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Hi.
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Welcome back.
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Today, we're going to do a fun
problem called the chess
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tournament problem.
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Now, it's a very long problem,
so I just want to jump
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straight in.
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Essentially, the problem
statement describes a very
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special chess tournament, which
involves players named
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Al, Bo, and Chi.
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Now Al is the current reigning
championship, and Bo and Chi
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are this year's contenders, and,
of course, they're vying
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with each other to beat out Al
and become the new champion.
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And so essentially,
the tournament is
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divided into two rounds--
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a first round, during which Bo
and Chi play against each
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other, and then a second
round, during which the
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surviving contender from the
first round plays against Al.
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And the problem statement also
gives you a bunch of
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information like what's the
probability that Bo beats Chi
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in a particular game,
et cetera.
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So without further ado, let's
get started on part a.
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In part a, the first thing we're
asked to compute is the
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probability that a second
round is required.
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Now, to save myself some
writing, I used the notation
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R2 to denote that event.
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So we are interested in
probability of R2.
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Now, I claim that this problem
is very sequential in nature
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so I would like to draw a tree
to describe what's happening.
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So in the first part of the
tournament, when Bo and Chi
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play their first game, exactly
one of two things can happen--
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either Bo can win
or Chi can win.
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And we're told by the problem
statement that Bo wins with
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the probability of 0.6 and,
therefore, Chi must win with
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the probability of 0.4, right?
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Because these two possibilities
must sum to 1,
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because either this must
happen or this happen.
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Now, let's imagine that the
first game has been played and
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that Bo won.
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Well, during the second game,
there's still two options for
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the outcome--
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Bo could win the second
game or Chi could
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win the second game.
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And because the problem
statement says that in every
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scenario Bo always wins
against Chi with the
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probability of 0.6, we can go
ahead and put a 0.6 along this
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branch as well.
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Similarly, 0.4 here.
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And similar logic, you've got
a tree that looks like this.
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And for those of you who haven't
seen trees before,
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it's just a structure that looks
something like this.
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And it helps us do better
accounting.
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It helps us keep straight in our
head what are the various
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outcomes, so that we
don't get confused.
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And so very quickly here, you
can see that there's four
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possible outcomes.
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So each node in this tree
corresponds to an outcome.
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And the leaves are those nodes
at the furthest stage.
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And it's convention to draw
the probability of a
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particular--
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so for instance, the probability
that Bo wins the
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first game-- it's just
convention to draw that
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probability over the
corresponding branch.
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And the reason why such diagrams
are so useful is
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because to compute the
probability of a particular
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outcome, if you've designed your
tree correctly, all you
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have to do is multiply the
probabilities along the
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branches that get into
that outcome.
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So let's see that in action.
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When is a second
round required?
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Well, a second round is
required here, right?
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Because in this case, Bo would
be the surviving challenger
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and he'd play the next
round against Al.
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It's also required here.
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But of course, it's not required
here or here, because
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no second round is played.
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And so these two outcomes
comprise the event R2.
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And now, to get the probability
of this outcome,
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you multiply along
the branches.
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So 0.6 times 0.6
give you 0.36.
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And 0.4 times 0.4
gives you 0.16.
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And we're almost done.
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We know that these two events
are disjoint, because if Bo
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won the first two games, then,
certainly, Chi could've won
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the first two games.
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And so you can just sum the
probabilities to get the
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probability of the reunion.
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So the probability of R2 is
equal to the probability that
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Bo won the first two games or
Chi won the first two games.
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And that's equal to 0.36
plus 0.16, which
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is equal to a 0.52.
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OK, now the second part
of part a asks for the
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probability that Bo wins
the first round.
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And this is first round.
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This is a very straightforward
one.
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So Bo wins the first round, that
correspondence only to
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this particular outcome.
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And we already know the
probability associated with
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that outcome is equal
to the 0.36.
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So we're done with that one.
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And now the last part is sort
of an interesting one.
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It asks for the probability
that Al retains his
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championship this year.
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So I'm going to just call
that A for short.
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A is the event that Al retains
his championship this year.
