G’day

- Seems to me the logic problem (I figure about 51% btw) is that “one daughter is a girl born on Friday” doesn’t preclude the other daughter from also being a daughter born on Friday. The other child’s birth day and sex is unconstrained. The probability any given child is born female is 51% if memory serves. (One coinflip was heads on a Monday what’s the chance that my other flip was tails?)
- Well, this may be a problem … but at least Mr Obama’s recent promise broken/lie isn’t at issue (after all he never promised his healthcare reforms would not cause job lose).
- Examining drug development.
- A boat and a typhoon.
- Bat-zappers.
- The car I’d really really like to own and drive when it comes out, but the boneheads in Washington most likely won’t let me. There is no excuse for this rampant stupidity.
- Art meets the modern Political Correctness.
- The iceberg tip.
- A science lesson for the global warming alarmists (and those think any science is settled on either side).

## 25 comments

Re #1.

I’ll take a stab:

First if you have two kids, there are 4 possible combinations:

MM

MF

FM

FF

If one of your kids is a girl born on Friday, then the possible combinations are:

MF

FM

FF

Odds that the ‘other kid’ is also a girl would seem to be 1/3. The moment we learn that one of your kids is a girl then the MM possibility is taken off the table.

The ‘coinflip’ answer here would be correct if the question was along the lines of “I have a girl, if we have another baby what’s the odds of it being a girl?”.

This sounds very much like the Monty Hall problem where it always makes sense to choose what’s behind the other curtain, does it not?

Re #2, strictly speaking you’re talking about job gains. If one person was working 40 hours a week and then a university turns that into two people working 20 hours a week you have two jobs, not one.

Assume in the short term the number of people teaching is fixed and cannot be easily increased (not a bad assumption, people may move into an area to fill an open job but to fill an open part time job?). You’re going to expect to get any given person working 40 hours a week but doing it split over two jobs.

(remember colleges can’t just make a 40 hour job a 20 hour job. Presumably before the ACA passed the reason they were paying someone for 40 hours was because they needed 40 hours of work done, not 20).

Boonton,

I see it as, if you have two kids, there are 4 possibilities. MM, MF, FM, FF. If one is a girl then your remaining possibilities are FM and FF (both MM and MF are out as the first kid considered is not a girl).

I’m not clear on the difference between the statement

“I have a girl, if we have another baby what’s the odds of it being a girl?”.and“I have a girl, what are the odds of my other child being a girl?”.I hold those two statements are equivalent.I added that to the comment on the blog you cited. The combos are:

MM

MF

FM

FF

In the question, you have one girl but you don’t know the order. So it could be MF or FM or FF.

But if you say “we had a girl, what’s the chance the next will also be a girl?”. Now you impose an order so MM and MF is eliminated. What’s left is FF or FM. 50-50.

so the Monty Hall problem, you get 3 choices. Two of the curtains conceal a dud prize, one a nice one. You choose and Monty let’s you see the dud behind one of the curtains you didn’t choose. He now let’s you keep your choice or go swap for the other curtain.

Initially the possibilities are:

Prize

Dud

Dud

So your initial choice had a 1/3 chance of winning and 2/3 chance of loosing.

Now he takes a dud off the table and let’s you choose between the two remaining options. Your original choice or the one you didn’t choose.

It may seem like swapping has a 50-50 shot of getting the winner so it doesn’t matter if you do or don’t. But in fact swapping has a 2/3 chance of scoring the winner.

Odds of the good prize being in the ‘set’ of curtains you rejected is 2/3 and there’s only a 1/3 odd your first choice was the winner. By switching you secure that 2/3 chance of winning hence you should always swap if given a choice.

Imagine if instead of showing you which curtain had the bum prize, Monty let you either keep the curtain you originally choose or swap it for the

twocurtains you didn’t choose. Clearly getting two curtains gives you a 2/3 chance of getting the good prize rather than the 1/3 chance by keeping your original choice. Monty is effectively doing that by letting you see which of the two unpicked options has the dud and letting you swap with the other one.I’m sure the girl problem is the same thing restated but I’m not sure how….

