The premise of the problem is that there is a 50% probability that any given child is a boy, and 50% probability of a girl, therefore the distribution of 2 child families is:
25% boy boy
25% boy girl
25% girl boy
25% girl girl
The father saying "Jamie is a boy does not say Jamie is the first child. It only says "at least one child is a boy, either the first or second, I'm not going to tell you which. However i can give you any number of useless details about the boy in order to conince you that you know him. Wanna know his name? I'll tell you regardless: it's one of an infinite number of possible names we could have chosen, and it's Jamie, btw. Wanna know which grandfather he takes after? He doesn't like mustard or cilantro. But sorry, I can't tell you whether he's the first child or second child. But I do have at least one boy. Jamie is his name. "
When the father makes the statement "Jamie is a boy" you are not specifying which child is a boy, because the name of the boy is an UTTERLY USELESS piece of information.
From a problem solving standpoint, the father saying "Jamie is a boy" is the equivalent of saying "I have a boy. His name is ~mumble~".
If I don't happen to catch the name, does it really affect the probability of the other kid being a girl? Was "~mumble~" information that helps me?
It's the same as the father saying "My child who likes pancakes more, who has blue eyes btw, here look at his picture, his favorite number is 4, and he is wearing a green shirt today, is a boy"
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Comparing it to the two color dice problem:
If I roll two dice, and YOU DONT KNOW THE COLORS.
I volunteer this information: "A beige dice shows 6. What does the other dice show?"
Do you think the probability of the other dice being a 6 is 1/6? It's not. It's 1/11. Because "beige" tells you nothing. Color information never tells us anything about the numbers unless we are going to consistently report the information for that color.
- - - - - - - -
(this actually reminds me of a really silly card trick, using red and black) but it also makes a cool CON too. I am not promoting gambling and betting is illegal. This is all hypothetical. I dont know if Ive stumbled on this variation on the problem new or if somebody already has thought of and presented this before. This MAY END UP CONFUSING a few people, so don't attack this until you've concluded that the OP's solution is 1/11, or 1:10. This might trip up some more people (or it might not) but here we go:
I set up a game, the red die, green die game. I have a perfectly balanced 6 sided red die, and a perfectly balanced 6 sided green die.
Whenever I roll just one 6, if it is red I will TRUTHFULLY announce: "The Red die shows 6! Want to bet $1 on whether i rolled 6-6? It pays off $8 on a $1 bet." and even show you the die. Whenever I roll just one 6, if it is green, I will TRUTHFULLY announce: "The green die shows 6! Want to bet $1 on whether i rolled 6-6? It pays off $8 on a $1 bet." and even show you the die.
Whenever I roll two 6s, I get to randomly choose whether to say red or green (I choose to flip a coin to decide), and EITHER TRUTHFULLY announce that "The red die shows 6! Want to bet $1 on whether i rolled 6-6? It pays off $8 on a $1 bet." & show you the red die, OR TRUTHFULLY announce that "The green die shows 6! Want to bet $1 on whether i rolled 6-6? It pays off $8 on a $1 bet." and show you the green die.
Let's play:
(1) so I roll a few times and then truthfully announce "The red die shows 6! Want to bet $1 on whether i rolled 6-6? It pays off $8 on a $1 bet." I pull back the curtain and show you the red die is clearly 6.
(2) I roll a few more times and then truthfully announce "The green die shows 6! Want to bet $1 on whether i rolled 6-6? It pays off $8 on a $1 bet." I pull back the curtain and show you the green die is clearly 6.
No tricks here.
Should you take these bets?
Is it a sucker bet?
Yes it's a suckers bet. The odds are still 1/11, yet I only pay you off at $8 on the $1 bet.
Why is it still 1/11 ? Because I'm not giving you usable information when I tell you the color there. Just the illusion of useful information.
Yes, the name is useless. It could be substituted for younger/older, etc.
However, the fact that the person who posed the question asked for the gender of the OTHER CHILD is not! The OTHER CHILD cannot refer to the child who was mentioned!
Jamie is a boy. What can my OTHER CHILD be? A boy or a girl.
Jamie is a boy. What can JAMIE be? A boy only, obviously.
"JAMIE" and my "OTHER CHILD" are two different entities! You can replace "Jamie" with "younger son" or whatever, but as soon as you reference your "other child", you're reducing the possibilities!
Look at the dice problem. "A beige die shows 6. What does the other die show?" Let's say the other die is purple. Here are the correct probabilities:
Beige 6, Purple not 6 - 5/6
Beige 6, Purple 6 - 1/6
Beige not 6, Purple 6 - 0
Beige not 6, Purple not 6 - 0
And not:
Beige 6, Purple not 6 - 5/11
Beige 6, Purple 6 - 1/11
Beige not 6, Purple 6 - 5/11
Beige not 6, Purple not 6 - 0
Lastly, your game isn't a sucker's bet. When I play the game, let's say you rolled a red 6 and announced it. There are only 2 possibilities:
1) You rolled a red 6, and a green 6 - 1/6
2) You rolled a red 6, and a green 1, 2, 3, 4 or 5 - 5/6
The same holds, with colors reversed, for you rolling & announcing a green 6.