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And for that we're going to need
a larger tree, because Al
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has a lot of activity in the
second round, and so far our
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tree only describes what happens
in the first round.
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Now, to save time, I've actually
drawn the rest of the
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tree over there up
in the corner.
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So let's get rid of
this one and let's
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look at the full tree.
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So let's see when does Al
retain his championship?
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Well, Al certainly retains his
championship here, right?
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Because no second round
is required.
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Similarly, here.
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Al retains his championship
here, because the second round
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was required, but Al beat Bo.
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And similarly, here Bo didn't
win both games in the second
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round against Al, so Al wins.
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Here, Bo is the new
championship.
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So we don't want to
include about one.
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And sort of by symmetry,
we also get this
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one and this one.
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So by my argument before, we
know that the outcomes that
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comprise our event of interest
are this one, this one, this
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one, this one, this
one, and this one.
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So we could multiply the
probabilities along each
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branch and sum them, because
they're disjoint, to get the
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total probability.
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But we're not going
to do that because
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that's a lot of algebra.
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Instead, we're going
to look at the
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complement of the event.
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So we're going to notice,
there's only two branches on
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which Al does not retain his
current championship.
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So P of A is, of course, equal
to 1 minus P of A. And we're
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going to get P of
A by inspection.
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I'm sorry, so P of
A compliment.
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I'm just testing you, guys.
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So P of A compliment corresponds
to here and to
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here, because those are the
outcomes where Al didn't win.
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And so again, you multiply along
the branches to get the
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probabilities.
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So you get 0.6 squared times 0.5
squared plus 0.4 squared
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times 0.3 squared.
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And if you do all the algebra,
you should get around 0.8956.
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So we're cruising through
this problem.
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Let's go to part b.
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Part b is a little bit less
straightforward than part a,
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because it starts asking you for
conditional probabilities,
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as opposed to a priori
probabilities.
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So in the first part--
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and again, I'm going to continue
my notation with R2--
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we want the probability that
Bo is the surviving
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challenger--
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so I'm just going to use
B to denote that--
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given R2.
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Now, by definition, you should
remember from lecture that
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this is equal to probability
of B and R2 divided by the
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probability of R2.
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And of course, we've already
computed this value right up
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here with part a.
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We know it's 2.5.
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So we don't have to do
any more work there.
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We only have to look
at the numerator.
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So we need to go and figure out
what nodes in that tree
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correspond to the event
B intersect R2.
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So let's use a new color.
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Let's see, Bo is the surviving
challenger here only, right?
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And R2 is automatically
satisfied, right?
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Because a second round is
required there and there, not
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on those two.
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But here Chi is the surviving
challenger, not Bo, so we're
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really only interested
in that node.
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And you multiply along the
branches to get the
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probabilities.
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So we have 0.36 over
0.52, which is
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approximately equal to 0.6923.
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OK, now, the next part wants
the conditional probability
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that AL retains his
championship, conditioned,
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again, on R2.
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So we already have A being
the event Al retains his
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championship.
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So we want the probability
of A, given R2.
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And let's just apply the
direct definition of
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conditional probability again.
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You get P of A and R2 divided
by a probability of R2.
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Of course, we have the
probability of R2 already, so
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we just need to find the node
in the tree that corresponds
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to A and R2.
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So where is R2?
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R2 is going to correspond to
every node to the right that
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is not one of these two.
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So a second round is required
here, here, here,
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here, here, and here.
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Now, where does Al retain
his championship?
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So Al retains his championship
here.
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He retains his championship
here.
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He retains his championship here
and here, but no second
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round is required, so these
guys don't belong in the
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intersection.
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But this does, and this does.
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So we can again multiply the
probabilities along the
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branches and then some them.
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So let's see, we get--
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this marker's not working very
well, so I'm going to switch
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back to the pink--
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so you get 0.6 squared
times 0.5.
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That gets rid of this one.
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And then we want 0.6 squared
times 0.5 squared.
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That gets rid of that one.
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And then plus--
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let's see--
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0.4 squared times 0.7, which
takes care of this one.
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And then lastly, 0.4 squared
times 0.3 times 0.7.
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And that is a long expression.
