Banner Logo
Home
The Real Kato
About Me
Twitter
Facebook
Frozen Lunches
Links
Dooce
Sweat Pants Mom
Secret Agent Josephine
Vindauga
Contact



Archives
Most Recent

2018 September
2007 September
2007 August
2007 July
2007 June
2007 May
2007 April
2007 March
2007 February
2007 January
2006 December
2006 November
2006 October


Categories
All Categories 

bloggers 
books 
commentary 
dating 
food 
funnyhaha 
interesting 
life 
movies 
music 
politics 
reviews 
science 
site-business 
sports 
style 
techwatch 
television 
theater 
travel 


Recent Comments
On New CBS Show Scorpion Riddled with Errors
e.* said:
also, 7a: disk-based backup targets don't work that way. you don't back up anything to one single so...
On New CBS Show Scorpion Riddled with Errors
Stephen J* said:
Wasn't planning on watching, now definitely won't.

19a: if the whole Ethernet cord to sportscar th...
On College Football 2013: Week 10 Preview
Ken said:
Update: Bryan Underwood is out with an injury. Umm, that's not good.
On It's Just (a) Lunch (Scam)?
kiki* said:
Run! Do not join this service! It is a waste of money. I was scheduled for several different dates. ...
On It's Just (a) Lunch (Scam)?
Informed* said:
Guys and women,

Amost everybody seems to be scammed by this company. Why then are they in business ...


<< Previous: College Football Rec... | Next: Amy's Birthday >>

Critical Thinking
Tuesday, 2007 September 18 - 11:41 pm
How would you do on a Google interview? Or... as President?

There's a widely-linked post making the rounds right now (I found it via Kottke). It lists interview questions that have been asked to Google job applicants. Google is well-known for screening job applicants tightly, using difficult interview questions.

The interesting thing about the post is the comments, where the vast majority of people are demonstrating exactly what Google doesn't want in their employees. See, most of the questions have answers that are both obvious and wrong. If you latch onto the first answer you think of because you haven't bothered to think things through carefully, then you're not the sort of person that Google wants.

I wanted to talk about four of the more interesting questions. Read the questions first and think about how you'd respond, then scroll down for a little analysis.

-----

1. How many golf balls can fit in a school bus?

2. You are shrunk to the height of a nickel and your mass is proportionally reduced so as to maintain your original density. You are then thrown into an empty glass blender. The blades will start moving in 60 seconds. What do you do?

10. In a country in which people only want boys, every family continues to have children until they have a boy. if they have a girl, they have another child. if they have a boy, they stop. what is the proportion of boys to girls in the country?

17. You have five pirates, ranked from 5 to 1 in descending order. The top pirate has the right to propose how 100 gold coins should be divided among them. But the others get to vote on his plan, and if fewer than half agree with him, he gets killed. How should he allocate the gold in order to maximize his share but live to enjoy it? (Hint: One pirate ends up with 98 percent of the gold.)

What do you think? How would you answer these questions during a job interview? Read on for some thoughts.

-----

1. How many golf balls can fit in a school bus?

An average non-mathematical person would say, "I dunno, a million?"

A narrow-thinking engineer answer would start calculating the volume of a golf ball (a sphere 1.68 inches in diameter), the approximate volume of the interior of a school bus (35 feet long, 8 feet wide, 6 feet high), and make a rough estimate of somewhere around 600,000.

An engineer with a little more savvy would ask, "How are the golf balls packed?" "What kind of school bus is it? Does it go to the 'special' school?" Or maybe, "How many students are already on the bus, and how many of them are carrying French horns?" (Disclaimer: I had to carry a French horn to school on the bus during junior high school, and nothing makes you a school bus don't-sit-next-to-me pariah like a French horn.)

The smart-aleck answer to this is, "all of them, if you carry them on one at a time". I actually like that answer; it points out the ambiguity in the question.

The point of this question is this: there are multiple interpretations to the question, and there are many parameters that have not been explicitly spelled out. If you answer the question quickly, without getting all the facts, you are not thinking critically.

2. You are shrunk to the height of a nickel and your mass is proportionally reduced so as to maintain your original density. You are then thrown into an empty glass blender. The blades will start moving in 60 seconds. What do you do?

