4 ms·

Bayes rule with odd ratios makes it pretty easy. base odds: 20:80 = 1:4 relative odds = (1 letter/6 letters) : (2 letters / 8 letters) = 2/3 po

Bayes rule with odd ratios makes it pretty easy. base odds: 20:80 = 1:4 relative odds = (1 letter/6 letters) : (2 letters / 8 letters) = 2/3 posterior odds = 1:4*2:3 = 1:6 Final probability = 1/(6+1) = 1/7 or roughly 14.2% Bayes rule with raw probabilities is a lot more involved.

thaumasiotes 5y ago
Odds are usually represented with a colon -- the base odds are 1:4 (20%), not 1/4 (25%).
- deleted 5y ago
  [deleted]
Aethylia 5y ago
Assuming that the algorithm is 100% accurate!
- mattkrause 5y ago
  I was also distracted by the fact that you can't (usually) hear the difference between English words written with one 'l' and those with two consecutive 'l's. "Voksal" and "Vauxhall" seem like they should each have six phonemes.
hervature 5y ago
I don't know about "a lot more". It is essentially the same calculation without having to know 3 new terms. Let: A = the event they are a spy B = the event that an l appears And ^c denote the complement of these events. Then, P(A) = 1/5 P(A^c) = 4/5 P(B|A) = 1/6 P(B|A^c) = 1/4 P(A|B) = P(B|A)P(A)/P(B) By law of total probability, P(B) = P(B|A)P(A) + P(B|A^c)P(A^c) Which is very standard formulation and really just your equation as you can rewrite everything I have done as: P(A|B) = 1/(1 + P(B|A^c)P(A^c)/P(B|A)P(A)) Which is the base odds, posterior odds, and odds to probability conversion all in one. The reason why this method is strictly better in my opinion is because the odds breaks down simply if we introduce a third type of person which doesn't pronounce l's. Also, after doing one homework's worth of these problems, you just skip to the final equation in which case my post is just as short as yours.
FabHK 5y ago
Hmm, a bit more involved maybe, but not that much. But your calculation sure seems short. With S = sleeper, and L = letter L, and remembering "total probability": P(L) = P(L|S)P(S) + P(L|-S)P(-S), (where -S is not S), we have by Bayes P(S|L) = P(L|S) P(S) / P(L) = P(L|S) P(S) / (P(L|S)P(S) + P(L|-S)P(-S)) = 1/6 * 1/5 / (1/6*1/5 + 1/4*4/5) = 1/30 / (1/30 + 6/30) = 1/7

thaumasiotes 5y ago

Odds are usually represented with a colon -- the base odds are 1:4 (20%), not 1/4 (25%).

deleted 5y ago
[deleted]

Aethylia 5y ago

Assuming that the algorithm is 100% accurate!

mattkrause 5y ago
I was also distracted by the fact that you can't (usually) hear the difference between English words written with one 'l' and those with two consecutive 'l's. "Voksal" and "Vauxhall" seem like they should each have six phonemes.

hervature 5y ago

I don't know about "a lot more". It is essentially the same calculation without having to know 3 new terms. Let: A = the event they are a spy B = the event that an l appears And ^c denote the complement of these events. Then, P(A) = 1/5 P(A^c) = 4/5 P(B|A) = 1/6 P(B|A^c) = 1/4 P(A|B) = P(B|A)P(A)/P(B) By law of total probability, P(B) = P(B|A)P(A) + P(B|A^c)P(A^c) Which is very standard formulation and really just your equation as you can rewrite everything I have done as: P(A|B) = 1/(1 + P(B|A^c)P(A^c)/P(B|A)P(A)) Which is the base odds, posterior odds, and odds to probability conversion all in one. The reason why this method is strictly better in my opinion is because the odds breaks down simply if we introduce a third type of person which doesn't pronounce l's. Also, after doing one homework's worth of these problems, you just skip to the final equation in which case my post is just as short as yours.

FabHK 5y ago

Hmm, a bit more involved maybe, but not that much. But your calculation sure seems short. With S = sleeper, and L = letter L, and remembering "total probability": P(L) = P(L|S)P(S) + P(L|-S)P(-S), (where -S is not S), we have by Bayes P(S|L) = P(L|S) P(S) / P(L) = P(L|S) P(S) / (P(L|S)P(S) + P(L|-S)P(-S)) = 1/6 * 1/5 / (1/6*1/5 + 1/4*4/5) = 1/30 / (1/30 + 6/30) = 1/7