Monday evening puzzle

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?

I think the puzzle here is trying to convince people that there is a different probability for each where there is none.

If the first and third questions explicitly stated ‘What is the probability that the youngest/eldest child is a boy?’, then we would have the 1/3 situation but it is not stated so 1/2 for all.

Cougar, That’s what I thought, alas the last isn’t a red herring, but almost conveyor belt levels of horribleness 🙂

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.

Nope.

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

As in opposed to the possibility of a 3rd sex? Yes.

I say 50% for each

You’re simplifying it so that you’re looking at one child in isolation – I did the same at first – but that’s not right. The two children are positional. The declaration of a son means the man has three possible configurations of kids – a son and daughter where the son is eldest, a son and daughter where the son is youngest, or two sons.

alas the last isn’t a red herring

Sods. Give me a minute to think about this then.

can you please give the answer? this is frustrating and I think I’m right and you haven’t said that I’m not

Gah, this is doing my head in now…

surely girl/boy = boy/girl where there is no stipulation of order?

I did already say..

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

I haven’t given the 3rd answer away yet, don’t want to spoil it for the rest of the class.

But

3) we know one gender so are only left with 2 options, so 1/2

Is wrong.

I’ve got it I think. Or rather, I’ve got the catch. Working it out is going to be a swine.

Torminalis, yes, it’s the difference between a permutation and a combination

ok, so 3) we know one is a son so 1/2

Gah, this is doing my head in now…

surely girl/boy = boy/girl where there is no stipulation of order?

Assuming you’re talking about question 1..

A man has two children.

So we know he can either have (youngest-oldest)

1.(B-B)
2.(B-G)
3.(G-B)
4.(G-G)

Then we find out he has at least one son, so 4 is no longer a possibility.

The probability of 1 is 0.5×0.5=0.25
The probability of 2 is 0.5×0.5=0.25
The probability of 3 is 0.5×0.5=0.25

Because all 3 outcomes have the same probability, and there is only one outcome that fits the outcome we want (2 boys), it must be 1/3.

I think I am starting to understand, we can either be talking about:

‘a 2 child family where one is known to be a boy’
or
‘randomly selecting one of two children and then noticing it is a boy’

Either will yield a different result. My head hurts now.

I’ve got it I think. Or rather, I’ve got the catch. Working it out is going to be a swine.

Only took me a few seconds when I was given the problem, try drawing a picture.

ok, so 3) we know one is a son so 1/2

Still no (that’s pretty much what you wrote before). The probability of 2 sons in question 3 is not 0.5.

His other child could also have been born on a Tuesday? Or any day of the week, and been either boy or girl….soooooo….

hang on, ‘cos 1 in 14 has already gone right?

Because all 3 outcomes have the same probability, and there is only one outcome that fits the outcome we want (2 boys), it must be 1/3.

That still doesn’t address my concerns about the lack of stipulation of the order. GB == BG surely? There is no stipulation of order of birth in the first or third so there are in fact only 3 options:

BB
GB
GG

The birth order is a red herring and I can’t understand how it could be that the answer to number 1 is anything but 1/2.

Okay, I have looked it up to save my own sanity, I shan’t spoil it but I think you are placing too much credence in the trickery of mathematicians.

Has anyone drawn a probability tree yet?

There is a difference between randomly selecting a child from two children and then noticing his sex and randomly selecting a two-child family with at least one son.

Apparently. Something to think about anyway.

Has anyone drawn a probability tree yet?

For the 3rd question? You’d be there for some time. Better off drawing a grid. If you do it in excel it will take you all of a minute.

logically, the day is irrelevant

Ultimately it all depends on the behaviour of the person asking the question (what if he was more proud of his son, what if he will always tell you about his eldest son first etc) which means that there are loads of different ways of establishing the probability. Given the constraints of the problem though and our limited understanding of the motivations of the questioner we have to find the most reasonable probability goven the info we have been given, not that which we have not, in these cases it is 1/2 for all to my mind.

If you imply the motivations of the questioner then you can probably make it come out as anything you wish.

Statistically the probability of two successive children both being born male is 0.25, regardless of the other factors mentioned in the question?? I can’t see how the answers are any different.

The probability of 1 is 0.5×0.5=0.25

You even say it there. The probability of outcome number 1 is 0.25. I don’t understand how or why you get 1/3 in your explanation. 😐

What do you mean when you say “logically”?

pocketshepherd, that would be true if the gender of the first child determined the gender of the second but it doesn’t

What do you mean when you say “logically”?

as in ‘ignoring statistical probability’ which is based on assumption and is only useful for actuaries, bookmakers and climatoligists 😀

You even say it there. The probability of outcome number 1 is 0.25. I don’t understand how or why you get 1/3 in your explanation.

But we want to know the probability of 1, GIVEN the fact he has at least one son.

