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.

3 comments:

Ranjeet said...

From Akash: http://en.wikipedia.org/wiki/Duckworth%E2%80%93Lewis_method

Mikael said...

I disagree with "Most people accept a bet when they think it is a game with at least a zero sum in their favor.", though less so than the original if 50-50...
That only works if the eventual losses can be offset by wins in multiple bets. This is a one time bet, so your logic doesn't apply.
Eg, I wouldn't take a one time bet of $1,000,000 to your $1,000,001 on a coin flip. My expected return on each coin flip is $0.50, but since we're only doing one flip that doesn't help me.

I also disagree with 10 fingers to 1 cadillac logic. At 50-50 odds of the lighter lighting, AB will only win 1 time in 1024, so the cadillac better be worth at least 1024 fingers. If the odds are 80% then he's expected to win 1 time out of 10 so the cadillac better be worth 10x the finger.
In other words, Q1 and Q2 are two inputs into the same function, you can't solve for one without knowing the other.

Q3 - AB is willing to go on, so he should get the Cadillac. MftS is not willing to upset his wife, that's his problem, not AB's.
AINAL, but from what I've read, gambling contracts aren't enforcable under the law, so AB doesn't have much recourse...

Q6 - This the wrong question. The fingers in and of themselves are unlikely to hold a monetary value for the MftS. This should be "How much does the MftS value watching someone cut off their finger vs the value of a cadillac." But even that's not really enough. Maybe it should really be: "How much does MftS value the game".

Q5 - This can't be answered without knowing the answer to Q6. We know the limit on what MftS is willing to pay, ie 47 fingers is worth at least 11 cars. He may very will have been willing to part with 22 cars to get the same 47 fingers, we can't know.

For MftS I believe that the real question is how often can I lose/win before the game gets boring.

Ranjeet said...

Okay, so I don't really think I disagree with Mikael's disagreements.

The "expectation value" approach seems to break down for one-time bets, and also for very large wagers, as his example shows.

The answers to the questions aren't independent of each other.

"Wrong question"? "Wrong question"? Oooh that hurts!

47/11 is a reasonable guess in the absence of more information.

I don't really care whether my answers are "correct", it has been loads of fun just thinking about the problem while sitting in traffic on 101 and 85.