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- Thread starter Howers
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the probabilities do add up if you count your chances of a boy girl family or a girl boy family... theres a .25 chance of both... then a .25 chance boy boy and .25 girl girlYour probabilities don't add up right. There is a .25 chance of boy boy, and a .25 chance of girl girl, and a .5 chance of boy girl. There is twice the chance of getting boy girl, than boy boy. Eliminating the chance for girl girl does not change this ratio.

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Green Zach fell into the trap of

The assumption of equiprobability is just that, an assumption. Whenever one applies this assumption one should also do some sanity check regarding the validity of making this assumption. The assumption makes sense in the case of a coin toss or the roll of a single die. The assumption does not make sense and is not valid when one looks at the problem of the sum of the rolls of two dice. There are 11 outcomes, the numbers 2 to 12. However, the 11 events are not equiprobable. The probability of rolling a seven is six times that of rolling snake eyes. Similarly, the assumption simply is not valid in the problem at hand.two children. 100% chance of one boy. 50% chance of another boy, 50% chance of a girl.

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CompuChip

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If I throw a die, I will get 6 or I won't. Therefore, the probability of getting 6 is 50%.

I don't think I'd agree :)

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Im pretty sure that is what he was implying

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so now we have boy boy and girl girl at 0.25 and boy girl at 0.5. becusae the families must have at least one boy, girl girl is eliminated. now the numbers dont add up to 1.0 becsaue one of the variables was eliminated. Now the probibilities are out of three. 1/3 = 0.25 or 25%.

therefore becasue we are left with 0.25 boy boy and 0.5 girl boy, that translates into a 1/3 chance of getting boy boy and 2/3 chance of getting boy girl.

please tell me if this makes any sence lol

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the way it is worded can lead you to a conclusion that it is 50% that is why i always hate probability... especially when i play my odds i always lose anyway so i tend to stay away from probability!

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This is how I think it is worded, firstly, a family with at least a boy is chosen, we should have factored in 75/25% with this selection, but this is wrong simply because the criteria is at least one boy, therefore it is not 75/25%. In fact there isn't any such probability simply because a boy MUST AT LEAST be chosen. Probability cannot be factored in. Only after asking of what probability is there a girl.Here is a very famous problem:

A random two-child family with at least one boy is chosen. What is the probability that it has a girl? An equivalent and perhaps clearer way of stating the problem is "Excluding the case of two girls, what is the probability that two random children are of different gender?"

So it is 2/3.

one simple way of thinking of this is, there is an equal chance of girls as there are boys for any slot. There are 4 combinations, if only 3 out of the 4 is used to count, then the

probability of finding one girl is 2 out of 3 while for a boy it is 3 out of a 3.

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Let's look at this random two child family.Here is a very famous problem:

A random two-child family with at least one boy is chosen. What is the probability that it has a girl? An equivalent and perhaps clearer way of stating the problem is "Excluding the case of two girls, what is the probability that two random children are of different gender?"

Child B: Boy

Child U: Unknown gender

The question is simply this, what is the probability that "Child U" is a girl?

A good estimate may rely upon population statistics. One way is to simply compare the birth rate of male children compared to the birth rate of female children in different countries. See http://www.photius.com/rankings/population/sex_ratio_at_birth_2007_0.html

The irony in the second form of the question is that if you assume "that (the) two random children are of different gender", "the case of two girls" is automatically excluded, and therefore, stating the problem as the following is redundant, "Excluding the case of two girls, what is the probability that two random children are of different gender?"

In there, the probability is clearly 2 out of 3 (ignoring the population data above), since 1 out of 4 of the possibilities is excluded. For example, the first child could be a boy and the second a girl. The second case is a first child who is female and a second child who is male. The third case is simply two boys.

The questions are obviously not equivalent. Yet possible arrangements are exactly the same. The paradox is well known, "There are lies, damned lies, and statistics."

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- #39

HallsofIvy

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If you mean exactly what you say, then the "possible arrangements" are NOT exactly teh same. For the first question, the possible arrangements are BB, GB, BG, GG. For the second, BB, GB, BG because we have been instructed to "Exclude the case of two girls".

When I saw the title to this thread, I thought you would be talking about the distinction between

"Of two children, at least one of them is a girl. What is the probability that the other child is a boy?" and

"Of two children, the older is a girl. What is the probability that the other child is a boy?"

It is easy to show that the first probability is 2/3 and that the second is 1/2. That is peculiar but still not a paradox. We have different probabilities based on different information.

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Let girls in the military and be done with it. In a while, they'll be suing the government for equal rights, anyway.I think there is another more-or-less equivalent problem formulated in a more interesting way.

A king of a certain distant country has decided that he wants more men in the population for military purposes. He thus decides of a new law enforcing that a couple is allowed to have only one girl.What is the boy/girl ratio resulting in the population ?(assuming we wait for several generations, after the last person born before the law has deceased, to get a stable asymptotic result for instance).

A family could have say 7 boys, the eighth kid being a girl preventing any further child in the family.

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I think we're saying the same thing...I hope ? I'm pointing out the fallacy of that argument...unless you really believe I believe there's only a 50% chance of the sun rising, in which case I need to build up my credibility

If I throw a die, I will get 6 or I won't. Therefore, the probability of getting 6 is 50%.

I don't think I'd agree :)

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Sorry to necro an old thread, but - as is usual with this "famous" problem - the exact wording is what makes it seem paradoxical, and that was overlooked. Compare these possible statements of the problem:A random two-child family with at least one boy is chosen. What is the probability that it has a girl?

