Mathematical Expectations: The Insider's Guide to Expected Values
Introduction
In probability theory, the expected value of a random variable, intuitively, is the longrun average value of repetitions of the experiment it represents.
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Understanding Expected Values
Let us understand the idea of expected values and intuition behind it by taking numerous examples.
Rolling a Dice
Let's take a sample of rolling a dice.
Outcome  Probability  Outcome \(\times\) Probability 

1  $${\frac{1}{6}}$$  $${\frac{1}{6}}$$ 
2  $${\frac{1}{6}}$$  $${\frac{2}{6} = \frac{1}{3}}$$ 
3  $${\frac{1}{6}}$$  $${\frac{3}{6}= \frac{1}{2}}$$ 
4  $${\frac{1}{6}}$$  $${\frac{4}{6}= \frac{2}{3}}$$ 
5  $${\frac{1}{6}}$$  $${\frac{5}{6}}$$ 
6  $${\frac{1}{6}}$$  $${\frac{6}{6}=1}$$ 
$${ Total = \frac{7}{2} = 3.5}$$ 
Expected value of rolling a dice
If we roll a dice 10,000 times using simulation and find average by summing up all the outcomes and divide by 10,000, it comes out to be 3.4733.
It's very close to the number 3.5 in the above example.
Playing Roulette
Let's take another example of Roulette. It's a betting game where we can bet on various parameters.
Let's say; we consider colour scenario alone for this example. Suppose, there are two colours in roulette to bet on 18 red numbers and 19 black numbers.
Let's say we bet on red for winning
\begin{align} Red & = net \space win \space of \space 1 \newline Black & = net \space loss \space of \space 1 \newline \end{align}
Consider Random Variable B for Win +1 and Lose 1\begin{align} P(B=1) & = \frac{18}{37} = 0.486486 \newline P(B=1) & = \frac{19}{37} = 0.513514 \end{align}
Expected Value of Bet results in loss:Outcome  Probability  Outcome x Probability 

1  $${\frac{18}{37}}$$  $${\frac{18}{37}}$$ 
1  $${\frac{19}{37}}$$  $${\frac{19}{37}}$$ 
$${ Total = \frac{1}{37} = 0.027027}$$ 
Expected value of betting in roullette
Car Rental Insurance
Let's take the reallife example of Car rental insurance.

Without insurance:
I have to pay $800 for loses. Loses above $800 to be born by Car rental company.We have two values with insurance
 Pay $800 with damage
 Pay $0
Expected Payments for accident
$${E(A) = 800 \times p + 0 \times (1p) = 800p}$$
where, p is the probability of the accident 
With paid insurance:
I have to pay $100 for loses with paid insurance of $80. Loses above $100 to be born by Car rental company.We have two values without insurance
 Pay $100 with damage
 Pay $0
Expected Payments for accident
$${E(A) + 80 = 100 \times p + 0 \times (1p) + 80 = 100p + 80}$$
where, p is the probability of the accident
Comparing With and Without Insurance
What is the probability that these two expected payments to be the same? We can equate and find out, the difference in the payments should be taken from the insurance amount.
\begin{align} Profitable \space Probability & = \frac{(Sum \space With \space Insurance  Sum \space Without \space Insurance)} {Insurance Amount} \newline & = \frac{800100}{80}\newline & = \frac{700}{80}\newline & = \frac{700}{80}\newline & = 8.75 \newline \end{align}
Hence, if your probability of an accident is higher than 8.75 %, you should get the insurance.