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Hey, everyone.

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Welcome back.

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Today, we're going to do another
fun problem that has

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to do with a random number
of coin flips.

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So the experiment we're going
to run is as follows.

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We're given a fair six-sided
die, and we roll it.

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And then we take a fair coin,
and we flip it the number of

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times indicated by the die.

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That is to say, if I roll a four
on my die, then I flip

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the coin four times.

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And then we're interested in
some statistics regarding the

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number of heads that show
up in our sequence.

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In particular, we want to
compute the expectation and

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the variance of the number
of heads that we see.

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So the first step of this
problem is to translate the

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English to the math.

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So we have to define
some notation.

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I went ahead and did
that for us.

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I defined n to be the outcome
of the die role.

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Now, since we flip the coin the
number of times shown by

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the die roll, n is equivalently
the number of

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flips that we perform.

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And n, of course, is a random
variable, and I've written its

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PMF up here.

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So Pn of n is just a discrete
uniform random variable

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between 1 and 6, because we're
told that the die has six

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sides and that it's fair.

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Now, I also defined h
to be the number of

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heads that we see.

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So that's the quantity
of interest.

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And it turns out that Bernoulli
random variables

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will be very helpful to
us in this problem.

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So I defined x sub i as
1 if the ith flip is

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heads, and 0 otherwise.

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And what we're going to do now
is, we're going to use these x

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sub i's to come up with
an expression for h.

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So if you want to count the
number of heads, one possible

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thing you could do is start with
0 and then look at the

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first coin flip.

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If it's heads, you add 1 to 0,
which I'm going to call your

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running sum.

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If the first flip is
tails, you add 0.

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And similarly, after that, after
every trial, if you see

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heads, you add 1 to
your running sum.

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If you see a tails, you add 0.

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And in that way, we can
precisely compute h.

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So the mathematical statement of
what I just said is that h

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is equal to x1 plus x2
plus x3, all the way

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through x sub n.

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So now, we are interested
in computing e of h, the

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expectation of h.

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So your knee jerk reaction might
be to say, oh, well, by

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linearity of expectation,
we know that this is an

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expectation of x1, et cetera
through the expectation of xn.

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But in this case, you would
actually be wrong.

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Don't do that.

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And the reason that this is not
going to work for us is

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because we're dealing
with a random

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number of random variables.

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So each xi is a random
variable.

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And we have capital n of them.

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But capital n is a
random variable.

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It denotes the outcome
of our die roll.

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So we actually cannot just
take the sum of these

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expectations.

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Instead, we're going to have
to condition on n and use

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iterated expectation.

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So this is the mathematical
statement of what I just said.

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And the reason why this works
is because conditioning on n

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will take us to the case that
we already know how to deal

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with, where we have a known
number of random variables.

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And of course, iterated
expectations holds, as you saw

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in lecture.

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I will briefly mention here that
the formula we're going

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to derive is derived
in the book.

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And it was probably derived
in lecture.

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So if you want, you can just
go to that formula

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immediately.

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But I think the derivation of
the formula that we need is

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quick and is helpful.

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So I'm going to go through
it quickly.

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Let's do it over here.

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Plugging in our running
sum for h, we get this

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expression--

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x1 plus x2 et cetera plus
xn, conditioned on n.

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And this, of course,
is n times the

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expectation of x sub i.

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So again, I'm going through
this quickly,

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because it's in the book.

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But this step holds, because
each of these xi's have the

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same statistics.

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They're all Bernoulli with
parameter of 1/2, because our

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coin is fair.

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And so I used x sub i to say it
doesn't really matter which

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integer you pick for i, because
the expectation of xi

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is the same for all i.

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So this now, the expectation
of x sub i, this is just a

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number, it's just some constant,
so you can pull it

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out of the expectation.

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So you get the expectation
of x sub i times the

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expectation of n.

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So I gave away the answer
to this a second ago.

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But x sub i is just a Bernoulli
random variable with

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parameter of success of 1/2.

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And we know already that the
expectation of such a random

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variable is just p, or 1/2.

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So this is 1/2 times
expectation of n.

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And now n we know is a discrete

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uniform random variable.

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And there's a formula that I'm
going to use, which hopefully

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some of you may remember.

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If you have a discrete uniform
random variable that takes on

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values between a and b--

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let's use w--

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if you call this random variable
w, then we have that

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the variance of w is equal to b
minus a times b minus a plus

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2 divided by 12.

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So that's the variance.

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We don't actually need the
variance, but we will need

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this later.

