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Yeah, adding a button to a codebase is an awful test. If someone made it to an onsite and you don’t know whether they can add a button in swift, something fail

Yeah, adding a button to a codebase is an awful test. If someone made it to an onsite and you don’t know whether they can add a button in swift, something failed in your screening process. If you’re testing how someone navigates a codebase, you can just look at it with them, and let them drive the chat. If you’re testing an engineer for a serious job, do an algorithms test. If you’re testing an engineer for a specific thing, test that specific thing. Both of those should be handled in the screen, not the onsite. IMHO, the onsite is about seeing how people think. Whether you can jam a button into a repo doesn’t tell me whether you can think or not. I guess it tells me whether you get flustered, but it’s pretty unfair to design things that are impossible just to see if people break. That candidate turned out to be awesome but I remember the interviewer telling me after they interviewed them “well, they couldn’t add the button, but as they were doing it, I realized I wouldn’t be able to add the button either so my interview was inconclusive”, and I replied “well it sounds like you need to design a better interview question”. The worst part was, the interviewer spent the first 20 minutes of the interview talking with the candidate before giving them 25 minutes to add the button!! There’s a question we gave candidates really early on which we no longer do because it biases towards math nerds that I absolutely loved. It’s based on a movie called 13 Tzameti. I’m probably screwing it up because it isn’t my question but it’s basically like this: You’re in a dark room after being abducted by a gang. The lights come on and there’s 12 other people in the room in a circle and everyone has a revolver with 1 bullet in it (6 slots in the chamber). Your instructions are to spin the chamber on the revolver and, when the lights go out, shoot the person to your right in order. What are the odds you make it to the next round? It’s a crazy question and, what’s even crazier is that for some reason, as you increase the number of people in the circle, the odds of making it to the next round converge on 1/e. No one has figured that out in the interview. Also no one has figured out why it converges on 1/e so if you have any ideas, let me know. I like this question because it shows you how free thinking people are. I dislike this question because it biases towards smartasses and probability nerds.

eschneider 7y ago
Your odds go way up if you repeatedly fire at the guy to your left. Being a senior engineer is knowing the difference between doing what your told and figuring out what needs to be done to achieve project success. In this case, getting to the next round. Next question?
- michaelt 7y ago
  He never said you had to spin the chamber by an unknown amount, aren't your odds even better if you make sure the bullet is in the next chamber before you shoot left? Of course, many game theory calculations assume all players know the payoff matrix and equilibrium strategies of the others; making sure the bullet isn't in the next chamber is the rational universal strategy if the game has a fixed number of rounds. Presumably the actual question has a bunch of provisos making sure you can't intentionally miss, or accidentally miss, or dodge, or shoot early, or fail to pull the trigger, or shoot the gang.....
  - eschneider 7y ago
    There are many 'optimizations' once the lights go out.
- richardhod 7y ago
  The odds might go even further up if you all get together and shoot the abductors instead
manwe150 7y ago
I'll bite, but how do you know it converges to 1/e if no one has figured out why? Your description sounds close to the wikipedia description of the first round of that movie (https://en.wikipedia.org/wiki/13_Tzameti https://en.wikipedia.org/wiki/13_Tzameti), so hopefully it's not just the wrong probability to analyze, but naively, the setup sounds like it would have slightly above 5/6 probabilty of survival for the first round. But some aspects of the question setup also almost remind me of the classic networking algorithm of slotted Aloha, which does yield an optimal utilization (packet survival) rate of 1/e in each round (for an altered question setup).
- josh2600 7y ago
  I was wrong, my engineer said that derangements are the answer to the convergence.
gjm11 7y ago
The murder question sounds quite interesting, but I'm not sure I understand it. You have n people (initially n=13) in a circle, firing shots that are fatal with probability p (here we apparently fix p=1/6, but I guess that for large n the final answer doesn't depend on p provided it's neither 0 nor 1?), and then 1 fires at 2, then 2 (if alive) or 3 (if not) fires at 3/4, and so on. And you're person 1 (this isn't stated explicitly, though...) so you're the last to have a chance to get shot at, and then you live if person n dies or if person n lives but misses. So, let's see. Suppose n is very large; then the probability that you live is approximately equal to the probability that your predecessor does; call that q. Then, as above, you live iff your predecessor dies (probability 1-q) or your predecessor lives but fails to kill you (probability q(1-p)). So q = 1-q+q(1-p) = 1-pq and 1 = 1/(1+p). That's a long way from 1/e, and a quick simulation seems to confirm this answer. It doesn't change a lot if we assume that a random person always fires first, instead of you, or if we assume that you're always last (which I think was the situation in that movie). If everyone has a revolver with n slots and one bullet, and they all fire at _you_, then you have a 1/e chance of survival for large n, but that sounds too different (and too easily found to be 1/e) to be the right thing. The probability p(n) that a permutation of n things is a derangement -- which tends to 1/e as n->oo -- satisfies the recurrence relation p(n) = [(n-1)p(n-1)+p(n-2)]/n; is it possible that the correct statement of the problem here leads to that same recurrence?
- josh2600 7y ago
  Ok, I talked to the engineer. I got it mostly right but everyone shoots to the right as soon as the lights go out. The question is "what's the probability you die?". Edit: You can also challenge people to think about the problem where everyone fires at exactly the same time OR random order since people have different reaction times. Edit 2: "The probability p(n) that a permutation of n things is a derangement -- which tends to 1/e as n->oo -- satisfies the recurrence relation p(n) = [(n-1)p(n-1)+p(n-2)]/n; is it possible that the correct statement of the problem here leads to that same recurrence?" <--- My engineer says that derangements are the correct key to the convergence.
  - gjm11 7y ago
ScottFree 7y ago
Haha, thanks for the response. Your question is worthy of being included in the nerd sniping box set. https://xkcd.com/356/ https://xkcd.com/356/

