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Comment by mamon | Hacker News Reader

"Problem 3. One is a boy born on Tuesday. What's the probability both are boys?" - this is not a problem at all. The fact that one boy is born on Tuesday is completely irrelevant as the sex of the child does not depend on day of the week. That information is given only to mess with readers mind. So, if you ignore that Tuesday part then you can reduce Problem 3 to Problem 2.

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Not really - like Norvig said, that's one interpretation, but there is another valid one.

Say you had a bunch of families with two children each. The children are evenly distributed in terms of gender and the the days of the week on which they were born. If you pick one parent from the crowd, the chance that they have at least one boy is 3 / 4, the chance that they have two is 1 / 4, and the chance that they have none is also 1 / 4:

      | B| G|
    --|--|--|
    B |BB|BG|
    --|--|--|
    G |GB|GG|

Each of the four squares on the above table is equiprobable. However, if the person says they have at least one boy, they must be in either the left column or the top row, so in one of three squares. There is only one square in those three with both boys, so the chance of that parent having two boys is 1 / 3.

Now, for the day-of-the-week problem. If you ask a parent if they have a male child born on Tuesday, it is not equiprobable that they're in any of those three squares. In the BG group, they all have male children, so the chance that any parent chosen has a male child born on Tuesday is 1 / 7. Similarly in the GB group. However, in the BB group, either one of their children may be born on a Tuesday to satisfy the condition. The chance that either child is born on a Tuesday is the same as the inverse of neither child being born on a Tuesday, or

    1 - (6/7 * 6/7) = 13/49

So the number of parents in the top left group (BB) who satisfy the condition is 13 / 49, whereas the number of parents in the top right (BG) is 1 / 7, and the bottom left (GB) is also 1 / 7. You're looking for the probability that a given parent in that satisfies the condition is in the top left group, which is

    (13/49) / (1/7 + 1/7 + 13/49) = 13/27

For me both problem 2 and problem 3 should also be 1/2 .

The variances are due to how people interpret the outcome space .

It is right that probability = favaroble out comes / total possible outcomes.

In problem 2, in my opinion the possible outcomes are not how it was suggested in the post but as below. When a family has 2 kids , the below are the only possible outcomes

1) Potential Outcome 1: Both are boys 2) Potential Outcome 2: One boy and one girl 3) Potential outcome 3: Both are girls

The 3rd one is not a legal potential outcome in our particular constraint of problem 2, since problem #2 statement already states 'at least one is a boy'

So Total possible legal outcomes = 2 Favorable out comes for our event (both boys) = 1

So probability for Problem 2 = 1/2

Similarly for problem #3, I think the post unnecessarily complicates the calculation of problem space. The fact that 'Tuesday' is mentioned is irrelevant in my opinon, if you state the problem #3 in a different way that is more clearly understood.

There are 14 baskets labelled as follows "Sunday Boy", "Sunday Girl", "Monday Boy", "Monday Girl",....."Saturday Boy", "Saturday Girl". A stork came and dropped 2 babies. One baby was dropped in "Tuesday Boy" basket. What is the probability that both are boys?

Now the total outcomes and favarable outcomes are :

Total possible outcomes = Number of ways second baby could have been dropped = 14 possible baskets = 14

Favorable outcomes = second baby dropped in 'boy' basket = 7 possible baskets = 7

Probability that both are boys = 7 / 14 = 1/2

If you understand problem 2, then you will get problem 3 as well, so let's focus on 2.

> It is right that probability = favaroble out comes / total possible outcomes.

No, not actually: It is only right if all outcomes are equally likely! (There's an old joke about the guy who has a 50% chance of winning the lottery, since either he will win it or he won't.)

In particular, you make that mistake here:

> 1) Potential Outcome 1: Both are boys 2) Potential Outcome 2: One boy and one girl 3) Potential outcome 3: Both are girls

These three outcomes are not all equally likely. Outcome 1 has probability 1/4, outcome 2 has probability 1/2, and outcome 3 has probability 1/4. (This is if you assume that each child has a half chance each of being a boy/girl.)

Norvig gets rid of this problem by listing out all four possible outcomes, which are all equally likely.

1. First child boy, Second child boy

2. First child boy, Second child girl

3. First child girl, Second child boy

4. First child girl, Second child girl

Again, both Norvig and you are messing with ordering of events. If you list 4 events like this, assuming that ordering of boys and girls matters, then you should stick with that. So, if ordering mattered when children were born it should also matter when you are doing "checks". So you shouldn't formulate problem as "one of the children is boy", you should formulate problem as "first child is boy" (with ordering in place), which eliminates possibility 3. Otherwise you are "solving" problem by listing sample space of completely different problem.

The event "at least one child is a boy" is well-defined on the 4-state sample space. It is the set of events (first boy/second boy, first boy/second girl, first girl / second boy). It has probability 3/4.

When you say "1) Potential Outcome 1: Both are boys 2) Potential Outcome 2: One boy and one girl 3) Potential outcome 3: Both are girls" you are right that those are three potential outcomes, but they are not equiprobable outcomes. One boy and one girl is twice as probable each of the others.

By that reasoning, in Problem 2 the fact that one child is a boy cannot affect the sex of his sibling so you can reduce Problem 2 to Problem 1. The point that Norvig is making is that there are other reasonable interpretations corresponding to different sample spaces (and therefore to other probabilities).

But if your sample space is just Cartesian of {B,G} and {1,2,3,4,5,6,7} then it is in fact just a noise. If you had some additional information, that, say, on Tuesdays there are two times as much boys born as girls, only then expanding sample space would make sense.

Also, for interpretation of problem 2a there is one fundamental flaw that I can see: The sample space listed as:'BB', 'BG', 'GB' is wrong, because 'BG' and 'GB' are in fact the same sample. Or, to put it in a different way: if you decide that ordering does not matter then you should list either 'BG' or 'GB', not both. And if order does matter then you should list 'BB' twice, for all the possible orderings. Both of which will make probability equal to 1/2.

EDIT: Moral of the story is: "If it looks like paradox you are doing it wrong". Which I think was the Norvig's point from the start.

"The sample space listed as:'BB', 'BG', 'GB' is wrong, because 'BG' and 'GB' are in fact the same sample."

Flip two coins. Exclude the case where both are tails. What is the probability the two coins are different? If 'BG' and 'GB' are "the same sample", is it also the case that 'HT' and 'TH' are the same sample?

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