MATH110 Sec 12.6 (F2019): Expectation

MATH110 Sec 12.6: Expectation

Today we'll cover section
12.6 on expectation.

Suppose I ask you to
play a game with me.

Here is the deal.

You roll a standard die.

If you roll an even number,
you will win $30 times a number

that you roll on that die.

If you roll an odd number, you
will lose $30 times the number

that you roll.

Or if you want to,
you can reverse it

so you'll win with the odd rolls
and lose with the even rolls.

Which one of those would
you choose and why?

The concept of expectation,
or sometimes it's call

"expected value,"
provides a formal way

of deciding issues like that.

So this is the topic
we're talking about today,

and we're going to illustrate
this idea of expectation

through basically
setting up games and also

some real-life situations
where it might come into play.

And let's go back to the
die-roll problem though.

I want to build a
chart that tells us

about winning and losing
based on what the die roll is.

And we want to use
that as a basis

of deciding whether to
choose the odd scenario

or the even scenario.

Let's go right now
though with the scenario

that I first posited, which
is that you win $30 time

the number that you roll if it's
even, and you lose $30 times

the number that you
roll that it's odd.

If we do that,
suppose you roll a 1.

Well, 1 is odd, so
you'd lose 1 times $30.

You would lose $30.

I'm going to indicate
that by putting

a negative sign in front of it.

If you roll a 2,
that's an even number.

You'll win 30 times 2.

You'll win $60.

If you roll a 3, that's odd.

So you lose 30 times
3, which is $90.

You see how this is working.

If you roll a 4, that's even.

You'll win 30 times
4 equals $120.

If you roll a 5, that's odd.

So you lose 30 times
5, which is $150.

If you roll a 6, that's even.

You win 30 times 6, $180.

In addition to how much
you can win or lose,

you've got to look at the
probability of each of those.

If you win on something
that rarely happens,

that's different than
winning on something

that happens a lot and
vice versa with losses.

So the probabilities
are important here.

Now, with a die roll,
if the die is fair,

it's actually going to
be the same every time

for this particular case.

There is a one sixth chance
of getting a 1, a 2, a 3, a 4,

a 5, or a 6.

So in this particular
problem, the probabilities

are the same all the way down.

But that in general does
not have to be true.

In this case, it is.

So the probability of
rolling a 1 is one sixth.

The probability of
rolling a 2 is one sixth.

The probability or
rolling a 3 is one sixth.

The 4 is one sixth.

The 5 is one sixth,
and the 6 is one sixth.

Now I want to do something
to merge those two

things together.

And I'm going to do that by
taking the first column, which

are the values either won
or lost, the dollar values,

and multiplying each by their
probabilities of occurring.

So across that, I will multiply
the first column number

times the second column number,
which is a dollar either gain

or lost times a probability.

So negative 30 times 1/6 is $5.

And it's a loss, so
it's negative $5.

60 times 1/6 is $10,
and it's positive,

so it's a positive $10.

Negative 90 times 1/6 is
$15, but it's negative.

120 times 1/6 is $20.

Negative 150 times
1/6 is negative $25.

And finally, 180
times 1/6 is $30.

So I've multiplied
across, and what

I've done by doing that is
created a third column which

merges the amount won or
lost with the probabilities

associated with each of
those things occurring.

It's kind of a blend
of those two things.

Are you with me?

Multiplying the value from
the first column, which,

in this case, is a dollar
amount either won or lost,

with its probability
of occurring.

And that's the value
in the third column.

Now I want to make this
into one number which

sort of encapsulates
the whole thing,

and I'm going to do
that by adding up

all those numbers
in the last column.

Now, you'll notice,
in this case,

if I get a positive
sum, that's good for me

because that means I overall
tend to win in this game.

If I get a negative
sum, that's bad for me

if I pick the even ones because
that would be a loss for me

on average.

So anyway, I add up those
last column numbers.

And if you add up all six of
those numbers, you get $15,

and it's positive.

So that means those even
numbers were the best choice.

It gives me, on average, a $15
win each time I play that game.

Now, doesn't mean I'll win
$15 every time I do it.

But if I play it a
lot of times, I'll

average, in general,
winning about $15 each time

I play that game.

So definitely the evens
are the best choice here.

That number at the end, which
is the sum of those products,

is called the
"expectation" of the game.

