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Jiro comments on Bayes' Theorem Illustrated (My Way) - Less Wrong

126 Post author: komponisto 03 June 2010 04:40AM

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Comment author: Jiro 22 December 2013 03:41:40AM *  -1 points [-]

No, that's not right. They don't assume that both boys can't be born on Tuesday. Instead, what they are doing is pointing out that although there is a scenario where both boys are born on Tuesday, they can't count it twice--of the situations with a boy born on Tuesday, there are 6 non-Tuesday/Tuesday, 6 Tuesday/non-Tuesday, and only 1, not 2, Tuesday/Tuesday.

Actually, "one of my children is a boy born on Tuesday" is ambiguous. If it means "I picked the day Tuesday at random, and it so happens that one of my children is a boy born on the day I picked", then the stated solution is correct. If it means "I picked one of my children at random, and it so happens that child is a boy, and it also so happens that child was born on Tuesday", the stated solution is not correct and the day of the week has no effect on the probability

Comment author: bigjeff5 22 December 2013 04:16:49AM *  0 points [-]

No, read it again. It's confusing as all getout, which is why they make the mistake, but EACH child can be born on ANY day of the week. The boy on Tuesday is a red herring, he doesn't factor into the probability for what day the second child can be born on at all. The two boys are not the same boys, they are individuals and their probabilities are individual. Re-label them Boy1 and Boy2 to make it clearer:

Here is the breakdown for the Boy1Tu/Boy2Any option:

Boy1Tu/Boy2Monday Boy1Tu/Boy2Tuesday Boy1Tu/Boy2Wednesday Boy1Tu/Boy2Thursday Boy1Tu/Boy2Friday Boy1Tu/Boy2Saturday Boy1Tu/Boy2Sunday

Then the BAny/Boy1Tu option:

Boy2Monday/Boy1Tu Boy2Tuesday/Boy1Tu Boy2Wednesday/Boy1Tu Boy2Thursday/Boy1Tu Boy2Friday/Boy1Tu Boy2Saturday/Boy1Tu Boy2Sunday/Boy1Tu

Seven options for both. For some reason they claim either BTu/Tuesday isn't an option, or Tuesday/BTu isn't an option, but I see no reason for this. Each boy is an individual, and each boy has a 1/7 probability of being born on a given day. In attempting to avoid counting evidence twice you've skipped counting a piece of evidence at all! In the original statement, they never said one and ONLY one boy was born on Tuesday, just that one was born on Tuesday. That's where they screwed up - they've denied the second boy the option of being born on Tuesday for no good reason.

A key insight that should have triggered their intuition that their method was wrong was that they state that if you can find a trait rarer than being born on Tuesday, like say being born on the 27th of October, then you'll approach 50% probability. That is true because the actual probability is 50%.

Comment author: Jiro 22 December 2013 04:24:10AM 0 points [-]

Here is the breakdown for the Boy1Tu/Boy2Any option:

Boy1Tu/Boy2Tuesda

Then the BAny/Boy1Tu option:

Boy2Tuesday/Boy1Tu

You're double-counting the case where both boys are born on Tuesday, just like they said.

A key insight that should have triggered their intuition that their method was wrong was that they state that if you can find a trait rarer than being born on Tuesday, like say being born on the 27th of October, then you'll approach 50% probability.

If you find a trait rarer than being born on Tuesday, the double-counting is a smaller percentage of the scenarios, so being closer to 50% is expected.

Comment author: bigjeff5 22 December 2013 04:55:28AM *  0 points [-]

I see my mistake, here's an updated breakdown:

Boy1Tu/Boy2Any

Boy1Tu/Boy2Monday Boy1Tu/Boy2Tuesday Boy1Tu/Boy2Wednesday Boy1Tu/Boy2Thursday Boy1Tu/Boy2Friday Boy1Tu/Boy2Saturday Boy1Tu/Boy2Sunday

Then the Boy1Any/Boy2Tu option:

Boy1Monday/Boy2Tu Boy1Tuesday/Boy2Tu Boy1Wednesday/Boy2Tu Boy1Thursday/Boy2Tu Boy1Friday/Boy2Tu Boy1Saturday/Boy2Tu Boy1Sunday/Boy2Tu

See 7 days for each set? They aren't interchangeable even though the label "boy" makes it seem like they are.

Do the Bayesian probabilities instead to verify, it comes out to 50% even.

Comment author: shinoteki 22 December 2013 05:25:35AM *  0 points [-]

What's the difference between

Boy1Tu/Boy2Tuesday

and

Boy1Tuesday/Boy2Tu

?

Comment author: bigjeff5 22 December 2013 05:35:18AM 0 points [-]

In Boy1Tu/Boy2Tuesday, the boy referred to as BTu in the original statement is boy 1, in Boy2Tu/Boy1Tuesday the boy referred to in the original statement is boy2.

That's why the "born on tuesday" is a red herring, and doesn't add any information. How could it?

