# bigjeff5 comments on Bayes' Theorem Illustrated (My Way) - Less Wrong

126 03 June 2010 04:40AM

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

What's the difference between

Boy1Tu/Boy2Tuesday

and

Boy1Tuesday/Boy2Tu

?

Comment author: 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: 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: 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: 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: 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: 22 December 2013 04:13:10PM 1 point [-]

No one is suggesting one flip informs the other, rather that when you say "one coin came up heads" you are giving some information about both coins.

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?

This is 1/2, because there are two scenarios, hh, ht. But its different information then the other question.

If you say "one coin is heads," you have hh,ht,th, because it could be that the first flip was tails/the second heads (a possibility you have excluded in the above).

Comment author: 22 December 2013 04:50:05PM 0 points [-]

No, it's the exact same question, only the labels are different.

The probability that any one child is boy is 50%. We have been told that one child is a boy, which only leaves two options - HH and HT. If TH were still available, then so would TT be available because the next flip could be revealed to be tails.

Here's the probability in bayesian:

P(BoyBoy) = 0.25 P(Boy) = 0.5 P(Boy|BoyBoy) = 1

P(BoyBoy|Boy) = P(Boy|BoyBoy)*P(BoyBoy)/P(Boy)

P(BoyBoy|Boy)= (1*0.25) / 0.5 = 0.25 / 0.5 = 0.5

P(BoyBoy|Boy) = 0.5

It's exactly the same as the coin flip, because the probability is 50% - the same as a coin flip. This isn't the monty hall problem. Knowing half the problem (that there's at least one boy) doesn't change the probability of the other boy, it just changes what our possibilities are.

Comment author: 22 December 2013 04:58:42PM 1 point [-]

No, it's the exact same question, only the labels are different.

No, it isn't. You should consider that you are disagreeing with a pretty standard stats question, so odds are high you are wrong. With that in mind, you should reread what people are telling you here.

Now, consider "I flip two coins" the possible outcomes are hh,ht,th,tt

I hope we can agree on that much.

Now, I give you more information and I say "one of the coins is heads," so we Bayesian update by crossing out any scenario where one coin isn't heads. There is only 1 (tt)

hh,ht,th

So it should be pretty clear the probability I flipped two heads is 1/3.

Now, your scenario, flipped two coins (hh,ht,th,tt), and I give you the information "the first coin is heads," so we cross out everything where the first coin is tails, leaving (hh,ht). Now the probability you flipped two heads is 1/2.

I don't know how to make this any more simple.

Comment author: 22 December 2013 04:43:07PM -1 points [-]

"The first coin comes up heads" (in this version) is not the same thing as "one of the coins comes up heads" (as in the original version). This version is 50%, the other is not.

Comment author: 22 December 2013 05:04:52PM *  0 points [-]

How is it different? In both cases I have two independent coin flips that have absolutely no relation to each other. How does knowing which of the two came up heads make any difference at all for the probability of the other coin?

If it was the first coin that came up heads, TT and TH are off the table and only HH and HT are possible. If the second coin came up heads then HT and TT would be off the table and only TH and HH are possible.

The total probability mass of some combination of T and H (either HT or TH) starts at 50% for both flips combined. Once you know one of them is heads, that probability mass for the whole problem is cut in half, because one of your flips is now 100% heads and 0% tails. It doesn't matter that you don't know which is which, one flip doesn't have any influence on the probability of the other. Since you already have one heads at 100%, the entire probability of the remainder of the problem rests on the second coin, which is a 50/50 split between heads and tails. If heads, HH is true. If tails, HT is true (or TH, but you don't get both of them!).

Tell me how knowing one of the coins is heads changes the probability of the second flip from 50% to 33%. It's a fair coin, it stays 50%.

Comment author: 22 December 2013 05:27:46PM 2 points [-]

Flip two coins 1000 times, then count how many of those trials have at least one head (~750). Count how many of those trials have two heads (~250).

Flip two coins 1000 times, then count how many of those trials have the first flip be a head (~500). Count how many of those trials have two heads (~250).

By the way, these sorts of puzzles should really be expressed as a question-and-answer dialogue. Simply volunteering information leaves it ambiguous as to what you've actually learned ("would this person have equally likely said 'one of my children is a girl' if they had both a boy and girl?").

Comment author: 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.