Boy/Girl probability in Monty Hall

Cecil's Columns/Staff Reports
1

Dear Cecil,

From column
http://www.straightdope.com/classics/a3_189.html

There is a family with two children. You have been told this family has a daughter. What are the odds they also have a son, assuming the biological odds of having a male or female child are equal? (Answer: 2/3.)

The possible gender combinations for two children are:

(1) Child A is female and Child B is male.

(2) Child A is female and Child B is female.

(3) Child A is male and Child B is female.

(4) Child A is male and Child B is male.

We know one child is female, eliminating choice #4. In 2 of the remaining 3 cases, the female child’s sibling is male. QED.

This is a fallacy. It remainds me of an old coin toss problem.

Let’s throw a coin three times. If it comes up all the same side you win, otherwise I win. The odds are fifty/fifty as you can see:

  1. All heads

  2. All tails
    =YOU WIN

  3. Two heads and a tail

  4. Two tails and a head
    =I WIN

Out of four options we each have two winners. But intuitively you’ve probably guessed there’s something wrong here. It’s not that easy to get three heads/tails. Our list of options should look like this (with letters for heads/tails):

1.H H H
2.T T T
=YOU WIN

  1. H H T
  2. H T H
  3. T H H
  4. T T H
  5. T H T
  6. H T T
    =I WIN

So the odds are actually 1/4 for you and 3/4 for me. The fallacy is assuming that T T H is the same as H T T, and T H T.

In the gender problem you list the options as:

  1. F M
  2. F F
  3. M F
  4. M M

While this is accurate as far as it goes, your final statement immediately shows up the fallacy. “In 2 of the remaining 3 cases, the female child’s sibling is male”

You seem to forget that in the real world the females in option 2. are in fact two different people, who BOTH have a sister. So let’s call the named child Debbie, in which case your options are:

  1. Debbie + M
  2. M + Debbie
    3 Debbie + F
  3. M M

But we’re leaving out the important fact that another option exists:

  1. F & Debbie

[and also in theory, 6. M M reversed]

Taking this into account, Debbie has potential for an older/younger brother OR an older/younger sister–making the odds 50/50, as nature intended it.

Thanks,
David, Ireland

2

Nope, Cecil is right. Of families that have two children, 1/4 will have two daugters, 1/4 will have two sons, and half will have a son and a daughter. So, of the families that have a daughter, 2/3 also have a son.

3

As soon as you choose one child (e.g., Debbie), the situation changes. But as long as all you know is “One child is a daughter,” the odds continue to be 2/3 that the other child is a son.

4

My way of looking at it:

(1) If you pick children at random, and find you have picked a girl witb one sibling, then it’s a 1/2 chance that her sibling is a boy.

(2) If you pick families at random, and find you have picked a family with two children, at least one of whom is a girl, then it’s a 2/3 chance that the family contains a boy as well as a girl.

So it depends on what your random selection process is.

5

I agree, BUT it’s important that phrasing the puzzle be clear. “One child” may mean there is AT LEAST one child (or at least one of whatever you may be talking about – For example: “This solid figure has one right-angle.”) That is, I believe, the way mathematicians speak.

But in common parlance it is often unclear. Rather than “one child” or “a child” perhaps it should be spelled out as “at least one.”

Even saying, “They are not both boys” wouold be clear enough, albeit an unusual way to say it.

But you are correct and onetimepad is not. My point here is one about semantics, not mathematics.

I found his example of all-the-same-way flipped coins interesting, though.

True Blue Jack

6

Isn’t it true that 1 and 3 are in fact the same scenario? 1 boy, 1 girl. If that’s the case, then the odds are 1/2, not 2/3.

If they’re not the same, and if child A is the first child mentioned (the girl), and B is the other child, then 3 is invalid because child A must be a girl.

In either case, the answer is 1/2, not 2/3.

7

1 and 3 are not the same. Lets list them as follows

  1. first born child girl second born child boy

  2. first born child girl second born child girl

  3. first born child male second born child female

  4. first born child male second born child male

case 1 the girl has a younger brother
case 3 the girl has an older brother

So they are clearly different families.

8

You’re adding an explanation that isn’t relevant to the puzzle. It doesn’t matter if child B is older or younger, because the puzzle only mentions the other child.

There only three ways of having two children:

  1. two boys
  2. two girls
  3. one of each

1 is invalid, and so the answer to the puzzle is 1/2.

There are four ways to order two children, or four different ways they can come to your house:

  1. boy then boy
  2. boy then girl
  3. girl then boy
  4. girl then girl

1 is invalid. 2 is also invalid, because the first child in the puzzle was a girl. Only 3 and 4 remain as possibilities, and so the answer to the puzzle is 1/2.

