Wednesday, November 28, 2012

Loud and Clear!

I love you sweetheart, I look forward to have the life I want with you. (I would never admit this to anyone, but know I think of you often, all the time!). I Love you :)))) and miss you and can't wait for our chats. When I come back I cannot wait to see you. :). You will come to the airport to pick me up, won't you, and take me back home? It will have been weeks, I can't imagine what it will feel like to be in your arms again  xoxo! Make that XOXO... :))

(Send the above to a real or imaginary actual loved one (for present purposes it doesn't matter which) and “accidentally on purpose” CC the one who doesn't seem to understand “N,O!”. Based on personal experience, I guarantee that it works like a charm. 

... Ouch!)

Monday, November 19, 2012

New Relics

We'd cleaned up the kitchen together after putting the girls and the baby to bed and I sat down at the table, wanting to be by myself before going upstairs. I telethought my spouse that I would come up in a bit. I felt the reassurance come back, and then turned the telepath off. My sister was asleep in one of the girls' rooms and Jess was downstairs in the study.

At the final moments of the atomization, I'd cried, my sister had cried, even Jess – his intimate from the last few years – had cried, and I'd found it strange that my mom hadn't. If there had been nothing else there, there was that bond through me, but perhaps she'd hardened herself long ago to potential loss.

I walked up to check on the girls. I never do that, but today, I just felt the need, to see them, with whom he'd spent more time lately than I'd been able to … neither with them nor with him. Some nights, if he was here visiting us when the girls got home with their father from swimming lessons or cello practice, he'd drop whatever we were talking about (math) or arguing about (math) to go be their toy. Surprisingly, he never once tried to teach them math, leaving that work to me and keeping his questions and inquisitiveness about my work for when the girls weren't around.

I'd wiped the tear stains off both the girls, but they were still a little puffy eyed. I am sure other parents, as the girls' aunt certainly did, would think me cruel for having taken them with us to the atomization. Now I found the younger one had crawled into her older sister's bed, pulled the blanket off her and cocooned herself. Just like my sister used to do to me. I spread the blanket out and covered them both up again - once they were sound asleep they would be fine, all the tussling took place early at night.

It struck me that as a child I'd never woken up cold. Did Dad cover me up at night when sis and I were asleep?

I suddenly wanted to hold him again, to feel his physical presence, to clutch him to myself - protection, protectiveness all jumbled together in this tangible vacuum that sat across from me as if forever. Milk seeped from my breasts and I realised I hadn't yet fed the “sleeps-through-the-night” baby but my body still felt too wracked to go up and do so immediately. I rebraided my hair, a single short braid I've worn since 5th grade, this time using generators of B_5, for which I'd had a soft spot ever since my discovery of their relation to conformal matrices had caused a small resurgence of interest in Strings.

On a sudden recollection I pulled open the wall behind me and there beneath the tabletop controller I found the only remaining physical vestiges I had of him. I lifted out the little plastic canister, the less-than-an-inch wide black body about the length of my pinky, the grey top with faded “IMLI SEEDS” written on it. Dad had had a few of those when I was a child - film-roll containers (from his childhood?) that he'd managed to hold on to through multiple international moves. I flipped open the top with my thumb and spilled the contents out on the kitchen table.

Nails. His nails. Both he and I had laughed when my sister had insisted on collecting these, “People collect hair and nails. Dad collected mine!”, she'd protested. “Only baby hair and nails, when they are soft!”, I'd sniggered. But she collected them anyway, that one time, and borrowed one of Dad's canisters which he used for taking spice mixtures on camping trips. She'd cried then because I'd made fun of her, but a few years ago she divided the contents with me, for safe-keeping. I opened the telepath and sent her a thank you, for morning delivery, and recalled her tying my half of his hair with a rubber band. I stuck my finger in the canister and sure enough there was a tuft of hair, with brittle pieces of broken rubber band sticking to it. The hair was a couple of inches long, grey. I sniffed it, painted my face with it, rubbed it between my fingers, desperately wanting to believe in re-incarnation, in a higher power, anything that would bring him back to me.

I carefully put the hair back in and looked at the nails. He'd been cutting his nails, we the three of us had all been cutting our nails before he took us climbing that day, and I remember that day because it was when I first climbed a 5.9, and then tried a 10a and he'd smiled and hugged me and said, “You already out-run and out-swim me, is there nothing you will leave me, not even climbing?”

.......................................

What have you left me, Dad?

