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.
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.