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And the expectation of w--

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actually, let's just
do it up here right

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ahead for this problem.

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Because we have a discrete
uniform random variable, the

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expectation is just
the middle.

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So you agree hopefully that the
middle is right at 3.5,

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which is also 7/2.

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So this is times 7/2, which
is equal to 7/4.

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So we are done with
part of part a.

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I'm going to write this answer
over here, so I can erase.

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And we're going to do something
very similar to

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compute the variance.

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To compute the variance,
we are going to also

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condition on n.

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So we get rid of this source
of randomness.

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And then we're going to use law
of total variance, which

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you've also seen in lecture.

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And again, the formula
for this variance is

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derived in the book.

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So I'm going to go through
it quickly.

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But make sure you understand
this derivation, because it

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exercises a lot of stuff
we taught you.

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So this, just using law of
total variance, is the

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variance of expectation of h
given n, plus the expectation

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of the variance of h given n.

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And now, plugging in
this running sum

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for h, you get this.

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It's a mouthful to write.

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Bear with me.

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x1 through xn given n--

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so I didn't do anything fancy.

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I just plugged this into here.

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So this term is similar
to what we saw

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in a previous problem.

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By linearity of expectation and
due to the fact that all

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of the x i's are distributed in
the same way, they have the

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same expectation, this
becomes n times the

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expectation of x sub i.

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And let's do this
term over here.

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This term--

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well, conditioned on
n, this n is known.

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So we essentially have
a finite known sum of

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independent random variables.

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We know that the variance of
a sum of independent random

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variables is the sum
of the variances.

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So this is the variance of x1
plus the variance of x2 et

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cetera, plus the
variance of xn.

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And furthermore, again, because
all of these xi's have

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the same distribution, the
variance is the same.

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So we can actually write this
as n times the variance of x

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sub i, where x sub i
just corresponds

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to one of the trials.

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It doesn't matter which one,
because they all have the same

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variance and expectation.

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So now, we're almost
home free.

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This is just some scaler.

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So we can take it out of
the variance, but we

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have to square it.

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So this becomes expectation
of xi squared times

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the variance of n.

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And then this variance is also
just a scalar, so we can take

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it outside.

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So then we get variance of x sub
i times expectation of n.

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Now, we know that the
expectation of x sub i is just

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the probability of success,
which is 1/2.

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So we have 1/2 squared, or 1/4,
times the variance of n.

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So that's where this formula
comes in handy.

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b is equal to 6, a
is equal to 1.

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So we get that the variance
of n is equal to 5 times--

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and then 5 plus 2 is 7--

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divided by 12.

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So this is just a formula from
the book that you guys

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hopefully remember.

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So we get 35/12.

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And then the variance of xi,
we know the variance of a

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Bernoulli random variable is
just p times 1 minus p.

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So in our case, that's 1/2
times 1/2, which is 1/4.

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So we get 1/4.

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And then the expectation of n,
we remember from our previous

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computation, is just 7/2.

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So I will let you guys do this
arithmetic on your own time.

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But the answer comes
out to be 77/48.

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So I will go ahead and put
our answer over here--

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77/48--

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so that I can erase.

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00:10:38,130 --> 00:10:40,490

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So I want you guys to
start thinking about

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part b while I erase.

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Essentially, you do the same
experiment that we did in part

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a, except now we use two
dice instead of one.

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So in part b, just to repeat,
you now have two dice.

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You roll them.

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You look at the outcome.

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If you have an outcome of four
on one die and six on another

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die, then you flip the
coin 10 times.

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So it's the same exact
experiment.

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We're interested in the number
of heads we want the

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expectation and the variance.

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But this step is now a
little bit different.

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Again, let's approach this by
defining some notation first.

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Now, I want to let n1 be the
outcome of the first die.

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And then you can let n2 be the
outcome of the second die.

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And we'll start with
just that.

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So one way you could approach
this problem is say, OK, if n1

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is the outcome of my first die
and n2 is the outcome of my

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second die, then the number of
coin flips that I'm going to

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make is n1 plus n2.

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This is the total coin flips.

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So you could just repeat the
same exact math that we did in

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part a, except everywhere that
you see an n, you replace that

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n with n1 plus n2.

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So that will get you to your
answer, but it will require

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slightly more work.

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We're going to think about
this problem slightly

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differently.

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So the way we are thinking about
it just now, we roll two

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dice at the same time.

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We add the results
of the die rolls.

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And then we flip the coin
that number of times.