Boonton,

The crux of the Monty Hall problem is you know that you have one prize and two duds. You have an established solution set (2 of one thing one of another). You don’t have the same situation here. The monty hall analogy would be to replace prize with girls. The initial M/H setup is girl, boy, boy. The initial girl problem is girl or boy and girl or boy. If the M/H setup was random Girl/Boy (prize/dud) then you wouldn’t gain information from his asking you if you want to switch. In the M/H problem he exposes one door after you choose, … he gives you information. This isn’t the case here.

Look at the two sentences I gave you, you need to tell me what the difference between them is. If there is none logically (which I claim is the case) then 51% is your answer.

well here you are given some information if you are told “I had a girl who was born on Friday”. That eliminates one out of four possibilities. Of the 3 possibilities that aren’t eliminated, the second child is a boy in 2 out of 3 of them so the chance that the second child is a girl is only 1/3.

In the case where you’re asking about the next child, you have a tossup between only two possibilities to make your 50-50 chance.

Now with Monty Hall, there’s a 2/3 chance that the winning door is among the set of two doors you didn’t pick. But Monty shows you one of those doors which means you can effectively swap a single unknown door for the set of two unknown doors.

Suppose Monty, though, offered you only a chance to swap your door for

oneof the two unpicked doorswithoutshowing you the dud? In that case the odds of the winner being among the two doors is still 2/3, but since you can only get one of those doors you have a 1/2 chance of getting it right. 1/2 * 2/3 = 1/3. So swapping one door for another door in that case is an even swap.So let’s say the boy is the winning prize and the girl is the dud. I tell you I have one dud, what are the odds that I won if I told you I have two kids? 2/3 since having one dud eliminates the possibility that I was a double winner with two boys. I might be a double loser with two girls but odds are I either had a boy before or after the girl to the tune of 2/3, not 49%.

Both problems seem to work by presenting a simple probablity problem (chance of having a girl is normally 50%, chance of picking the right door out of 3 is 33.3%) but confuse the issue by presenting a bit of information that alters the possible states of the universe. In the first you know at least one out of two children is a girl, in the other you learn at least one of the dud doors after you’ve picked another door.

Boonton,

No. That’s the problem. It eliminates

twoof the possibilities, not one. The date gives you no information. I had a girl who was born on Friday. When was your other daughter born. Friday is a possible answer.The M/H problem logically resolves itself when you realize Monty gives you information on the state of the system when he opens a door. No information is given by the girl on Friday revelation, except that two of the possiblities are eliminated (first child considered is a girl eliminates MM and MF).

Boonton,

Again you say with the coin flip information about first coin yields nothing about the second. You had (HH HT TH and TT). Heads on first eliminates the first two. Look at those two sentences. This is not (you have three coin tosses, to tails, one heads … pick one … I reveal another to be a tail … pick the head). The boy/girl problem gives you no information about the second child or the distribution.

“I have a girl, if we have another baby what’s the odds of it being a girl?”. and “I have a girl, what are the odds of my other child being a girl?”

If they are the same logically 51% is your answer. If not, how are they different?

Boonton

Let me put it another way. You flipped a coin twice. On Friday you had heads. What is the probability the other flip was a tail?

Boonton

Now what if i said, woops I haven’t flipped the 2nd time, give me a moment and I’ll flip it. Same difference logically, coins behave the same if flipped in the past as future.

No. That’s the problem. It eliminates two of the possibilities, not one. The date gives you no information. I had a girl who was born on Friday. When was your other daughter born. Friday is a possible answer.Reading the blog carefully, two questions are asked:

•I have two children. One of my children is a girl who was born on Friday. What’s the probability I have two girls?

•I have two children. One of my children is a girl. Before you came in, I selected a daughter at random from the set of all my daughters, and this daughter was born on Friday. What’s the probability I have two girls?

