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Lets take what we've learned about averages, moments, centroids, and put

Â them to use in an entirely different setting.

Â That of probability we're going to begin our introduction to probability with a

Â geometric treatment, that is both visual and visceral.

Â Our probability begins with counting. Consider something as simple as asking,

Â what happens when you roll a pair of dice?

Â What's the probability that the sum of the numbers you get adds to greater than

Â seven? Well, in this case, you can simply make a

Â table with all of the possible outcomes for the first and the second die.

Â Add up their face values and identify which of those are greater than 7.

Â When you do so, the probability is a ratio of the number of outcomes that are

Â greater than 7 to the total number of outcomes.

Â In this case, the numbers work out to 15 over 36.

Â That means you have to have a little bit less than a 50-50 chance of getting

Â something bigger than 7. Now, this is simple enough but do you see

Â an integral hiding in there somewhere. Stay tune.

Â Consider a simple situation of spinning a dial.

Â Some outcomes, you win, others you lose. You could compute the probability of a

Â win by counting sectors as before but there's another way to approach this.

Â One can think of the random variable as being an angle where some angles connote

Â a win. Spinning the dial corresponds to choosing

Â an angle at random between 0 and 2 pi. In this case, how would you compute the

Â probability? Well, one could think of it as a ratio

Â between the number of winning angles to the total number of angles.

Â But we're really not counting angles, are we?

Â We're really computing length in this simple example, one would get an answer

Â that is the same as counting segments. But in other examples, we would really

Â need to compute lengths. For example, what's the probability that

Â a randomly chosen angle on the circle has sine larger than one half.

Â Well, we would compute this probability as a ratio, as before.

Â The total number of angles has length to pi, the length of the circle.

Â If we look at those angles that have sine larger than one half, well, that length

Â is 2 pi over 3. Taking the ratio gives us a probability

Â of 1 3rd. We can do the same thing with area.

Â With what probability does a randomly chosen point in a square lie within the

Â inscribed circle. So, take the square, consider the

Â inscribed circle and then choose a point in the square at random.

Â Think of it as throwing a dart. What are the odds that you land inside

Â that circular region, well in this case it's going to be an area fraction.

Â We would take the area of the disc and divide by the area of the square that is

Â our probability. We can compute that easily enough if the

Â radius is r, than the disc has area pie r squared and the square has area 2r

Â quanity squared. That leads to pie over 4 or about 79%.

Â Those are your odds. In some context volume is the appropriate

Â tool. Consider the following, with what

Â probability does a randomly chosen point in a ball lie within the outer 1%.

Â That is within 1% of the boundary as measured by radius.

Â So, we have a solid ball radius r. It's the point that is chosen at random,

Â not the radius. Given a random point in that ball.

Â What are the odds that it's radial coordinate is within the outer 1%?

Â Well, we have to think in terms of volumes since this is a three dimensional

Â ball. We know the volume of a ball of radius r

Â is 4 3rds pie r cubed. So, to compute the probability in this

Â case, what kind of computation should we do?

Â Well, given our formula for volume, it's really easy to compute the volume of a

Â ball, if we pick a point at random in the ball, then the total volume of the

Â possibilities 4 3rds pi r cubed. Now, what are the odds that a randomly

Â chosen point lies within the inner 99%? Well, that too is a ball of radius .99 r.

Â And so, if you look at this fraction, this ratio of volumes this gives us the

Â odds of not being within the outer 1%. So, to compute the odds of being in the

Â outer 1%, we take one minus this ratio. What you'll notice is that there's some

Â convenient cancellation that goes on. The 4 3rd pi cancels.

Â And the r cubed cancel and we're left with 1 minus .99 cubed.

Â That works out to about .0297 et cetera, that means there's about a 3% probability

Â of being within that crust, that outer shell.

Â Notice, it's not 1% because it's the point that is random and not the radius.

Â Now, having done length, area, volume, of coarse, you know what's coming next.

Â That is, high dimensional volume or hypervolume.

Â In this case, let's repeat the problem with a ball of dimension n.

Â In this case, the volume of an n-dimensional ball, radius r, is some

Â constant v sub n times r to the n. You don't have to remember what that

Â constant v sub n is, because if we follow the exact same steps as before, computing

Â the probability of being within that outer 1%.

Â We get the ratio, the volume of that inner 99% ball, which is v sub n times

Â .99r to the nth, divided by the volume of the full n-dimensional ball v sub n r to

Â the n. Just as before the sub ends cancel, the r

Â to the n's cancel and we are left with 1 minus .99 to the n.

