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Who lays off the layers-off? HR departments look after themselves.
Executives?
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you'd get 98% of the value with 10% of the time invested. Hard to say. The economics of hiring aren't intuitive - a bad hire can do a lot of damage, esp. if it's a management position (which this wasn't, I know). It's probably better to nicely turn down 10 people (and nicely ask them to re-apply in 6 months) than to let in 1 charming, but conniving and free-riding bullshitter into your team.
I dunno-- I'm a big fan of "fire fast". If you can't detect a charming, conniving, free-riding bullshitter in 1-2 weeks of working with them, how are you going to detect 'em with 10 man-hours worth of interviews?
in addition firing in a company like google is expensive: there's three months of training for a position like SRE (given they're scale at which google operates), during which they're paid a competitive salary (as usual), provided free food, flown out to mountain view (if from remote office).
i can understand why google would have the requirements they would: they want to maintain the start-up culture (everyone is working there by choice, there is a perfect culture fit) at a much bigger place and for a longer duration.
for what it's worth they're also open to letting people interview multiple times as well, for the same or different position. to this experience definitely help the original poster. he could come back within a year to interview (say) for a munich office position, where the competition wouldn't be as intense.
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That was my second thought too (my first thought was to upvote the comment), but at least in the 2-die case, the modulus takes care of it. If you work out the probabilities, the chance of getting 0mod2 = 1/36 (2) + 3/36 (4) + 5/36 (6) + 5/36 (8) + 3/36 (10) + 1/36 (12) = 18/36 = 1/2. Same goes for getting 1mod2. So you end up with a fair result even though the chances for each individual outcome are biased. I didn't…
experimental evidence says that it is in fact a uniform distribution: r = random.Random() def one_to_five(): return r.randint(1, 5) def mod_seven(): return (sum(one_to_five() for x in xrange(7)) % 7) + 1 def test_dist(lst): return [(x, lst.count(x)) for x in [1,2,3,4,5,6,7]]
Google isn't hiring anymore out of top management or very specific exceptions.
Citation needed.
Similar leaks are being posted every other day.
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I asked my recruiter specifically if they were in a hiring freeze (I'm in the middle of the application process now), and she was very adamant that they were not. If they were, they would lay off the whole HR department, except for a skeleton crew needed for benefits and such. After all, why spend ~40M/year on recruiters if you're not recruiting?
Who lays off the layers-off? HR departments look after themselves.
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Execute f 7 times, add all numbers, call that x. Then you do mod(x,7)+1, and you get a random number between 1 and 7. If the original function was unbiased, this one is going to be unbiased too.
This won't work. f gives 1 to 5. 7 * f gives 7 to 35. But 7 to 35 will not be evenly generated. Think about it: There are more ways to get a 20 than there are to get a 7 or a 35. Same thing with rolling 2 die.
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Execute f 7 times, add all numbers, call that x. Then you do mod(x,7)+1, and you get a random number between 1 and 7. If the original function was unbiased, this one is going to be unbiased too.
This won't work. f gives 1 to 5. 7 * f gives 7 to 35. But 7 to 35 will not be evenly generated. Think about it: There are more ways to get a 20 than there are to get a 7 or a 35. Same thing with rolling 2 die.
I've seen this guys articles get posted on reddit/hacker news over the course of the past few years and he just seems like a tome of knowledge that keeps expanding at a rapid pace Buddy, you have a lot of time on your hands and you're using it well, keep chugging along, looks like you're close to your dream job
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i've tried it out experimentally in google docs and it looks really good for big numbers. No real proof tho and it might be that the errors counter each other by chance: http://spreadsheets.google.com/ccc?key=pq4tB7LQWN03gF7ImGhIP...
Have a look at the distributions for each sum here: http://brianpmearns.com/ext/dice/dice.html edit: Oops bad addition on my part, could be uniform, but you'd have to work out the actual number exactly.
1 -> 11177
2 -> 11172
3 -> 11158
4 -> 11144
5 -> 11144
6 -> 11158
7 -> 11172
...not quite
The "I got interviewed by Google" articles are getting old for me. I just don't care, and their process isn't that interesting anymore.
First basic recruiter basic questions.
Second part, call from Mountain View to London. They couldn't organize a VOIP line that didn't sound underwater. They tried calling again and the line was still bad. They proceeded anyway.
I spent 5 minutes answering a basic question about deploying some changes across a thousand machines, then the interviewer mentioned he'd heard part of it, but how would I deploy it across a thousand machines?
I talked about things stored in inodes, including POSIX attributes, he asked for an example, I mentioned immutable. Google: WHAT? Me: IMMUTABLE. Google: WHAT? Me: IMMUTABLE. INDIGO MIKE MIKE UMBRELLA ...
The whole call went along those lines and was generally a massive waste of time.
Apparently they were going to organize a second interview with a working phone line. They haven't gotten back to me in the two weeks since.