[Closed] Monday evening puzzle

I should know this... 😉

/has flashbacks.

im hard pressed to think of why it's not 50% for all three, but Im sure there's a dastardly deviant reason why it's not.

(and Im assuming we're not going into any odd days of the week gender distributions are we...)

his eldest son

that should be either "elder son", implying two sons (100%) or he actually has 3+ sons (in which case it is also 100%)

1) 1/3
2) 1/2
3) 13/27 unless he's standing on a conveyor belt.

0%?

Smith and Johnson have one son each but Jones has two sons.

Which of the sons is standing on the conveyor belt?

Depends on the random distribution of conveyor belts in the local area relative to the sons, I'd assume. Impossible to say, I guess.

(and Im assuming we're not going into any odd days of the week gender distributions are we...)

Not sure what you mean by that?

75%
100%
75%

One quarter

Not sure what you mean by that?

Im wondering if there's some freakishness of nature about the probability of boys or girls being born on various days of the week not being equal.

Assuming the likelyhood of a boy or a girl is even,

Smith - probability is 1 in 2, the presence of the first son makes no difference.

Jones - "eldest son" implies two sons, otherwise it'd be "eldest child" - probability 1 in 1.

Johnson - 1 in 2 again, what difference does the birthday make?

Im wondering if there's some freakishness of nature about the probability of boys or girls being born on various days of the week not being equal.

Not that I know of, but in this question you can take the probability of being born on a certain day to be the same for each gender.

Slightly reworded question 2, sorry for any confusion.

Mr Jones definately has 2 sons...not sure about the other two.

Just seen the update...haven't got the foggiest for any of them.

cougar and I think the same way. Probably both wrong then 🙂

"eldest son" implies two sons

superlative is used for 3+...so it implies three sons at least,
Now edited by realman.

[i]Slightly reworded question 2, sorry for any confusion.[/i]

I wouldn't worry, I've reworded my answer about four times so far 🙂

"elder child"!

what kind of cackhanded editing do you call that? The only one I was certain I knew the answer to...and now I dont! pah!

Smith - probability is 1 in 2, the presence of the first son makes no difference.

It does make a difference.
If he says he had 20 children what are the chances of him having 20 sons?
He then introduces 19 sons (all called Dave)....Now what are the odds of having 20 sons?

Its the old Goats and Gameshows.

Is it my imagination or is John doing the rounds a bit!

It has to be 50/50 for all of them. The presence, age or recent birth of a son has no bearing at all on the sex of the other child

Jones - "eldest son" implies two sons, otherwise it'd be "eldest child" - probability 1 in 1.

It does make a difference

No it does not. How can the sex of one child have ANY bearing on the sex of another?

I'm going to take a couple of Nurofen.....

😕

You have to define 'one child' more carefully.
'one of my children' or 'this one child, ere'

You aren't convinced PeterPody ?

I dont think this is the same as Monty Hall.

cougar and I think the same way

You have my sympathies.

It does make a difference.
If he says he had 20 children and introduced 19 sons all called Dave....what are the odds of having 20 sons.

From a logic puzzle perspective, the odds are the same, 1 in 2.

From a genetics point of view, the odds are somewhat higher I expect. I'm assuming it's not that sort of a puzzle; if it is, I'm oot.

Nope. Re-read it!

You've changed it, you toe-rag.

Ian, would you care to explain your working..? 😉

What, my real workings, or the workings I'd present if this was a presentation to a client? 🙂

I wonder if one of the other two are grandparents, given the presence of Mr "John's son" in the third one...

Assuming we are working on the basis that the male/female demographic split is 50/50 all of these have a 50% probability of having 2 boys.

The day on which birth occurs has no bearing whatsoever so in each case we have 4 possibilities

1) BB
2) BG
3) GB
4) GG

Question 1 omits possibility 4. 2 & 3 are effectively the same (in this case having no information on the order of birth) and are of equal probability which leaves us with 2 options, a boy and a girl or 2 boys. 2 options = 50%.
Question 2 omits possibilities 2 & 4 which means it has to be 50%.
Question 3 works out the same as question 1.

What am I missing?

I wonder if one of the other two are grandparents, given the presence of Mr "John's son" in the third one...

I like the thinking, shame its wrong. All separate questions, no one is related to anyone from a different question.

What, my real workings, or the workings I'd present if this was a presentation to a client?

I just want to know if you worked it out yourself or just googled it lol.

if you have two children, you will either have 2 sons, 2 daughters or one of each - 3 options
therefore:
1) we know John is a son so two options left so 1/2
2) we don't know any genders so 1/3
3) we know one gender so are only left with 2 options, so 1/2

(can I add that we can't be sure that John is a boy)

2/1
4/1
2/1

I was hoping for an opportunity to say 'Burlington Berty' here...

2/1
4/1
2/1

No, no, and no (and not just because 2/1=2, 4/1=4 etc).

I seem to remember reading somewhere that there are more boys born than girls. So it's not a 50:50 ratio. But I don't know what the figure is.

I seem to remember reading somewhere that there are more boys born than girls. So it's not a 50:50 ratio. But I don't know what the figure is.

You can assume that for any random child there is a 0.5 probability of it being a boy, and 0.5 for it being a girl.

So it's not a 50:50 ratio. But I don't know what the figure is.

Men:Women 48:52 IIRC

1) BB
2) BG
3) GB
4) GG

Question 1 omits possibility 4. 2 & 3 are effectively the same (in this case having no information on the order of birth) and are of equal probability which leaves us with 2 options, a boy and a girl or 2 boys.

Ah, 2 and 3 are not effectively the same, they are still two distinctly different configurations, so you have 3 options(1,2,3) and only 1 where the answer is BB so the odds are 1/3. Were you do say the eldest was a boy, then you have two options(1,2) (aasuming the eldest is listed first), so the odds are 1/2.

The tuesday one makes my head hurt, so I'm not going into that 🙂

I see where this is going.

Looking at the first question: You have four possibilities,

Daughter, daughter
Daughter, son
Son, daughter
Son, son

So when you know that one child is male, you rule out the first option. That leaves you with one out of three chances that both kids are boys.

It's all 0% surely as it's about probability of lying and nothing to do with 49/51 gender split or age. Unless the questioner wants to tell us to assume the men are truthful?

In the second one, as Ian says (simultaneous edits), you rule out two possibilities, so probability is 1/2.

2 and 3 are not effectively the same

Yeah they are because in this case there is no mention of the order of birth so BG and GB are the same.

I think. Still waiting to be told what I am missing. why would day of the week have any bearing on the other child whatsoever? I say 50% for each and stand by it until I hear proof otherwise.

What do I win?

On the third one, it's a red herring - day of week has no bearing on gender. This is the first question again, only going "look, over there" as you shuffle the deck. It's 1 in 3 again.

It's all 0% surely as it's about probability of lying and nothing to do with 49/51 gender split or age. Unless the questioner wants to tell us to assume the men are truthful?

Bingo, all 3 men have no sons, they're just lying to distract you while their mates nick your bike.

Yes, first one is 1/3, and the second one is 1/2.

OK so assuming 50:50.

Can we also assume that by "children" we mean directly as in son or daughter?