And I've already
alluded to this.

But what that number means
is if you play this game over

and over again, in
the long run, you

can expect on average to
win about $15 per game.

If the expected value
or the expectation

had come out to
be negative, that

would mean on average
you would lose money,

so that would be a bad choice.

Had that come out
negative, you'd

been better off flipping your
choice and choosing the odds.

Let's look at some
more examples.

In the simplest case,
for example here, it

says the outcomes
of an experiment

and the probability of each
outcome are given in the table

below.

Compute the expectation
for this experiment.

They give you a column of
outcomes, 5, 6, 7, 8, 9, 10,

and the probability
of each outcome.

In any case, if they give
you a problem like this,

all you have to do is add
a column for the product.

Remember, we have a
column for the outcome.

In the previous example,
it was dollar amounts.

And then you have a
probability column.

And all you need to do is
multiply across and then add

down.

So in this particular case,
the expectation is 6.23.

A problem like this is
strictly giving you practice

doing the calculations.

How about this one?

One of the wagers in
the game of roulette

is to place a bet that the ball
will land on a black number.

Now, on this wheel,
you can count them.

But I'll tell you that
there are 18 black numbers.

There are 18 red numbers.

And then, as you can see,
there are two green numbers.

If the ball lands
on a black number,

the player wins the
amount of his bet.

If the player bets $3, find
the player's expectation,

and round your answer
to two decimal places.

So we're looking for the
player's expectation.

So what you have to do
is build your own chart.

You want a column for how
much the player wins or loses

with various scenarios.

Then you're going to get a
column for probabilities,

and you will multiply
across and add down.

That's your strategy
here every time.

So let's start with
the won/lost situation.

The player either wins or loses,
so there are only two rows

to this table.

If you land on
black, you're going

to win the amount of your bet.

So if a player bets $3 and lands
on black, the player wins $3.

On the other hand, if the
ball does not land on black,

that would mean landing
on either red or green

because there are
only two other colors.

Then the player loses his or
her bet, which is, in this case,

$3.

So it's a negative $3.

Now we have to figure
the probabilities

of those two things happening.

Notice that if
you land on black,

you've got to count the
number of black numbers

divided by the total
number of numbers.

Well, if you look up
here, there are 18 blacks,

so it's 18 out of, and then
you've got to count them all.

And if you count them
all, you get 18 plus 18

plus 2, which happens to be 38.

So the probability of
landing on a black is 18/38.

So what's the probability of
landing on a red or green?

Well, if there are 38
and 18 of them are black,

the rest of them
are red or green,

which means 18 and 2 is 20,
or you can say 38 minus 18.

In any case, you'll find out
that there are 20 nonblack

numbers.

So there's a probability
of 20 out of 38

that you land on
a nonblack number.

So now it's just a matter
of doing the arithmetic.

The last column is the
product of the won/lost column

and the probability column.

So if you multiply across--

and this time, they're asking us
to round to two decimal places.

So you can go ahead.

And when you multiply
3 times 18, you get 54.

So that's 54/38.

But since you're going to
round to two decimal places,

it's not a bad idea
to go ahead and change

that 54/38 to a decimal.

Now, I will say
though, if you're

going to round in the end
to two decimal places,

you do not need to round
to two decimal places

in the intermediate
calculations.

They need to be
taken out further.

So I went ahead and showed
five decimal places.

But certainly, go beyond two.

And it comes out to
be about 1.42105.

And then when you multiply
across the second row,

remember, that's a negative
$3, so the answer is negative

3 times 20.

That's negative 60/38.

And if you change
that to a decimal,

it comes out to be
about negative 1.57895.

And then if you add down the
column, that last column,

you get about negative 0.1579.

And that's what they told you
to round to two decimal places.

So that would be negative 0.16.

So on average, if you played
this game over and over again,

you would lose about $0.16 every
time you played, on average.

Sometimes you'd win.

Sometimes you'd lose.

But if you averaged it out, you
would tend to lose about $0.16

every time you played this game.

Let's play another
roulette game.

One of the wagers in roulette is
to bet that the ball will stop

on a number that's
a multiple of 3,

and that's not counting
the green numbers,

which are labeled 0 and 00.

They're not included in
that being a multiple of 3.

If the ball stops
on such a number,

the player wins
double the amount bet.

