Philosophy

# Lesson Two Redux: More Mysticism

Is it written into sport announcers’ contract that they shall speak in nothing but cliché?

Since there is always great confusion about why premises about “fairness” or “randomness” are not needed, we had better cover it in a main post.

Premises:

(P1) I have a six-sided object, just one side of which is labeled ‘6’.
(P2) Upon tossing, only one side will show

————————————————————————————————————–
(C) A ‘6’ will show.

The conclusion is not certain with respect to the premises. But we can say

Pr( C | P1 & P2 ) = 1/6.

Many would like to add this premise:

(P3) The die is “fair.”

But that is equivalent to

(P3) Each side of the die is equally likely to show,

which is the same as saying

(P3′) The probability of a ‘6’ is 1/6.

I can write

Pr( A ‘6’ will show | P1 & P2 & The probability of a ‘6’ is 1/6) = 1/6.

and since I do not now need (P1) or (P2), I can write

Pr( A ‘6’ will show | The probability of a ‘6’ is 1/6) = 1/6.

Or in plain English, “The probability a ‘6’ will show given the probability a ‘6’ will show is 1/6 is 1/6.”

This is a fine argument. It is valid. That means the conclusion has been deduced with certainty from the premises. Arguments which have conclusions that are deduced are the strongest there is, and isn’t that glorious.

But the argument is circular: its premises contain the conclusion, the thing we wanted to prove. Aristotle showed us that this argument is the same as saying, “A, therefore A,” where A is any statement.

So give up on (P3), because it is not needed. Instead, think carefully about (P1) and (P2); think hard about what they do not tell us. What words are there in (P1) that tell us that the sides are symmetric, that they are weighted equally, that they are of the same substance, that they have the same friction, or the same of any other probative factor?

Not one word. Too, there is not a shred of evidence that any of these things—and of an infinite number of other things—are such that we should consider them.

There isn’t any word in (P2) about the gravitational field in which the six-sided object—we do not know it is square, nor that it is a die—will be tossed, nor about the amount of spin or other force imparted to the object, nor about viscosity of the air in which it will be tossed, nor do we know that there will even be air! We do not know about the elasticity of the surface upon which the object will land, nor about that surface’s roughness. Nor do we know about an infinite number of other things, each of which could influence the object showing a ‘6’.

If there are 12 Johnstons and 18 Freihaufs in a room and you will select the one closest to the door, then, given this evidence and no other, there is a 60% chance that you will select a Freihauf. Just as above, we do not need to add any evidence that the people in the room are, perhaps by polkaing, distributed “randomly.”

Given only the evidence we have—and no other tacked on imaginatively —we have no idea where anybody in the room is. It is the lack of information that makes the outcome “random”, which is to say, unknown.

The problem we have with these types of arguments is that we cannot help but add premises: the stated ones are never enough; we are always greedy for more. Adding (unstated) premises is like a mathematician changing axioms mid-proof to suit him so that he gets the result he desires. This behavior might be because the above arguments sound like situations in which we have a lot of experience. But here is another example, adapted from Stove, which shows that we should consider arguments as given.

(P1) Just half of all winged horses are yellow.
(P2) Bob is a winged horses.
————————————————————————————————————–
(C) Bob is yellow.

Here, everybody always agrees that the probability that “Bob is yellow” given just (P1) and (P2) is 50%. We don’t feel the need to talk about “opaque bags” of winged horses, with half of an infinite amount of them yellow, and with the other half some other color or colors. We don’t have the gut feeling that if we were to know that that probability is 50%, that we need to repeat a “random” experiment with winged horses an infinite number of times.

This, obviously, is because there are no winged horses that we know of, yellow or not. Thus, we are able to tackle the argument in its intended sense.

“OK, Briggs. Maybe. But what about real dice?” Well, as to that, stick around.

Categories: Philosophy, Statistics

### 15 replies »

1. Rich says:

This is about information. Doesn’t your introduction of natural language vocabulary necessarily introduce more information than you’re allowing in the conclusion? You use the word “die” (in P3). We can look it up in a dictionary and find: “die: a small cube with 1 to 6 spots on the six faces; used in gambling to generate random numbers”. At which point we cry, “Aha! Our ignorance of the symmetry of this cube precludes even probability conclusions beyond “between 0 and 1″!”

Even Bob the winged horse is a horse, about which we fancy we know something especially that he doesn’t exist.

2. JH says:

My daughter asked me how to measure the hardness of thinking. She said she liked to think soft. ^_^ Think hard they donâ€™t tell us = Imagine what they donâ€™t tell us?!

â€œAll men are mortal. Therefore, Socrates is mortal.â€ We can suppress the premise of â€œSocrates is a manâ€ since itâ€™s a known fact, in a way, itâ€™s redundant. However, are we certain that a die is fair? I am not. You may assume/guess/estimate so, but the additional premise of â€œfair dieâ€ is necessary for me to conclude that the probability of rolling a six is 1/6.