The game would only be a sucker bet if you said "I rolled a 6, but I'm not going to say whether it's the red or green die that came up 6. I'll take you on 7-to-1 that the green die is 6."
Yes, the name is useless. It could be substituted for younger/older, etc.
However, the fact that the person who posed the question asked for the gender of the OTHER CHILD is not! The OTHER CHILD cannot refer to the child who was mentioned!
Jamie is a boy. What can my OTHER CHILD be? A boy or a girl.
Jamie is a boy. What can JAMIE be? A boy only, obviously.
"JAMIE" and my "OTHER CHILD" are two different entities! You can replace "Jamie" with "younger son" or whatever, but as soon as you reference your "other child", you're reducing the possibilities!
you've chosen to eliminate girl-boy from the list without offering a valid reason. The correct list of probabilities is this:
Boy boy
Boy girl
Girl boy Girl girl
Look at the dice problem. "A beige die shows 6. What does the other die show?" Let's say the other die is purple. Here are the correct probabilities:
Beige 6, Purple not 6 - 5/6
Beige 6, Purple 6 - 1/6
Beige not 6, Purple 6 - 0
Beige not 6, Purple not 6 - 0
And not:
Beige 6, Purple not 6 - 5/11
Beige 6, Purple 6 - 1/11
Beige not 6, Purple 6 - 5/11
Beige not 6, Purple not 6 - 0
you don't know the other die is purple. How do you know it's not beige? Certainly you acknowledge that if they're both beige, that the probability is 1/11. If I am trying to beat you at this game, then whenever a single die comes up 6, I will say "A X colored die came up 6." because that apparently offering useless color or name information is sufficient to con you into thinking that the other color has 1 in 6 change of being a 6.
It's would be like me saying: "At least one die shows a 6. How likely is the other die a 6... But wait! This die showing 6 was made in Indonesia... Oh don't answer yet, I want to give you more useful information! This die was carved by a guy named Jamie! He's a boy! Now, tell me what is the chance that the die not made by Jamie, not made in Indonesia, shows 6?"
Lastly, your game isn't a sucker's bet. When I play the game, let's say you rolled a red 6 and announced it. There are only 2 possibilities:
1) You rolled a red 6, and a green 6 - 1/6
2) You rolled a red 6, and a green 1, 2, 3, 4 or 5 - 5/6
The same holds, with colors reversed, for you rolling & announcing a green 6.
The game would only be a sucker bet if you said "I rolled a 6, but I'm not going to say whether it's the red or green die that came up 6. I'll take you on 7-to-1 that the green die is 6."
thank you. You've been suckered.
You see in your case (1), whenever I roll a red 6 AND a green 6, only half the time I will say "red 6", and the other half of the time I will say "green 6". Yeah there are 5 losers for green 6, and 5 losers for red 6, but red 6 and green 6 half to SPLIT the WINNER TWO WAYS!
It's an awesome sucker bet.
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Here let's break It down by rolling the dice 360 times:
50 chances out of 360 I will roll a red 6 and green non-6. I announce "I rolled a red 6" (Loser gimme $1)
50 chances out of 360 I will roll a green 6 and a red non-6. I announce "I rolled a green 6" (Loser gimme $1)
10 chance out of 360 I will roll a green 6 and a red 6, and 50% of these cases I choose to randomly announce red 6, and 50% of the time I will choose to announce a green 6. Therefore:
5 times out of 360 I announce "I rolled a red 6" (Winner collect $8)
5 times I announce "I rolled a green 6" (Winner collect $8)
You lose. After we play this 360 times, I have $50+$50 of your money, and you have $40+$40 of mine. I do believe that nets me $20. Thanks for playin' you've been suckered! Welcome to Reno! Please, pleeeeeeaaaase come again!
1) Because Jamie is not a girl?
When you say stuff like "boy boy" or "boy girl", have you ever thought what those refer to? Here are the clear, definite possibilities:
Jamie boy, Andy boy
Jamie boy, Andy girl
Jamie girl, Andy boy
Jamie girl, Andy girl
Eliminate all those with "Jamie girl", because it is established that Jamie is a boy.
2) So you're saying that the two dice might be beige? Then you MUST take away the die that you referred to as the beige 6, so there can be no ambiguity as to which one is "the other".
My point is not that irrelevant information can alter the probabilities. My point is, as soon as you use the words "THE OTHER", you have made the two dice/kids distinct. You cannot play around ambiguities.
