Don't Ask Marilyn a Physics Question

Marilyn needs a physics class. A chemistry class would help, too.  In this morning's Parade magazine*, Marilyn vow Savant answered the question "Is it true that if water is 100 percent pure, it will not conduct electricity?" with "Yes."

Uh, no.

Electricity is the motion of electrically charged particles.  Most commonly, it is the motion of electrons in metal wires that most people think of as electricity.  However, in batteries and other electrochemical applications, both positive and negative ions are moving.  How well a substance conducts electricity depends on a number of factors such as the number of charged particles, the distance the charges have to travel, the area through with they move, the temperature of the material, complex interactions with the atoms and molecules, and the applied voltage.

Tap water is a poor conductor; copper is about 10 billion times better**.  Adding ions makes it somewhat better.   Copper is only 10 million times better than sea water. In other words, sea water is around 1000 times better at conducting electricity than drinking water.

So it makes some sense to think that if one were to remove all impurities from water than it wouldn't conduct at all.  Except…

Water autodissociates.  Even in pure water, there are hydronium ions (H₃0⁺) and hydroxide ions (OH^-).

Chemists have a way to express how much hydronium ions are in solution; it's called pH, the logarithm (base 10) of the H₃O⁺ concentration.  Pure water at room temperature has a pH 0f 7.  

Therefore, pure water will conduct**.

There are other substances that one might think don't conduct electricity - air and glass, for example.  Under normal conditions they don't, but if I were to apply enough of a voltage, even these materials will conduct.  Remember this the next time you watch a lightning storm.

NOTES:

*As of noon on Nov. 27, neither Parade.com nor www.marilynvossavant.com have a link to this column.  When one appears, I will update the blog.  UPDATE: The link to the column is http://www.parade.com/askmarilyn/2011/11/Sundays-Column-11-27-11.html.

** Pure water has a conductivity of 5.5 X 10^-6 S/m.  For comparison, copper has a conductivity of 5.96 X 10^7 S/m.  

Bartender! Another round for me and my women friends.

Want to prevent osteoporosis?  Drink a beer.  An ale is better than a lager.

A nutritionist at Cambridge University, Jonathon Powell, has found that while ethanol in beer helps to prevent bone loss, the presence of silicon in the form of orthosilicic acid promotes bone growth.  Guess what?  Beer is an excellent source of this dietary silicon.

A good ale will contain about 1/3 of the daily recommended amount of silicon, and for some reason the absorption of the silicon is enhanced with the beer.

It's even more important for older women, because silicon combines with estrogen, so it may be more important for post-menopausal women to raise a pint.

So treat the important women in your life.  L'Chaim.

p.s.  An old beer advertisement.

Bartender! Beer me.

One of my favorite activities is to order a Boddingtons, a creamy British ale.  As good as it tastes, I always anticipate the visual splendor that is the head.  As it forms, tiny bubbles flow downwards.  If you're used to beers and sparkling wines where the bubbles go up, it's entrancing.

Boddingtons and Guinness are two examples of a beer that foams because there is both carbon dioxide and nitrogen dissolved in the beer.  The CO2 results from the fermentation process, but the nitrogen must be introduced artificially.  In a bar where these beers are served on tap, nitrogen is pumped into the beer at high pressure.  In a can or bottle, nitrogen has already been added, but a widget is used to introduce pressurized nitrogen to initialize the bubbling process.

Have you ever wondered why bubbles form at all?  In order for a bubble to form there must a small gas pocket.  Gas then diffuses into the pocket, and when it reaches some critical size, it detaches and  - voila!  A bubble.  I have always thought that the bubbles formed on the surface of the glass where small imperfections created the gas pockets.  I remember my father taught me an important lesson - always use the same glass.  He thought that in a freshly washed glass there would be enough soap (surfactant for you science-y folk) left behind to coat the sides of the glass and inhibit bubble formation.  

Then in 2002, French scientists studying champagne (of course) found that bubbles form on small cellulose fibers.  These fibers are thought to be added either by the cloth used to dry the glass or by falling from the air.

Two Irish mathematicians - who I'm sure enjoy a pint of Guinness now and then - have shown that cellulose is an efficient method to promote bubbling in these 'nitrogen' beers and have proposed that the widget could be replaced by a coating of cellulose on the side of a bottle or can.

Please enjoy your beer responsibly and scientifically.

What goes "Quack"? Dr. Oz?

URGENT!  On tomorrow's Dr. Oz show, he investigates features claims that magnets are a cure for chronic pain.  I won't be able to watch it, but I hope to catch either a rerun or find it on the web.  If any reader sees it, leave a comment.