Friday, November 16, 2012

Review - The Wind-up Bird Chronicles

Review of “The Wind-up Bird Chronicles”, by Haruki Murakami

Murakami seems to be a very popular author with women on the dating websites, at least with the women who interest me. As preparation for the battle between the sexes, I wanted to read what they are reading. I could have armed myself by reading “Think like a man, act like a woman”, but while the movie was funny, I don't know if it was particularly edifying and none of the other BASPs with whom I watched it found it worth commenting on afterwards. Besides, having been a full-time male mom for 8 years, I'd already thought like a man and acted like a woman, and well, here I am.

So Murakami it is. Now, if you want a regular old book review, look for one on the web.

A few lines I wish I would have written: “Which is not to say that I didn't have any distinguishing characteristics. I owned a signed copy of Miles Davis' Sketches of Spain. I had a slow resting pulse rate: forty-seven normally and no higher than seventy with a high fever. I was out of work. I knew the names of all the brothers Karamazov. But none of these distinguishing characteristics was external.”

Now for the feelings the book aroused in me: mostly it left me very disturbed, there is palpable occult-like evil, but it is so unexplained and unmotivated, and only tangentially personified, that I could never really grasp what it was. This is mostly carried out against women, and the violations are mostly more mental than physical. There is gristly, bloody, gruesome, slow, pit of the stomach violence and detailed empathetic pain. This is mostly against men. The scenes seem gratuitous, except as mysterious and near mythical events in the past that explain future senses of deep connection. The women survive, the men don't.

I suspect that one of the reasons women like this book is because it is a singular example of a book with anticipated, prolonged, detailed descriptions of infernal physical violence on men and of violations of males bodies on a mass scale. Does it make women who like this book man-haters? Perhaps, but only in the sense that men who like Paul Auster's “The Book of Illusions” (in which 8 women are gratuitously and violently killed off) or indeed who like any of the numerous books and movies in which gratuitous or non-gratuitous violence against women is lovingly portrayed, are woman-haters.

As the accompanying diagram shows, the book has numerous loose ends, vestigial characters with no development that play a repeating, non-minor role but remain completely disconnected from any other character other than the protagonist. I was never even able to understand whether they were the shades of some other characters. These are the dangling characters in the diagram. They seem superfluous in the sense that the book would not have been essentially different without them.

Now for the diagram. It is a map of the main characters, a few key non-human objects and their interconnectedness. I had hoped that drawing it would show me the center, and was motivated to draw it when the protagonist sketches out some of the interconnections (as a circle) and wonders what is at the center. However, circles needn't have centers in the space they belong to: while the rim of a flat dish has a center, what is the center of the rim of your coffee cup? Or the center of a circular orbit around a black hole? And indeed, I haven't found any obvious center in the Wind-Up bird network.

In the diagram, green characters are founts of positive energy, red characters of evil. Green arrows indicate positive, life-enhancing acts, passive or active. Red arrows indicate life-draining actions. Light blue characters are carriers of energy of undefined sign. However, almost all the characters are in some way contaminated by this “energy”, whether positive or negative, except for May, who in the end seems the most real of all the characters.



Other comments or unanswered questions:
  1. Nutmeg's father has a blue stain of the same shape and color on his cheek that Mr. Wind-Up Bird acquires and loses. But there is no other connection: the former is a birthmark and brings no special powers or heat, the latter is acquired on a parallel world excursion, is connected with and is a source of this mysterious energy and is lost towards the end of the narrative.
  2. What happens to Noboru Wataya the cat, and what is his connection to Malta Kano, if any? What is the significance of his tail and whether it has the same bend or not as before?
  3. Why is the wind-up bird audible only to males? Is there any associated mythology with this ill-omen?
  4. WTF with the connection between Lt. Mamiya and Creta? This is an instance of the author getting tired of writing the book towards the end, and puts these two unrelated characters together to tie up two ends.


Sunday, November 4, 2012

Answers for "Man from the South" probability questions

Links to Dahl's  "Man from the South"

The questions are repeated in this post.

Preliminary answers

Q1. What does American Boy think is the probability of his lighter lighting during any single attempt?


Most people accept a bet when they think the odds are at least 50-50. Since Man from the South and American Boy have had time to negotiate the terms and think about it, both think they can win it. So as neutral observers we can consider the probability that the American Boy will win the bet to be 0.5.

