All of Dan_Nelson's Comments + Replies

But There's Still A Chance, Right?

Let's say you and your friend Suzie bumped into a guy on the street. This guy is holding a red marble in his right hand, and a velvet bag in his left. You and Suzie ask the man what is in the velvet bag. You realize very quickly that he doesn't speak your language. You take the velvet bag to see for yourself what is inside. It contains 19 blue marbles. In fact, it has a sticker on the outside of the bag that says "Contents: 1 red marble, 19 blue marbles". Suzie wonders out loud whether the man looked in the bag and specifically pulled out the re... (read more)

0pnrjulius10y
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.