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## coin tossing, unknown gendered siblings and zero x zero is zero - part II

(836 Posts)
ZingDollyChops Sat 12-Oct-13 00:04:39

carry on folks.

we'll get there one day!

BlackeyedSusan Sat 12-Oct-13 00:06:51

its cold. do i need to take my socks off to do counting?

ZingDollyChops Sat 12-Oct-13 00:07:54
Sat 12-Oct-13 00:10:27

The answers are still 10 and 1/3

ZingDollyChops Sat 12-Oct-13 00:26:52

but you forget that if the coins land on a table either or both of them could land their Edges!

is it probable? how likely to happen? what are the chances? is tat even possible?

well, it is certainly possible. I have done that a few times.
and if you practice a lot to try and land a coin on its edge the probability of having Tail/Edge, Head/Edge, Edge/Tail, Edge/Head & Edge/Edge all increase.

so only what you assume to be possible is what you can expect to be probable!

Sat 12-Oct-13 02:18:01

I knew I hated (and failed!!) STatistics in school for a reason!

raggedymum Sat 12-Oct-13 07:13:25

Okay, I've woken up early in the morning pondering the Tuesday Boy thing -- what have you all done to me???

At the risk of making things worse for everyone else, can I please ask those who understand the boy-boy & Tuesday boy to tell me if I've got this right. I actually think that if I understand the Tuesday boy it will make boy-boy easier for me, just like the 100-doors makes more sense than the 3-doors.

Okay, so the question is: "I have two children. One of them is a boy born on a Tuesday. What is the probability I have two boys?"

The first statement ("I have two children") says the speaker is a member of the set of individuals who have two children. We can stop at this point (right??) and enumerate all the possible ways of having two children of either sex on different days of the week. We assume the probability of having a boy or girl is 1/2 each and of any given day of the week is 1/7 (ignoring planned C-sections and other things which shift the distribution into working days).

So that is:
GMon-GMon GMon-GTue GMon-GWed GMon-GThu GMon-GFri GMon-GSat GMon-GSun
GTue-GMon GTue-GTue GTue-GWed GTue-GThu GTue-GFri GTue-GSat GTue-GSun
GWed-GMon GWed-GTue GWed-GWed GWed-GThu GWed-GFri GWed-GSat GWed-GSun
GThu-GMon GThu-GTue GThu-GWed GThu-GThu GThu-GFri GThu-GSat GThu-GSun
GFri-GMon GFri-GTue GFri-GWed GFri-GThu GFri-GFri GFri-GSat GFri-GSun
GSat-GMon GSat-GTue GSat-GWed GSat-GThu GSat-GFri GSat-GSat GSat-GSun
GSun-GMon GSun-GTue GSun-GWed GSun-GThu GSun-GFri GSun-GSat GSun-GSun

GMon-BMon GMon-BTue GMon-BWed GMon-BThu GMon-BFri GMon-BSat GMon-BSun
GTue-BMon GTue-BTue GTue-BWed GTue-BThu GTue-BFri GTue-BSat GTue-BSun
GWed-BMon GWed-BTue GWed-BWed GWed-BThu GWed-BFri GWed-BSat GWed-BSun
GThu-BMon GThu-BTue GThu-BWed GThu-BThu GThu-BFri GThu-BSat GThu-BSun
GFri-BMon GFri-BTue GFri-BWed GFri-BThu GFri-BFri GFri-BSat GFri-BSun
GSat-BMon GSat-BTue GSat-BWed GSat-BThu GSat-BFri GSat-BSat GSat-BSun
GSun-BMon GSun-BTue GSun-BWed GSun-BThu GSun-BFri GSun-BSat GSun-BSun

BMon-GMon BMon-GTue BMon-GWed BMon-GThu BMon-GFri BMon-GSat BMon-GSun
BTue-GMon BTue-GTue BTue-GWed BTue-GThu BTue-GFri BTue-GSat BTue-GSun
BWed-GMon BWed-GTue BWed-GWed BWed-GThu BWed-GFri BWed-GSat BWed-GSun
BThu-GMon BThu-GTue BThu-GWed BThu-GThu BThu-GFri BThu-GSat BThu-GSun
BFri-GMon BFri-GTue BFri-GWed BFri-GThu BFri-GFri BFri-GSat BFri-GSun
BSat-GMon BSat-GTue BSat-GWed BSat-GThu BSat-GFri BSat-GSat BSat-GSun
BSun-GMon BSun-GTue BSun-GWed BSun-GThu BSun-GFri BSun-GSat BSun-GSun

