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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,535

http://www.maa.org/devlin/devlin_04_10.html

"I tell you I have two children and that (at least) one of them is a boy, ... your right answer to my question is not 1/2 but 1/3." Agree!

"I tell you I have two children, and (at least) one of them is a boy **born on a Tuesday**. What probability should you assign to the event that I have two boys?" he gives the answer **13/27** ... Really?

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,705

Bellos wrote:

Probability strikes fear into even the most brilliant of mathematicians, since it is so full of bonkers counter-intuitions.

He's right, it is easy to make errors, and difficult to check.

Although it is conterintuitive:

They treed it.

There is the 27 ways he is talking about and you can count the 13 winners.

Tree-ing is what Marilyn did in Parade magazine for that famous Monty Hall problem.

Based on how they define the problem I can't see any error. Could be that it is not defined well enough.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,535

So if he had specified Tuesday at 11:23:38 the odds would approach 1/2.

Giving more un-related details of one boy changes the odds ... very counter-intuitive indeed.

**Don't let him speak! He is changing the probabilities!**

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,705

That is interesting, it does appear the more things he specifies the closer the prob. will get to 1/2???

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**zayzay199****Member**- Registered: 2010-05-21
- Posts: 5

bobbym wrote:

That is interesting, it does appear the more things he specifies the closer the prob. will get to 1/2???

hi bobbym whats up.

r u the owner of this website too cos its a webside fab

*Last edited by zayzay199 (2010-05-21 07:33:43)*

hiya

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,705

Hi zayzay199;

No I am not. MathsisFun is the one you are thinking of. Welcome to the forum!

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**ZHero****Real Member**- Registered: 2008-06-08
- Posts: 1,889

DA@maa.org wrote:

So now you know the possible gender combinations are BB, BG, GB. Of these three possibilities, in two of them I have a boy and a girl, and in only one do I have two boys, so you should calculate the probability of my having two boys to be 1 out of 3, namely 1/3.

Out of the stated three combinatons BB, BG, GB, I think that BG and GB are identical!!

And hence, instead of 1/3, 1/2 should be correct!

If two or more thoughts intersect with each other, then there has to be a point.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Try this page here.

Try to read it with an open mind. They are both mathematicians so...

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**ZHero****Real Member**- Registered: 2008-06-08
- Posts: 1,889

I haven't yet thought about "a boy born on Tuesday" one...

But for the first one, the only possibilities are...

1. Both children are boys. (√)

2. Both children are girls. (x)

3. One of the children is boy and the other one is girls. (√)

I agree that the question is not framed properly since nothing is mentioned about who's younger/elder.

The question is analogous to possessing (order doesn't matter) two balls of different colors (say PINK and BLUE) and not to drawing two balls from two sacs of two different colored balls wherein the order of ball drawn matters!

If two or more thoughts intersect with each other, then there has to be a point.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Probably not, no pun intended. Apparently they have used a tree to answer the questions , in other words they counted them all manually. Best to try to absorb what they did for the next time. That's the way probability works.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

ZHero wrote:

I haven't yet thought about "a boy born on Tuesday" one...

But for the first one, the only possibilities are...

1. Both children are boys. (√)

2. Both children are girls. (x)

3. One of the children is boy and the other one is girls. (√)I agree that the question is not framed properly since nothing is mentioned about who's younger/elder.

The question is analogous to possessing (order doesn't matter) two balls of different colors (say PINK and BLUE) and not to drawing two balls from two sacs of two different colored balls wherein the order of ball drawn matters!

Yes, those are the only possiblities. But they're not equally likely.

You can try it yourself, by flipping coins. Flip 2 coins at a time and the possible results are:

- Two heads

- Two tails

- One of each

The third result should come up roughly twice as much as the other two.

It's an interesting question though, MIF. I've half got my head around it now.

The two extremes of the question are:

"I have two children. This one here, who I'm directly pointing at with no possiblility for confusion, is a boy. What's the probability that I have two boys?"

Answer 1/2.

"I have two children. One of them, and I'm not giving you any information about which, is a boy.

What's the probability that I have two boys?"

Answer 1/3.

The rest of this type of question form a 'sliding scale'. The probability depends on how likely it is that the given information could describe either child.

So, if he says "One of them likes ice-cream and is a boy", then the answer is close to 1/3, since clearly everyone loves ice-cream.

With the Tuesday clue it's less likely that the two children could be confused, and so the answer is closer to 1/2.

If you said that one child was born on Tuesday while football was on TV and it was a bit cloudy and in the room next door somebody was listening to Lady Gaga, and is a boy, then you're almost certainly specifying the child you mean. So the probability there is 1/2 - ε

Why did the vector cross the road?

It wanted to be normal.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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You know Leibnitz got into trouble in the same way. By stating that the odds of throwing an eleven were the same as throwing a twelve in two rolls of a dice. He said there is one 5 and one six, one way, period.