If the player bets $3, compute
the player's expectation,

and round your answer
to two decimal places.

This time, we're
building the same column.

We want a won or lost column.

We want a probability
column, and then we

want a column for the prize.

But this time, the winning
condition is different.

You win if you roll a number
that's a multiple of 3.

Well, the multiples of 3 are 3,
6, 9, 12, 15, 18, 21, 24, 27,

30, 33, 36.

If you notice, I
colored them differently

and bolded them up above so
you can easily count them.

So that is the
winning condition,

rolling a multiple of 3.

So you need a column
for winning, which is

you do land on a multiple of 3.

And you need to figure out
how much you win or lose.

Well, it says, if the
ball stops on a number,

you win double that amount.

So if a player
bets $3, the player

wins double that, which is $6.

However, if the player does
not land on a multiple of 3,

the player loses the
bet, which was $3.

So you have to read
this carefully.

And now we want to do the
probabilities just like before.

What is the probability of
landing on a multiple of 3?

Well, you've got to
count the multiple of 3s.

And if you count
those multiples of 3s,

you see that are 12
of them, and there

are 38 numbers just as in
the other roulette problem.

So the probability of landing
on a multiple of 3 is 12/38.

If you don't land
on a multiple of 3,

notice there are 38 numbers,
and there are 12 multiples of 3.

So if you subtract, you find
out there are 26 numbers

that are not multiples of 3.

So the probability of not
getting a multiple of 3

is 26 out of that same 38.

So at this point, we need
a column for the product.

We multiply across.

6 times 12 is 72,
so there's 70--

72/38.

And again, we want to
divide it out and round it

to two decimal places.

But we do not want to round
off to two decimal places

until we get to the end.

So I carried a
few more decimals,

and I got 1.8947
on the first row,

and I got 3 times 26 is
78, but it's negative.

So it was negative 78/38.

And again, that comes out
to be about negative 2.0526.

And if you add down
the last column,

you get about negative 0.1579.

You get about negative $0.16.

It turns out to be about the
same as the last problem.

It doesn't have to be
the same each time.

That's just the way
the numbers came out.

But the point is, if you're
playing roulette in a casino,

you can expect to have
an expectation that's

negative from your point of view
because the casino would go out

of business if they did not on
average cause you to lose money

every time you play, on
average, not every single time.

Sometimes you win.

Sometimes you lose.

But on average, you
can expect to lose

about $0.16 on
this game as well.

Before we leave
the casino, let's

play one more of these
games involving a wheel.

It says, many casinos have a
game called the Big Six Money

Wheel, which has
54 slots in which

are displayed a Joker, a
casino logo, and various dollar

amounts as shown in this table.

Players may bet on the Joker,
the casino logo, or one

or more dollar denominations.

The wheel is spun.

And if the wheel stops on the
same place as the player's bet,

the player wins that
amount for each dollar bet.

So you have to read
that enough so you

understand what's happening.

If a player bets $3 on
the Joker denomination,

find the player's expectation,
and round the answer

to two decimal places.

So this is very similar to the
roulette problems we've worked.

You have to build a
table, but you also

have to count how many
of everything there are.

They told you that
there's one Joker slot.

There was one casino slot.

And there are various numbers
of dollar slot, and one Joker,

and there are 53 things
that aren't Jokers.

And the reason I
broke it up that way

is because the player's
going to bet on the Joker,

so I need to know whether
you get a Joker or not,

which means I need those counts.

Also remember that this
information is for a $1 bet.

It says that the player wins
this amount for each $1 bet.

So if the player makes a $3 bet,
then that $40 changes to $120

because 40 times 3 is 120.

Build your table just as before.

You want a won/lost column.

You want a probability column.

And you want a column
for the product.

You win if you land on a Joker.

You lose if you get
something that's not a Joker.

So you need a Joker row
and a non-Joker row.

Now, what happens if
you do land on a Joker?

Well, we just said you win $120.

So we'll put that in there.

What happens if you
don't land on a Joker?

You lose your bet.

It was a $3 bet,
so you lose the $3.

Now we need to do
the probabilities.

Well, there's only
one Joker out of 54.

There's one Joker and 53
things that aren't Jokers.

So there's one chance out of
54 of getting a Joker, which

means there's 53 chances out of
54 that you do not get a Joker.

Once you get here,
you've got it made.