BMon-BMon BMon-BTue BMon-BWed BMon-BThu BMon-BFri BMon-BSat BMon-BSun
BTue-BMon BTue-BTue BTue-BWed BTue-BThu BTue-BFri BTue-BSat BTue-BSun
BWed-BMon BWed-BTue BWed-BWed BWed-BThu BWed-BFri BWed-BSat BWed-BSun
BThu-BMon BThu-BTue BThu-BWed BThu-BThu BThu-BFri BThu-BSat BThu-BSun
BFri-BMon BFri-BTue BFri-BWed BFri-BThu BFri-BFri BFri-BSat BFri-BSun
BSat-BMon BSat-BTue BSat-BWed BSat-BThu BSat-BFri BSat-BSat BSat-BSun
BSun-BMon BSun-BTue BSun-BWed BSun-BThu BSun-BFri BSun-BSat BSun-BSun

That's a total of 49x4=196 possibilities.

The second statement ("One of them is a boy born on a Tuesday") narrows this down to only those possibilities which contain a Tuesday boy:

GMon-BTue
GTue-BTue
GWed-BTue
GThu-BTue
GFri-BTue
GSat-BTue
GSun-BTue

BTue-GMon BTue-GTue BTue-GWed BTue-GThu BTue-GFri BTue-GSat BTue-GSun

BMon-BTue
BTue-BMon BTue-BTue BTue-BWed BTue-BThu BTue-BFri BTue-BSat BTue-BSun
BWed-BTue
BThu-BTue
BFri-BTue
BSat-BTue
BSun-BTue

That's now 27 possibilities. 13 of them (the last block) are the ones that contain two boys. So the final part of the question ("What is the probability I have two boys?") is answered 13/27.

Have I got this right? Even now that I've typed it all out, I am really struggling to BELIEVE it. How could mentioning the day of the week on which the boy was born suddenly make it so much more probable (up from 1/3) that there are two boys? But I think the above calculation is consistent with the calculation that gives 1/3 for boy-boy. Therefore if someone can convince me the above is correct (if it is!), I think boy-boy = 1/3 will make sense to me.

(And to the person who asked about studying permutations, combinations, and probability in school, I learned in the US and I can say that we did NOT study these, and I went all the way through to AP-calculus. I had to pick up a lot of it on my own, later. I eventually took a couple graduate-level stats classes which helped a lot, but has left some strange holes in my knowledge.)

marcopront Sat 12-Oct-13 07:43:48

Ignoring the Tuesday issue, some conditional probability.

p(the woman has two boys given that she has a least one boy)
= p(two boys) / p(at least one boy)
= 0.25 / 0.75
= 1/3

PeteCampbellsRecedingHairline Sat 12-Oct-13 07:58:28

What hell have I stumbled across?

PenguinsDontEatPancakes Sat 12-Oct-13 08:05:00

Right, I literally lost sleep trying to think of an example that would explain this to people without somehow translating in their minds as implying birth order (or selection order) is relevant. I have come up with something I think both works and is seasonal! Here goes.

You are trick or treating and see an old school friend up ahead who you haven't seen for years. She has two children with her.

Scenario one

You walk up to her. One child is a boy in a devil costume. The other child is dressed as a ghost, covered head to toe in a sheet.

Clearly, the chance she has two boys is 50%. The child under the sheet could be a boy or a girl.

Scenario two

You walk up to her. One child is dressed as a ghost, covered head to toe in a sheet. The other is dressed as a gruffalo, covered head to toe in a costume.

During the course of a quick chat she mentions 'my son'. (Assume the two children she has with her are her only ones and the son isn't stashed at home!).

We do not care or know about the identity of the son. There are three possibilities:

- Ghost is a boy, Gruffalo is a girl
- Ghost is girl, Gruffalo is a boy
- Ghost is boy, Gruffalo is a boy

The chance that both costumes contain a boy is 1/3

This should explain as well why GB is different to BG. Ghost is a girl and Gruffalo is a boy is obviously different to Ghost is boy, Gruffalo is a girl without implying any order.

Scenario three - the one of with the confusion

This is the one I think a lot of people are getting stuck into seeing and why they can't see scenario two.

You walk up to her. One child is dressed as a ghost, covered head to toe in a sheet. The other is dressed as a gruffalo, covered head to toe in a costume.

During the course of a quick chat she mentions 'Freddie'. (Assume the two children she has with her are her only ones and Freddie isn't stashed at home!).

So you think that there are four possibilities:

- Ghost is Freddie, Gruffalo is a girl
- Ghost is Freddie, Gruffalo is a boy
- Ghost is girl, Gruffalo is Freddie
- Ghost is a boy, Gruffalo is Freddie.