In the first we are not told there’s ‘another daughter’. We are first told there’s two children (that gives us 4 possible combinations), then we are told one is a daughter (we are down to 3 combinations since boy-boy is eliminated).

In the second, the question seems to be asked using terminology from set theory. Again out of 4 possible combinations, we learn the ‘set of daughters’ is not an empty set. But nothing says the set of ‘all my daughters’ must be greater than one. Note the ‘set of all my daughters’ is a subset of ‘all my children’ and the set of ‘all my children’ is set equal to 2.

In terms of conventional language, you may answer the second question at 100% because usually when someone says something like “I’m going to pick one at random from this set” it’s assumed you are NOT talking about a set of just one member. “A daughter selected at random” means you have more than one daughter and if you only have 2 kids then chances are 100% you have two daughters.

Let me put it another way. You flipped a coin twice. On Friday you had heads. What is the probability the other flip was a tail?Flip two coins, possibilities are:

HH

HT

TH

TT

But you tell me one of your coins came out heads. So TT is off the table.

HH

HT

TH

Probability the other flip was a tail? 2/3.

Now re-ask the question. I flipped a coin once and got heads, what’s the chance the next flip will be a tail?

Of the original possibilities, getting heads on the FIRST flip eliminates the possibilities I marked with a *

HH

HT

TH*

TT*

What’s left? HH and HT. That means the chance of a tail is 50-50

Note that in the case of the daughter, all the writer tells you is that she has a daughter, she doesn’t say the daughter is the first or second child!

Boonton,

This is your mistake. Look at the coin flip. I flip two coins. One flip was heads. What is the chance the other flip was heads. Doesn’t matter which flip. The chance of the *other* flip being heads is 50/50. Doesn’t matter which child is first or second. You still only have one child left and no information. Order isn’t important.

See this is wrong

TH is also off the table. You have one possibility left, one coin which has two states. One coin remains it has two possible values.

Boonton,

Look when monty opens a curtain he

gives you informationyou have no information related to the remaining child. There is no Car/Goat/Goat constraint on number.And btw one can talk of the set of daughters born on Friday of that my parents had … it just happens to be the empty set. If you have a set of something it can be 0,1,2, … of them. Sets can be uncountable, but one might reasonably assume that the set of children is not an uncountable set. The point being the “set of my daughters” is something one can talk about just as reasonably (mathematically speaking) as the set of my parents daughters (which as note is empty for they had two sons). It may actually be the case that “a set of” in casual English means more than two. I cannot speak to that, as if you say “set” my brain switches to math-glish or math-ese.

Boonton,

That’s akin to “or”, which I automatically treat as the mathematical conjunction not the computer science exclusive or which is the common parlance. That is to say, Bob has hair or Bill has hair mathematically speaking allows both to have hair, but is often understand to mean “Either” Bob or Bill have hair. When I was in High School apparently (as we noted last winter when my kids were reviewing old year books) a lot of mileage from my cutting short a History class discussion in which the teacher had get a argument going asking an ambiguous questions was X this *or* the other and I promptly responded “Both!”.

Anyhow that was something of a digression on your set of daughters remark => 100% remark which I think is correct only if your interpretation of set is not mathematically sound.

TH is also off the table. You have one possibility left, one coin which has two states. One coin remains it has two possible values.Why is TH off the table? If you said “I flipped two coins, one was heads” then TH would be a possibility which means the odds two heads becomes 1/3. Only if you said “I flipped two coins, the FIRST one was heads” would TH be removed (or if you said the second one was heads then HT would be removed).

Boonton,

Why is “first” the first flipped, not the first considered. Again. Look at those two sentences. Tell me the difference.

So we have 4 possibilities. First we have two doubles (HH, TT). Order doesn’t matter for those. Second we have one of each (TH, HT).