You multiply across.

You get 120/54 on the top row.

Multiply across on the bottom.

You get negative 159/54.

And then if you
change each of them

to decimals, carrying more
than two decimal places

to start with, you'll add
those last two columns

after you change
them to decimals,

and it comes out to be about
negative $0.72 by the time

you've rounded it to
two decimal places.

So again, this is a losing
game for the average player.

On average, if you played
this game over and over again,

you would tend to lose about
$0.72 each time you play.

That's an average.

If a pair of regular
dice are tossed once,

use the expectation
formula to determine

the expected sum of the numbers
on the upward faces of the two

dice.

That's a lot of words.

But all it's saying is you're
going to roll the two dice

and look at the sums.

So anytime they give me
a problem with two dice,

I'm going to build my
6-by-6 table that lists

all the possible two die rolls.

I want the sum, so
I'm going to make

a table where the first column
gives me every possible sum.

And then I'm going to have
a probability column to see

how likely that is.

And I'm going to
use my 6-by-6 table

to decide how
likely it is, which

will allow me to calculate
the probability of getting

every particular sum.

Now, the lowest sum is a 2.

That's a 1 and a 1.

And the biggest sum is a
12, which is a 6 and a 6.

So you can get anything--

any sum between 2 and 12.

So you can get a 2, or 3, or 4,
or 5, or 6, or a 7, or 8, or 9,

or 10, or 11, or 12.

I just need to know how
likely those sums are.

Those sums are not
equally likely.

Unlike just rolling one die,
when you're rolling two dice

and adding up and
getting a sum, those sums

are not equally likely.

For instance, looking
at the 2, there's

only one way to get a 2, and
that's if you get a 1 and a 1.

So the probability of
getting a 2 is just 1/36.

But looking at the 3s,
there's two ways, 2-1 and 1-2,

so that's 2 out of 36.

For a 4, there are three ways.

That's 3 out of 36.

For 5, there are four
ways, which is 4 out of 36.

For 6, there are five ways,
so that's 5 out of 36.

For 7, there are six ways,
which is 6 out of 36.

And then by the
time you get to 8,

you can see this kind of go
down again, the probability.

If you look at the 8s,
there are only five ways

to get a sum of 8, so
that's 5 out of 36.

When you look at the
9s, you get four ways.

So that's 4 out of 36.

When you look at the
10s, you get 3 out of 36.

When you get to 11,
there are only two ways.

So that's 2 out of 36.

And by the time you get to
12, there's only one way,

so that's 1 out of 36.

Once you get the value
column, which, in this case,

are the sums, and the
probability of those sums

column, which is the second
column, all you have to do

is multiply across and add down.

So let's do that.

If you multiply 2 times
1/36, you get 2/36.

If you multiply 3 times
2/36, you get 6/36.

And if you follow
that all the way down,

the very last one is 12
times 1/36, which is 12/36.

So the expectation is the
sum of all those numbers.

And they're all 36th,
so you can add them up

by adding the numerators and
putting them all over 36.

So really all you have to do
is add up 2, and 6, and 12,

and 20, and 30, and 42, and 40,
and 36, and 30, and 22, and 12,

and put that sum over 36.

It turns out to be 252 over 36.

If you divide out, you'll
see that that comes out

to be exactly 7.

So the expectation for rolling
two dice is a sum of 7.

In other words, that's
the most likely roll.

If someone ever asks you
to make a bet on a sum

and roll when you're rolling
two dice, always bet on 7.

That's the most likely sum.

I alluded to it earlier, but
we did a lot of these things

where you're just
playing gambling games.

But there are some actual
business applications

of expectations.

In fact, there are tons
of real-life applications

of expectation.

But we're going to talk
about one in particular.

This problem says, when
an insurance company

sells a life insurance
policy, the premium, which

is the cost of
buying the policy,

is based largely
on the probability

that the insured person will
outlive the term of the policy.

Such probabilities are found
in mortality tables, which

give the probability that
a person of a certain age

will live one more year.

But what the insurance
company wants to know

is the expectation
on a policy that it's

going to sell, that
is, how much it'll

have to pay out on average
for every policy it writes.

That's an extremely important
thing for that company to know.

Let's look at a
couple of examples.

According to mortality tables
in the National Vital Statistics

Report, the probability
that a 21-year-old

will die within a year
is about 0.000962.

