Now at the heart of the test that we are proposing
is a consideration of a string of coin tosses.
And asking questions about the likelihood of a run of successes,
a subsequence of successes somewhere in that sequence.
To begin to make progress, we're going to have to understand how to describe
success runs in a neat and compact mathematical formulas.
So, let's begin, by taking a look at a made-up sequence of coin tosses.
1 represents a head or a success.
0 represents a tail or a failure.
And what I've shown you on the screen is a sequence of 30 trials.
Of course, we are assuming that every trial is an independent trial.
And so we are dealing with repeated independent trials, and
underlying all of this is a product space and a product measure.
Now in the long string, those 1s and 0s are alternating around,
they're bouncing around.
But in the middle and at the end,
we see strings of repeated ones, repeated successes.
And so clearly, here's an example of a sequence of 30 tosses,
where there are sequences of five or more successes at one or more places.
To keep at accurate accounting of when success runs occur,
we will need to do two things.
First, figure out what length of success run is of immediate importance.
In this case,
let us say we're going to look at success runs of length five, five and more.
The second, and this is important, is as a convention
we imagine that we start the sequence of tosses at, on the left,
work our way towards our right, looking for five successes in a row.
The first time it happens we say here is a first success run.
Here, here visibly, now marked in red, is an example of what I mean.
The first success run of length five occurs at trial 9.
Now, and this is critical.
We reset the the clock, if you like, at trial 9.
We say we have a renewal.
And it's as if we begin the coin tossing anew with trial 10.
And then we proceed all the way up to trial 16 where
we discover another string of files.
And say, okay, hold it.
We have a new renewal.
Of course, on trial 16 was the second success run.
Now notice there was a success immediately after trial 16, but
that is part of the renewal of the next stage of the new sequence of tosses
that has started after the success run.
We now begin anew at trial 17.
Proceed all the way until trial 27 where we have the next success run.
And we proceed in this fashion.
So, at the heart of this idea is that we start from the left,
with the first toss, proceed to the first occurrence of the success run
of length five that you're interested in, consider a renewal to have taken place,
a new sequence of coin tosses to have been initiated from that point on,
what will the next success run, a new renewal, the next success run, and so on.
Of course, when you've tied the success run of length five,
it is possible that it is part of a larger success run.
But for our purposes right now, granting length six or
seven are not germane, we are worrying about runs of length five.
And of course eventually we'll want to take a step back and
tackle runs of any length.
But for the time being, let us focus on runs of length five.
Here's one more example of a sequence of 30 tosses and
in red I have outlined where the success runs have terminated, have occurred.
Starting from the left again on the 10th trial we have the first success run of
length five.
The clock starts again immediately with trial 11.
But here, we actually have a very long string of consecutive successes.
So, from 11 to 15 gives you an immediate new success run.
The clock starts again.
We have a renewal, from trial 16, a new run starts.
And we find that we have consecutive successes, culminating in trial 20,
with yet another success run.
Trial 21 happens to be a success, but we don't quite complete the success run, so
we move on now all the way up to trial 29,
when the next success run comes into place.
So this is at the heart of a convention.
We go from left to right and each time we have a success run, we have a renewal.
It's as if the coin tosses start anew.
Now we understand.
We come to a right understanding of what a success run is in our setting.
Now to make probabilistic progress on this,
we have to be able to identify what the probability space in question is.
And so let's go back to first principles.
So here's a sample run of tosses with various success runs identified,
the first success run at trial 9, the next success run at trial 16,
the next success run at trial 27 and so on.
Our very first question, naturally, what is the sample space for this problem?
What is the space of idealized outcomes that represents this chance experiment?
Now, there are a couple of ways we could go about this.
But, the most general and extensible fashion is
to think of a Gedankenexperiment where a coin is
tossed repeatedly indefinitely often.
The outcomes of the tosses, call them x1, x2, x3,
x4, x5, and so forth, represents successes and failures, 0's and 1's.
And this entire sequence of 0's and
1's, will represent a sample point or an idealized outcome for this experiment.
Of course, as we've seen in tableau six, part two, that is a dangerous
Penn tableau and if the listener has, has had the fortitude to walk through it,
she would have seen that this leads to actually a very rich and complex space.
It actually leads to the continuum.
Now the subtleties of the continuum are not going to be germane for our purposes
but this is just to understand that even simple settings could read,
lead to very complex problems.
But in principle we now have an understanding of what
an outcome in this experiment is.
Practical simplicity is going to be a turn for
us because we are going to essentially terminate the experiment after n trials
and ask questions about what happened in n trials.
In which case we'll deal with repeated independent trials
over the financed domain.
But we're jumping ahead of ourselves.
Our next quest naturally should be in this space of sample points,
what are the evens of interest?
So let us say that the basketball player in question has taken in shorts and
we're interested in whether in that sequence of n active to the basket,
there is a string of five or more successes, consecutive.
So naturally, the event of interest to
us is event that there is at least one success run
of length five at or before the end of the trial.
Let us call this event S sub n.
Of course, in our sample space,
this corresponds to all sequences of trials where in the first n trials,
you discover consecutive subsequences of five or more ones.
Now, connected to this natural event S sub n is the event
Q sub n, which is the complement of S sub n, which is the event that,
in the first n trials, there is no success run of length five, homo.