3) EDIT: OK, it seems like a sucker bet now, if you're forced to say "the red is 6" every time you roll 6-6 instead of choosing randomly, I lose every single time on "the green is 6". Still not sold on the other 2 though.
You like sucker bets? How about this:
If you flip 3 coins, at least 2 land the same way:
HHH, HHT, HTH, THH, TTT, TTH, THT, HTT
So the last one is either gonna match them, or it isn't. 1/2 chance, right?
But look at the possibilities - only 2 out of 8 have all 3 coins match up (HHH and TTT). That's not 1/2, that's 1/4.
The tricky bit here is, which is supposed to be the third coin?
Similarly with the dice/kids...which is supposed to be "the other"?
If you flip 3 coins, at least 2 land the same way:
HHH, HHT, HTH, THH, TTT, TTH, THT, HTT
So the last one is either gonna match them, or it isn't. 1/2 chance, right?
But look at the possibilities - only 2 out of 8 have all 3 coins match up (HHH and TTT). That's not 1/2, that's 1/4.
The tricky bit here is, which is supposed to be the third coin?
Similarly with the dice/kids...which is supposed to be "the other"?
So, we have the following possible states:
HHH
HHT
HTH
THH
HTT
THT
TTH
TTT
That gives us 8 equiprobable states. We konw that the current state (result of three flipped coins) is one of the states involve *at least* two matching states. Which, gives us no information, since *every* possible state involves two matching sets.
It becomes a simple matter then to identify that HHH and TTT are the two states that "win". Odds -- 2/8 (1/4).
How you can understand *this* and not understand the Jami problem I don't konw, since they are mathematically exactly the same principle.
The main catch is: any of the coins can be the "other" coin. just as either of the kids, or either of the dice can be the "other" dice.
The "third" coin can only be identified when you know the states of all coins. You cannot pick out one coin (as the "third") and say "the other two are going to land the same way" with certainty.
The boy/girl problem: Jamie has been identified, so when you say "the other", you are not referring to him. I can pick out Jamie the boy, and there is no way in hell that "The other kid is a boy, and Jamie is a girl".
Here's the question from TomCat26, requoted:
"I have two children. Billy (or some other male name) just turned 12 the other day. What are the odds that the other one is a girl?"
Tell me, can "the other one" refer to Billy?
Because if it can't, then "girl boy" needs to be struck off, because "girl" refers to Billy. And the answer is 1/2.
The game can only start if you roll a 6, AND you must specify which one (or a random die, if both are 6) landed on 6.
Once you state that, you might as well show me the die that landed on 6. Upon seeing this, I have RELEVANT INFORMATION, which will alter the probabilities.
do you actually read the responses carefully, or do you just throw the canned response back? fine let's show you the red dice
Let's make the game exactly the same, and you will BET every time I show you the red die 6. You will always bet on red.
I roll red 6:
(a) 5/6 times, red 6, green is not 6. I show you red die 6 (you must bet $1 and LOSE)
(b) 1/6 time, red 6, green is 6, I am supposed to show you red die and let you bet $1 and pay you off for a winner... But HOLD UP! We still have to flip a coin to decide whether I'm going to show you the red 6 or the green 6! (So Only 50% of the time when I roll 6-6, I will show you the red die you must bet $1 and win $8) the other 50% of the time when I roll 6-6, I show you green and you don't bet.
Get it now?
Somebody betting green die 6 gets the same con job.
You see, I show the red die 5.5 times per 36 rolls and I show the green die 5.5 times per 36 rolls, but I PAY OUT ONE TIME PER 36 rolls. A green better wins .5 out of 5.5, a red better wins .5 out of 5.5, and a guy betting red and green both wins 1 time out of 11. Cant you see it?
Surely you see that we will show a red or green die 11 times out of 36.
Surely you see that I will show red as often as green, meaning I show each 5.5 out of 36.
Surely you see that red has to win as often as green.
Surely you see that there is only one payout per 6 rolls.
Yeah you see a red die 6, but I've essentially skimmed half your winnings off the top before you ever see that red die.
Nath'd. I edited my post at 7:56 and you posted at 8:13.
I'm still not sold on the boy/girl problem, as stated this way:
"I have two children. Billy (or some other male name) just turned 12 the other day. What are the odds that the other one is a girl?"
If you're saying that the answer is 2/3, I want you to look at your workings and confirm that you did not include "Billy is a girl and the other one is a boy" as one of the possibilities.
Let's go back to my lotto example earlier, which I'm sure you didn't read, but fine, let's go.
Lets use families with 10 kids. There are 1024 families on the island, each with equally likely combination and order of boy-girl distribution.
There is exactly one family with bbbbbbbbbb
There is exactly one family with bbbbbbbbbg
There is exactly one family with bbbbbbbbgb
There is exactly one family with bbbbbbbbgg
...
