4 ms·

Seeing the headline, and remembering that 50% of employees were laid off, my first thought was: "what're the odds?" I assume I'd have to use NCR or something t

Seeing the headline, and remembering that 50% of employees were laid off, my first thought was: "what're the odds?" I assume I'd have to use NCR or something to calculate that. I know the upper (rarest) bound is 1/2^40, but it's likely much more likely, and not random.

thaumasiotes 3y ago
You want the binomial distribution. ( https://support.microsoft.com/en-gb/office/binom-dist-function-c5ae37b6-f39c-4be2-94c2-509a1480770c https://support.microsoft.com/en-gb/office/binom-dist-functi... , https://help.libreoffice.org/6.2/en-US/text/scalc/01/04060181.html https://help.libreoffice.org/6.2/en-US/text/scalc/01/0406018... if you want to use Excel; https://help.libreoffice.org/6.2/en-US/text/scalc/01/04060181.html https://help.libreoffice.org/6.2/en-US/text/scalc/01/0406018... ("B") for LibreOffice Calc.) 1/2^40 doesn't come into it. There are 67 relevant employees. Given that 50% of all employees were laid off, what were the odds of at least 40 employees out of 67 randomly-selected employees getting laid off? =B(67;0.5;40;67) > 0.07103569 [This is the LibreOffice formula: "in 67 trials which each have a 0.5 chance of success, what is the probability of seeing between 40 and 67 successes?"] One-tailed p-value of 0.07, two-tailed p-value of 0.14. So the odds are about 1 in 7, or 1 in 14. You might want to ask yourself "Do I really believe that, if I thought I might be about to get fired, that would have no influence on whether I tried to join a union?" [Answering this question in the negative means you can justify using the one-tailed p-value, but it also means the odds of getting fired should be substantially above 50%.] Note also that this number shifts substantially as you move the odds of getting laid off away from 50%. At 53%, you have a one-tailed value of 16.44%. At 47%, it's 2.5%.
- freeAgent 3y ago
  There’s an implicit assumption that the layoffs were evenly distributed and that the union bargaining team was also evenly distributed within the organization. If those aren’t true, this analysis is going to give deceiving results (could be either direction).
  - mjburgess 3y ago
    And also that the union employees were identical to others -- which is kinda demonstrably false. People who self-select as union leaders are likely to be different from the other employees in various relevant ways. Modelling any of these assumptions yields an arbitrary probability of your own choosing.
  - vanderZwan 3y ago
    Isn't this a statistical catch-22? If being in a union increases the odds of getting fired… well, then there's nothing to discuss because that's illegal. And if it should decrease it because of fear of legal retaliation, wouldn't that make it less likely that this happened randomly. This feels like the wrong conclusion to draw (especially the second part), but I suck at statistics so someone please explain where I'm off.
    - eru 3y ago
      > Isn't this a statistical catch-22? If being in a union increases the odds of getting fired… well, then there's nothing to discuss because that's illegal. Extreme example: suppose a company has 100 employees. Four of them are in the union bargaining team and they just happen to all work for the widgets department because of an accident of history or randomness etc. The widgets department has 10 people overall. Now the company decides for business reasons entirely unrelated to unions, that thanks to GPT they don't need so many people in the widgets department. So they fire 8 people from widgets. Assume further that they went out of their way to protect the union bargainers: of those 8 people fired, only 2 are in the union bargaining team. (That means that the remaining 2 widgets people are both union bargainers.) Even though the company went out of their way to protect the union bargainers, statistically they only laid off 8% of the company, but 50% of union bargainers. Would that be illegal or even unfair? (In the real world, I am fairly sure that union bargainers are concentrated in some departments. Mostly because it attracts certain kinds of people, and different departments also attract certain kinds of people. But, of course, I have no clue what the reality inside of Bandcamp was like. I'm just speculating about hypothetical examples.)
      - 3y ago
- vitus 3y ago