Â Now, what happens as n gets large, .99 to the n goes to 0.

Â And we're left with the somewhat surprising conclusion that when the

Â dimension is high enough, for example, if it is 459 or bigger then 99% of the

Â volume of this ball is in the crust. If you pick a point in the ball at

Â random, your odds of being within 1% of the boundary are greater than 99%.

Â That comes as a bit of a surprise. But not if you know how high dimensional

Â volume works. These examples illustrate the basis of

Â what we call fair probability or uniform probability.

Â For a uniform probability distribution over some domain D, whether it's a, a

Â ball or a circle or a square. Then, the probability of a random point

Â being in some subset A, within D is given by a fraction, by a volume fraction.

Â The probability of being within A is the ratio.

Â The volume of A to the volume of D. In this case, what we mean by volume is

Â dependent on the dimension. When we were spinning a dial, we looked

Â at length. When we were choosing a point at random

Â in the square, then we considered area fractions.

Â When we were choosing a point on a ball. Then, we were considering volume, either

Â three dimensional or higher, depending on the dimension of D.

Â And lastly, when we simply rolled dice, we were counting.

Â But that counting is in itself zero dimensional volume.

Â The volume associated to a discreet set. Now, that one example had some unique

Â features, in that there were two variables, that two die and they were

Â independent. And so we could compute probabilities in

Â terms of multiplication. For an example it involves both

Â independent variables and integrals. We're going to consider the classical

Â buffon needle problem. This problem goes as follows.

Â Consider a collection of parallel lines in the plane, separated by some distance,

Â l. Then drop a needle on the plane.

Â Let's say the needle also has length l. Then, what are the odds that you have a

Â crossing that the needle crosses one of the lines.

Â Well, one way to answer this question would be to drop a whole bunch of

Â needles, count how many of them crossed the line and divide by the total number

Â of needles that you dropped. That would be an approximation.

Â But we can do better that. If as stated, we let l denote the

Â distance between these parallel lines and the length of the needle, then what

Â variables do we use to characterize where a randomly tossed needle has fallen?

Â We'll let h denote the horizontal distance from the leftmost tip of the

Â needle to the rightmost, nearest line. But that doesn't completely characterize

Â the needle. We also need to know theta.

Â That is the angle that the needle makes with this line.

Â Now, what are the bounds on h and theta? H can vary between 0 and l.

Â Theta doesn't vary from 0 to 2 pi, but rather from 0 to pi by means of how we've

Â defined h in term of the left tip. Now, given this, how can we say, whether

Â or not, a needle crosses a line. Well, if we set up a right triangle based

Â on that needle, then the hypotenuse is of length l.

Â And we know the angle theta. Therefore, the horizontal width of this

Â triangle is equal to l times sine beta. If that quantity is greater than or equal

Â to h, then we have a crossing of the line.

Â So, if in this theta h plane we graph l times sine theta then, dropping a

Â collection of needles is the same thing as sampling this rectangle at random.

Â And any random point that obeys this inequality, that falls under the curve

Â can note a crossing. And now we can do some calculus because

Â these variables are independent. Changing h doesn't change theta, and

Â changing theta doesn't change h. Then we can compute the probability of a

Â crossing as an area fraction. In this case, taking the ratio of the

Â area under the curve to the total area of the rectangle.

Â What's the area under the curve? Why that's simply the integral.

Â This data goes from 0 to pi of l sin data D theta.

Â What's the area of the entire domain? It's the area of the rectangle, pi times

Â l. Now, I think you can do this integral in

Â your head noticing that the l's cancel, one obtains a probability of 2 over pi.

Â That's very interesting. But maybe more interesting than you might

Â suspect because of the following fact, this probability has pi within it.

Â And we have an interpretation in terms of dropping needles on say a sheet of paper.

Â This principle means that you should be able to approximate the value of pi

Â simply by dropping needles on a field of lines, counting how many times the needle

Â crosses the line. And then dividing by the total number of

Â needles that you dropped. That ratio should come closer and closer

Â to 2 over pi by taking the reciprocal, multiplied by 2.

Â You can, in principle, approximate pi. Maybe you should try it.

Â If you do, one of the things that you'll find is that probability is a somewhat

Â mysterious subject. You'll also learn a thing or two about

Â convergence. Our treatment of probability won't end

Â with volumes and needles and Pi. In fact, we've hardly begun.

Â In our next lesson, we're going to consider what happens when the world

Â isn't fair, or at least when the underlying probability distribution isn't

Â fair. To do that, we'll need to wed our

Â volumetric approach with a more mask based approach from our previous lessons.

Â