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“A quick, elegant derivation of Bayes’ Theorem” by RohanS
I'm glad I know this, and maybe some people here don't, so here goes.
_P(A text{ and } B) = P(A) cdot P(B mid A)_
_P(B text{ and } A) = P(B) cdot P(A mid B)_
Order doesn't matter for joint events: "A and B" refers to the same event as "B and A". Set them equal:
_P(B) cdot P(A mid B) = P(A) cdot P(B mid A)_
Divide by _P(B)_:
_P(A mid B) = frac{P(B mid A) cdot P(A)}{P(B)}_
And you're done! I like substituting hypothesis (H) and evidence (E) to remember how this relates to real life:
_P(H mid E) = frac{P(E mid H) cdot P(H)}{P(E)}_
You might also want to expand the denominator using the law of total probability, since you're more likely to know how probable the evidence is given different hypotheses than in general:
_P(H_i mid E) = frac{P(E mid H_i) cdot P(H_i)}{sum_j P(E mid H_j) cdot P(H_j)}_
---
First published:
January 22nd, 2026
Source:
https://www.lesswrong.com/posts/GjkqijXHakMyDxF9e/a-quick-elegant-derivation-of-bayes-theorem
---
Narrated by TYPE III AUDIO.
View all episodes
By LessWrongI'm glad I know this, and maybe some people here don't, so here goes.
_P(A text{ and } B) = P(A) cdot P(B mid A)_
_P(B text{ and } A) = P(B) cdot P(A mid B)_
Order doesn't matter for joint events: "A and B" refers to the same event as "B and A". Set them equal:
_P(B) cdot P(A mid B) = P(A) cdot P(B mid A)_
Divide by _P(B)_:
_P(A mid B) = frac{P(B mid A) cdot P(A)}{P(B)}_
And you're done! I like substituting hypothesis (H) and evidence (E) to remember how this relates to real life:
_P(H mid E) = frac{P(E mid H) cdot P(H)}{P(E)}_
You might also want to expand the denominator using the law of total probability, since you're more likely to know how probable the evidence is given different hypotheses than in general:
_P(H_i mid E) = frac{P(E mid H_i) cdot P(H_i)}{sum_j P(E mid H_j) cdot P(H_j)}_
---
First published:
January 22nd, 2026
Source:
https://www.lesswrong.com/posts/GjkqijXHakMyDxF9e/a-quick-elegant-derivation-of-bayes-theorem
---
Narrated by TYPE III AUDIO.
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