"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?

This problem, as stated, is ambiguous. In most fields of Mathematics, that makes it unanswerable. But Probability can handle it - in fact, you could say that the probability is a quantitative measure of the ambiguity in the scenario. You just need to make reasonable assumptions about the ambiguity.

Unfortunately, most people who answer problems like this don't recognize all of the ambiguities. Without even realizing it, they make assumptions that avoid the ambiguity, but affect the answer. But when you don't know that you are making an assumption, you can't defend whether or not it is a reasonable one.

In this case, the ambiguity is that we need to know why this person told you "I have two children and that (at least) one of them is a boy (born on a Tuesday)."

Case 1: The person was required to tell you that fact, if it was true. If it wasn't, other people would have been chosen until one who could truthfully make the required statement was found.

In this case, the answers as presented are correct: 1/3 (without "Tuesday") and 13/27 (with).

Case 2: Just one person was selected, and with free will offered to tell you (1) the number of his or her children, (2) the gender of (at least) one, and maybe (3) the day of the week a child of that gender was born.

In this case, **it is possible** for the father of a boy born on a Tuesday to tell you he has a boy born on a Thursday, or for the mother of a boy born on a Tuesday to tell you she has a girl born on a Friday. The previous answers are wrong, because they count every who parent could tell you about the boy born on a Tuesday, not the ones who would.

If, instead, you count only half of the parents who have a boy born on a Tuesday and another child that fits some other description, you get 1/2 for every possible permutation of the facts.

One of these cases is reasonable to assume, when you aren't told the parent's motivation. The other is not. And your intuition is a good guide to tell which.

BTW, this problem originated at a conference honoring Martin Gardner, the long-time author of the Mathematical Games column in Scientific American. He originally answered 1/3 for the simpler problem, which is all most people recall. They somehow forget that he withdrew that answer, and said that either 1/3 or 1/2 could be correct based on the very ambiguity I described.

]]>"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 answer13/27... Really?

Assume that boy/girl are equally likely and each day of the week is equally likely for each child.

Case 1: Both children were born on the same day of the week (1/7).

This day could be Tuesday (1/7) or NotTuesday (6/7).

Genders could be BB (1/4), BG (2/4), GG (1/4).

Case 2: The children were born on different days of the week (6/7).

Case 2a: Exactly one of the children was born on a Tuesday ((6 choose 1)/(7 choose 2) = 2/7).

Gender of the Tuesday child could be B (1/2) or G (1/2).

Gender of the other child could be B (1/2) or G (1/2).

Case 2b: Neither child was born on a Tuesday ((6 choose 2)/(7 choose 2) = 5/7).

Genders could be BB (1/4), BG (2/4), GG (1/4).

This is the sample set. Applying the restriction "one of the children is a boy born on a Tuesday" leaves us with

3/196 from Case 1, including 1/196 that both are boys.

24/196 from Case 2, including 12/196 that both are boys.

The probability of both being boys given the restriction is thus (1/196 + 12/196)/(3/196 + 24/196) = 13/27.

To put it another way, saying the boy mentioned was born on a Tuesday has a 12/13 chance of identifying the boy mentioned (1/2 chance of the other being a boy), and a 1/13 chance of leaving the boy unidentified (1/3 chance of both being boys).

]]>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.

]]>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.

]]>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.

]]>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.

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.

]]>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.

______________________________________________________________________________

]]>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.

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%!

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.

]]>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.

]]>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!?

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