I am a little, but how does that relate to the probability of winning or losing when you need to beat the hand of an opponent?
We use probability as the weight to calculate the weighted average or the expected outcome of an event. If the expected outcome is favourable, we should proceed. Otherwise, we shouldn't.
Let's use a simple example. Let's say you're rolling a fair dice. If you roll a six, you win $5 and anything else you lose $5. Now let P be the profit from rolling the dice once.
[math]P = \begin{cases} +5 &\text{if roll a 6} \\ -5 &\text{if roll 1, 2, 3, 4, 5 } \end{cases}[/math]Now, we're going to use probability as weights.
\(\displaystyle \Pr(6)=\frac{1}{6}\)
\(\displaystyle \Pr(\text{anything else})=\frac{5}{6}\)
So we have:
[math]P = \begin{cases} +5 &\Pr(6)=1/6 \\ -5 &\Pr(\text{anything else)}=5/6 \end{cases}[/math]The expected profit:
[math]E(P)=\$5\left( \frac{1}{6}\right)-\$5\left(\frac{5}{6}\right)\approx -\$3.33[/math]Since the expected outcome is not favourable, you shouldn't play the game. If you were to play this game indefinitely, on average, you would expect to lose $3.33 every game. It should make intuitive sense. Since if you win, you'll win $5, but when you lose, you'll also lose $5. However, the chance of getting a 6 is lower than getting anything else, so you should expect to lose money.