â€œIf x+1=3, then x=2.â€ To many of us, â€œx+1=3â€ is equivalent to â€x=2â€. Doesnâ€™t the premise contain the conclusion here? So I still donâ€™t get it.

The conclusion is not certain with respect to the premises. But we can say Pr( C | P1 & P2 ) = 1/6.

But you are certain that Pr( C | P1 & P2 ) = 1/6. Yes, you have a six-sided object, and it can be hexahedron. So, again, with P1 and P2, I am not certain that Pr( C | P1 & P2 ) = 1/6.

3. SteveBrooklineMA says:

In the dice example, it seems to me that P3 and P3′ are not the same. P3 tells us that the probability of a ‘5’ is 1/6, but P3′ does not.

4. Ari says:

I did a project on probability in college where I studied the probabilities of rolls with 20-sided dice. I learned two things.

1. There are a lot of discussions of winged horses, scaly dragons, and enraged barbarians when it comes to 20-sided dice.

2. The mere mention of “20-sided dice” is enough to label you unfit to breed almost instantaneously.

5. costanza says:

Speaking of sports cliches…one of my personal favorites happens in PGA tournaments: after a shot, one of the announcers will utter “That was a great golf shot.” What the hell kind of shot would it have been?

6. Swade016 (Wade Michaels) says:

I’m with Steve, P3 implies P3′, it is not logically equivalent.

As for the 1/6th issue, I tried proving you wrong by drawing up a diamond looking thing with one side much much larger than the others, but since nowhere did you mention in P1 that “6” had to be labelled on a specific side. So no matter how high the probability that any one side has of showing (or how disproportionate) the probabilities end up washing out back to 1/6th.

7. DEBEE says:

WB your blog audience is way too imaginative. Llike all the winged yellow horses, I know, they refuse wear blinders.

8. Bernie says:

Under what circumstances would P3 not be the same as P3′ ? Also, Matt indicated that one side was labelled with a “6” – there is no information on the labels on the other sides – if they even have labels.

Speaking of “fair” – is the new ball being used at the World Cup “fair”? Given the masterful display by the Germans in their first game, has Addidas given the Germans an advantage?

9. TomVonk says:

William

I think this has been already discussed here .
So I submit that in :

(P1) I have a six-sided object, just one side of which is labeled â€˜6â€².
(P2) Upon tossing, only one side will show

â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€“
(C) A â€˜6â€² will show.

The conclusion is not certain with respect to the premises. But we can say

Pr( C | P1 & P2 ) = 1/6.

I can prove that the probability that the conclusion Pr( C | P1 & P2 ) = 1/6 is TRUE is exactly 0 .
Of course you can say it . But if the probability for the conclusion being TRUE is 0 , then it is FALSE .

1) You have an OBJECT with SIDES . So clearly you refer to an object which can be described geometrically . The sides are surfaces with an equation in an adequate coordinate system . The object is what is inside the volume delimited by the surfaces . This volume defines among others the density of matter distribution . The use of the word object suggests that this density is non 0 .

2) Upon TOSSING only ONE side will SHOW . From this premise several highly non trivial conclusions can be drawn . As the object can be tossed , it confirms that the matter density distribution is indeed not 0 . As only one side shows , ALL the surfaces must meet at exactly 90Â°or less . For any bigger angle several sides would show .
I make here the reasonable assumption that by “show” is meant “can be seen” .
Let’s restrict the surfaces to planes without loosing generality .
The object you defined with your premises is a cuboid , a cubic die is just a particular case of a cuboid .

OK now we will take the simplest assumption – the matter distribution density is a non zero constant .
To go farther and finish the proof , we must define “toss” .
We’ll say that “tossing” is “providing momentum in a constant gravitationnal field” .
It could be more complicated but we take the simplest case again . The gravitational field must be non zero because otherwise the object would never show only one face .
Then it can be shown easily that the probability for any of the surfaces to show is proportional to the fraction of its surface to the total surface . As we deal with a cuboid , 4 faces have the same surface S1 and the remaining 2 have the same surface S2 .
It is a 2 dimensional problem and the probabilities for a given object are represented by a point in a plane (S1,S2) . There is one degenerated case (first diagonal) where all surfaces are equal and then both probabilities are equal to 1/6 (this is the classical “fair die”) .
Now as the probability to pick randomly a point with coordinates (x,x) in a plane is 0 , follows that the probability for the statement “Pr( C | P1 & P2 ) = 1/6” to be true is zero.
QED
Beyond an amusing logical exercice that I hope you’ll appreciate William , I think there is a deeper insight .
And it is that it is impossible to disconnect any probabilistic statement from the physical reality .
For me probability is just a specific mathematical language that can be used (in some cases)to describe reality .
But if one speaks about baseball what matters is knowing about baseball and not whether one speaks German or English .