Anyway it is twice as likely to throw an 11 as it is to throw a 12. (5,6) is not the same as (6,5). Neither is (h,t) and (t,h) the same when discussing coin tossing.

Enumerating the possibilites:

(G,G) (B,B) (B,G) (G,B) , so knowing that I have at least 1 boy, it is 1 in 3 that the other kid is a boy. Funny thing is on Devlins page some guy used Bayes Theorem to get the wrong answer! That's probability for ya.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**ZHero****Real Member**- Registered: 2008-06-08
- Posts: 1,889

mathsyperson wrote:

Flip 2 coins at a time and the possible results are:

- Two heads

- Two tails

- One of eachThe third result should come up roughly twice as much as the other two.

Got it since the Sample Space is {(HH), (HT), (TH), (TT)} and there are Two favorable outcomes for "One of each"! (I wouldn't go about the crazy idea of trying flipping a coin myself though )

But this is fairly coz of one main reason... Coin-1 and Coin-2!!

Your next few words explain it More Clearly........

mathsyperson wrote:

The two extremes of the question are:

"I have two children. This one here, who I'm directly pointing at with no possiblility for confusion, is a boy. What's the probability that I have two boys?"

Answer 1/2."I have two children. One of them, and I'm not giving you any information about which, is a boy.

What's the probability that I have two boys?"

Answer 1/3.

Nice explication of the effect of distinguishableness.

As i mentioned earlier.....

ZHero wrote:

I agree that the question is not framed properly since nothing is mentioned about who's younger/elder.

1/3 is possible for the case Child-1 (Younger) and Child-2 (Elder).

I'd set it that the probability is 1/2 if they are twins!?

If two or more thoughts intersect with each other, then there has to be a point.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Hi;

Doesn't matter, throwing a die one at a time or both at once , same probabilities. The tree structure is the correct way to think about this.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**ZHero****Real Member**- Registered: 2008-06-08
- Posts: 1,889

Hi;

Throwing a die one at a time → Throw-1 and Throw-2

Throwing tow dies at once → Die-1 and Die-2

Think distinguishability of the events plays an important role here!

Both the experiments are Similar because the events can be Differentiated!

I Can't think of any suitable example resembling "Twins" right now.... Hope there must be.

*Last edited by ZHero (2010-05-21 21:44:49)*

If two or more thoughts intersect with each other, then there has to be a point.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Nope, don't worry, about that. The two dice are twins and so are the two coins and B and B. Twins already. Hope not , Vegas would come apart. Probability needs to work.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**ZHero****Real Member**- Registered: 2008-06-08
- Posts: 1,889

Nope, don't worry, about that. The two dice are twins and so are the two coins and B and B. Twins already. Hope not , Vegas would come apart. Probability needs to work.

Consider the following experiments..

1. Throwing two identical unbiased dies and getting a sum of 11.

CASE-I: you get a 5 on dies-1 and 6 on dies-2

CASE-II: the numbers swap.

Both the cases are different and acceptable.

2. Throwing a dies twice and getting a sum of 11.

CASE-I: you get a 5 on throw-1 and 6 on throw-2.

CASE-II: a 6 on throw-1 and 5 on throw-2.

The cases, again, are different and acceptable too!

Now consider the experiment dealing with babies exactly identical in > height, weight, date/time of birth, choices (if that matters) but with different possible sexes except of course for the case that both are girls}...

CASE-I: both are boys.

CASE-II: **one** (which one?) is a boy and the **other** is not.

This is in no manner similar to stating that the probability of drawing a card from a pack of 52 cards and its being an A is 1/2 (either it **is** an A or **it is not**) because if it ain't an A then there are 12 other possibilities.

On the other hand, in case of the problem in context, no other outcomes are possible.

Normally, it doesn't have ANY business (whatsoever) with Vegas.

Probability works only on papers. Probability that you'r an Asian is 40%!

If two or more thoughts intersect with each other, then there has to be a point.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,705

Nope, probability also works empirically. Vegas is the best example of that. There is a nice motto here. It goes like this: If you want to make money, find someone who knows less probability than you do.

Fact is this the tree:

(G,G) (B,B) (B,G) (G,B) this is the pertinent examples. (B,B) (B,G) (G,B) . 2 to 1 against or 1 in 3. Ya just can't argue with counting.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**gurthbruins****Member**- Registered: 2010-05-09
- Posts: 157

What a lot of baloney! If you have two children, and you tell me at least one is a boy, this can in no way influence what the other is. That is not only intuitive, it is obviously true. Regardless of whether the mentioned boy is one born on a Tuesday, having blue eyes, being a runt or anything else, the properties of the boy have no bearing on the fact that it is a boy.

So any reasoning to the contrary must be fallacious. In fact all Devlin's reasoning contains numerous non-sequiturs (unwarranted assumptions).

The tree is also fallacious, since it assumes cases to be equally probably which are no longer equally probable. The tree, however, does highlight the inaccuracy of the 1 in 3 assumption.

In all cases, the other child is equally likely to be boy or girl, there is no information given which can affect this.