When you have one of each, if you DO NOT specify the order then you have two possibilities. If you DO specify the order then you have only one. If I say I got a head then it could be TH or HT. If I say I rolled a head first then it can only be HT.

Returning to the original questions:

**

•I have two children. One of my children is a girl who was born on Friday. What’s the probability I have two girls?

•I have two children. One of my children is a girl. Before you came in, I selected a daughter at random from the set of all my daughters, and this daughter was born on Friday. What’s the probability I have two girls?

***

I’m not seeing any evidence that she is giving us a birth order. She is saying she has at least one (possibly two) girls. So of the 4 possible combinations, only Male Male is taken off the table. We are left with 3 combinations; MF, FM, FF.

The probability that the other child is also a girl would be 1/3, not 1/2.

Now if you read these questions as providing some type of birth order, if she is saying the girl was the first child then the combination MF is eliminated leaving you with just FM and FF as possibilities (in which case you would say the odds of the unknown child being a girl is 50-50).

I’m not seeing a difference between the two sentences above. Both are saying the writer has two children (4 combinations possible), at least one of which is a girl (3 possible combinations), so what’s the odds it’s both girls (1 out of 3 possible combinations).

You can’t get to a 50-50 combination unless you can justify eliminating one of the MF/FM possiblities. If she specified an order that would do it but I don’t see that she did.

Boonton,

I think you are wrong on this. You have 4 possiblies. HT,HH,TH and TT. You select a coin … it is heads. You now have two possiblies left. HT and HH. Not three. What you call “first” and I call first is irrelevant. I call “first” the one you indicated.

Again, if you don’t see a difference in the sentences inidicated you in some way realize that who defines order is unimportant.

I think the “born on Friday” here is important if you had two boys. If that wasn’t said then you haven’t actually excluded the two boys example, i.e., “I selected a daughter at random from the empty set …” may logically possible but (elements of an empty set can be anything … you just don’t get a result, but that set can’t be empty if you have a birth date).

The point is once you have told me you have a daughter and you have two children. I know

nothingabout the other child. It is unconstrained by any statement you have made. It has a 51% chance of being a girl. That is all there is to consider. I think your confusing lies in the “three choices, one eliminated” way of thinking about is that the selected daughter is represented still in the choices remaining, they are not all equally likely. Yes you’ve excluded the TT/MM possibility, but the consequences of what has been revealed is not yet exhausted. Again, remember, the final child has no constraints. You have been given 0 information on the second child.I think you are wrong on this. You have 4 possiblies. HT,HH,TH and TT. You select a coin … it is heads. You now have two possiblies left. HT and HH. Not three. What you call “first” and I call first is irrelevant. I call “first” the one you indicated.Imagine you write on 4 index cards: HH, HT, TH, TT and put them in a box. Pull one out, how many are left?

Boonton,

That is a different problem, I’ve been telling you all along that isn’t relevant.

Let me explain it this way.

1. You have two children. One is in Europe. What is the chance that the remaining child is a girl. (51)

2. You have two children. One is in Europe and is a girl. What is the chance that the remaining child is a girl. Same problem same answer. (51).

3. You have two children. One is in Europe and was born on Friday and is a girl. What is the chance that the remaining child is a girl. Same problem same answer. (51).

4. You have two children. One was born on Friday and is a girl. What is the chance that the remaining child is a girl. Same problem same answer. (51).

It makes no difference in probability of “first” or the sex whether of the child was born on any given day. The sex of the remaining child remains unconstrained. This is

notthe same as the Monty Hall problem, and the crucial difference is that Montygives you informationthat you didn’t have.(4 added. note, no difference between any of these)

Boonton,

The salient point is that there is not distinction no change to the situation on the child in the US whether I choose to reveal or not reveal the gender of the child sent overseas.

Boonton,

The point is I’ve by saying “one daughter is a girl” you’ve pulled

twonot one card from your deck, you’ve pulled TT and HT .. you’ve selected position one completely.Damm, I think you’re right.