Note to faithful readers - Sorry it's been a while since a post.  My beer post is coming soon.  Life at Rio Hondo College has been hectic.  I wish I could fill you in, but one day, I promise to regale you with details either here or at River Deep Faculty.

Don't Ask Marilyn - Part 4 or Let's Do Math!

I want to conclude this series on the Marilyn von Savant's die roll problem (see the original column from Parade magazine and my part 1, part 2, and part 3) with a proper probability calculation, even though my analysis using entropy is spot on.  When I'm done, I will be writing about beer.  Here's a brief highlight of what's  to come.  Beer bubbles are cool, and drinking beer may aid women in preventing osteoporosis.

My reader thinks that Marilyn's die roll problem is one of conditional probability.  I disagree.

Conditional probability means "what is the probability that an event occurs (let's call this event B) if we already know that another event (let's call this event A) has already occurred."  To use the phrase of my reader and one that is used in conditional probability, "what is the probability that event B occurs given that event A has already occurred."

Two examples to illustrate:

• I throw a die, and I roll a 4.  What's the probability that the next roll is a 1 given that the first roll is a 4?  Since these two events are independent, the probability of rolling a 1 is 1/6.  In other words, knowing I threw a 4 does not affect the next roll.
• I throw a die, but don't tell you what I roll, except I do tell you it's not a 5 or a 6.  What's the probability that it's a 4 given that it's not a 5 or a 6?  The answer - 1/4.

MATH ALERT!   In chapter 4 of "Introduction to Probability" by Grinstead and Snell, conditional probability is calculated with the formula

In order to calculate the conditional probability P(B|A) [the probability that B occurs given that A has already happened], we need to calculate P(A) [the probability that A happened] and P(A and B) [the probability of A and B; the upside down U is the mathematical symbol for union and can be understood to mean 'and.']

Let's examine the two examples from above.

• P(A) = 1/6.  P(A and B) = 1/36.  Then P(B|A) = 1/6.
• P(A) = 4/6, since knowing that the roll is not 5 or 6 is the same as knowing that it is a 1, 2, 3, or 4.  P(A and B) = 1/6.  Then P(B|A) = 1/4.

Now I examine Marilyn's problem.  There are two possible outcomes. Roll (a) 11111111111111111111 and roll (b) 66234441536125563152.  

So let's assign events and be careful.  Event B is roll (b), since I'm interested in knowing what the probability of rolling (b) given that a die has been rolled 20 times.  Then event A is rolling a die 20 times.

P(A and B) = the probability of rolling (b) and rolling a die 20 times = 2.7 x 10^-16.

P(A) = the probability of rolling a die 20 times = 1.

Therefore, Then P(B|A) = the probability of rolling (b) given that a die has been rolled 20 times = 2.7 x 10^-16.

See.  George Alland, the math teacher that challenged Marilyn, is correct.  Marilyn is wrong in insisting "It was far more likely to have been that mix than a series of ones."  Marilyn corrects her mistake when she writes that "a jumble of numbers" is more likely.  This was my point in part 1, part 2, and part 3.

There is one more issue I'd like to address. Something that might help clarify the difference between 'that mix' and 'a jumble.'  There is in math the concepts of permutations and combinations.

A permutation is a arrangement of things in which order of the things matters.  Suppose you buy a raffle ticket an your number is 407.  Do you win if the number called is 074?  No, because the order of those digits matter.  Now suppose you buy a lottery ticket. and your numbers are 1, 13, 25, 26, 33, and 42.  Do you win when you see the ping pong balls come up 13, 26, 42, 33, 1, 25?  Yes you do, because the only thing that matters is the combination of numbers, not the order in which they are drawn.  When Marilyn wrote 'that mix' she - perhaps inadvertently - specified a particular order, one permutation.  When she wrote 'a jumble' - perhaps she caught her previous mistake - she now highlights the combination, not the order.

UPDATE:
How does the probability change when when we consider the combination rather than the permutation?

Event B is roll (b), since I'm interested in knowing what the probability of rolling (b) given that either (a) or (b) is rolled. Event A is now rolling (a) or (b).

P(A and B) = 2.7 x 10^-16.
P(A) = 5.4 x 10^-16.
Therefore, Then P(B|A) = 1/2.

But if I change the event B to rolling three 1s, three 2s, three 3s, three 4s, four 5s, and four 6s,  then

P(A and B) = 0.239.

P(A) = 0.239+ 2.7 x 10^-16.

Therefore, Then P(B|A) = 1.

I hope this settles the matter. I need a beer.

p.s. Many thanks to my reader. He truly highlights the need for all of us to be clear in our writing and our mathematics. I hope that all my readers hold me to such standards.

Don't Ask Marilyn - Part 3 or I Get Email

UPDATED!  See below.
My correspondence concerning the Ask Marilyn column with a reader continues.  The emails are copied below.  I have removed the reader's name and have only deleted some friendly asides and such. I have more comments about the Marilyn vos Savant column after the emails.

10/30/2011

READER: In any event, I'm not sure what you're saying.  In your 1st blog

entry, you say Marilyn is incorrect.  In your 2nd blog entry, you

seem to say Marilyn actually is correct.

10/30/2011

VP: In her first answer, she writes "It was far more likely to have been that mix [emphasis added] than a series of ones."  In my view, when she writes 'that mix', she is referring to that one specific roll, and that roll has the same probability as all 1s.  That's why I claimed "Marilyn's first answer is wrong" in my first post.

However, in her answer to George Alland, she changes her answer to '"It was far more likely to be (b), a jumble [emhasis added] of numbers."  She has changed the conditions of the problem from considering one particular throw to a roll that is jumbled.  That's why I wrote in my first post "Her second answer (the "jumble of numbers" is more likely) is correct..."

As you wrote, Marilyn may sometimes be ambiguous and being confined to one small column in Parade magazine, that can be all too easy. 

10/30/2011

READER: You are correct that when Marilyn writes "that mix", she is talking

about the specific series (b).

However, what you are omitting in your analysis of Marilyn's answer,

is that "that mix", viz., (b), is indeed "far more likely" GIVEN THAT

the rolled series must be either (a) or (b).

The "GIVEN THAT" clause is crucial in determining likelihood.  I had

stated this key point in my first response to your blog entry, along

with the other key point that the writing down of the series occurs

after the 20 die rolls.

Changing Marilyn's problem by omitting the "GIVEN THAT" clause

constraint, would make your probability analysis correct and your

ambiguity complaint reasonable.

BTW, I'm not a die-hard Marilyn fan.  When she messes up, e.g., when

she claimed that Wiles' proof of Fermat's Last Theorem was invalid,

I'm the first to throw a stone.

11/2/2011

VP: I must admit that I'm at a loss.  I fail to grasp how the phrase 'given that' affects the probabilities.  Could you explain further?

The reader points to the original problem as stated by Marilyn.  I reread it and I see that there's an even more egregious error.  Marilyn writes 'It’s (b) because the roll has already occurred.'  This implies there is some conditional probability.

As far as my understanding of probability goes, there's three issues here.  (1) What is the probability of rolling a die twenty times and getting one out of 3,656,158,440,062,976 possible outcomes?  (2) What is the probability of rolling a die twenty times and getting a particular mix of 1s, 2s, 3s, 4s, 5s, and 6s?   And (3) this problem does not involve any conditional probabilities.

Are there any readers who can find some oversight, misconception, and/or goof on my part?

UPDATE  11/2/2011  Email

First note that Marilyn doesn't explicitly use the words "given that".  However, the meaning of her wording involves the same idea, viz., conditional probability.

You can google something like:  "given that" probability to find numerous examples using the phrase "given that" in this conditional probability context, e.g.,    http://www.mathgoodies.com/lessons/vol6/conditional.html
OK, let's move to Marilyn's article.  I've carefully chosen wording and formatting to make what's going on easier to understand.

The 1st half of Marilyn's article basically says:

    The specific mix of numbers (b) 66234441536125563152
    is as likely to appear next, as
    the specific series (a) 11111111111111111111,
    GIVEN THAT
    I've already written down (a) and (b).

Hopefully, you agree with this wording and the correctness of the statement, so far.

The 2nd half of Marilyn's article basically says:

    The specific mix of numbers (b) 66234441536125563152
    is more likely to have been the rolled series, than
    the specific series (a) 11111111111111111111,
    GIVEN THAT
    I wrote down (a) and (b) after I finished rolling the die,
    AND
    the series I rolled is indeed either (a) or (b).

Please take a moment to confirm that this captures the meaning of the 2nd half of Marilyn's article.

Now, do you also see how the "given that" clause for the 2nd half fundamentally changes the likelihood of (a) vs. (b), even though Marilyn still compares explicitly "that mix", 66234441536125563152, with the all ones series?

Note that Marilyn is NOT saying that, if we run the entire experiment again, that 66234441536125563152 would again be the series written down on the piece of paper.