Hi.
Today we're going to do another fun problem that
involves rolling two dice.
So if you guys happen to frequent casinos, this problem
might be really useful for you.
I'm just kidding.
But in all seriousness, this problem is a good problem,
because it's going to remind us how and when to use the
discrete uniform law.
Don't worry, I'll review what that says.
And it's also going to exercise your understanding of
conditional probability.
So quick recap.
The discrete uniform law says that when your sample space is
discrete, and when the outcomes in your sample space
are equally likely, then to compute the probability of any
event A, you can simply count the number of outcomes in A
and divide it by the total number of possible outcomes.
OK, so coming back to our problem.
The problem statement tells us that we roll two fair
six-sided die.
And it also tells us that each one of the 36 possible
outcomes is assumed to be equally likely.
So you know alarm bell should be going off in your head.
Our sample space is clearly discrete.
And it says explicitly that all
outcomes are equally likely.
So clearly, we can use the discrete uniform law.
And again, this is helpful because it reduces a problem
of computing probabilities to a problem of counting.
OK, and before we go any further, I just want to review
what this graph is plotting.
You've seen it a few times, but just to clarify, on one
axis, we're plotting the outcome of the first die roll,
and on the second axis, we're plotting the outcome of the
second die roll.
So if you got a 4 on your first die, and you get a 1 on
your second die, that corresponds to this point over
4 and up 1.
OK, so part a asks us to find the probability
that doubles our rolls.
So let's use some shorthand.
We're going to let D be the event that doubles are rolled.
And we want to compute the probability of D.
I argue before we can use the discrete uniform law.
So if we apply that, we just get the number of outcomes
that comprise the event "doubles rolled" divided by
36, because there are 36 possible outcomes, which you
can see just by counting the dots in this graph.
Six possible outcomes for the first die, six possible
outcomes for the second die.
That's how you--
6 times 6 is 36.
So I've been assuming this entire time that you know what
doubles are.
For those of you who don't know, doubles is essentially
when that number on the first die matches the number on the
second die.
So this outcome here 1-1 is part of the event "doubles
rolled." Similarly, 2-2, 3-3, 4-4, 5-5, and 6-6--
these six points comprise the event "doubles rolled."
So we can go ahead and put 6 over 36,
which is equal to 1/6.
So we're done with part a.
We haven't seen any conditioning yet.
The conditioning comes in part b.
So in part b we're still interested in the event D, in
the event that "doubles are rolled." But now we want to
compute this probability conditioned on the event that
the sum of the results is less than or equal to 4.
So I'm going to use this shorthand sum less than or
equal to 4 to denote the event that the role results in the
sum of 4 or smaller.
So there's two ways we're going to go about
solving part b.
Let's just jump right into the first way.
The first way is applying the definition of conditional
probability.
So hopefully you remember that this is just probability of D
intersect sum less than or equal to 4, divided by
probability of sum less than or equal to 4.
Now, sum less than or equal to 4 and D intersect sum less
than or equal to 4 are just two events.
And so we can apply the discrete uniform law to
calculate both the numerator and the denominator.
So let's start with the denominator first because it
seems a little bit easier.
So sum less than or equal to 4, let's figure this out.
Well, 1-1 gives us a sum of 2, that's less
than or equal to 4.
2-1 gives us 3.
3-1 gives us 4.
4-1 gives us 5, so we don't want to include this or this,
or this point.
And you can sort of convince yourself that the next point
we want to include is this one.
That corresponds to 2-2, which is 4, so it makes sense that
these guys should form the boundary, because all dots
sort of up and to the right will have a bigger sum.
3-1 gives us 4.
And 1-2 gives us 3.
So these six points--
1, 2, 3, 4, 5, 6--
are the outcomes that comprise the event sum less than or
equal to 4.
So we can go ahead and write in the denominator, 6 over 36,
because we just counted the outcomes in sum less than or
equal to, 4 and divided it by the number
of outcomes in omega.
Now, let's compute the numerator.
D intersect sum less than or equal to 4.
So we already found the blue check marks.
Those correspond to sum less than or equal to 4.
Out of the points that have blue check marks, which one
correspond to doubles?
Well, they're actually already circled.
It's just these two points.
So we don't even need to circle those, so we get 2 over
36, using the discrete uniform law.
And you see that these two 36s cancel each other.
So you just get 2/6 or 1/3.
So that is one way of solving part b, but I want to take
you, guys, through a different way, which I think is
important, and that make sure you really understand what
conditioning means.
So another way that you can solve part b is to say, OK, we
are now in the universe, we are in the conditional
universe, where we know the sum of our
results is 4 or smaller.
And so that means our new sample space is really just
this set of six points.
And one thing that it's worth noting is that conditioning
never changes the relative frequencies or relative
likelihoods of the different outcomes.
So because all outcomes were equally likely in our original
sample space omega, in the conditional worlds the
outcomes are also equally likely.
So using that argument, we could say that in our sort of
blue conditional universe all of the
outcomes are equally likely.
And therefore, we can apply a conditional version of the
discrete uniform law.
So namely, to compute the probability of some event in
that conditional world.
So the conditional probability that "doubles are rolled", we
need only count the number of outcomes in that event and
divide it by the total number of outcomes.
So in the conditional world, there's only two outcomes that
comprise the event "doubles rolled." These are the only
two circles in the blue region, right?
So applying the conditional version number
law, we have two.
And then we need to divide by the size of omega.
So our conditional universe, we've already said, has six
possible dots.
So we just divide by 6, and you see that we get the same
answer of 1/3.
And so again, we used two different strategies.
I happen to prefer the second one, because it's slightly
faster and it makes you think about what does conditioning
really mean.
Conditioning means you're now restricting your attention to
a conditional universe.
And given that you're in this conditional universe where the
sum was less than or equal to 4, what is then the
probability that doubles also happened?
OK, hopefully you, guys, are following.
Let's move on to part c.
So part c asks for the probability that at least one
die roll is a 6.
So I'm going to use the letter S to denote this, the
probability that at least one die roll is a 6.
So let's go back to our picture and
we'll use a green marker.
So hopefully you agree that anything in this column
corresponds to at least one 6.
So this point, this point, this point, this point, this
point, and this point your first die landed on a 6, so at
least one 6 is satisfied.
Similarly, if your second die has a 6, then we're also OK.
So I claim we want to look at these 11 points.
Let me just check that, yeah, 6 plus 5--
11.
So using the discrete uniform law again, we get
11 divided by 36.
OK, last problem, we're almost done.
So again, we're interested in the event S again, so the
event that at least one die roll is a 6.
But now we want to compute the probability of that event in
the conditional world where the two dice land
on different numbers.
So I'm going to call this probability of S. Let's see,
I'm running out of letters.
Let's for lack of a better letter, my name is Katie, so
we'll just use a K. We want to compute the probability of S
given K. And instead of using the definition of conditional
probability, like we did back in part b, we're going to use
the faster route.
So essentially, we're going to find the number of outcomes in
the conditional world.
And then we're also going to compute the number of outcomes
that comprise S in the conditional world.
So let's take a look at this.
We are conditioning on the event that the two dice land
on different numbers.
So hopefully you agree with me that every single dot that is
not on the diagonal, so every single dot that doesn't
correspond to doubles, is a dot that we care about.
So our conditional universe of that the two dice land on
"different numbers", that corresponds to these dots.
And it corresponds to these dots.
I don't want to get this one.
OK, that's good.
So let's see, how many outcomes do we have in our
conditional world?
And I'm sorry I don't know why I didn't include this.
This is absolutely included.
I'm just testing to see if you,
guys, are paying attention.
So we counted before that there are six dots on the
diagonal, and we know that there are 36 dots total.
So the number of dots, or outcomes to use the proper
word, in our conditional world is 36 minus 6, or 30.
So we get a 30 on the denominator.
And now we're sort of using a conditional version of our
discrete uniform law, again.
And the reason why we can do this is, as I argued before,
that conditioning doesn't change the relative frequency
of the outcomes.
So in this conditional world, all of the outcomes are still
equally likely, hence we can apply this law again.
So now we need to count the number of outcomes that are in
the orange conditional world, but that also satisfy at least
one die roll is a 6.
So you can see--
1-- we just need to count the green circles that are also in
the orange.
So that's 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.
So we get a 10, so our answer is 10 over 30, or 1/3.
So now we're done with this problem.
As you see, hopefully, it wasn't too painful.
And what are the important takeaways
here for this problem?
Well, one is that whenever you have a discrete sample space,
in which all of outcomes are equally likely, you should
think about using the discrete uniform law, because this law
lets you reduce the problem from computing probabilities
to just counting outcomes within events.
And the second takeaway is the way we thought about
conditioning.
So we talked about one thing, which is that in your
conditional world, when you condition, the relative
likelihoods of the various outcomes don't change.
So in our original universe, all of the outcomes were
equally likely.
So in our conditional universe, all of the outcomes
are equally likely.
And we saw it was much faster to apply a conditional version
of the discrete uniform law.
So that's it for today.
And we'll do more problems next time.