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But another way you can think
about this is, you roll one

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die, and then you flip the coin
the number of times shown

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by that die and count
the number of heads.

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And then you take the second
die and you roll it.

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And then you flip the coin that
many more times and count

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the number of heads
after that.

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So you could define h1 to be
number of heads in the first

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n1 coin flips.

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And you could just let h2 be
the number of heads in the

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last n2 coin flips.

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So hopefully that terminology
is not confusing you.

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Essentially, what I'm saying
is, n1 plus n2 means you'll

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have n1 flips, followed by
n2 flips, for a total

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of n1 plus n2 flips.

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And then within the first n1
flips, you can get some number

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of heads, which we're
calling h1.

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And in the last n2 flips, you
can get some number of heads,

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which is h2.

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So the total number of heads
that we get at the end--

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I'm going to call it h star--

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is equal to h1 plus h2.

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And what part b is really
asking us for is the

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expectation of h star and
the variance of h star.

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But here's where something
really beautiful happens.

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h1 and h2 are independent,
and they are

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statistically the same.

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So the reason why they're
independent is because--

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well, first of all, all of our
coin flips are independent.

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And they're statistically the
same, because the experiment

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is exactly the same.

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And everything's independent.

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So instead of imagining one
person rolling two die and

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then summing the outcomes and
flipping a coin that many

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00:14:55,390 --> 00:14:58,670
times and counting heads, you
can imagine one person takes

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00:14:58,670 --> 00:15:00,720
one die and goes
into one room.

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00:15:00,720 --> 00:15:02,710
A second person takes a
second die and goes

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00:15:02,710 --> 00:15:04,500
into another room.

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00:15:04,500 --> 00:15:06,100
They run their experiments.

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00:15:06,100 --> 00:15:08,520
Then they report back
to a third person

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the number of heads.

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And that person adds them
together to get h star.

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00:15:13,460 --> 00:15:16,660
And in that scenario, everything
is very clearly

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independent.

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So the expectation of h star--

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you actually don't need
independence for this part,

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because linearly of expectation
always holds.

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But you get the expectation of
h1 plus the expectation of h2.

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And because these guys are
statistically equivalent, this

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is just two times the
expectation of h.

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And the expectation of h we
calculated in part a.

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So this is 2 times 7 over 4.

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Now, for the variance,
here's where the

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independence comes in.

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00:15:50,930 --> 00:15:54,238
I'm actually going to write this
somewhere where I don't

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have to bend over.

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00:15:55,200 --> 00:16:00,690
So the variance of h star is
equal to the variance of h1

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00:16:00,690 --> 00:16:03,820
plus the variance of
h2 by independence.

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00:16:03,820 --> 00:16:07,020
And that's equal to 2 times
the variance of h, because

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they are statistically
the same.

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And the variance of h
we computed already.

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00:16:12,810 --> 00:16:18,650
So this is just 2 times
77 over 48.

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00:16:18,650 --> 00:16:23,390
So the succient answer to part
b is that both the mean and

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00:16:23,390 --> 00:16:28,830
the variance double from part
A. So hopefully you guys

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00:16:28,830 --> 00:16:30,060
enjoyed this problem.

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00:16:30,060 --> 00:16:32,510
We covered a bunch of things.

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00:16:32,510 --> 00:16:37,870
So we saw how to deal with
having a random number of

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00:16:37,870 --> 00:16:38,830
random variables.

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00:16:38,830 --> 00:16:41,840
Usually we have a fixed number
of random variables.

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00:16:41,840 --> 00:16:44,670
In this problem, the number of
random variables we were

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00:16:44,670 --> 00:16:47,350
adding together was
itself random.

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00:16:47,350 --> 00:16:49,410
So to handle that, we
conditioned on n.

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00:16:49,410 --> 00:16:53,110
And to compute expectation, we
use iterated expectation.

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00:16:53,110 --> 00:16:57,450
To compute variance, we used
law of total variance.

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00:16:57,450 --> 00:17:03,020
And then in part b, we were
just a little bit clever.

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00:17:03,020 --> 00:17:07,270
We thought about how can we
reinterpret this experiment to

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00:17:07,270 --> 00:17:08,859
reduce computation.

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00:17:08,859 --> 00:17:12,869
And we realized that part b is
essentially two independent

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00:17:12,869 --> 00:17:14,150
trials of part a.

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00:17:14,150 --> 00:17:15,900
So both the mean and the
variance should double.

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00:17:15,900 --> 00:17:19,567