# Expected Value

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

## Example 1

Case 1: Head

If the 'head' is shown, you get 3 points.

So *x*_{h} = +3.

The probability of 'head' shown, *p*_{h}, is 1/2.

Case 2: Tail

If the 'tail' is shown, you lose 2 points.

So *x*_{t} = -2.

The probability of 'tail' shown, *p*_{t}, is 1/2.

The expected value, E(*X*), is the sum of *x*⋅*p*

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])

Case 1: 0 'head's

There's 1 way that can happen: [T, T].

And (total) = 4 ways.

So *p*_{0} = 1/4.

Case 2: 1 'head'

There are 2 ways that can happen: [H, T] and [T, H].

And (total) = 4 ways.

So *p*_{1} = 2/4.

Case 3: 2 'head's

There's 1 way that can happen: [H, H].

And (total) = 4 ways.

So *p*_{2} = 1/4.

The expected value, E(*X*), is the sum of *x*⋅*p*

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, *A*_{tot}, is 9*π*.

Area of a circle

Case 1: 3 points (Blue)*x*_{3} = 3

The area of the blue part, *A*_{3}, is *π*.

And *A*_{tot} = 9*π*.

So *p*_{3} = *π*/9*π* = 1/9.

Case 2: 2 points (Green)*x*_{2} = 2

The area of the green part, *A*_{2}, is 4*π* - *π* = 3*π*.

And *A*_{tot} = 9*π*.

So *p*_{2} = 3*π*/9*π* = 3/9.

Case 3: 1 point (Brown)*x*_{1} = 1

The area of the brown part, *A*_{1}, is 9*π* - 4*π* = 5*π*.

And *A*_{tot} = 9*π*.

So *p*_{1} = 5*π*/9*π* = 5/9.

The expected value, E(*X*), is the sum of *x*⋅*p*

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 *x*_{win} = 5.0⋅10^{8} - 2.

-2 is very small than 500⋅10^{6}.

So -2 is negligible.

So *x*_{win} ≈ 5.0⋅10^{8}.

There are _{69}C_{5}⋅_{26}C_{1} ways

to pick 6 numbers.

Combinations (_{n}C_{r})

And there's only 1 way

to pick the 6 winning numbers.

So the probability of winning the jackpot, *p*_{win}, is

1/_{69}C_{5}⋅_{26}C_{1} = 1/(2.9⋅10^{8}).

Case 2: Lose

One pays $2 and gets nothing.

So *x*_{lose} = -2.

If one doesn't win the jackpot,

then one loses the lottery.

So *p*_{lose} = 1 - *p*_{win}

= 1 - 1/(2.9⋅10^{8}).

Probability of (not *A*, Complement)

1/(2.9⋅10^{8}) is very small than 1.

so 1/(2.9⋅10^{8}) is negligible.

So *p*_{lose} ≈ 1.

The expected value, E(*X*), is the sum of *x*⋅*p*

for each case.

So E(*X*) = (5.0⋅10^{8})⋅[1/(2.9⋅10^{8})] + (-2)⋅1.

E(*X*) = -8/29 means

each time one buys a lottery ticket,

one expects to lose $8/29. (= $0.276...)