Having one son means he can either have 3 different sets of children, one of which is two boys.

If you draw a probability tree of all the possible combinations, the scrub out the paths that don’t have at least one boy, you’ll see there are only 3 possibilities.

Admittedly the tree will show that each of those possibilites has only 25% of being true, but you have more info, you know that one of them has to be true, so (With your Bayesian head on) you adjust your probabilites.

I know you still want to say that the numbers show that it is only 25%, but the numbers also show you that there is a 25% chance of it being 2 girls and you know that ain’t the case

There is a difference between randomly selecting a child from two children and then noticing his sex and randomly selecting a two-child family with at least one son.

Apparently. Something to think about anyway.

eh? this is a troll right?
you haven’t worded the original question that way. effectively all we’re being asked is what is the probability of 1 child being male.

Anyway onto the fun one 🙂
We know one boy is born on a Tuesday
The possible BG/BB/GB combos with a Tuesday Boy (TB) are

1) tbmb
2) tbtb
3) tbwb
4) tbtb *
5) tbfb
6) tbsb
7) tbsb

8 ) tbmg
9) tbtg
10) tbwg
11) tbtg
12) tbfg
13) tbsg
14) tbsg

15) mgtb
16) tgtb
17) wgtb
18) tgtb
19) fgtb
20) sgtb
21) sgtb

22) mbtb
23) tbtb *
24) wbtb
25) tbtb
26) fbtb
27) sbtb
28) sbtb

Days of the week are abbreviated to M,T,W,T,F,S,S

Note the *, i’ve accidently listed the same combination twice, so taking that out we have 27 possibilities

1) tbmb
2) tbtb
3) tbwb
4) tbtb
5) tbfb
6) tbsb
7) tbsb

8 ) tbmg
9) tbtg
10) tbwg
11) tbtg
12) tbfg
13) tbsg
14) tbsg

15) mgtb
16) tgtb
17) wgtb
18) tgtb
19) fgtb
20) sgtb
21) sgtb

22) mbtb
23) wbtb
24) tbtb
25) fbtb
26) sbtb
27) sbtb

I’ve struck out girl combos, leaving the 13 BB ones.
So that’s 13 viable options out of 27, i.e. 13/27 odds.

you haven’t worded the original question that way. effectively all we’re being asked is what is the probability of 1 child being male.

Sorry, that’s not got anything to do with the original question. That was just for Torminalis, as he was asking the question “why?”. In a real life situation, it may turn out that the probability would be 0.5. In the question, the answer is 1/3 though.

In a real life situation, it may turn out that the probability would be 0.5. In the question, the answer is 1/3 though.

!!!???

that would be true if the gender of the first child determined the gender of the second but it doesn’t

flipping a coin, the probability of getting two heads in a row: 0.5 x 0.5 = 0.25. The first flip doesn’t affect the second but the overall probability is changed. the same goes with gender of successive children (assuming a 0.5 probability of either gender).

Right.

Looking just at winning conditions:

He can be the eldest child born on a Tuesday with a younger brother born on any other day. Seven permutations.

He can be the youngest child born on a Tuesday, another seven permutations, except we’ve just counted him once in “both on Tuesday” in the previous step, so we need to knock that off then we don’t count it twice. 7 + 6 = 13 permutations.

Now, looking at losing conditions we have a similar situation:

He can be the eldest child born on a Tuesday with a younger sister born on any other day. Seven permutations.

He can be the youngest child born on a Tuesday, another seven permutations, however the “both on tuesday” gotcha doesn’t apply here as the two are distinct, it’s girl/boy or boy/girl. 7 + 7 = 14 permutations.

So the answer is (winning permutations) out of (total permutations), 13 / (13 + 14), which is 13/27.

Assuming you’re talking about question 1..

A man has two children.

So we know he can either have (youngest-oldest)

1.(B-B)
2.(B-G)
3.(G-B)
4.(G-G)

Then we find out he has at least one son, so 4 is no longer a possibility.

The probability of 1 is 0.5×0.5=0.25
The probability of 2 is 0.5×0.5=0.25
The probability of 3 is 0.5×0.5=0.25

Because all 3 outcomes have the same probability, and there is only one outcome that fits the outcome we want (2 boys), it must be 1/3.

Surely this all comes down to how you interpret the question? If you take it as “he has one son, what’s the probability he has another” then you’d scrap both hypotheses 3 and 4 because your prior knowledge dictates that he has a son already. That then makes it a 50/50 jobby. In this case B-G == G-B as they’re the same scenario, the ordering makes no difference.

I’m writing this down for my own sanity, this is what I believe:
1) we know he has one son, so there are 3 possible remaining combinations of children, one of which is two sons, so 1/3
2) assuming that John is a son, there are 2 remaining combinations (BB and BG), so 1/2
3) is the same as 1) as the day is logically irrelevant