- A family is selected at random from all two-child families. An observer picks one of the children at random, and tells you that child is a boy.
- A family is selected at random from all two-child families that have at least one boy.
- A two-child family is known to have at least one boy.
- A family is selected at random from all two-child families. An observer looks at both children, and decides to tell you that at least one is a boy.

- 1/2. The unselected child's gender is independent of the selected child's gender.
- 2/3. 1/4 of all two-child families have two boys, 1/2 have a boy and a girl, and 1/4 have two girls. The two-girl families are eliminated from consideration, but all of the remaining ones are possible. P(BG)/[P(2B)+P(BG)] = (1/2)/[(1/4)+(1/2)]=2/3.
- It is ambiguous.
- It is still ambiguous, but less so. The best answer, 1/2, can be defended where no other can.

The last statement is also ambiguous. It is an

The problem is famous because not everybody recognizes the ambiguities, even while misinterpreting some of the clearer versions. Both answers have been given to most version, by intelligent and educated people. Some will take the minimalist approach, you only know about one child and you can't assume that both were observed. Others will take the maximalist approach, that the implication in "at least" means both children must have been used to create the condition.

The problem is old, originating at least as long ago as 1959, in Martin Gardner's

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Haha, this (the original problem) reminds me of the Monty Hall Problem. The answer is indeed 2/3.

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There is no ambiguity. In cases 1 and 4, if the randomly selected family is a two family, the observer has two choices: lie about one of the children being a girl, or pick another family. The implicit assumption here is that the report of a family having at least one boy is truthful. All bets are off of course if the report is untrustworthy.Compare these possible statements of the problem:

- A family is selected at random from all two-child families. An observer picks one of the children at random, and tells you that child is a boy.
- A family is selected at random from all two-child families that have at least one boy.
- A two-child family is known to have at least one boy.
- A family is selected at random from all two-child families. An observer looks at both children, and decides to tell you that at least one is a boy.

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CompuChip

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Which is why you can't treat it as a lie (any more than you should consider the possibility the family has 5 children, or that the selection wasn't done at random.)The implicit assumption here is that the report of a family having at least one boy is truthful. All bets are off of course if the report is untrustworthy.

The "implicit assumption" you are making here is that the observer is required to say "at least one is a boy." There is no such requirement. The observer merely states a fact.In cases 1 and 4, if the randomly selected family is a two (did you mean to say "two-girl" here?) family, the observer has two choices: lie about one of the children being a girl, or pick another family.

Case 1 is trivial to solve by a frequentist approach: There are four possible outcomes for the random selections, each with probability 1/4: the family has two boys and you pick a boy, the family has a boy and a girl and you pick the boy, the family has a boy and a girl and you pick the girl, or the family has two girls and you pick a girl. The last two cases, where you picked a girl, are eliminated by the observation (which can't be considered to be a lie), so the answer is (1/2)/(1/2+1/2)=1/2.

Since statements 3 and 4 could result from either 1 or 2, they are ambiguous.

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And this is why there is always a controversy about these problems, no matter how they are asked. :)

First off, birth order is only important for counting the various possibilities. It sometimes is easier if you consider a smaller number cases with

Here's the solution for problem #1, "collapsing" the birth order: There are eight possible cases, based on birth order and which child you picked: BBO, BBY, BGO, BGY, GBO, GBY, GGO, and GGY. Since you actually picked a boy, you can eliminate the four cases where a girl would have been picked; leaving BBO, BBY, BGO, and GBY. Of these four equally-likely cases, in two of them the family also has a girl. The probability is therefore 2/4=1/2. (This solution is a varaition of the one used for another famous "paradox," called either Bertrand's Box Paradox or Gardner's Three-Card Swindle. The only difference is that those have three equally-likely sets-of-two, where we have four.)

Essentially, my problem #1 picks a child at random. Along the way it also picked a family at random, but the information you got was only about the child (just as in Bertrand's Box, you picked a coin at random, and for the There-Card Swindle, you picked one side of a card). In particular, the observer may never even have seen the "other" child.

In my problem #2, the information is about the family. In particular, whatever observer gave you the information (1) had to look at both children in the family, and (2) had pre-decided to tell you that at least one child was a boy whenever that was true. (As CompuChip pointed out, there was a "requirement that he will tell you that [at least one is a boy] whenever there is at least one boy.")

In my #3 and #4, it is not clear whether the observer saw one or two children; or whether there was a "requirement that he will tell you that [at least one is a boy] whenever there is at least one boy." In fact, I made a point of leaving such a requirement out. And that is what makes the two cases ambiguous.

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Let me assure you that I choose the wordings of problems #1 and #2 very carefully, so that there could be no ambiguity other than by misperception. The answers are 1/2 and 2/3, respectively. You can easily write a computer simulation to verify them, if you are so inclined.

But we have not been trained properly to see what factors are important to establishing what the probability experiment is, and what is part of the smoke and mirrors. It hasn't helped that ambiguous versions of this problem have been presented, and answered as if they always had to be similar to #2 by people who should know better. That's why I pointed ou the one person, Martin Gardner, who recognized his mistake after making it

What you need to do is figure out what the information you have is "connected to," and what possibilities exist that would allow you to get that information. In my problem #1, "boy" is connected with one individual. In my problem #2, it is connected with the most liberal observation of the family. In #3 and #4, it is not clear what it is connected to.

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