_____________________________________________________________________________________________

I posted the above before, then deleted it thinking it mistaken, so I have thought a lot before coming to the following rather earth-shaking conclusions.

The set of probabilities BB BG GB GG can only be considered valid or applicable if we have no information about the possibility of any one of these 4 possibilities. As far as we know, they are all equally possible.

The moment it is stated that one or two are boys, the applicability of the set is destroyed. We cannot simply delete the possibility GG and keep the other three, because the whole set DEPENDS on GG being a possibility.

Therefore we'll have to look at the problem afresh, and only apply probabilities to uncertain facts. We can't mix certain and uncertain data in one equation.

So we can say: there is at least one boy; that is a certainty. So we put one boy in a cage. That leaves one other child about whom we know absolutely nothing. Therefore the chances are equal that the second child will be B or G.----------------QED.

The second problem is exactly the same: the probability that the boy was born on a Tuesday is immaterial, because it is no longer a probability but a certainty.

______________________________________________________________________________

*It's the activity of the intelligence above all that gives charm to existence.*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,705

Hi gurth;

Good topic. You only have these possibilities:

BB BG GB GG. They are all equally likely 1 /2 * 1/2 = 1 / 4. You know you have at least 1 boy. that rules out GG.

You are left with BB BG GB , still equally likely. There is only one them that gives you 2 boys BB , BG and GB don't. It is 2 to 1 against or the odds are 1 in 3 or 1 / 3.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**gurthbruins****Member**- Registered: 2010-05-09
- Posts: 157

No, bobby, you miss my point entirely. The possibilities BB BG GB GG are only equally likely if you know nothing. Once you know something, they no longer apply. Your reasoning is the old reasoning, which I also accepted before I thought about it more deeply.

*It's the activity of the intelligence above all that gives charm to existence.*

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**gurthbruins****Member**- Registered: 2010-05-09
- Posts: 157

FURTHER CONSIDERATIONS

To restate the problem:

"Consider a pair of siblings. Given that at least one is a boy, what is the probability that both are boys?"

This "given that" is too glib. We need to ask HOW this CAN be given, or KNOWN.

Let's consider a city with a million pairs of siblings. We know that the ratio of boys:boy-girl:girls

will approximate 1:2:1. This means that if we take any pair AT RANDOM, (not a SELECTED pair) then we

can say the chances of boys:boy-girl:boys are 1:2:1.

Now we are confronted with a single pair and told that at least one is a boy. This implies that the

pair has NOT been taken at random but SELECTED from the million pairs in the city. Therefore the

chances 1:2:1 are no longer applicable. We are not informed by what process the pair has been

selected.

We are entitled to consider the possibility that the pair may have been SELECTED by some PERSON. This

person may have SELECTED a pair of boys, in which case the probability of 2 boys is 1 in 1. Or he may

have SELECTED a boy-girl pair, in which case the probability of two boys is 0 in 1.

We are entirely in the dark about how the pair has been SELECTED. We cannot do any better than confess

our complete ignorance of the probabilities.

*It's the activity of the intelligence above all that gives charm to existence.*

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**mathsyperson****Moderator**- Registered: 2005-06-22
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I'm not sure that knowledge implies intent like you suggest.

I could throw a dice and say "Oh look. It's not a 6." But that doesn't mean I meant for it to not be a 6. It's just what happened.

Why did the vector cross the road?

It wanted to be normal.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,705

Hi gurthbruins;

You have a programming background. Try this, don't think about it, program a simulation. Or draw a tree. When in programming mode I don't need mathematical certainty, just empirical evidence. Program it honestly, no putting your own opinions into it, make it exactly model the problem like they explain it. Might help.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**aawood****Member**- Registered: 2010-06-25
- Posts: 2

gurthbruins wrote:

No, bobby, you miss my point entirely. The possibilities BB BG GB GG are only equally likely if you know nothing. Once you know something, they no longer apply. Your reasoning is the old reasoning, which I also accepted before I thought about it more deeply.

You're right in that given a set of options, once you know something that limits that set, those possibilities no longer apply. But as all you know is that 1) there are two siblings, and 2) one of these at least is a boy, then all you've done is reduce it to 3 possibilities; BB, BG, or GB. At this point, only one possibility leaves you with two boys, and two with a boy and a girl; 1/3 chance of there being two boys. Your assumption is that once you know a boy is there that you've removed one of the ELEMENTS of the puzzle, and that's not quite true.

The problem with your reasoning, and is is a very understandable one, is you've taken an element out of the equation without knowing which element it is you've removed, and that's important to know. If it was the first kid, you know it's either BG or BB. If it was the second, it's either GB or BB. If it is BB, the chances of a given boy being taken is only half, so you're left with a certainty of kid A being taken if it's BG (1/3 chance), a certainty of kid B being taken if it's GB (1/3 chance), or a 50/50 split of either kid being taken if it's BB (1/3 chance). So, 1/3 chance of it being two boys, 2/3 chance of it being a boy and a girl.

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