9

True enough. If we assume that girls and boys are equally likely then the probabilities for these families are

1/4 two boys
1/4 two girls
1/2 one of each

In a population of 1000 families

250 have two boys
250 have two girls
500 have one of each

There are 750 families that have at least one chiled a girl
250 have two gilrs
500 have one of each

500 out of these 750 have a boy or 2/3 of them.

10

I should expand on the probabilities a little bit.
The probabilities can be shown as follows again the same 1000 families

500 families have the first child a boy of those families 250 have the second child a boy and 250 of them have the second child a girl.
250 are two boys and 250 one of each

500 families have the first child is a girl of those families 250 have the second child a boy and 250 of them have the second child a girl.
250 are two girls and 250 one of each

This makes
250 families have two boys
250 families have two girls
500 families have one of each

11

The point you are missing is that case #3 (one boy, one girl), is TWICE as common as the other two cases. 25% of two-child families will have two girls, 25% will have 2 boys, and 50% will have one of each. So in the first puzzle (the probability that a pair of siblings, of which at least one is a girl, also includes a boy) is, as stated, 2/3.

The second problem (a girl with one sibling visits, what is the probability that she has a brother) is a little trickier. What many people have forgotten to take into acount is that a family with two girls has two girls who each have a sister, whereas a familiy with one of each only has one girl who has a brother.

To make following this easier, let us consider these families:

  1. Aaron and his brother Adam (2 boys)
  2. Betty and her brother Bill (1 girl, 1 boy)
  3. Cedric and his sister Cecilia (1 boy, 1 girl)
  4. Daisy and her sister Dot (2 girls)

You are expecting one set of siblings to visit, but you don’t know which set. You hear a knock at the door. Your roomate answers the door and says there is a girl looking for you. What is the probability that she will have a brother?

There are four possible girls (Betty, Cecilia, Daisy and Dot). Two of them have brothers (Betty and Cecilia) and two of them have sisters (Daisy and Dot). Therefore the probability that the girl at the door has a brother is 2/4, or 1/2. QED.

JRB

12

Incidentally, onetimepad’s explanation is the solution to the second puzzle, not the first.

13

I wrote a lengthy amount of posts on this puzzle in the several other threads and I won’t bore everyone with rehashing all of stuff again. This puzzle appears to have been around at least 60 years, and it was shot down by both *Scientific America *and the guru of puzzles, Martin Garner. Oddly enough, Martin Garner was use by Marilyn as the judge of her work in the Monty Haul puzzle.
The 2/3 answer is just bad logic…its as simple as that.

This are the facts:

The four possible family combinations:
Column A ….Column B
1)……Girl………… Girl
2)……Girl………… Boy
3)……Boy………… Girl
4)……Boy………… Boy

Statements of Equality

*If the families are equal, then the children are not. If we are given a girl and the families are equal, then each families (#1, #2, #3) are each likely 33.3% of the time… making each girl in family #1 worth half the value of the girls in families #2 and #3.

*If the girls are equal, then the families are not. If we are given a girl and the girls are equal, then each girl has an equal chance… making family #1 50% likely to be the home of the girl we are talking about.

Now if we are given a girl, then that girl is either the girl from column A or the girl from column B. From this we can conclude that there isn’t two boys in the family because… AGAIN we know either column A is a girl or Column B is a girl but we don’t know that both are girls. If there are no boys in column A then neither families #3 and #4 are possible; If there are no boys in column B then neither families #2 and #4 are possible. The idea that we can just get rid of family #4 without throwing out either of the corresponding families(#2 or #3) with it , is poor logic.

Before we meet a child, the families are equal, but the childrem are not. Once we meet a child, the children become equal and the families become unequal.
We are twice as likely to meet a girl from family #1 than we are in either family #2 or #3. In fact, half the time with #2 and #3 we will meet a boy. The reverse of this is turn as well, if we meet a boy then that boy is twice as likely to be from family #1 than he is from families #2 or #3. These events are true because the children are equal not the families.

Having meet a girl then the possible combinations are:
Known girl and another girl
Known girl and another boy
A girl and the known girl
A boy and the known girl

In the four combinations, the other child is a boy 50% of the time and a girl 50% of the time.

14

It’s actually Martin Gardner.

15

This may not be relevant, as the original problem did not mention meeting a child. But it can point toward the following line of thinking.

There are two possible problems here; they are related and can sound as if they’re the same, but they are different. Here’s one way to state them:

From a population of two-child families:

  1. A family is chosen at random and found to include at least one daughter. What is the probability that the family also includes a son?