So what does this tell us about what he thinks is the probability of the lighter lighting on any one attempt? Let this probability be p. The probability that he will win the bet, i.e. that it will light N times out of N trials, is pN. We just decided that this is ½, for N=10. So,
p10 = ½
I was driving and couldn't very well use my laptop (my phone is not a very smart phone), leave alone the log tables, slide rule or abacus in my back seat. And I can't do powers of 1/10 in my head, not in general. But I can calculate powers of numbers near 1! So let's calculate q=1-p instead, which is bound to be small since the boy is so sure of success.
½ = (1-q)N ~ 1-qN, so for N= 10, q=0.5/10 = 0.05.
Which gives us our first answer: p = 95%, which we know is a bit of an overestimate. (The exact answer is 93.3%.)

But this is like focusing on conversion ratio and not on cost-per-action.

Q4. Were there to have been a 9th attempt, and were American Boy to have failed in it, what would he have lost?


A: His left pinky, those were the terms of the bet. So American Boy is staking his pinky every time! All but the last time, if he wins, all he wins is the right to stay in the game, if he loses, chop-chop (and the right to play!)

So this is kinda sorta like conversion attribution: every bid-request we (RFI) win and then every impression we serve only gives us the right to stay in the game, in the sense that if we don't serve the impression we certainly won't get attribution credit.

Q2. How many fingers to a Cadillac?

On the face of it, it would seem to be one finger to a Cadillac, C = F, since if he loses, he loses a finger, and if he wins, he wins a Cadillac. But as we saw above, the American Boy stakes his one finger 10 times and thus faces 10 opportunities to lose his finger vs. one opportunity to win the Cadillac. So,
10F = C


After the eighth successful attempt, 

Man from the South's wife surprises them and ends the game. She sends American Boy away empty handed.

Q3. How should he have been compensated, if at all?

Aha! “How to distribute the spoils in an interrupted game?” The very question that my colleague Jack pointed out was the leading cause of the rise of probability theory!
The way I think of this is in terms of “vesting”. Each time they play and American Boy wins, he wins a 10th of a Caddy, but, BUT, he only gets to keep his winnings (the entire Cadillac) if he keeps his head (figuratively) and finger (literally) for 10 trials. American Boy can't get cold feet and walk out after say 5 trials and demand half a Cadillac – so really he can get cold feet and walk out but he can't drive off. So there are two possibilities. Under one, the Man from the South gets cold feet and decides not to play anymore. In this case he forfeits his Cadillac, which was held in escrow by the narrator. Under the other possibility, the “authorities” intervene. Since neither party has broken their terms, in this case I think American Boy gets to keep his “unvested” winnings, which would be 8/10 of a Cadillac. Now since the Cadillac wasn't the possession of the Man to begin with, but belonged to the “authorities”, … this isn't a math problem anymore. (By the way, who intervened and broke up Cardano and Pascal's poker game?)

Legal opinions? Aji, Joanne?

Q1 (Re-evaluated). What does American Boy think is the probability of his lighter lighting during any single attempt?


Most people accept a bet when they think it is a game with at least a zero sum in their favor. Since Man from the South and American Boy have had time to negotiate the terms and think about it, both think they can win it. So as neutral observers we can consider the game to be a zero-sum game.

What does it mean for this to be a zero-sum game? Winnings * probability of winning – losses * probability of losing = 0! Which yields:
p = 1/(W/L + 1). (Check: if W are high, p is low; if W are nearly 0, p is nearly 1 and if W=L, p =1/2.)

In our case the winnings are the Cadillac C, the losses are the finger F and the probability P of winning the entire game is P = pN, where p is the probability of the lighter lighting in a single trial. Combining things we have
C*P – F*(1-P) = 1, or P = 1/(C/F + 1)
Putting in C/F = 10,
p10 = 1/11

1/11 = (1-q)N ~ 1-qN, so for N= 10, q=1/11 = 0.1.
Which gives us 78.7%.

Really? Would you play that game with a lighter which only lights less than 80% of the time? I think that the Man from the South has fuddled the American Boy into undervaluing his finger, by making him think he is wagering a finger vs. a Cadillac, whereas really he is wagering a finger against a 10th of a Cadillac.

Some preliminaries

If you aren't interested in the nuts and bolts, skip them, but this is so anyone can check my work.
Notation: Sum[i, 0, Infinity] f(i) is to be interpreted as the sum of the function or series f(i) over the index i from i=0 to i= Infinity. Then,
Sum[n,0,Infinity] pn = 1/(1-p)
Sum[n,0,N] pn = (1-pN+1)/(1-p)
and
Sum[n,1,N] pn = (1-pN)*p/(1-p)

Look at the following table of outcomes of consecutive tosses and the overall probabilities
W ← 1 → L
1: p (1-p) (End)
2: p2 p(1-p) (End)
3: p3 p2(1-p) (End)
So after n trials, the probability of winning all is pn and the probability of losing any is
(1-p)* Sum[i,0,n-1] pi = (1-p) * (1-pn)/(1-p) = 1-pn = 1- prob(Winning). Which is good since it indicates I can still sum correctly.