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But it happens to
be about 0.7992.
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OK, so we are done with
part b and we can move
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along to part c.
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And I am, since we're running
out of room, I'm actually just
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going to erase this.
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And hopefully you guys
have had a chance to
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copy it down by now.
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If not, you can always pause
the video and go back.
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So let's see, part c asks us
given that the second round is
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required and that it comprised
of one game only.
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So let's denote I. So let's I
be the event that the second
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round was one game only.
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So essentially, in math
conditioned on R2 and I, what
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is the probability that it was
Bo who won the first round?
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So let's let B be the event that
Bo won the first round.
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OK, so again translating the
English to math, we just want
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the probability of B given R2
and I. Now, I am once again
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going to use the definition of
conditional probability.
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You might be concerned that we
haven't defined explicitly yet
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the definition of conditional
probability, when what lies
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behind the conditioning bar is
not a single event, but it's
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rather an intersection
of an event.
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And so my claim to you is that
it doesn't matter and that the
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same exact definition applies.
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But we'll go through
it slowly.
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So R2 is an event, I is an
event, and we know that the
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00:14:18,710 --> 00:14:21,880
intersection of two events
is itself an event.
248
00:14:21,880 --> 00:14:25,760
So I'm going to make up a new
letter, and I'm going to call
249
00:14:25,760 --> 00:14:33,030
this event W. So just using
the new notation, this is
250
00:14:33,030 --> 00:14:37,950
equal to probability
of B, given W.
251
00:14:37,950 --> 00:14:40,280
Now, this is the normal
definition that we know.
252
00:14:40,280 --> 00:14:44,460
We know that this is probability
of B intersect W
253
00:14:44,460 --> 00:14:51,120
over probability of W. And then
we just resubstitute what
254
00:14:51,120 --> 00:14:53,710
the definition of W was.
255
00:14:53,710 --> 00:14:57,680
And so if you do that over here,
you get probability of B
256
00:14:57,680 --> 00:15:06,480
and R2 and I divided by
probability of R2 and I.
257
00:15:06,480 --> 00:15:10,230
So hopefully, jumping from here
ahead to here, you see
258
00:15:10,230 --> 00:15:14,170
that the definitions act
exactly the same way.
259
00:15:14,170 --> 00:15:17,570
But these are two very short
intermediate steps that should
260
00:15:17,570 --> 00:15:21,120
help you convince yourself
that the same
261
00:15:21,120 --> 00:15:23,880
definition still works.
262
00:15:23,880 --> 00:15:27,550
So let's start with the
denominator, because the
263
00:15:27,550 --> 00:15:30,040
denominator looks a
little bit easier.
264
00:15:30,040 --> 00:15:34,680
Where is R2 and I in our tree?
265
00:15:34,680 --> 00:15:37,530
Well, let's see.
266
00:15:37,530 --> 00:15:39,970
Here, a second round was
required, but it
267
00:15:39,970 --> 00:15:41,830
comprised two games.
268
00:15:41,830 --> 00:15:43,540
Same with this one.
269
00:15:43,540 --> 00:15:46,830
Here, a second round was
required and it was comprised
270
00:15:46,830 --> 00:15:48,060
only of one game.
271
00:15:48,060 --> 00:15:49,132
So this is good.
272
00:15:49,132 --> 00:15:53,370
This is one of the outcomes
that we're looking for.
273
00:15:53,370 --> 00:15:55,410
Here, no second round
was required.
274
00:15:55,410 --> 00:15:56,590
So this doesn't qualify.
275
00:15:56,590 --> 00:15:58,020
Same with this one.
276
00:15:58,020 --> 00:16:01,530
Here, a second round was
required, and there was only
277
00:16:01,530 --> 00:16:03,430
one game, so that's good.
278
00:16:03,430 --> 00:16:05,810
And then these don't qualify
for the same reasons
279
00:16:05,810 --> 00:16:07,370
as we set up there.
280
00:16:07,370 --> 00:16:09,010
So we just have to multiply the
281
00:16:09,010 --> 00:16:12,800
probabilities along those branches.