10. Briggs says:

Tom,

You are too good a physicist; you are misleading yourself. Think instead about Johnstons and Freihaufs in a room. No sides, no gravitational fields, not QM, no nothing except people in a room. That probability is just a matter of information is easier to see, here.

Incidentally, the premises about the six-sided object do not say anything about a cube. There is no information about symmetry of any kind, for that matter.

And I’ll bet I can get you to agree with me that probability can be separated from physics: think of the winged horse examples. No physics at all there; yet we have probability.

11. TomVonk says:

Incidentally, the premises about the six-sided object do not say anything about a cube. There is no information about symmetry of any kind, for that matter.

This is exactly the point I have been making William .
The premises are overloaded with information about symetries and matter density .
I wouldn’t believe that you did not carefully read the post but when the premise is saying that “ONLY ONE FACE SHOWS” , it is logically equivalent to say that the considered object is a cuboid with plane or concave faces .
A cuboid being a 6 faced object where 2 opposing faces are rectangles .
So yes , your premise actually says that the object is a cuboid (note that a cube is a particular case of a cuboid) 🙂

And this premise coupled to the premise of “tossing” logically implies that there IS a gravitational field too .
Note that this part , even if it is clearly contained in the premises is not necessary to prove that the statement “Pr( C | P1 & P2 ) = 1/6” has probability 0 to be true .
The only necessary part is that the premises imply that the object is a cuboid what I have proven .
If somebody tells something else , then he stands in contradiction with the premises what we cannot have , can we ?

The example with winged horses is good . It is also physics – a winged horse is a horse with wings also called Pegasus antineus and quite common in the Proxima Centauri region .
But the statement about the winged horses is very different from the statement about the 6 sided object and is actually true .
And I can even explain why it is so different 🙂
It is because the yelowness of the horse depends only on its horsitude .
So the implicit (and NECESSARY !) “fair horsitude” is already present in the premise .
If you were not sure whether you can make the difference between a flying horse and a flying wharg (who is very similar to a horse) then the conclusion about the probabilities would be wrong .
This is exactly what happens with the 6 sided object – it is because the sides of a cuboid are not equivalent (symetric , invariant) that it really matters on which side is the 6 and the probabilities vary accordingly .

I hope that I am clear but what I say here is universal .
You can make any probabilistic statement you wish and I bet you what you want that I will be able to show that in each and every statement you make , there is an explicit but generally implicit assumption about symetries .
By symetry I mean an invariance in a transformation so not only simple geometrical symetries .

12. Swade016 (Wade Michaels) says:

Bernie, P3 is not LE to P3′ because P3′ does not imply P3. To prove this, I will show P3′ is true, then I only have to find a single object in the entire universe where when tossed each side of the die is NOT equally likely to show.

Let my object be the ace of hearts from a bicycle deck. It has six sides (this isn’t flatland afterall), and the probability that a 6 being randomly labelled on a side and showing is 1/6. We’ll just guess here, but the probability either face shows is 99.99994%. The P(ace) = 49.99998% and P(red bycicle face) is also 49.99998%. The probability of any edge = .00001%. (.4999998+.4999998+.0000001+.0000001+.0000001+.0000001) = 1.

Since the probability that the 6 will be labelled on any edge is 1/6 (6 equally likely choices to be labelled), we multiply our probabilities to get the expected outcome and they all wash out at 1/6. P3′ = (Sum prob of face up * prob of label being 6) = (.083333+.083333+1.67e-08+1.67e-08+1.67e-08+1.67e-08) = .16666 = 1/6

So, P3′ holds because he probability 1/6., however, the ace of spaces is NOT a “fair die”, i.e. “each side of the die is equally likely to show”, which is what P3 states. Therefore, P3′ does not imply P3, which means they are not logically equivelant.

If I made a mistake, please correct…I’m not 100% sold on my solution.

13. Swade016 (Wade Michaels) says:

Based on my last post, I’m also with Tom on this, P3 states that “each side of the die is equally likely to show”. It says nothing of any labelings showing, only sides. Since only a fair die of sides 6 has each side equally likely to show, and Tom shows us that only a regular cube with angles 90 degrees and all face areas equal and the center of mass being the center of the object, AND that there are an infinite number of 6-sided objects in the real 3-d plane, the Probabiliy that P3 is true is 0, just as P(X = x) in a continuous distribution = 0.

14. JH says:

(P3) The die is â€œfair.â€â€¦, which is the same as saying (P3â€²) The probability of a â€™6â€² is 1/6.

Building upon SteveBrooklineMAâ€™s comment, as long as the probability of NOT observing a 6 is 5/6, then (P3â€™) is correct. For example, P(1)=1/12, P(2)=1/12; P(3)=2/12, P(4)=3/12, P(5)=3/12 and P(6)=1/6â€¦ not fair! This can bee readily extended to a six-sided object. Furthermore, since P3 and P3â€™ are not the same, so the claim o f circularityâ€¦

15. TomVonk says: