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## Probability using combinatorics

# Getting exactly two heads (combinatorics)

## Video transcript

I'm going to start
with a fair coin, and I'm going to
flip it four times. And the first question I want to
ask is, what is the probability that I get exactly
one head, or heads? This is one of those
confusing things, when you're talking about what
side of the coin. I know I've been not
doing this consistently. I'm tempted to say,
if you're saying one, it feels like you should do the
singular, which would be head. But I've read up a little
bit of it on the internet, and it seems like when
you're talking about coins, you really should say one
heads, which is a little bit, it seems a little
bit difficult for me. But I'll try to go with that. So what is the probability
of getting exactly one heads? And I put that in
quotes to say, you know, really, we're just to
talking about one head there. But it's called heads when
you're dealing with coins. Anyway, I think you get
what I'm talking about. And to think about
this, let's think about how many different
possible ways we can get four flips of a coin. So we're going to have one
flip, then another flip, then another flip, then another flip. And this first flip
has two possibilities. It could be heads or tails. The second flip has
two possibilities. It could be heads or tails. The third flip has
two possibilities. It could be heads or tails. And the fourth flip
has two possibilities. It could be heads or tails. So you have 2 times
2 times 2 times 2, which is equal
to 16 possibilities. 16 possible outcomes when
you flip a coin four times. And any one of the possible
outcomes would be 1 of 16. So if I wanted to
say, so if I were to just say the
probability, and I'm just going to not talk
about this one heads, if I just take a, just
maybe this thing that has three heads right here. This exact sequence of events. This is the first flip, second
flip, third flip, fourth flip. Getting exactly
this, this is exactly one out of a possible
of 16 events. Now with that out of
the way, let's think about how many possibilities,
how many of those 16 possibilities, involve
getting exactly one heads? Well, we could list them. You could get your heads. So this is equal
to the probability of getting the heads
in the first flip, plus the probability of getting
the heads in the second flip, plus the probability of getting
the heads in the third flip. Remember, exactly one heads. We're not saying at least
one, exactly one heads. So the probability in
the third flip, and then, or the possibility that you
get heads in the fourth flip. Tails, heads, and tails. And we know already
what the probability of each of these things are. There are 16 possible
events, and each of these are one of those
16 possible events. So this is going to be 1 over
16, 1 over 16, 1 over 16, and 1 over 16. And so we're really
saying the probability of getting exactly one
heads is the same thing as the probability of getting
heads in the first flip, or the probability
of getting heads-- or I should say the
probability of getting heads in the first flip, or
heads in the second flip, or heads in the third flip,
or heads in the fourth flip. And we can add the probabilities
of these different things, because they are
mutually exclusive. Any two of these things cannot
happen at the same time. You have to pick one
of these scenarios. And so we can add
the probabilities. 1/16 plus 1/16 plus
1/16 plus 1/16. Did I say that four times? Well, assume that I did. And so you would get 4/16,
which is equal to 1/4. Fair enough. Now let's ask a slightly
more interesting question. Let's ask ourselves
the probability of getting exactly two heads. And there's a couple of
ways we can think about it. One is just in the
traditional way. [? We ?] know the
number of possibilities and of those equally
likely possibilities. And we can only use
this methodology because it's a fair coin. So, how many of the
total possibilities have two heads of the total of
equally likely possibilities? So we know there are 16
equally likely possibilities. How many of those
have two heads? So I've actually, ahead
of time so we save time, I've drawn all of the 16
equally likely possibilities. And how many of these
involve two heads? Well, let's see. This one over here has two
heads, this one over here has two heads, this one
over here has two heads. Let's see, this one
over here has two heads, and this one over
here has two heads. And then this one over
here has two heads, and I believe we
are done after that. So if we count them, one,
two, three, four, five, six of the possibilities
have exactly two heads. So six of the 16 equally likely
possibilities have two heads. So we have a-- what
is this-- a 3/8 chance of getting exactly two heads. Now that's kind of what
we've been doing in the past, but what I want to do
is think about a way so we wouldn't have to write
out all the possibilities. And the reason why
that's useful is, we're only dealing
with four flips now. But if we were
dealing with 10 flips, there's no way that
we could write out all the possibilities like this. So we really want a different
way of thinking about it. And the different
way of thinking about it is, if we're
saying exactly two heads, you can imagine we're
having the four flips. Flip one, flip two,
flip three, flip four. So these are the flips,
or you could say, the outcome of the flips. And if you're going to
have exactly two heads, you could say, well,
look, I'm going to have one head in
one of these positions, and then one head in
the other position. So how many, if I'm
picking the first, so I have kind of a heads
one, and I have a heads two. And I don't want you to
think that these are somehow the heads in the first flip or
the heads in the second flip. What I'm saying is
we need two heads. We need a total of two
heads in all of our flips. And I'm just giving one
of the heads a name, and I'm giving the
other head a name. And what we're going to see in a
few seconds is that we actually don't want to double
count, we don't want to count the
situation, we don't want to double count this
situation-- heads one, heads two, tails, tails. And heads two, heads
one, tails, tails. For our purposes, these are
the exact same outcomes. So we don't want to
double count that. And we're going to have
to account for that. But if we just think
about it generally, how many different spots,
how many different flips can that first head show up in? Well, there's four
different flips that that first head
could show up in. So there's four possibilities,
four flips, or four places that it could show up in. Well, if that first head takes
up one of these four places, let's just say that first head
shows up on the third flip, then how many different places
can that second head show up in? Well, if that first head is
in one of the four places, then that second head can only
be in three different places. So that second
head can only be-- I'm picking a nice
color here-- can only be in three different places. And so, you know, it could
be in any one of these. It could maybe be
right over there. Any one of those three places. And so, when you think about
it in terms of the first, and I don't want to say
the first head, head one. Actually, let me
call it this way. Let me call it head A
and head B. That way you won't think that I'm
talking about the first flip or the second flip. So this is head A, and this
right over there is head B. So if you had a particular,
I mean, these heads are identical. These outcomes aren't
different, but the way we talk about it
right now, it looks like there's four places that
we could get this head in, and there's three places where
we could get this head in. And so if you were to multiply
all of the different ways that you could get, all
of the different scenarios where this is in four
different places, and then this is
in one of the three left over places, you get
12 different scenarios. But there would only be
12 different scenarios if you viewed this as
being different than this. And let me rewrite it with
our new-- So this is head A, this is head B,
this is head B, this is head A. There would only be
12 different scenarios if you viewed these two things as
fundamentally different. But we don't. We're actually double counting. Because we can
always swap these two heads and have the
exact same outcome. So what you want to do is
actually divide it by two. So you want to divide it by
all of the different ways that you can swap
two different things. If we had three heads here,
you would think about all of the different ways you could
swap three different things. If you had four heads here, it
would be all the different ways you could swap four
different things. So there's 12
different scenarios if you couldn't
swap them, but you want to divide it by all
of the different ways that you can swap two things. So 12 divided by
2 is equal to 6. Six different scenarios,
fundamentally different scenarios, considering
that you can swap them. If you assume that head A and
head B can be interchangeable. But it's a completely
identical outcome for us, because they're
really just heads. So there's six
different scenarios, and we know that there's a total
of 16 equally likely scenarios. So we could say that the
probability of getting exactly two heads is 6
times, six scenarios and-- Or there's
a couple of ways. You could say there
are six scenarios that give us two heads,
of a possible 16. Or you could say there are
six possible scenarios, and the probability of each
of those scenarios is 1/16. But either way, you'll
get the same answer.