Back to the problem, to get a handle on 

what if anything AB deserves when the game is stopped.

From making the last, Nth, trial a zero-sum game, we know that
pN = 1/(C/F + 1). We've assumed that AB wins 1/N th of a Cadillac (virtually) when his lighter lights. So assuming the first trial is also a 0-sum game, we have:
p*C/N = (1-p) *F, which resolves as
p = (C/NF + 1) ^(-1).

Can these two equations be solved simultaneously for both C/F and p?
Yes, but the solution is i) independent of N and ii) meaningless:
We have (1+C/NF)^N = (1 + C/F), but the RHS is simply the first two terms in the binomial expansion for the LHS, so the equality holds only when C/F = 0 and p =1.

So one of our assumptions above is wrong. 

Let's try another tack. 

Assume that at the nth trial, AB wins some unknown portion of the Cadillac a(n)*C. (When Ari and I were talking about this last week, Ari guessed, “Wouldn't it be some quadratic or increasing portion that he wins?” Ari's motivation was to take into account the wearing out of the flint, the gas running out, the thumb getting tired, AB getting nervous etc. Dahl, spends an entire paragraph describing the care and attention to detail taken by AB, after each light, he blows on the lighter, closes the lighter, waits a few seconds perhaps for gas pressure to build up again, re-opens it and then flicks it once. ) What we know is that
Sum[n,1,N] a(n) = 1, over the course of the entire game, if he survives, he wins the entire Cadillac. So at every trial, AB stands to lose not just his finger and the right to play, but also the “won but not vested” portion of the Cadillac, and he stands to win some portion of the remaining.

At the last, Nth trial: W : L
p : (1-p)
a(N)C : F + (1-a(N))C
Using the 0-sum equation, we get:
a(N) = (1 +F/C)(1-p).
Great, so now we have ONE equation and 3 unknowns: a(N), F/C and p. But we also know that if the game is 0-Sum over all: p^N * C = F*(1-p^N),
which yields:
0 < p = (1+C/F)^(-1/N) < 1. So now at least we have two equations for three unknowns, and the solution for p is valid. This doesn't guarantee that 0


If we knew C/F, we could solve the problem. However, clearly, C=F is no longer valid. C = 10F could be used for AB's assumption. We also have another source of information: The Man from the South's wife explains that he has lost eleven cars and taken forty-seven fingers. Assuming that he considers these equivalent, we have 11C = 47F
so,

The fraction C/F is the ratio of the value of one Cadillac to the value of one Finger.


But all this hasn't answered the question of 
what AB deserves when the game is interrupted. 
Also, we've assumed that the entire game is 0-sum and that the last trial is 0-sum. Can't we make use of the assumption that 

every trial is also 0 sum 

to see if we can figure out the intermediate non-vested winnings?

Recall that at the nth trial, AB stakes his previous winnings and his finger for a chance to win a(n)*C portion of the Cadillac. So the 0-Sum equation for the nth trial is:
a(n)*C = (1/p -1) *(F + C* Sum[i, 0, n-1] a(i)). The resulting recursion relation is for a geometric series!
a(n+1) = (1/p)* a(n),
whose solution is
a(n) = a(0)/p^n.
Note immediately that a(0) != 0, so AB has to stake (even if only virtually) something more than just his finger. 

We find a(0) 

by using the fact that the total portion of the Cadillac gained over 10 trials is 1:
1 = Sum[i,1,N] a(i) = a(0) * Sum[i,1,N] (1/p)^i = a(0) * (1/p^N -1)/(1-p), or

a(0) = (1-p)/(1/p^N – 1)

For the first trial (note that this is independent information since so far we have used the recursion relation and established a(0) using the “normalization”, but we haven't yet used any 

“initial conditions”):

a(1)*C = (1/p – 1) * ( F + a(0)*C), which yields
C/F = ((1/p)N -1)/p, which is a different relationship between C/F, N and p than we had previously. I am not sure I can invert this to yield p(C/F), but it can certainly be numerically solved.

Let us also calculate the virtual winnings after each trial:

W(n) = C*Sum[i,1,n]a(i) = C*a(0)*(1/p^n -1)/(1-p), which simplifies to
W(n) = ((1/p)n - 1)/((1/p)N - 1)

So given p we could calculate C/F (or vice versa), a(0), a(n) and W(n)


Working on the “wife's numbers”, we see that the American Boy should be compensated with 66% of the Cadillac when the wife interrupts the game after the 8th trial.