This was the question that appeared on Kottke. The most common answer was to either duck under the blades, or try to ride out the wind created by the blades. But those are the answers that don't demonstrate a lot of thought. I especially don't like the "wind" answer, because the fact is, blender blades do not blow air like a fan would, and even if they did, they'd be more likely to create swirling vortices instead of a steady upward draft. Here's an experiment for you: put a wad of paper in a blender, and see if the blender gently blows it out when you turn the power on.

If you suppose that it were somehow physically possible to shrink yourself to that size without diminishing your normal human capabilities, then you should consider that your strength (a function of the area of your cross-section, proportional to the square of your height) has been reduced about 4900 times, but your mass (a function of your volume, proportional to the cube of your height, given that your density remains the same) has been reduced 343000 times. If you currently have a 24-inch vertical leap, you'd still have a 24-inch vertical leap regardless of how small you became. So given those facts, you could simply jump out of the blender.

Another interesting thing to note: you'd actually have what seems to be 70 minutes to come up with the solution, because time would travel 70 times slower for you at that size. The speed of light and the speed of electrons is constant, but when you're tiny, photons and electrons have less distance to travel.

If you were clever, you might ask how, exactly, did someone manage to shrink you to the height of a nickel? Are your cells the same size, but you just have fewer of them? In that case, you probably wouldn't be very smart, as you'd only have 1/343000 as many brain cells as you have now. Do you have the same number of cells, but fewer molecules per cell? In that case, the cells probably wouldn't function correctly: DNA strands would be shorter, and cell wall integrity would be compromised.

So really, the clever answer is, "Unless you change the laws of physics... you'd die." (But it's only clever if you talk about molecules and cells, and not so clever if you just think the rotating blades would chop you to death.)

10. In a country in which people only want boys, every family continues to have children until they have a boy. if they have a girl, they have another child. if they have a boy, they stop. what is the proportion of boys to girls in the country?

There are two different poorly-thought-out answers: "There'd be more boys because every family has to have a boy"; or, "There'd be more girls because families might have lots of girls but only at most one boy."

A mathematically-inclined person might work out the probabilities: There's a 1-in-2 chance that a family would have exactly one boy, and stop. There's a 1-in-4 chance that a family would have one girl, then one boy. There's a 1-in-8 chance that a family will have two girls, then one boy. If you continue down this line, then you realize that the expected number of boys in a family is the sum of the infinite series (1/2 + 1/4 + 1/8....), which is 1. The expected number of girls is (0 + 1/4 + 2/8 + 3/16 + 4/32...) which is also 1. So the number of boys and girls is, non-intuitively, exactly the same. You could even note that the average family would have two children.

A clever person would note that it doesn't matter how many kids a family has: the odds of having a boy or girl are always the same. It doesn't matter when the parents stop having kids. Even if all parents stopped having kids after they've had 2 boys and 5 girls, and the answer would still be that there are an equal number of boys and girls. This is an example of how logic can produce an answer more simply than math... but that "logic" is not necessarily a matter of just jumping at the first thought that comes to mind.

Some people might note that all those probability calculations are based on the assumption that the odds of having a boy and a girl are equal. In reality, as far as humans go, there's a slightly higher chance of having a boy baby than a girl baby, by about 51% to 49%. You could also throw out factors like the mortality rates of men versus women in the country, and the odds that a couple might be more likely to abort a pregnancy if they knew they were having a girl. But then again, if you nit-picked the question to that degree, you just might be considered a geek.

17. You have five pirates, ranked from 5 to 1 in descending order. The top pirate has the right to propose how 100 gold coins should be divided among them. But the others get to vote on his plan, and if fewer than half agree with him, he gets killed. How should he allocate the gold in order to maximize his share but live to enjoy it? (Hint: One pirate ends up with 98 percent of the gold.)

Oh, so many people get this question wrong. This is an interesting one. The poorly-thought-out answer says that Pirate 5 has to offer some sort of bribe to any two other pirates in order to get them to go along with his plan. But that's not really a complete answer.

Part of the problem people have is that they don't get a full explanation of the parameters of the problem. Now, you could form an answer that's based on that ambiguity, but the real challenge is to understand the game theory and inductive reasoning that can lead you to a truly meaningful answer. There's a complex thought process involved, and unless you've seen this kind of question before, there's a very high chance you won't know how to figure it out.