  You've computed the odds that at least 40 of the 67 union members get laid off, bargaining team member or not. You actually want the odds of a specific set of 40 individuals getting laid off. It looks like Bandcamp had 118 employees and 60 layoffs per various news outlets (and on the wikipedia page [0]). Assuming each employee was equally likely to get laid off given the target retention [1], you could compute this from the binomial distribution using "number of ways to predetermine those 40 are getting laid off, plus 20 from the remainder, divided by number of ways to select 60 individuals out of 118". Or, more concisely, (78 choose 20) / (118 choose 60), which is actually closer to 10^-16 (2 orders of magnitude more unlikely than the simple approximation of 2^-40, which assumes an infinitely large pool). [0] https://en.wikipedia.org/wiki/Bandcamp https://en.wikipedia.org/wiki/Bandcamp : "Songtradr stated that only 60 of Bandcamp's previously 118 workers had been offered a contract." [1] This is by no means guaranteed.
  - thaumasiotes 3y ago
    > You've computed the odds that at least 40 of the 67 union members get laid off, bargaining team member or not. Well, close. I've computed the odds that at least 40 of a group of 67 special people get laid off. There is another comment in this thread suggesting that 67 is the number of bargaining team members. I don't know, because I haven't read the article. So my calculation might or might not be right, depending on whether you interpreted the article correctly or the handful of people taking the other position did. My interpretation was that 40 out of 67 union members got fired, which appears to have been wrong. However... > You actually want the odds of a specific set of 40 individuals getting laid off. I definitely don't want that. You never want to make a comparison against a specific outcome when you ask "what are the odds?" like this. All specific outcomes are rare, so that question will never tell you anything informative. (I almost wrote "will never tell you anything useful", but if what you're looking for is a scapegoat, you might find the calculation you propose useful for that. It's not useful for anything else, and frankly it's a disgrace that you suggested it.) edit: This is what the article says: > “On Monday, October 16, 2023 over half of Bandcamp was laid off as a result of Epic Games’ divestiture to Songtradr,” Bandcamp United said in a statement. “Of those laid off, 40 were in the union bargaining unit out of a total 67 members. None of the eight (8) democratically elected bargaining team members received a job offer.” So it looks like there are these groups: - bandcamp employees (number unknown) - union members (number unknown) - union members on the bargaining team (67) - union members on the bargaining team who got laid off (40) - union members elected to the bargaining team (8) - union members elected to the bargaining team, who got laid off (8) This suggests that the calculation I gave was the same one that was sought, and also that the base rate, as far as we believe in it, was over 50%. Remember that shifting the odds of being fired from 50% to 53% more than doubled the odds of seeing the pattern we did see.
bananaquant 3y ago
As an HR would say, those people are "a bad cultural fit", which legitimately raises the probability of them being laid off to 100%. Simple.
- hkt 3y ago
  "legitimately"? I'd be surprised..
  - deleted 3y ago
    [deleted]
  - banannaise 3y ago
    That's the argument HR would make, as a means of laundering the actual reason.
- pastacacioepepe 3y ago
  You might need an /s, this is a perfectly plausible comment on HN.
  - taejavu 3y ago
    I believe it’s a sincere comment actually.
    - jacquesm 3y ago
      The fact that you're not 100% sure is what worries me.
- dathinab 3y ago
  still not legitimate, because then any laying of on union members/leaders could be legitimized that way in any country which properly enforces labor protection law laying of any union leaders without the agreement of the union is extremely hard and requires missteps of the members like e.g. stealing. Or really unusual situations like you lay of half of the members and over half of the members have young children (or e.g. are disabled, project leaders etc.) but non of the union members have any of that. The likelihood of which is basically 0 in practice.
hatsix 3y ago
It clearly wasn't random, in fact, they said as much. It's a cost-cutting effort, and US salaries are generally much higher than non-US salaries. If you want to know the odds, you'd have to know their criteria. You can't assume that it was random, because it definitely wasn't. (Though, I've seen the aftermath of a random layoff, I'd take criteria'd layoff every day, even if I'm the one with the pink slip.)