Comment author: Jiro 22 December 2013 05:51:45AM *  0 points [-]

This sounds like you are trying to divide "two boys born on Tuesday" into "two boys born on Tuesday and the person is talking about the first boy" and "two boys born on Tuesday and the person is talking about the second boy".

That doesn't work because you are now no longer dealing with cases of equal probability. "Boy 1 Monday/Boy 2 Tuesday", "Boy 1 Tuesday/Boy 2 Tuesday", and "Boy 1 Tuesday/Boy 1 Monday" all have equal probability. If you're creating separate cases depending on which of the boys is being referred to, the first and third of those don't divide into separate cases but the second one does divide into separate cases, each with half the probability of the first and third.

doesn't add any information. How could it?

As I pointed out above, whether it adds information (and whether the analysis is correct) depends on exactly what you mean by "one is a boy born on Tuesday". If you picked "boy" and "Tuesday" at random first, and then noticed that one child met that description, that rules out cases where no child happened to meet the description. If you picked a child first and then noticed he was a boy born on a Tuesday, but if it was a girl born on a Monday you would have said "one is a girl born on a Monday", you are correct that no information is provided.

Comment author: bigjeff5 22 December 2013 06:06:39AM *  0 points [-]

The only relevant information is that one of the children is a boy. There is still a 50% chance the second child is a boy and a 50% chance that the second child is a girl. Since you already know that one of the children is a boy, the posterior probability that they are both boys is 50%.

Rephrase it this way:

I have flipped two coins. One of the coins came up heads. What is the probability that both are heads?

Now, to see why Tuesday is irrelevant, I'll re-state it thusly:

I have flipped two coins. One I flipped on a Tuesday and it came up heads. What is the probability that both are heads?

The sex of one child has no influence on the sex of the other child, nor does the day on which either child was born influence the day any other child was born. There is a 1/7 chance that child 1 was born on each day of the week, and there is a 1/7 chance that child 2 was born on each day of the week. There is a 1/49 chance that both children will be born on any given day (1/7*1/7), for a 7/49 or 1/7 chance that both children will be born on the same day. That's your missing 1/7 chance that gets removed inappropriately from the Tuesday/Tuesday scenario.

Comment author: EHeller 22 December 2013 07:21:43AM 1 point [-]

I have flipped two coins. One of the coins came up heads. What is the probability that both are heads?

1/3 (you either got hh, heads/tails,or tails/heads). You didn't tell me THE FIRST came up heads. Thats where you are going wrong. At least one is heads is different information then a specific coin is heads.

This is a pretty well known stats problem, a variant of Gardern's boy/girl paradox. You'll probably find it an intro book, and Jiro is correct. You are still overcounting. Boy-boy is a different case then boy-girl (well, depending on what the data collection process is).

If you have two boys (probability 1/4), then the probability at least one is born on Tuesday (1-(6/7)^2). ( 6/7^2 being the probability neither is born on Tuesday). The probability of a boy-girl family is (2*1/4) then (1/7) (the 1/7 for the boy hitting on Tuesday).

Comment author: bigjeff5 22 December 2013 10:12:57AM 0 points [-]

Lets add a time delay to hopefully finally illustrate the point that one coin toss does not inform the other coin toss.

I have two coins. I flip the first one, and it comes up heads. Now I flip the second coin. What are the odds it will come up heads?

Comment author: Jiro 22 December 2013 04:39:33PM -1 points [-]

. There is still a 50% chance the second child is a boy and a 50% chance that the second child is a girl.

No there's not. The cases where the second child is a boy and the second child is a girl are not equal probability.

I have flipped two coins. One of the coins came up heads. What is the probability that both are heads?

If you picked "heads" before flipping the coins, then the probability is 1/3. There are three possibilities: HT, TH, and HH, and all of these possibilities are equally likely.

I have flipped two coins. One I flipped on a Tuesday and it came up heads. What is the probability that both are heads?

If you picked "heads" and "Tuesday" before knowing when you would be flipping the coins, and then flipped each coin on a randomly-selected day, and you just stopped if there weren't any heads on Tuesday, then the answer is the same as the answer for boys on Tuesday. If you flipped the coin and then realized it was Tuesday, the Tuesday doesn't affect the result.

The sex of one child has no influence on the sex of the other child, nor does the day on which either child was born influence the day any other child was born.

If you picked the sex first before looking at the children, the sex of one child does influence the sex of the other child because it affects whether you would continue or say "there aren't any of the sex I picked" and the sexes in the cases where you would continue are not equally distributed.

Comment author: bigjeff5 22 December 2013 04:48:30AM *  0 points [-]

Which boy did I count twice?

Edit:

BAny/Boy1Tu in the above quote should be Boy2Any/Boy1Tu.

You could re-label boy1 and boy2 to be cat and dog and it won't change the probabilities - that would be CatTu/DogAny.