So let's make some clarifying assumptions.

First, assume that if the Pirate 5 (the top pirate) is voted down and gets killed, then the remaining pirates retain their rankings and continue the game, with Pirate 4 now in charge. If Pirate 4 is killed, then Pirate 3 is in charge, and so on.

Second, assume that any vote includes the person who proposed the plan (the top pirate), and a tie vote is enough to carry the plan.

Third, assume that all pirates are acting rationally to maximize the amount of gold they receive, and are not motivated by emotion or vindictiveness.

Fourth, assume that pirates are ruthless and cannot be trusted to collaborate with each other.

Okay. So consider if there were only two pirates involved. In that case, the top-ranked pirate (Pirate 2) would get 100 gold coins, and Pirate 1 wouldn't get any, because Pirate 1 has no way to vote down the plan.

Let's work backwards from there. Clearly, Pirate 1 doesn't want it to come down to just two people. So consider if there were three pirates. Then, Pirate 1 would accept even just one coin to avoid letting things get down to two people. Thus, if there were three pirates involved, Pirate 3 should offer Pirate 1 a single coin; Pirate 1 would agree because one coin is better than zero. Pirate 2 could not overrule this plan.

Now if there were four pirates, Pirate 2 would accept even just one coin to avoid letting things get down to three pirates; if it got down to three, as we have shown above, Pirate 2 would get nothing. Thus, Pirate 4 could offer to keep 99 coins and pay Pirate 2 a single coin; Pirate 1 and Pirate 3 would get nothing, but they don't have enough votes to override the plan.

If there were five pirates, then Pirate 1 and Pirate 3 are anxious to avoid letting things get down to four pirates, because then they'd get nothing. So if Pirate 5 offered Pirate 1 and Pirate 3 a single coin each, then they'd vote in favor of the plan, since one coin would be better than nothing. So this is how Pirate 5 keeps 98 coins for himself.

You can actually keep this line of reasoning going: Pirate 6 would have to offer a coin to Pirates 2 and 4 to satisfy everyone's self-interests. Pirate 7 would have to offer a coin to Pirates 1, 3, and 5.

There are actually many, many interesting things about this question. You might consider that if two or more low-ranking pirates decided to collaborate and act towards their mutual benefit rather than their individual benefit, then they could greatly alter the outcome of the game. You might also consider that if the top-ranking pirate is cognizant of game theory himself, he might try to fool the other pirates (via something like a dollar auction) to take advantage of them.

-----

The last problem is particularly interesting to me. Besides its illustration about critical thinking, the problem shows how a group might arrive at a non-optimal solution to a problem when everyone unilaterally acts in their own self interests. (This solution is called the Nash equilibrium, which you might remember from the movie "A Beautiful Mind", starring Russell Crowe. And by the way, Russell Crowe is a big Michigan football fan. Ahh, see how everything ties together?)

Acting against one's own unilateral self-interests might not seem like right thing to do, if you look at things in a shallow way, without thinking critically... but as I'll discuss in an upcoming article, that's exactly what we have to do when it comes to setting governmental policy.

Maybe Google should pick our next President.
Permalink  9 Comment   Bookmark and Share
Posted by Ken in: interesting

Comments

Comment #1 from Brett (Guest)
2007 Sep 19 - 11:40 pm : #
I used to ask people how many toilets there are in the U.S. and why they think that, when I interviewed people. And yes, one person actually answered "1 million...just...because"...they were not hired.
Comment #2 from John C (Guest)
2007 Sep 20 - 9:21 pm : #
I am part of the problem, and not the solution. I am far too lazy to spend time thinking about how pirates distribute teasure.
Comment #3 from dazy po (Guest)
2007 Sep 21 - 3:00 pm : #
The phrasing of question 17 does not include the top pirate in the sum of voting pirates: "...the others get to vote on his plan, and if fewer than half agree with him...".

Therefore, the last coordinated clause in this statement is untrue: "...Pirate 4 could offer to keep 99 coins and pay Pirate 2 a single coin; Pirate 1 and Pirate 3 would get nothing, but they don't have enough votes to override the plan."
Comment #4 from Ken (realkato)
2007 Sep 21 - 3:43 pm : #
Very good point, dazy po (if that IS your real name).

