## Weekend combinatorics work problem (IV)

10 Dec, 2022 at 15:46 | Posted in Statistics & Econometrics | 10 Comments

When my daughter (who studies mathematics) and yours truly solve a combinatorics problem together it takes 12 minutes. If my daughter tries to solve the problem herself it takes her 10 minutes more than it takes when I solve it alone. How long does it take me to solve the problem?

1. Why do you get to cherry-pick away negative solutions? What if the professor starts with an answer and it takes 6 minutes to come up with a problem?
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Why do you pretend that throwing away half of your math-model predictions is just fine?

12/(t+10) + 12/t = 1 =>
12t + 12(t+10) = t(t+10) =>
24t + 120 = t^2 + 10t =>
14t + 120 = t^2 =>
t^2 – 14t -120 = 0 =>
(t-20)(t+6) = 0
Only applicable root positive, so it takes the professor 20 (minutes) 🙂

• Lars,
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I think your starting premise is totally conjectural. 🙂
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I suggest the way to proceed is to speculate about who is the dominant personality, the professor or daughter.
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Let X = time for professor to complete, therefore time (Y) for daughter to complete is Y = X +10.

If the professor is the dominant personality and totally gets his way then X = 12 and Y = 22.
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If the daughter is the dominant personality and totally gets her way then Y = 12 and X = 2.

• Interesting that the “fudge” necessary to solve this problem is the nature of the interaction between the father and daughter. You assumed the interaction was additive, I assumed it was multiplicative (efficiency through interaction), and Henry assumed that the interaction was negative (inefficiency through interaction – or interference).
What is so essential illustratively is that the entire problem solution depends on the starting assumption. This represents the object lesson for economic modeling. The result is defined by the starting assumption that only Henry stated explicitly. You and I made our assumptions implicitly, and the assumptions could be slipped by many an unwary reader.
This is illustrative of many economic models: Choose the assumptions that get the desired result.
— John Lounsbury

• John,
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If Lars’ specification of the problem is based on actual experience, that is, he knows it takes him 20 minutes to solve the problem and his daughter takes 30 minutes and it takes them 12 minutes co-operatively, then his starting premise models the empirical results accurately.
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If he is speculating about his time to completion, then it is anybody’s guess as to how to proceed.
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If the results are based on his experience then, in this case, it could be said that two heads are better than one.
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But as we all know from experience, this is not necessarily always the case. 🙂

• Yes, I agree. My previous statement about your solution was not correct. Your assumption about the effect of a dominant personality led to proof that cooperation produced no improvement, not that the interaction effect was negative. The interactive effect was nil.
The dominant personality effect has been seen in some group case studies I have read over the years. Example: https://pubmed.ncbi.nlm.nih.gov/19159145/

3. Let me address this as a theoretical economist would.

This problem is a simple one involving solving a quadratic equation.

Assume perfect efficiency of interaction between father (Y) and daughter (Z). Then YxZ = 12.
Since Y = Z+10, then
(Z+10)xZ =12.
Z^2 + 10Z -12 = 0
Z = {-10 +/- (10^2 + 4x1x12)^(1/2)}/2
= {-10 +/- (148^(1/2)}/2
= {-10 +/- 12.2}/2
= 1.1 or -11.1
-11.1 is not physically possible.
Therefore, the father alone solves the problem in 1.1 seconds and the daughter in 11.1 seconds.

However, there are few, if any, children who can work perfectly with a parent. Thus perfect efficiency is an imperfect assumption, and frictions must be considered.
Assume frictions amount to 50% for each. Then
(Z-0.5Z)x(Y-0.5Y) = 12
(0.5Z)x(0.5Y) = 12
0.25ZxY = 12
and Y = Z+10 as before.
Substituting and expanding
0.25Zx(Z+10) = 12
0.25Z^2 + 2.5Z -12 = 0
Z^2 + 10Z – 48 = 0
Now
Z = {-10 +/- (10^2 + 4x1x48)^(1/2)}/2
= {-10 +/- (296^(1/2)}/2
= {-10 +/- 17.2}/2
= 3.6 or -13.6, which can be discarded
With frictions, the resulting times for both the daughter and the father increase.

It is left to the econometric technician to determine the observed values of the frictions.

QED

— John Lounsbury

• Correction: The father’s and the daughter’s times have been reversed in the problem statement above. Should be:
“Assume perfect efficiency of interaction between father (Z) and daughter (Y).”
Cross-checking solutions using starting equations show imperfect results due to “rounding errors”.

4. The philosopher’s daughter, a mathematician, deduced that there were numerous theoretical possibilities for x given that z =12 and y = x +10:
If z = x then x = 12
If z = y then x = 2
If z = x + y then x = 1
If z = x + w then x = 12 – w
etc.
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The philosopher king derided the abstract “axiomatic deductivity” of his mathematician daughter.
However, he himself was utterly incapable of providing any empirical light on this matter or anything else.
Perched in his academic ivory tower high above the real world of practical decisions, he pontificated that:
– The world is dominated by “fundamental uncertainty”.
– No evidence about the past can ever give any guidance regarding the future because “real-world social systems are not governed by stable causal mechanisms or capacities”.
– Science is incapable of providing guarantees that it has found “real causes”. There is always the possibility of other vital and perhaps unobservable causes.
– Science is incapable of providing explanations in terms of “mechanisms, powers, capacities, or causes”. These exist in a “deeper reality” beyond the reach of everyday experience and science. Scientific observation and induction is merely concerned with the measurable aspects of reality.
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Then a humble econometrician addressed the problem. He carefully observed and studied the philosopher and his daughter doing “combinatorics” both separately and together. He formulated probabilistic hypotheses. Given the paucity and poor quality of the data presented in this post he concluded that there is a 99% probability that ln(x) = 2.5 +/- 2.5, ie
2 hours > x > 1 minute.

• I apologize for leaving your cleverly constructed scenario out of my discussion. It simply did not fit in the narrow construct.
It was hilarious, nonetheless. (Barbed Bayesian)
— John

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