[EM] Richard's frontrunners example

[MIKE OSSIPOFF]

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

In the first row after the header, the probabilites for each pair of
candidates of being the two front-runners are given; they must add up
to one (I'm ignoring the remote possiblity of a three-way tie).

The next row gives the probability of a tie involving each pair. The
sum of these probabilities is the probability of any tie occurring.

The last row is the probability, given a tie, that it is between the
specified pair. These numbers are derived by dividing the probability
of a tie between the two candidates (from the second row) by
the probability of any tie (0.14). The sum in the third row is one
(ignoring the rounding error after the division).

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 shouldn't have asked for the example, because it proves nothing.
You didn't show how the probability that A & B are the 2 biggest
votegetters can be different from the probability that if there's
a tie for 1st place it will be between A & B. You merely wrote them
different. Well, I got the example that I asked for, but it doesn't
mean anything. I asked for an example where the probabilities are
different, and so you said "Ok, they're different".

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.

So I'll ask one more time: Which part of the following argument do
you disagree with?:

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.

I've demonstrated that those 2 probabilities must be the same.
I repeat: If you disagree with that conclusion, then which part of
the above argument do you disagree with?

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

Instead of those specific statements a) & b), let's talk about two
unspecified statements A & B. Maybe one of them is an if-then
statement, maybe both are, or maybe neither is. It makes no difference.

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.

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.

Instead of just expounding about Bayes' theorem, I suggest that
you look it up.

Do you disagree with any part of my argument above, or don't you?
If so, which part?

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.

Mike Ossipoff




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