Actually the phrasing is ambiguous... it says "fewer than half" but that could mean either "fewer than half of the others" or "fewer than half of all the pirates". I agree that the wording makes it sound very much like "fewer than half of the others"... but I worked out that solution and ran into a few problems.

If you think of the problem that way, Pirate 2 is in the worst position; if it gets down to two pirates remaining, then Pirate 1 pirate will always vote to kill him and take all the gold for himself.

So if Pirate 3, as top pirate, offers a plan that involves keeping all the gold himself, Pirate 2 would vote in favor of that plan, as it would allow him to avoid death. Pirate 1 would get nothing either, but can't override Pirate 2's vote.

Pirate 4 could offer Pirate 2 a single coin, and Pirate 1 a single coin, to secure their votes.

Pirate 5 would need to offer Pirate 3 a single coin, and also offer *two* coins to either Pirate 2 or 1, leaving him with 97 coins. Or would he? It's conceivable that Pirate 2 would be comfortable with a single coin, since there are no scenarios that allow him to get any more than that. And the question explicitly states that one pirate would get to keep 98 coins for himself, not 97.

It's that last bit of ambiguity that made me question this interpretation of the problem. It led me to believe that the transcription of this question wasn't accurate, and that's why I put in my second clarifying assumption.

And anyway, there are really two points that are more important than the actual solution: first, that the answer to this question can be derived without assuming that the pirates make completely unjustified decisions; and second, that when the pirates act entirely in their own self-interests, then the best interests of the collective group are not necessarily well-served, even when there are democratic processes involved.
Comment #5 from Chip (Guest)
2007 Sep 25 - 8:31 am : #
How awesome! Thanks for sharing.

The blender question got me. I thought about trying to stay alive by trying to move toward the bottom center (under the blades) and trying to attach myself. Although I would be sick because of the movement, I felt I had the best chance.

I really did not consider "jumping" out of the blender. Hmmmm...mmmm.

I was going down the right path with the other questions.
Comment #6 from Anthony Borelli (Guest)
2007 Sep 25 - 12:34 pm : #
First of all, the equation clearly states the "others" get to vote on the plan. The word "others" is exclusionary (this is not ambiguios langauge), so the top pirate isn't voting. I understand why you assumed he does vote, because following the rest of the line of your reasoning, you can only come to the hinted 98% if the top pirate did have a vote, but you have made a second error as well. LIVES are at stake.

I think your approach works only if the top pirate just gets cut out of the money when voted down. When you consider that his life is at stake, though, it changes the practical consideration. I'm 100% certain that pirates 1 through 3 (whose lives, I will show you later will never be at stake) cannot be bought for just 1 gold coin, when they know the top pirate's life is in their hands. They clearly have the real leverage and would know it.

So let's work through it again, remembering lives are at stake and that the top pirate doesn't get a vote.

Let's assume;

1. A pirates LIFE is his PRIMARY concern (can't spend gold if your dead).

2. We must also assume that pirates are vindictive (since they are) and self serving in about equal parts. So if a pirates life is not at stake, getting his vote will require a fair distribution of the gold considering the relative balance of power in a given circumstance. If treated unfairly (in relation to the hand dealt them) a pirate will turn down an incremental self interest (1 coin) in favor of their vindictive nature. They might accept 1 coin, however, if it is slightly better than they would get otherwise to save their life, and if others in similar circumstance are not getting more.

If it came down to Pirates Two and One, the top ranked pirate (who cannot vote) would die for certain. Pirate One has the only vote, and can kill Pirate Two for the treasure if he chooses. Even if Pirate Two gave him the whole treasure, since there is still no self interest to balance a pirate's natural vindictiveness, Pirate One will definitely kill Pirate 2. So Pirate Two definitely doesn't want this to happen.

If it came down to Pirate Three at the top, Pirate Two will vote for him to save his own life, so Pirate Three can take it all, as he has Pirate Two's vote guaranteed. Clearly there is no chance of going lower than Pirate Three, if it got that far. Since Pirate Two MUST Guarantee Pirate Three's safety by voting for him, we can see that Pirates One through Three will never really have their lives at stake.