If the premium for
a one-year, $25,000

life insurance policy
for a 21-year-old is $32,

what is the insurance company's
expectation for the policy?

You've got to start
out thinking just

like you did with the roulette
problems and all the others.

You're going to build a table.

So what we're looking at
in that table basically

is whether that 21-year-old
person dies or not.

So you could say, let
S sub 1 be the event

that the person
dies within a year.

That's the term of the policy.

And S sub 2 is the event
that the person doesn't die.

In other words,
however you phrase it,

you're looking at
whether the person will

die within a year or not.

If the person dies
within the year,

the insurance company has
to pay out the policy.

If the person doesn't
die within a year,

then the company gets to
make a profit on that person

because that person had
to pay for that policy.

The probability that the person
dies within a year is 0.000962.

But if it does
happen, the company

has to pay out $25,000
to whoever the policy

beneficiary is.

However, keep in mind that
the policy wasn't given free.

The company charged
$32 for the policy.

So even if they pay out, they
actually don't lose 25,000.

They lose 25,000 minus the
$32 that the policyholder paid

to get the policy.

So 25,000 minus 32 is 24,968.

So it's close to 25,000,
but it's not quite 25,000.

So you have to adjust for that.

The probability that the person
does not die within a year

is 1 minus the probability
that that person does die.

So if you take 1 minus
0.000962, you get 0.999038.

And if that happens, the company
keeps that entire premium

of $32, and that's the company's
profit for issuing that policy.

And that's, of course, what
the company hopes will happen.

And I'm pretty sure it's
what the person that

took out the policy hopes
will happen as well.

Any case, you have
to build a table.

One row is the 21-year-old
person dies within a year.

The second row is that
that person does not

die within the year.

The first column gives
you the amount of profit,

and we're looking at it from
the company's point of view.

And then the second
column is the probability

of that happening.

So we'll start off
just the same as we did

for the roulette wheel problem.

We look and see that
the amount of profit

that the company makes if that
person does die within a year

is actually a loss, so
it's a negative 24,009.

It's the $25,000 policy
adjusted for the fact

that they actually charged
$32 for the policy.

That's what the
company will end up

losing if that 21-year-old
person dies within a year.

However, the probability
of that is very small.

It's only 0.000962.

So it's highly unlikely
that that 21-year-old would

die, not impossible, obviously.

And what happens if the person
doesn't die within a year?

The company keeps
that $32 premium.

So that's a $32 profit
for the company,

and that's highly likely.

We already calculated
that that's

going to be 1 minus 0.000962.

It comes out to be 0.999038.

So it's highly unlikely that
the company will pay out

that big amount, and
it's highly likely

that the company will keep a
small amount, but not 100%.

Now all you have to
do is multiple across,

taking more than two decimal
places and rounding at the end,

and then adding down
that last column.

And you'll find out that
the expectation is $7.95,

and it's positive.

The company needs a
positive expectation.

If that number comes out
negative, on average,

the company's going to
lose money on every policy.

And a company that loses money
on every policy on average

is not going to
stay in business.

So on average, that
company issuing

that particular
type of policy will

tend to average $7.95
per policy issued.

You'll find one variation on
this problem in your homework.

And I want to show you how
to solve it algebraically,

mathematically, however
you want to phrase that.

And then I want to show you a
quicker way to get the answer,

but I do want to go through
the mechanics of it first.

It says, the probability that
a 30-year-old male in the US

will die within one
year is about 0.00142.

An insurance
company is preparing

to sell a 30-year-old male
a one-year, $50,000 life

insurance policy.

How much should it
charge for its premium

in order to have a positive
expectation for the policy?

And then round your answer
to the nearest dollar.

Do you see how this
problem is different

than the previous one?

Because, last time, we knew
what the cost of the policy was.

This time, we don't.

We need to figure that out.

That's what they're
asking, how much

should they charge a policy to
have a positive expectation?

So it's the same idea, but
we don't know the premium

this time.

In the last problem, we worked
it with a certain premium.

I think it was $32.

This time, we don't know
what the premium is.

Since we don't know
what the premium is,

we'll just make up
a variable for it.

You can call it x.

I decided to call it P here.

But P stands for the premium.

In the previous problem,
we knew what it was.

In this problem, we don't, so we
have to have a variable for it.