eschneider 7y ago

Your odds go way up if you repeatedly fire at the guy to your left. Being a senior engineer is knowing the difference between doing what your told and figuring out what needs to be done to achieve project success. In this case, getting to the next round. Next question?

michaelt 7y ago
He never said you had to spin the chamber by an unknown amount, aren't your odds even better if you make sure the bullet is in the next chamber before you shoot left? Of course, many game theory calculations assume all players know the payoff matrix and equilibrium strategies of the others; making sure the bullet isn't in the next chamber is the rational universal strategy if the game has a fixed number of rounds. Presumably the actual question has a bunch of provisos making sure you can't intentionally miss, or accidentally miss, or dodge, or shoot early, or fail to pull the trigger, or shoot the gang.....
- eschneider 7y ago
  There are many 'optimizations' once the lights go out.
richardhod 7y ago
The odds might go even further up if you all get together and shoot the abductors instead

manwe150 7y ago

I'll bite, but how do you know it converges to 1/e if no one has figured out why? Your description sounds close to the wikipedia description of the first round of that movie (https://en.wikipedia.org/wiki/13_Tzameti https://en.wikipedia.org/wiki/13_Tzameti), so hopefully it's not just the wrong probability to analyze, but naively, the setup sounds like it would have slightly above 5/6 probabilty of survival for the first round. But some aspects of the question setup also almost remind me of the classic networking algorithm of slotted Aloha, which does yield an optimal utilization (packet survival) rate of 1/e in each round (for an altered question setup).

josh2600 7y ago
I was wrong, my engineer said that derangements are the answer to the convergence.

gjm11 7y ago

The murder question sounds quite interesting, but I'm not sure I understand it. You have n people (initially n=13) in a circle, firing shots that are fatal with probability p (here we apparently fix p=1/6, but I guess that for large n the final answer doesn't depend on p provided it's neither 0 nor 1?), and then 1 fires at 2, then 2 (if alive) or 3 (if not) fires at 3/4, and so on. And you're person 1 (this isn't stated explicitly, though...) so you're the last to have a chance to get shot at, and then you live if person n dies or if person n lives but misses. So, let's see. Suppose n is very large; then the probability that you live is approximately equal to the probability that your predecessor does; call that q. Then, as above, you live iff your predecessor dies (probability 1-q) or your predecessor lives but fails to kill you (probability q(1-p)). So q = 1-q+q(1-p) = 1-pq and 1 = 1/(1+p). That's a long way from 1/e, and a quick simulation seems to confirm this answer. It doesn't change a lot if we assume that a random person always fires first, instead of you, or if we assume that you're always last (which I think was the situation in that movie). If everyone has a revolver with n slots and one bullet, and they all fire at _you_, then you have a 1/e chance of survival for large n, but that sounds too different (and too easily found to be 1/e) to be the right thing. The probability p(n) that a permutation of n things is a derangement -- which tends to 1/e as n->oo -- satisfies the recurrence relation p(n) = [(n-1)p(n-1)+p(n-2)]/n; is it possible that the correct statement of the problem here leads to that same recurrence?

josh2600 7y ago
Ok, I talked to the engineer. I got it mostly right but everyone shoots to the right as soon as the lights go out. The question is "what's the probability you die?". Edit: You can also challenge people to think about the problem where everyone fires at exactly the same time OR random order since people have different reaction times. Edit 2: "The probability p(n) that a permutation of n things is a derangement -- which tends to 1/e as n->oo -- satisfies the recurrence relation p(n) = [(n-1)p(n-1)+p(n-2)]/n; is it possible that the correct statement of the problem here leads to that same recurrence?" <--- My engineer says that derangements are the correct key to the convergence.
- gjm11 7y ago

ScottFree 7y ago

Haha, thanks for the response. Your question is worthy of being included in the nerd sniping box set. https://xkcd.com/356/ https://xkcd.com/356/