Suzie’s suspicion is correct in general, though the two could work out the same in certain cases.
We know the probability P(draw red | choose random) = 1⁄20
What we need to know is P(choose random) where P(~choose random) is the prior probability of cheating. We also need to know P(draw red | ~choose random), the probability of drawing red if you cheat (presumably 1, but not necessarily—maybe it’s an unreliable cheating method). From all those, we can solve the system and compute P(chose random | drew red).
What you’re asking is whether P(chose random | drew red) = P(draw red | choose random); and in general this is not the case.
Indeed, we get to be so Bayesian we actually use Bayes’s Theorem explicitly:
P(A|B) = P(B|A) * P(A)/P(B)
Unless the priors are equal P(A) = P(B) [P(draw red) = P(choose random)] those two conditional probabilities will be distinct.
Suzie’s suspicion is correct in general, though the two could work out the same in certain cases.
We know the probability P(draw red | choose random) = 1⁄20 What we need to know is P(choose random) where P(~choose random) is the prior probability of cheating. We also need to know P(draw red | ~choose random), the probability of drawing red if you cheat (presumably 1, but not necessarily—maybe it’s an unreliable cheating method). From all those, we can solve the system and compute P(chose random | drew red).
What you’re asking is whether P(chose random | drew red) = P(draw red | choose random); and in general this is not the case.
Indeed, we get to be so Bayesian we actually use Bayes’s Theorem explicitly: P(A|B) = P(B|A) * P(A)/P(B)
Unless the priors are equal P(A) = P(B) [P(draw red) = P(choose random)] those two conditional probabilities will be distinct.