How do the winnings increase as the trials proceed:

Winnings in units of "Cadillacs"
 
which look like
Levenfeld curve


Conclusion and final answers:

Q1. What does American Boy think is the probability of his lighter lighting during any single attempt?
About 80%.

Q2. How many fingers to a Cadillac?
In American Boy's valuation based on his behaviour, 10 fingers to a Cadillac.

After the eighth successful attempt, Man from the South's wife surprises them and ends the game. She sends American Boy away empty handed.

Q3. How should he have been compensated, if at all?
With 60 or 66% of a Cadillac. I would go with 66%, which is based on the Man from the South's experienced equivalence between Cadillacs and fingers.

Q4. Were there to have been a 9th attempt, and were American Boy to have failed in it, what would he have lost?
Ah, his finger, of course, and, his virtual stake, which is 3.9 % of a Cadillac. How could he have lost something he never had to begin with? Well, for the bet to proceed, AB would have had to ask the narrator to spot him 4% of a Cadillac, or its cash equivalent, or its (OUCH) finger equivalent, which is 17% (3.9% * 4.3 F/C).
If AB's lighter fails during the game, he loses 1.17 fingers since he would have to sell 0.17 fingers to the Man from the South to pay off the debt to the narrator. If AB's lighter doesn't fail during the game, he simply returns the cash or Cadillac equivalent from whoever he borrowed it, and is ahead one Cadillac. 

Why did I ever think of approaching the problem this way, with a “virtual stake”? In particle physics, one can borrow virtual particles from the vaccuum in order to simplify calculations. It is all halal as long as the virtual particles don't violate any conservation laws for quantum numbers and the mass-energy of the particles exists for a short enough duration of time that Heisenberg's Uncertainty principle is not violated. The really interesting thing is that these virtual particles have real effects: A pair of uncharged conducting plates will attract each other because a virtual charged particle – anti-particle pair will come into existence from the vacuum for a brief time, and the effective dipole and its images will cause the plates to experience an attractive force. Don't believe me, look up the Casimir Effect.

What happens if you grab those particles and forcibly separate them from each other and prevent them from annihilating each other as any decent particle-anti-particle pair should do? You end up creating a Black Hole-White Hole pair, which you can then use for superluminal transportation and as a time-machine! (Okay, I just made that up, but is it really any crappier than “The Secret” or Deepak Chopra?)

Back to reality: The Man from the South's wife explains that he has lost eleven cars and taken forty-seven fingers.

Q5. What does Man from the South think is the probability that American Boy's lighter will light during a single attempt?
86%.

Q6. How many fingers to a Cadillac does Man from the South figure?
Th ratio of the values is C/F = 47/11.

Q7. Do your answers to Q3 and Q4 change?
Yes.

Added on 16th Nov. 2012
What was Fermat and Pascal's approach? Instead of looking backwards, they looked forward and calculated the probability (on the condition of the current circumstances) of winning or losing the game and divided the spoils accordingly. So if AB has a probability of p of lighting the lighter and has already done so 8 times, the probability that he will then do so 10 times is simply p^2 and the probability that he will lose is (1-p^2). According to this approach, AB wins p^2 of the Cadillac and loses 1- p^2 of his pinky. I think this is close to Jon's suggestion, who strongly felt that since the game hadn't finished AB would have to lose part of his finger in exchange for part of the Cadillac.

Thursday, November 1, 2012

Probability questions from "Man from the South"

The Man from the South (see earlier post) and the American Boy negotiate terms and make a bet: If the American Boy lights his lighter 10 times in a row he gets Man from the South's Cadillac. If the lighter fails to light during one of those 10 attempts Man from the South immediately cuts off American Boy's left pinky, and keeps it.

Think about the following questions in any order. You can make any assumptions, come up with an approximation, an estimate, a minimum or maximum.

Q1. What does American Boy think is the probability of his lighter lighting during any single attempt?

Q2. How many fingers to a Cadillac?

After the eighth successful attempt, Man from the South's wife surprises them and ends the game. She sends American Boy away empty handed.

Q3. How should he have been compensated, if at all?

Q4. Were there to have been a 9th attempt, and were American Boy to have failed in it, what would he have lost?

The Man from the South's wife explains that he has lost eleven cars and taken forty-seven fingers.

Q5. What does Man from the South think is the probability that American Boy's lighter will light during a single attempt?

Q6. How many fingers to a Cadillac does Man from the South figure?

Q7. Do your answers to Q3 and Q4 change?

I don't necessarily have answers to all the questions, but here they are.