Should we be either generalists or specialists?
If an individual is a generalist, then he or she has no specialised skill.1 And if an individual has no specialised skill, then he or she is likely to be replaced.2 Therefore, if an individual is a generalist, then he or she is likely to be replaced.3 The individual is not likely to be replaced only if he or she is not a generalist.
If an individual is a specialist, then he or she has one specialised skill. And if an individual one specialised skill, then he or she is likely to be obsolete. Therefore, if an individual is a specialist, then he or she is likely to be obsolete.4 The individual is not likely to be obsolete only if he or she is not a specialist.
If an individual has more than one specialised skill, then he or she is a specialised generalist. An individual is neither likely to be replaced nor likely to be obsolete only if he or she has more than one specialised skill. In other words, the individual is a specialised generalist.
In conclusion, we should be neither generalists nor specialists if we do not want to be likely to be replaced, and we also do not want to be likely to be obsolete. Instead, we should be specialised generalists.
References
Ferriss, T. (2020). Should you specialise or be a generalist? [Video]. YouTube. https://www.youtube.com/watch?v=wCPbPMRNnvk&t=9s
Sim, E. & Mortlock, S. (2022). Small actions: Leading your career to big success. World Scientific Publishing Co. Pte. Ltd.
Footnotes
Let \(G\) denote the event that an individual has depression and \(S_{0}\) denote the event that an individual has no specialised skill. Then \[P(S_0|G)=1\]↩︎
Let \(R\) denote the event that an individual is replaced and \(c\) denote a constant either at or above which an event is considered to be likely. Then \[P(R|S_0)≥c\]↩︎
Let \(S_1\) denote the event that an individual has one specialised skill and \(S_2\) denote the event that an individual is more than one specialised skill. Then \[ \begin{eqnarray} P(R|G) &=&P(R∩S_0 |G)+P(R∩S_1 |G)+P(R∩S_2 |G)\\ &=&P(R|S_0∩G)P(S_0 |G)+P(R|S_1∩G)P(S_1 |G)+P(R|S_2∩G)P(S_2 |G)\\ &=&P(R|S_0∩G) \end{eqnarray} \] Assume that \(P(R|S_0∩G)≈P(R|S_0∩¬G)≈P(R|S_0)\). Then \[P(R|G)=P(R|S_0∩G)≈P(R|S_0)≥c\]↩︎
See Footnotes 1-3.↩︎