{align}
P(N_\text{hit} = n) = (1-p)^{n-1} \cdot p.
\end{align}
While we could use this, we can use the same shortcut to find the expected value as before.
\begin{align}
\mathbb{E} N_\text{hit} = \sum_{n=0}^\infty P(N_\text{hit} > n)
\end{align}
The probability that we haven't gotten a hit in $n$ dice, $P(N_\text{hit} > n)$ is the probability that we roll a miss all $n$ times.
\begin{align}
P(N_\text{hit} > n) = (1-p)^n
\end{align}
Thus, the expected value is,
\begin{align}
\mathbb{E} N_\text{hit} = \sum_{n=0}^\infty (1-p)^n .
\end{align}
This is a familiar infinite series, although it might not be immediately recognizable in this form. If we instead substitute $a = 1-p$ and notice that $a < 1$ for $p > 0$, then it becomes more familiar.
\begin{align}
\mathbb{E} N_\text{hit} &= \sum_{n=0}^\infty a^n \\
&= a^0 + a^1 + a^2 + a^3 + \ldots \\
& = \frac{1}{1-a}
\end{align}
This is very similar to the analysis done related to rolling for actions in First Martians.
Finally, we substitute back in that $a = 1-p$.
\begin{align}
\mathbb{E} N_\text{hit} &= \frac{1}{1-a} \\
&= \frac{1}{1 - (1-p)} \\
& = \frac{1}{p}
\end{align}
Combining this result for a single figure with the previous analysis we find that indeed $\mathbb{E} N = f/p$.

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