thaumasiotes 3y ago

You want the binomial distribution. ( https://support.microsoft.com/en-gb/office/binom-dist-function-c5ae37b6-f39c-4be2-94c2-509a1480770c https://support.microsoft.com/en-gb/office/binom-dist-functi... , https://help.libreoffice.org/6.2/en-US/text/scalc/01/04060181.html https://help.libreoffice.org/6.2/en-US/text/scalc/01/0406018... if you want to use Excel; https://help.libreoffice.org/6.2/en-US/text/scalc/01/04060181.html https://help.libreoffice.org/6.2/en-US/text/scalc/01/0406018... ("B") for LibreOffice Calc.) 1/2^40 doesn't come into it. There are 67 relevant employees. Given that 50% of all employees were laid off, what were the odds of at least 40 employees out of 67 randomly-selected employees getting laid off? =B(67;0.5;40;67) > 0.07103569 [This is the LibreOffice formula: "in 67 trials which each have a 0.5 chance of success, what is the probability of seeing between 40 and 67 successes?"] One-tailed p-value of 0.07, two-tailed p-value of 0.14. So the odds are about 1 in 7, or 1 in 14. You might want to ask yourself "Do I really believe that, if I thought I might be about to get fired, that would have no influence on whether I tried to join a union?" [Answering this question in the negative means you can justify using the one-tailed p-value, but it also means the odds of getting fired should be substantially above 50%.] Note also that this number shifts substantially as you move the odds of getting laid off away from 50%. At 53%, you have a one-tailed value of 16.44%. At 47%, it's 2.5%.

