[EM] Richard's frontrunners example

[Richard Moore]

MIKE OSSIPOFF wrote:

> Richard said:
>
> Granted. For a three-way race,
> probability that:      AB     AC     BC     Sum
> _________________________________________________
> both are front
> runners               0.5    0.3    0.2    1.0
> _________________________________________________
> a tie occurs
> involving             0.1    0.03   0.01   0.14
> both
> _________________________________________________
> if a tie occurs,
> both are front        0.71   0.21   0.07   1.00
> runners

I left out another row because it's not essential to the example,
but it may help the discussion that follows:

if both are
front runners,             0.2    0.1   0.05
a tie occurs

There is no fourth column because the sum of these numbers isn't
particularly meaningful. The significance of these numbers is that,
by multiplying them with the first row numbers, you get the second
row numbers. I'll talk more about these new numbers in a bit.

> I reply:
>
> Richard's calculation of his last row seems reasonable.
>
> Richard continues:
>
> Note that the numbers on the first row do not equal those in the
> last row.
>
> I reply:
>
> I noticed that you wrote the 1st row different from the last row.
>
> Right after I asked you for an example, I realized that this is what
> I should expect. You believe that the 1st row can be different from
> the last row, and so you wrote it different.

I wrote the first row first. I wrote the second row with numbers of
descending probability -- if "AC are front runners" is less likely
than "AB are front runners" then we can expect "AC are tied" to
be less likely than "AB are tied". Then I calculated the last row
from the second.

Note: You can put any positive numbers you like in the second row,
so long as each is less than the numbers in the first row of the same
column. That is because the first row contains single-event probabilities
and the second row contains joint probabilities. There is no magic
formula that gets you from row 1 to row 2. There is no constant of
proportionality that keeps the numbers in row 2 to the same ratios
as the numbers in row 1.

I know you would like to claim that the numbers in the "missing row"
should be the same all the way across. That will give the example
you would believe it should. But wishful thinking doesn't make it so.

Here's one example of why the second row can't be assumed to
be uniform. Note that "BC are front runners" is fairly unlikely at 20%.
Let's assume the probabilities are derived from a poll with a known
random sampling error, and no other error sources. So if the unlikely
outcome "BC are front runners" happens, then we are experiencing
an extreme excursion of the random sampling error. The sample was
skewed, and that means the probability of a tie changes because of
that skew.

Here's a more extreme example. Suppose B drops out of the race
the day before the election. Now "AC are front runners" is at 100%
probability. All the B voters must now vote for A or C. Let's say
that the B voters prefer A to C. A is already ahead of C, so the gap
widens. A tie becomes less likely. On the other hand, if the B voters
prefer C to A, the gap narrows. So a 100% probability of two candidates
being front runners cannot be used to determine the probability of a
tie without additional information. Likewise, you cannot determine
the second row from the first, without the additional information in
the "missing" row.

So please don't try to tell me that you can derive the joint probabilities
from the single-event probabilities without some knowledge of the
conditional probabilities. And please don't try to tell me the numbers
on the "missing" row must all be the same. Each column represents a
different set of circumstances and it is rare that you will find the same
conditional probability across the board.

> But when I showed you why those 2 probabilities can't be different,
> I asked you which part of my argument you disagree with. You didn't
> answer tha question.

You didn't show they can't be different. Instead you wrote this:

> 1. We agree that if i & j are the 2 top votegetters in the election,
> then, if there's a tie for 1st place, it will be between them.
>
> 2. We agree that if i & j are not the 2 biggest votegeters in the
> election, then, if there is a tie for 1st place, it will not be between
> them.
>
> 3. That means that either both of the following 2 statements are true,
> or neither of them are true:
>
> a) i & j are the 2 biggest votegeters in the election.
> b) If there's a tie for 1st place, it will be between i & j.
>
> 4. Since, after the election, those 2 statements are either both true
> or both false, then the probability, at a time before the election
> that a) will be true after the election must be the same as the probsbility,
> at that same time before the election, that b) will be
> true after the election.

You see, the problem with this argument is as follows:

b is a conditional statement of the form, if X then Y. Now b is true whenever
X and Y are both true. But b is also true if X is false and Y is true. What's
more, it's true if X and Y are both false. The only way statement b can
be false is if X is true and Y is false.

The consequence of this is that b can be true at times when a is not. b will
be true whenever a is true; it will also be true whenever X (there is a tie)
is false. If the probability that there is no tie is 99%, then b will be true
at least 99% of the time.

We can't determine the probability of b from the probability of a.

> You know, I have to say that you've disappointed me here, Richard.
> This is simple & obvious.

If it were so simple and obvious, a light would have turned on in your
head by this time. I've got to admit, step 3 of your fallacy is pretty
subtle.

> Say we agree that if A is true, then B is true. And that if A is
> false, then B is false. Then either A & B are both true, or they're
> both false. At some particular time, the probability that A will
> be true at a certain later time must be the same as the probability
> that B will also be true at that same later time. That's because
> either they're both true or they're both false.

If we agree that A implies B, and B implies A, then we have to agree
that p(A) = p(B). But you need to establish that A and B are equivalent.
If the equivalence is not in evidence to begin with, you can't use that
equivalence to prove the probabilities are equal.

> What kind of statements A & B are is irrlelevant. Say B is an if-then
> statement that says: "If x is true, then y is true". Your Bayes
> theorems are about the relationship between the probability that x
> is true, the probability that y is true, and the probasility tha B is
> true.
>
> But please note that neither Bart nor I said anything about the
> probability that i & j are in a 2-way tie for 1st place. And neither
> Bart nor I said anything about the probability that there will be
> a tie for 1st place. So this just isn't a Bayes problem.

How can you deduce whether or not Bayes applies from what
you or Bart never said? Let's look instead at what was said. Here's
what Bart said:

Slight correction/clarification:  The precise meaning of Pij is usually
taken to mean the probability, given a tie exists for first place, that
i and j will be involved in that tie.

This *is* a conditional probability. You've been talking about the
probability of a simple hypothesis: A and B are front runners. The
prior hypothesis of Bart's conditional is a different simple hypothesis:
there is a tie. Multiplying Bart's prior probability and his conditional
probability gives a joint probability. Now your hypothesis is related
to the same joint probability by a second conditional probability.
You've got two simple hypotheses, two conditional probabilities, and
one joint probability tying them all together. And yet you say Bayes
doesn't apply.

> It's obvious now, to everyone but you, that you're way off when you
> say that the probability that Bart named Pij can be  different than the
> probability that I named Pij. You can answer my question about which
> part of my argument you disagre with, or not, but I there's no justification
> for me to spend any more time trying to explain that
> to you.
>
> But if you decline to answer that question, that will mean that you
> can't find a part of that argument that you dsiagree with.

Well, I didn't decline, but that's a very funny statement just the same.

 -- Richard