282
00:16:12,800 --> 00:16:18,350
And we see that it's 0.4 squared
times 0.7 plus 0.6
283
00:16:18,350 --> 00:16:20,680
squared times 0.5.
284
00:16:20,680 --> 00:16:22,200
OK, we're almost done.
285
00:16:22,200 --> 00:16:26,930
We just need to look at the
intersection of R2 and I. So
286
00:16:26,930 --> 00:16:29,570
R2 and I are the ones we've
already circled.
287
00:16:29,570 --> 00:16:33,990
But now, we want to add one more
constraint, which is that
288
00:16:33,990 --> 00:16:36,930
Bo had to have won
the first round.
289
00:16:36,930 --> 00:16:41,100
And so we see here that Chi won
the first round, if we're
290
00:16:41,100 --> 00:16:42,015
looking at this outcome.
291
00:16:42,015 --> 00:16:44,050
And so he's no good.
292
00:16:44,050 --> 00:16:45,960
Let's use a different color.
293
00:16:45,960 --> 00:16:47,840
Let's see, maybe this one.
294
00:16:47,840 --> 00:16:51,540
But here Bo did win
the first round.
295
00:16:51,540 --> 00:16:57,160
So we're going to get 0.6
squared times 0.5.
296
00:16:57,160 --> 00:16:59,370
And I got that, of course,
just by multiplying the
297
00:16:59,370 --> 00:17:03,180
probabilities along the
right branches.
298
00:17:03,180 --> 00:17:11,069
And this, if you're curious,
comes out to be about 0.6164.
299
00:17:11,069 --> 00:17:15,819
OK, so I know that was a lengthy
problem, but you
300
00:17:15,819 --> 00:17:20,910
should feel really comfortable
now doing sort of basic
301
00:17:20,910 --> 00:17:22,819
probability manipulations.
302
00:17:22,819 --> 00:17:27,109
One thing that this problem
emphasized a lot was your
303
00:17:27,109 --> 00:17:30,150
ability to compute conditional
probabilities.
304
00:17:30,150 --> 00:17:33,190
So you saw me apply the
definition of conditional
305
00:17:33,190 --> 00:17:35,550
probability twice in part b.
306
00:17:35,550 --> 00:17:39,010
And then you saw me apply the
definition again in part c in
307
00:17:39,010 --> 00:17:42,600
a sort of slightly
modified way.
308
00:17:42,600 --> 00:17:44,840
So that's one thing that
you should have
309
00:17:44,840 --> 00:17:46,340
gotten out of this problem.
310
00:17:46,340 --> 00:17:50,050
And then another thing is that
hopefully, you noticed that by
311
00:17:50,050 --> 00:17:54,870
using a tree diagram, we made
the problem much easier.
312
00:17:54,870 --> 00:17:57,210
We almost didn't even have
to think about computing
313
00:17:57,210 --> 00:17:58,750
probabilities anymore.
314
00:17:58,750 --> 00:18:02,450
We reduced the problem to just
saying, OK, what are the
315
00:18:02,450 --> 00:18:06,260
outcomes that comprise our
event of interest?
316
00:18:06,260 --> 00:18:08,620
And then once you select
those, to compute their
317
00:18:08,620 --> 00:18:10,130
probability you multiply the
318
00:18:10,130 --> 00:18:13,400
probabilities along the branches.
319
00:18:13,400 --> 00:18:17,100
You have the right to just add
those together, because if you
320
00:18:17,100 --> 00:18:21,200
draw your tree correctly, all
of these guys should be
321
00:18:21,200 --> 00:18:23,200
disjoint from one another.
322
00:18:23,200 --> 00:18:25,970
So you have to be careful, of
course, to set up your tree
323
00:18:25,970 --> 00:18:26,800
appropriately.
324
00:18:26,800 --> 00:18:30,670
But once you do set up your tree
appropriately, your life
325
00:18:30,670 --> 00:18:32,270
is much simpler.
326
00:18:32,270 --> 00:18:33,520
So that's it for today.
327
00:18:33,520 --> 00:18:34,870
And we'll see you next time.
328
00:18:34,870 --> 00:18:36,234