# Expected Value

How to find the expected value from the probability problems: examples and their solutions.

## Example 1

If the 'head' is shown, you get 3 points.
So xh = +3.

The probability of 'head' shown, ph, is 1/2.

Case 2: Tail

If the 'tail' is shown, you lose 2 points.
So xt = -2.

The probability of 'tail' shown, pt, is 1/2.

The expected value, E(X), is the sum of xp
for each case.

Sigma Notation

So E(X) = 3⋅(1/2) + (-2)⋅(1/2).

E(X) = 1/2 means
each time the coin is tossed,
you expect to get 1/2 points.

## Example 2

For each toss, there are 2 ways: 'head' or 'tail'.

The coin is tossed 2 times.
And each toss is an independent event.

So (total) = 2⋅2 = 4 ways.
([H, H], [H, T], [T, H], [T, T])

There's 1 way that can happen: [T, T].
And (total) = 4 ways.

So p0 = 1/4.

There are 2 ways that can happen: [H, T] and [T, H].
And (total) = 4 ways.

So p1 = 2/4.

There's 1 way that can happen: [H, H].
And (total) = 4 ways.

So p2 = 1/4.

The expected value, E(X), is the sum of xp
for each case.

So E(X) = 0⋅(1/4) + 1⋅(2/4) + 2⋅(1/4).

E(X) = 1 means
if a coin is tossed 2 times,
you expect to see 1 'head'.

## Example 3

The total area of the dart, Atot, is 9π.

Area of a circle

Case 1: 3 points (Blue)

x3 = 3

The area of the blue part, A3, is π.
And Atot = 9π.

So p3 = π/9π = 1/9.

Case 2: 2 points (Green)

x2 = 2

The area of the green part, A2, is 4π - π = 3π.
And Atot = 9π.

So p2 = 3π/9π = 3/9.

Case 3: 1 point (Brown)

x1 = 1

The area of the brown part, A1, is 9π - 4π = 5π.
And Atot = 9π.

So p1 = 5π/9π = 5/9.

The expected value, E(X), is the sum of xp
for each case.

So E(X) = 3⋅(1/9) + 2⋅(3/9) + 1⋅(5/9).

E(X) = 14/9 means
for the given dart,
you expect to get 14/9 points per trial.

## Example 4

Case 1: Win

One pays \$2 and gets \$500 million.
So xwin = 5.0⋅108 - 2.

-2 is very small than 500⋅106.
So -2 is negligible.
So xwin ≈ 5.0⋅108.

There are 69C526C1 ways
to pick 6 numbers.

Combinations (nCr)

And there's only 1 way
to pick the 6 winning numbers.

So the probability of winning the jackpot, pwin, is
1/69C526C1 = 1/(2.9⋅108).

Case 2: Lose

One pays \$2 and gets nothing.
So xlose = -2.

If one doesn't win the jackpot,
then one loses the lottery.

So plose = 1 - pwin
= 1 - 1/(2.9⋅108).

Probability of (not A, Complement)

1/(2.9⋅108) is very small than 1.
so 1/(2.9⋅108) is negligible.

So plose ≈ 1.

The expected value, E(X), is the sum of xp
for each case.

So E(X) = (5.0⋅108)⋅[1/(2.9⋅108)] + (-2)⋅1.

E(X) = -8/29 means
each time one buys a lottery ticket,
one expects to lose \$8/29. (= \$0.276...)