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Mathematical Expectations: The Insider's Guide to Expected Values

Introduction

In probability theory, the expected value of a random variable, intuitively, is the long-run average value of repetitions of the experiment it represents.
- Wikipedia

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 real-life 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
    1. Pay $800 with damage
    2. Pay $0
    Expected Payments for accident

    $${E(A) = 800 \times p + 0 \times (1-p) = 800p}$$

    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
    1. Pay $100 with damage
    2. Pay $0
    Expected Payments for accident

    $${E(A) + 80 = 100 \times p + 0 \times (1-p) + 80 = 100p + 80}$$

    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{800-100}{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.

Mathematical Expectations: The Insider's Guide to Expected Values
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