Your iPhone just got less secure. Blame the FBI

Re:Your iPhone just got less secure. Blame the FBI by logfromblammo on Wednesday March 30, 2016 @04:20PM (#11392646)

After first datum (there are two children):

  Presuming A and B to be independent events
  (first child is a girl, second child is a girl)
  P(A) = 1 - P(!A) = 1/2
  P(B) = 1 - P(!B) = 1/2

  P(A and B) = P(B|A) * P(A) = 1/2 * 1/2 = 1/4
  P(A and !B) = P(!B|A) * P(A) = 1/2 * 1/2 = 1/4
  P(!A and B) = P(B|!A) * P(!A) = 1/2 * 1/2 = 1/4
  P(!A and !B) = P(!B|!A) * P(!A) = 1/2 * 1/2 = 1/4

Good so far.

But the probability later on depends on how you find out that at least one of your friend's children is a girl. If your friend tells you a random fact about her whole family, the probability is 1/3. If she tells you a fact about a randomly chosen child, the probability is 1/2.

You can presume from "they're not both boys" that the statement is equivalent to "there is at least one girl". When she tells you the name, you now know "a specific child of mine is a girl".

"Not both boys" gives you the precondition "!(!A & !B)" which equates to "A or B".

  P(A or B) = P(A and B) + P(A and !B) + P(!A and B) = 3/4
  P(A and B|A or B) = P(A or B|A and B) * P(A and B) / P(A or B) = 1 * 1/4 / 3/4 = 1/3

"Mary is a girl" gives you either the precondition "A" or the precondition "B"

  P(A and B|A) = P(A|A and B) * P(A and B) / P(A) = 1 * 1/4 / 1/2 = 1/2
  P(A and B|B) = P(B|A and B) * P(A and B) / P(B) = 1 * 1/4 / 1/2 = 1/2

These are both the same number, so that works out fine.

[ Parent | on HN ]

Re:Your iPhone just got less secure. Blame the FBI by logfromblammo on Wednesday March 30, 2016 @04:33PM (#11392763)

Now see if you can work it out if your friend told you this beforehand:

Between pregnancies, I had an accident that resulted in an odd medical condition. The doctors said that after it happened, 50% of boy fetuses that otherwise would have resulted in pregnancy would simply fail to implant, and I wouldn't even know about it.

Now there's an all new ambiguity. How do you decide whether your friend was planning on having two children all along, or if she has two children because that's how many times she got pregnant?

[ Parent | on HN ]

Re:Your iPhone just got less secure. Blame the FBI by pklausler on Wednesday March 30, 2016 @04:45PM (#11392854)

Here's an epistemological head-scratcher: do the odds still change if you don't hear the daughter's name accurately? Why?

[ Parent | on HN ]

Re:Your iPhone just got less secure. Blame the FBI by logfromblammo on Thursday March 31, 2016 @12:27PM (#11398386)

Remember, those odds are the level of certainty you have that a given statement may be true based only on what you already know. You can also assign a level of certainty to the facts you think you know.

Given the number of ways to mishear "Mary" and the dialectical variant in pronunciation characterized by the Mary-marry-merry merger, I'd tentatively assign the following values:

  25% One child is a girl.  (Heard correctly)
  25% One child is a boy.   (Heard incorrectly)
  50% No new information.   (Incomprehensible or ambiguous)

  In the first case P(A and B|A) = P(A and B|B) = 1/2
  In the second, P(A and B|!A) = P(A and B|!B) = 0
  In the third, P(A and B|A or B) = 1/3

In any case, you're just guessing at the probability you might mishear something, so there's no definitive answer. The mere possibility that I might have heard a boy's name means that the overall probability could be anywhere between 0 and 1/2. By the above, my new odds are 5/12.

So the odds do change, based on your confidence in what you heard (and your confidence that your friend isn't the kind of parent to name her son "Meriadoc" or something similar).

[ Parent | on HN ]