If it came down to Pirate 4, he cannot possibly get Pirate Three's vote (since Pirate Three automatically gets 100 coins if the top spot comes to him), and his ONLY hope of saving his own life is insuring both Pirates One and Two vote for him. They will clearly demand 50 coins apiece and Pirate Four cannot be 100% sure of saving his own life unless he gives them 50 each. (I'm not saying he definitely would, only that this is probable, and that this is the scenario everyone must consider as the most likely next step when Pirate Five makes his choice and the others vote on it).

So Pirate Five, as a practical matter, cannot buy votes from Pirates One and Two and still have money left for himself, as they can expect as much or more from Pirate Four. He must buy Pirate Three and Four's Votes. Pirate Three's life will never be in danger, but Pirate Four's life will be (to some degree) if he seeks any coins for himself once he is Top Pirate. So, unlike Pirates One, Two, and Three who could turn down an offer of just one coin without risk to their lives, Pirate Four must be inclined to take an offer of a single coin, as he would have to risk his own life to get more than zero if he becomes Top Pirate.

Pirate Three, however, will never have his life at risk (Pirate Two must vote for Three's plan to save his own life, remember), as a result Pirate Three can demand maximum return. Pirate Three gets nothing if the top spot passes to Pirate Four, so he doesn't want that to happen, but since that is balanced against Pirate Five's actual LIFE, Pirate three will definitely require at least more coins than Pirate Five or become vindictive, but a 49/50 split between Five and Three would leave Pirate Four in a vindictive mood.

Pirate Four recognizes Pirate Three has no risk to his life, and is therefore in an enormously better position, and Pirate Four can therefore accept that Pirate Three will get much more than him. But Pirate Five, whose life is still at risk, is in no better position than Pirate Four, and so Pirate Five cannot take more than 1 coin for himself without tipping Pirate Four's vindictive vs. self-interest scales.

So the split is;

Pirate Five: 1
Pirate Four: 1
Pirate Three: 98

That's how I see it, anyway. (actually, it isn't. I think he could PROBABLY offer Pirate Four as much as 24 - maybe 25 - and thereby raise his own share without risking tipping Pirate Three's vindictive vs. self-interest scales, but this is the safest scenario for Pirate Five and its as close as I can get without the Top Pirate voting and the language clearly precludes that - though perhaps it precluded it by accident).

How big a nerd am I?
Comment #7 from Ken (realkato)
2007 Sep 26 - 12:04 am : #
I think that assuming pirates are vindictive and self serving in "about equal parts" puts an unacceptable amount of vagueness into the problem. How does one determine how many coins vindictiveness is worth, in your "vindictive vs. self-interest scale"? Consider your analysis of the four-pirate scenario. Why would Pirate One vote in favor of a 50/50 split with Pirate Two, when Pirate Two's only alternatives are zero coins (in the three-pirate scenario) or death (in the two-pirate scenario)? Maybe Pirate One would demand 75 coins. Or 100. Is there any plan that Pirate Four can really offer that would absolutely ensure his own survival? Under your rules, I don't think so. Pirate Three will always vote against him, and Pirate One can vote against him out of spite, no matter how many coins he's offered.

If you take that line further, then Pirate Four would have to vote for any plan that Pirate Five proposed; that'd be the only way to ensure his own survival. But how would Pirate 5 gain anyone else's vote? If the other pirates are vindictive enough, they'd vote against him anyway. So Pirate Five is at a loss, and there is no solution to the problem. Somehow, that's unsatisfying to me.

I believe the most reasonable assumption is that the transcription of the problem, as originally provided on the web site, is incorrect and incomplete. I believe that the problem was supposed to have been worded such that the top pirate DOES get a vote, and that the accumulation of coins is the sole motivation among the pirates. Or perhaps, the idea was to ask the questioner to clarify all these points before providing an answer.
Comment #8 from A-Kato (Guest)
2007 Sep 26 - 12:13 am : #
pirates distribute treasure arrrrbitrarily
Comment #9 from Anthony Borelli (Guest)
2007 Sep 26 - 4:26 pm : #
Ken,

As I said, I don't really see it that way, that's just the closeest I can get to the hinted answer given the question as it is presented.

I completely agree with your last paragraph.

I also like A-Kato's answer.

Tony

Comments are closed for this post.
Login


Search This Site
Powered by FreeFind