  2. A child is chosen at random and found to be female. What is the probability she has a brother?

You are giving the correct answer to Problem #2: 1/2. But the problem posed in the OP is #1, and Cecil has given the correct answer there: 2/3

16

And note that J.R. Brown is examining the same two (distinct) problems.

17

Look, whether you count four possibilities or three doesn’t matter. Whether or not you consider birth order doesn’t matter. Either way, Cecil is right.

If the odds of having one girl are 50%, then the odds of having two girls must be 25%. (This assumes the sex of the two kids is independent.) This is because 50% times 50% is 25%.

Similarly, the odds of having two boys must be 25%.

So we have:

  1. Odds of two girls: 25%
  2. Odds of two boys: 25%

The only remaining possibility is that the couple had one boy and one girl (in some order). Since the total odds of all possibilities must add up to 100%, then the odds of this possibility must be 50%.

  1. Odds of one boy and one girl: 50%

Now, we’ve all agreed that the sex of one kid won’t effect the sex of the other – no one is saying it does. But if a couple waits until they have two kids, and then tells you “One of our kids is a girl,” then they’re effectively telling you “We’re not in group 2 (two boys)”

That means they must be in group 1 (two girls), or group 3 (one of each). Since having one of each was twice as likely as having two girls, there’s a two thirds chance that that’s the group they’re in.

I said nothing about birth order, I treated “one of each” as a single group, and the answer is still the same.

18

Yet another way to look at it: there’s a 50% chance of both children being the same sex. That’s because whatever sex you get for the first kid, you have a 50% chance of the second one matching it.

That means 50% of the parents have one kid of each sex. The other 50% are split evenly between having two boys and having two girls.

So there are more parents with a boy and a girl than with two girls. Thus, of the parents with at least one girl, most have a boy and a girl.

This all assumes that the sex of each kid is random, and one doesn’t affect the other. No one is saying having one girl causes the other child to be a boy. It’s just that there are more parents with one of each than with two girls, so if the parents have one girl they’re more likely to fall into the “one of each” group. Correlation does not imply causation.

Or, two put it in a less cliche way: The odds don’t change because having a girl directly effects the other kid’s sex – the odds change because the parents gave you more information to help guess which group they’re in.

19

The correct answer is 50%. We’ve been conditioned with brain teasers that the less obvious answer is always correct, but in this case it really is the simpliest answer.

Both sides are overcomplicating the problem. One child’s gender has already been determined. Only one child’s gender remains to be “randomly generated” . The fact that there are two children total is irrelevant. The order they were born in is irrelevant. The only fact that matters is there is ONE child left up to the 50/50 gender assignment. The odds that the child is female (making both children daughters) is 50%.

To go by the combination charts people have been posting (overcomplicating the simple answer):

  1. M,M
  2. M,F
  3. F,M
  4. F,F

The first (first as in first to have their gender “generated”, not first born) child is locked into being female, so options 1&2 are impossible. Only the second child can be either gender. Since only one child’s gender remains subject to statistics, there are only 2 possibilities, or 50/50.
Other examples

Coin toss-

Question: A coin is tossed. What are the odds that it will be tails?

Answer: 1/2

Q: You flip a coin. Another coin is already lying tails side up on the table. What are the odds that both coins will end up tails?

A: 1/2. The fact that there is a second coin is irrelevant to determining the outcome of one single toss. The first toss’ outcome has already been locked and only the second is still up to chance.
Die roll-

Q: What are the odds of rolling a 12 with two six sided dice?
A: 1/36 (1/6 for a ‘6’ on the first toss multiplied by 1/6 for the second)

Q: One die is on a table showing a ‘6’. You roll a second die. What are the odds that the dice will add up to 12?
A: 1/6. Only one die is subject to chance.
I’m really surprised Cecil debunked the game show prize door question and got the correct 50/50 answer instead of the incorrect 2/3, but then went on to get this similar question wrong with an answer of 2/3 instead of the correct 50/50.

Of course, the wording of the question is a bit unclear. “they have a daughter” could mean they have AT LEAST one daughter or that they have ONLY one daughter. In the latter case, the answer would be a 100% chance that they have a son. In either case, the answer still would never be 2/3.

20

I believe some folks are confusing independent events in statsitics from this word problem.

In a truly independent event world, the answer is indeed 1 in 2. No matter how many kids walk through a door, the next one has a 50/50 shot of being a boy.

We’re not in an independent world here. There are four cases, each with a 25% probability:

1- FF
2 - FM
3 - MF
4 - MM

Someone TELLS you that one is a girl (ruining total independence). I would assume they know exactly what the family has. They have essentially eliminated case 4. Assuming this person picked the family at random (A big assumption), 2 of the 3 remaining cases has a boy. Therefore, if some person TELLS you a family has one girl, the odds of a boy for the other is 2 in 3.