freeAgent 3y ago
There’s an implicit assumption that the layoffs were evenly distributed and that the union bargaining team was also evenly distributed within the organization. If those aren’t true, this analysis is going to give deceiving results (could be either direction).
- mjburgess 3y ago
  And also that the union employees were identical to others -- which is kinda demonstrably false. People who self-select as union leaders are likely to be different from the other employees in various relevant ways. Modelling any of these assumptions yields an arbitrary probability of your own choosing.
- vanderZwan 3y ago
  Isn't this a statistical catch-22? If being in a union increases the odds of getting fired… well, then there's nothing to discuss because that's illegal. And if it should decrease it because of fear of legal retaliation, wouldn't that make it less likely that this happened randomly. This feels like the wrong conclusion to draw (especially the second part), but I suck at statistics so someone please explain where I'm off.
  - eru 3y ago
    > Isn't this a statistical catch-22? If being in a union increases the odds of getting fired… well, then there's nothing to discuss because that's illegal. Extreme example: suppose a company has 100 employees. Four of them are in the union bargaining team and they just happen to all work for the widgets department because of an accident of history or randomness etc. The widgets department has 10 people overall. Now the company decides for business reasons entirely unrelated to unions, that thanks to GPT they don't need so many people in the widgets department. So they fire 8 people from widgets. Assume further that they went out of their way to protect the union bargainers: of those 8 people fired, only 2 are in the union bargaining team. (That means that the remaining 2 widgets people are both union bargainers.) Even though the company went out of their way to protect the union bargainers, statistically they only laid off 8% of the company, but 50% of union bargainers. Would that be illegal or even unfair? (In the real world, I am fairly sure that union bargainers are concentrated in some departments. Mostly because it attracts certain kinds of people, and different departments also attract certain kinds of people. But, of course, I have no clue what the reality inside of Bandcamp was like. I'm just speculating about hypothetical examples.)
    - 3y ago
vitus 3y ago
You've computed the odds that at least 40 of the 67 union members get laid off, bargaining team member or not. You actually want the odds of a specific set of 40 individuals getting laid off. It looks like Bandcamp had 118 employees and 60 layoffs per various news outlets (and on the wikipedia page [0]). Assuming each employee was equally likely to get laid off given the target retention [1], you could compute this from the binomial distribution using "number of ways to predetermine those 40 are getting laid off, plus 20 from the remainder, divided by number of ways to select 60 individuals out of 118". Or, more concisely, (78 choose 20) / (118 choose 60), which is actually closer to 10^-16 (2 orders of magnitude more unlikely than the simple approximation of 2^-40, which assumes an infinitely large pool). [0] https://en.wikipedia.org/wiki/Bandcamp https://en.wikipedia.org/wiki/Bandcamp : "Songtradr stated that only 60 of Bandcamp's previously 118 workers had been offered a contract." [1] This is by no means guaranteed.
- thaumasiotes 3y ago
  > You've computed the odds that at least 40 of the 67 union members get laid off, bargaining team member or not. Well, close. I've computed the odds that at least 40 of a group of 67 special people get laid off. There is another comment in this thread suggesting that 67 is the number of bargaining team members. I don't know, because I haven't read the article. So my calculation might or might not be right, depending on whether you interpreted the article correctly or the handful of people taking the other position did. My interpretation was that 40 out of 67 union members got fired, which appears to have been wrong. However... > You actually want the odds of a specific set of 40 individuals getting laid off. I definitely don't want that. You never want to make a comparison against a specific outcome when you ask "what are the odds?" like this. All specific outcomes are rare, so that question will never tell you anything informative. (I almost wrote "will never tell you anything useful", but if what you're looking for is a scapegoat, you might find the calculation you propose useful for that. It's not useful for anything else, and frankly it's a disgrace that you suggested it.) edit: This is what the article says: > “On Monday, October 16, 2023 over half of Bandcamp was laid off as a result of Epic Games’ divestiture to Songtradr,” Bandcamp United said in a statement. “Of those laid off, 40 were in the union bargaining unit out of a total 67 members. None of the eight (8) democratically elected bargaining team members received a job offer.” So it looks like there are these groups: - bandcamp employees (number unknown) - union members (number unknown) - union members on the bargaining team (67) - union members on the bargaining team who got laid off (40) - union members elected to the bargaining team (8) - union members elected to the bargaining team, who got laid off (8) This suggests that the calculation I gave was the same one that was sought, and also that the base rate, as far as we believe in it, was over 50%. Remember that shifting the odds of being fired from 50% to 53% more than doubled the odds of seeing the pattern we did see.

bananaquant 3y ago

As an HR would say, those people are "a bad cultural fit", which legitimately raises the probability of them being laid off to 100%. Simple.

hkt 3y ago
"legitimately"? I'd be surprised..
- deleted 3y ago
  [deleted]
- banannaise 3y ago
  That's the argument HR would make, as a means of laundering the actual reason.
pastacacioepepe 3y ago
You might need an /s, this is a perfectly plausible comment on HN.
- taejavu 3y ago
  I believe it’s a sincere comment actually.
  - jacquesm 3y ago
    The fact that you're not 100% sure is what worries me.
dathinab 3y ago
still not legitimate, because then any laying of on union members/leaders could be legitimized that way in any country which properly enforces labor protection law laying of any union leaders without the agreement of the union is extremely hard and requires missteps of the members like e.g. stealing. Or really unusual situations like you lay of half of the members and over half of the members have young children (or e.g. are disabled, project leaders etc.) but non of the union members have any of that. The likelihood of which is basically 0 in practice.

hatsix 3y ago

It clearly wasn't random, in fact, they said as much. It's a cost-cutting effort, and US salaries are generally much higher than non-US salaries. If you want to know the odds, you'd have to know their criteria. You can't assume that it was random, because it definitely wasn't. (Though, I've seen the aftermath of a random layoff, I'd take criteria'd layoff every day, even if I'm the one with the pink slip.)