Given there are four options, two of which have two boys, it looks like the answer is 50%.

The problem with this approach is that it is treating the identity of the boy as relevant and therefore introducing additional permutations which the question didn't reference and which mess up the probabilities.

Please, please if this doesn't explain it to any of you, can I at least get points for trying

AuntyEntropy Sat 12-Oct-13 08:06:32

Going back to the original BG thing. I think the people who say that 1/3 is "too theoretical" might find it useful to make it really concrete.
There are a million MNers. For the sake of argument, 400,000 of them have exactly two children. For well rehearsed reasons on the previous thread, that's 100,000 with GG, 100,000 with BB, and 200,000 with one of each.

If, in a discussion about the high cost of football boots, I mention that I have a boy, and in a discussion about food budgets I mention that I have 2 DC, then you know that I am one of the 300,000 that have one of each or two boys. And since the former group is twice as large as the latter, then it's twice as likely that I'm one of them.

But there are subtleties about the exact way you pose the question, as the Wikipedia page attests.

PenguinsDontEatPancakes Sat 12-Oct-13 08:09:37

OOh, I like your example Aunty. I did try and do that waaaaay back on the old thread about people in a hall but no one liked it!

prism Sat 12-Oct-13 08:31:16

It's simple really. The chances of having a boy are 50%. It doesn't matter whether you already have a boy, already have a girl, have no other children, or have dozens of them. Each time, it's 50%, or 1/2. To get the probability of one particular outcome, you multiply the probabilities of the steps involved. This is easy with the 2 boys scenario as there's only one way to do it, whereas if you were to have a one of each there are two ways. To have two boys, you have a 50% chance of success first time, and a 50% chance the second time, and obviously you're no longer in the running if you had a girl first. Multiply the probabilities, and your chances of having two boys are 1/4. Likewise the chances of having 3 boys (with 3 attempts) are 1/8, and 4 boys, 1/16.

Anyone who thinks that the chances of having two boys is 1/3 is extremely welcome to come to Prism Towers and play cards with me for money (I will send a car).

Sat 12-Oct-13 08:34:15

Like I said, a lovely tree diagram helps. They're visual and you can just walk along the path. Looks like people were busy till very late last night ��

PenguinsDontEatPancakes Sat 12-Oct-13 08:38:09

Prism - Yes, the chances of having two boys out of two children are 1/4. That isn't what people are discussing. They are arguing, if you know there is at least one boy, what the odds are of two. Your example still factors in the option of two girls to make 1 (1/4 two boys, 1/4 two girls, 1/2 one of each) whereas the 1/3 option removes two girls from the outset.

marcopront Sat 12-Oct-13 08:39:38

But prism the question is what is the probability of two boys when you know that one is a boy. You have ignored the knowing one is a boy.

NumTumRedRum Sat 12-Oct-13 08:45:34

Oh gawd. I wanted Mumsnet, seem to have stumbled accross Mathsnet. I don't understand . I sort of want to, but I think it might be beyond me this morning. Carry on, I'll lurk.

BeckAndCall Sat 12-Oct-13 08:47:09

OMG we've got time on our hands!

But, that said, kim - I want a tree diagram!!!

And yes agree that the conditional probability changes the simple 50/50 question. ( doesn't matter if I agree or not, does it, it just IS)

(And thank you to zing for the lovely flourish finish to the old thread)

Sat 12-Oct-13 08:53:30

Penguin's ghost & gruffalo example is very good. Read Scenario 2 until you understand it.

Alternatively, just flip the flipping coins and see.

PenguinsDontEatPancakes Sat 12-Oct-13 08:56:09

Why thank you Cote. I am not sure it was worth still being awake at 12.30 thinking it up. Lying in bed, am pregnant, getting over a bug and have two small children. Can I sleep? Nope. I'm thinking about maths

Sat 12-Oct-13 08:57:41

It took me a while to get to sleep, too.

All this for coins. I shudder to think how many threads we'll go through when we get to dice

Sat 12-Oct-13 08:57:52

Wonders if she should make a tree diagram and put it on her profile.

Sat 12-Oct-13 09:00:06

Here's a good question. If you get a coffee at Costa, how many different ways can they make it? Think of size, type, milk choice etc.

Sat 12-Oct-13 09:00:09

Do it, kim. Share the pain

prism Sat 12-Oct-13 09:00:37

Apologies. The chances of having a boy if you already have one (or indeed if you don't) are 50%. The chances of a pair of siblings being boys, given that one is a boy, are 1/3.

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