Now we should pause.
This event Sn is what is germane for us.
And we would like to say something eventually about its probability.
But that looks like a very complicated endeavor.
There are very, very many ways in which one could put in
a consecutive sequence of file ones somewhere in a long sequence of n tosses.
And tabulating and
capturing all of them systematically looks to be a very difficult task.
Naturally, we will try to break this task up into simple pieces,
partition it so that the individual pieces become transparent and
we can glue things together.
In other words, we're going to try to use additivity
by finding proper partitions of the space.
And little thought will tell us that we're worrying about sequences
of five heads, five successes somewhere in the sequence,
a natural event of interest then is a question of, well,
what is the first run of five successes?
Does that all come?
And so accordingly, let's promptly define a new event, anticipating that this
just might be useful in stitching together a bigger picture.
For each value n, 1, 2, 3, 4, 5 and so
forth let F sub n represent the event that the first
success run of length five occurs at trial n.
Now there are other possibilities.
If you're at the 8th trial and a success run occurs at it,
it could be the first success one.
But it could be a 2nd, 3rd or subsequent success run.
Let's promptly introduce a little more notation to capture this richer
possibility.
Let U sub n represent the event that there is some success run,
the first or subsequent, which occurs at trial n.
All right, now we have three events at hand.
Sn, its complement Fn, and Un.
n of course is an index, and be allowed to,
to run freely ov, over the positive integers.
We'd like to say something about the probabilities of these things, n.
And of course, naturally, the question is,
what is the underlying probability measure that's going to fuel these probabilities?
And as I've mentioned, the underlying sample space is very rich.
It actually expands the continuum of possibilities.
But for our purposes we can afford to look at finite strings of tosses at a time.
And what comes to our rescue immediately is the fact
that the probability measure is implicitly specified
by the fact that we're dealing with a setting of repeated independent trials.
For any given n, we're effectively dealing with a product space and
a product measure.
The computations, the probabilities are going to become almost transparent
in consequence of this key observation.
And immediately, with this at the back of our mind,
let's introduce a little notation for the probabilities of the events in question
just to keep our notation a little more compact to prevent it from burgeoning.
So let's use lowercase letters for the probabilities of the associated events,
lowercase s then for the probability of the event big S of n,
lowercase q of n for the probability of the event big Q sub n,
little f sub n and little u sub n respectively for
the probabilities of the events big F sub n and big U sub n respectively.
So before we go forward into a full-fledged analysis of these
probabilities, we should begin by exploring the ground, seeing the layout,
understanding what we can say about these probabilities in simple settings.
What are the boundary conditions that are known to us?
So accordingly, let's start by looking at the first few trials, trials 1 through 5.
Our immediate observation is the following.
We are interested in sections one of length five.
Therefore it is manifestly impossible that any success
run of length five concludes on or before the 4th trial,
and therefore the probabilities f1, u1,
s1, f2, u2, s2, f3, u3, s3, and
f4, u4, s4 are all identically 0.
What about the possibility of a success run at trial 5?
Well, of course, this is possible.
And there's only one way in which this can be accomplished, and
that is, if one starts with a sequence of five successes, five ones in a row.
And the moment this happens, we'll have triggered a first success run,
in fact the only possible success run at trial 5.
And the probability of this, because we're dealing with repeated independent trials,
is naturally, one half times one-half times one-half times one-half
times one-half, or 2 to the power minus 5, nice and simple.
What about the next trial?
What can we say about f6, u6, and s6?
Pause the lecture while you take this in,
and think about what the problem
setting gives you in this context.
Resume the lecture when you're ready.
You're back.
Let's take a look at a picture.
Let us consider the first six trials.
If there is a success run occurring on trial 5,
then in the first trial of the table, you will see, we must have 1's in a row.
The probability of that is 2 to the power minus 5.
But what if a success run terminates on the 6th trial.
Now if that happens, it is manifestly the case that
trials to prove six must give you five consecutive 1's.
But it also constrains what must happen on trial 1.
The 1st trial can not be 1 itself because if that were
the case then a success run would already have concluded at trial 5.
By a convention, remember we are counting from left to right and
starting a new, the renewal every time we get a success run.
So if there is to be a success run at trial 6,
we must start the sequence off with a failure on the first trial.
But this now gives us a setting where in the first 6 trials,
I start with a failure, and then I have five consecutive successes, successes and
failures are equally likely.
The coin is fair.
They each have a probability of one-half, and therefore, this particular run of six,
failure with five successes has probability one-half times one-half
times one-half times one-half times one-half times one-half,
or f6 and u6 are both 2 to the power minus 6.
What can we say about s6?
The probability that there is a success run of length five at or
before their 6th trial.
And a little introspection tells us that this can only happen if
either there is a first success run at trial 6,
or there is a first success run at trial 5,
which means that s6 is simply f5 plus f6.
Because these two events are disjoint, or they are mutually exclusive.
You can have a success run at, at trial 5, but
then you will not have a success run at trial 6, or vice versa.
And putting together the algebraic values we find that we get 3
times 2 to the power minus 6.
What do we appeal to?
A universal, subtle and
marvelous principle of additivity, the whole here was a sum of the two parts.