But once you use the letter
P, the process works the same.

It's just that you've got a
variable in the process which

generates some unknowns,
which generate something

you have to solve for.

But otherwise, the
process is the same.

What happens if the person dies?

Again, it's all about whether
the person dies within the year

or doesn't.

If the person dies--

actually, this is a male.

So if he dies, the
company loses $50,000

but keeps the policy
premium, which is P dollars.

So the company's actual
loss is 50,000 minus P

just like in the last problem.

It was, I think,
25,000 minus $32,

or whatever that
previous premium was.

We don't know what
the premium is here.

That's why the P is in there.

But the process is the same.

So if we build our
table, the first row

is that this 30-year-old
male dies in a year.

The second row is that that
30-year-old male doesn't die.

The first column is how much
the company gains or loses,

and the second column is the
probability of that happening.

So we just proceed
just like before.

The gain or loss for the
company of the person dying

is actually a loss.

That's why there's a
negative sign in front of it.

It's a loss of 50,000
minus P. And you

need those parentheses
because it's the 50,000

minus P is the amount of money.

And to make it negative, you've
got to negate the whole thing.

So it's negative and then, in
parentheses, 50,000 minus P.

If the person doesn't
die, the company

keeps that premium,
which is P dollars.

So the second position in
that column is just P dollars.

What are the probabilities?

Well, the probability that the
person dies within one year

is 0.00142.

The probability that
that male does not die

will be 1 minus 0.00142.

And if you do that
calculation, you'll

find out that that's 0.99858.

Now you need to multiply
across and then add down.

If you multiply across, and
being very careful about that

minus sign, you get negative
0.00142 times, in parentheses,

50,000 minus P on the first
row, and you get 0.99858P

for the second row.

Now you want to
add those together.

As you're adding,
you've got to be

careful about those
parentheses and signs.

So if I multiply the negative
0.00142 times the 50,000,

that product will be negative.

But when I multiply the negative
0.00142 times the negative P,

that will make a
positive 0.00142P.

And then, of course, the
0.99858P is just itself.

And I'm setting it equal to
0 because that's where you--

where expectation changes
from positive to negative.

We want a positive expectation.

So you find out we're at 0.

And if you get any more than
that, it'll be positive.

Anything greater than 0
is a positive expectation.

So we just set it equal to 0 and
realize that anything greater

than that will be a
positive expectation.

It looks awful.

But if you multiply out that
negative 0.00142 times 50,000,

it comes out to be negative 71.

And also, 0.00142P plus 0.99858P
comes out to just be 1P or just

P. So what looked like a
terribly messy equation came

out very simple.

It just came out to be
negative 71 plus P equals 0.

If you add a 71 to
both sides, you'll

see that P is equal to $71.

So if you set that
premium $71, that's

the dividing line between
a positive expectation

and a negative expectation.

So the company needs to set
that policy price above $71

in order to have a
positive expectation.

It's a lot of work.

The process is exactly
like the previous problem

except you've got
a variable in it.

But it turns out to be
quite a bit of work.

And I did promise you I was
going to show you a shortcut

way to do this, and I'm going to
follow through on that promise.

Here's the shortcut.

If you look at that
equation near the bottom,

there are two terms that
involve that 0.00142P.

The other term was 0.99858P.

When you add those together,
you remember, you got 1.

That was not a coincidence.

So you got 1P,
which is P. That's

going to happen every time.

That was not coincidental.

So what that means
is you really don't

have to go through all
this work every time.

Every time, the two decimals
will always add to 1.

That means you'll always get
something with a P in it.

And the number that
eventually winds up

on the other side
of the equation

is nothing more than the
product of the policy

value times the probability
that that 30-year-old male dies,

or whoever the policyholder is.

So every time you do a problem
like this, all you have to do

is look at the policy value,
and find the probability

of that person dying, and
multiply those two numbers

together.

And if you multiply those
two numbers together,

you will get $71 in this case.

And when the numbers change and
you have a different problem,

you'll get that
product, whatever it is.

The product of the policy value
and the probability of death

will give you the cutoff
for the policy price

that will give you a
positive expectation.

Even though that
problem looks nasty,

if you look at
the shortcut, it's

actually the easiest problem
you could possibly have.

You just have to
search out the policy

value, and the
probability of death,

and multiply them together.