# Expected value VS. probability for the purpose of classification

Say that we have two gun shooter teams, $t_1$ and $t_2$, that will compete against each other tonight to see which of them is a better gun shooter.

The rules of this shooting game are very simple:

• Each team will shoot 50 shots.
• If a shot lands on the center dot, the team is given a score of 5 points.
• If a shot lands in the 1st outer circle, the team is given a score of 2 points.
• If a shot lands in the 2nd outer circle, the team is given a score of 1 points.
• Else, they get 0 points.
• Match ends when each one of the teams shoots 50 shots.
• Winning team is one that gets the largest number of points once the match ends.

Our task in this scenario is to predict whether team $t_1$ will win against $t_2$.

The only input that we are given in order to predict the winning team is only the following:

• Two random variables, $S_1$ and $S_2$, that take values in the set of scores that teams $t_1$ and $t_2$ can achieve, respectively. For example, the expected value $\mathbb{E}[S_1]$ reflects the expected number of scores team $t_1$ will have in this match.

• Number of scores that each team will achieve is perfectly independent of any external factor, including (but not limited to), weather and time. So we need not worry about the performance degradation or improvement of any of the teams, as their performance is constant (independent of any factor)

In order to predict whether $t_1$ shall win, we have the following methods:

Prediction method 1: $$m_1(t_1,t_2) = \begin{cases} \text{predict t_1 will win} & \text{ if } \Pr(S_1 > S_2) > 0.5\\ \text{predict t_2 will win} & \text{ otherwise}\\ \end{cases}$$

Another way to rephrase that (maybe more correctly) is as follows. Let $\mathcal{S} = \{0,1,2,...,250\}$ be the set of possible scores, and let $\mathcal{T} = \{(a,b):(a,b) \in \mathcal{S}\times\mathcal{S}, a > b\}$ be the set of score tuples where the first component is larger than the 2nd. Then, we can represent method 1 as follows: $$m_1(t_1,t_2) = \begin{cases} \text{predict t_1 will win} & \text{ if } \left(\sum_{(a,b)\in\mathcal{T}}\Pr(S_1=a, S_2=b)\right) > 0.5\\ \text{predict t_2 will win} & \text{ otherwise}\\ \end{cases}$$

Prediction method 2: $$m_2(t_1,t_2) = \begin{cases} \text{predict t_1 will win} & \text{ if } \mathbb{E}[S_1] > \mathbb{E}[S_2]\\ \text{predict t_2 will win} & \text{ otherwise}\\ \end{cases}$$

## Question:

• Q1: How do the methods $m_1, m_2,$ compare against each other from the perspective of prediction accuracy?

For completeness, for any method $m_i$, we measure its predictive accuracy as follows: $$\text{acc}\big(m_i(t_1,t_2)\big)=\frac{\#\text{of times method m_i predicted correctly}}{\#\text{of times method m_i was used to predict}}$$

• What is the criterion of winning? The greatest total score? – Tim Nov 16 '16 at 10:47
• Yes (I edited my question to incorporate that). – caveman Nov 16 '16 at 10:51
• What do you mean by $\Pr(S_1 > S_2)$? Btw methods 2,3,4 are identical so I don't get what do you mean? – Tim Nov 16 '16 at 16:19
• Since $\Pr(S_1 \gt S_2)$ is the chance that team 1 will win outright and each method always predicts the same team will win, then either they are the same method or they always predict opposite results. Why, then, isn't the answer perfectly obvious? – whuber Nov 17 '16 at 15:37
• They are quite different. There is no required mathematical relationship between the chance of the event $S_1\gt S_2$ and the expectations of those two variables. It's easy for $m_1$ and $m_2$ always to give opposite predictions. As a tiny example, consider a situation where only one shot is taken. Let $S_1$ always score $2$. Let $S_2$ score $1$ with probability $2/3$ and $5$ with probability $1/3$. The expectations are $2$ and $7/3$, so $S_2$ has the higher expectation. However, $S_1$ will score higher than $S_2$ in two-thirds of all shots. – whuber Nov 17 '16 at 18:28