There is exactly one family with gggggggggg
If you ask a dad, do you have at least 9 boys? And dad say "yes, their names are jimmy, jonny, jacob, joseph, jodi, jack, jed, jeremy, and jackson, what are the chances my OTHER child is a girl?"
What's your answer?
There are 10 families on the island with nine boys and a girl, and 1 family with 10 boys.
Does the fact that the father named 9 boys AND used your "magic word" ( "OTHER" ), mean that the other child has a 50% chance of being a girl?
The answer is "NO, there is a 10/11 possibility of the OTHER kid being a girl. The fact that you learned all their names means nothing."
Nath'd. I edited my post at 7:56 and you posted at 8:13.
I'm still not sold on the boy/girl problem, as stated this way:
"I have two children. Billy (or some other male name) just turned 12 the other day. What are the odds that the other one is a girl?"
If you're saying that the answer is 2/3, I want you to look at your workings and confirm that you did not include "Billy is a girl and the other one is a boy" as one of the possibilities.
The identifying information "Jamie is a boy" only eliminates the last state because we do not know which child is Jamie. Jamie could be child 1 OR child 2.
Thus, there are three possible (and equiprobable) states, two resulting in a "loss" and one in a "win" for the goal of two boys.
Where you are getting hung up, is you are assuming that the identifier "jamie is a boy" identifies one specific result (IE child 1), but it doesn't.
I see it like this: Although it's true that there are two possible events--Andy is either a boy or a girl--the likelihood of one of these events and the likelihood of the other are not the same. You first have to "create children in the womb" before naming them. Imagine taking an ultrasound. Ruling out GG, three possibilities are equally likely from an ultrasound image from right to left: BB, BG, and GB.
Now, the children are named. If Jamie is a boy, then it is still twice as likely that Andy is a girl than not. This is because there is one BB, but there are two permutations of boy-girl.
You can't necessarily assume that probabilities are equal between two TYPES of different things--e.g., Andy's being or girl and Andy's being a boy. You have to FIRST figure out how likely those types are to show up in the first place, which means treating the children as if they were first "dice in the womb."
The other magic word is "at least".
If "at least" 9 kids are boys is general, and I can accept that the answer is 1 in 10. Like how Ral Zarek used "at least one die is 6" and the answer was 1/11.
Let's say you ask a dad, he says "My 9 youngest children are all male, what's the probability of my eldest child being female?" There are 2 families whose first 9 children were boys. They only differ in their last child.
bLatch: In two cases, Jamie is child 1. In one case, Jamie is Child 2. How do you reconcile this inconsistency?
My thought process right now:
I flip 2 coins, both hidden from you. If both land tails, I show them to you and start over.
If at least 1 lands heads, I cover the other one up. You guess that the other lands tails.
HH - 1/3, you lose
HT - 1/3, you win
TH - 1/3, you win
Let's say I name the revealed coin Billy. It doesn't change the chances.
Suppose I named the coin Billy before flipping it? And I only start the game if Billy lands on heads? The probability is 1/2.
Suppose I name my first kid Billy? And I only have a second kid if Billy is a boy? The probability of the second kid being a girl is 1/2?
Or do I have 2 kids, start over if they're both girls.
Then if at least 1 is a boy, I name him Billy and you guess that the other is a girl?
WTF? Who waits until their second child is born to name their first?
Is that settled? People name their kids when they're born. Not at the moment when the second is born, or when you tell a stranger that they just turned 12.
So the prof's question is poorly worded...right?
Suppose I name my first kid Billy? And I only have a second kid if Billy is a boy? The probability of the second kid being a girl is 1/2?
But since it's not assumed that the first kid is named Billy, the fixed order of the kids is either Billy-Andy or Andy-Billy, and if Billy is a boy then you can say the order is Boy-Andy or Andy-Boy.
And if you fill in the sex of Andy as "Girl," you get Boy-Girl or Girl-Boy (=2).
If you fill in the sex of Andy as "Boy," you can only have Boy-Boy (=1).
So it is twice more likely that Andy is a girl.
(1) Non-relevant information posing as relevant information. I hope I can help with that.
(2) Emphasis on the concept of using the word "OTHER" somehow changing things. I cant help you with this because I still have no idea why you think saying "other child" is reasonable.
As for the issue of (1), and Jamie. The name doesnt matter.
The father has two boys, Jamie and Andy, he would have said "I have a boy named Andy" 50% of the time, instead of announcing "I have a boy named Jamie"
If the farmer is required to truthfully reveal that he has a boy, and he has two, then just like in the red dice / green dice sucker game, he is prefiltering his answer.
He says "I have a boy named Jamie" only half as often if he has two boys, as he does if he has a boy named Jamie and a girl.
If I flip the first coin and it lands tails, I can still flip the second and have it land heads. Then I name the second coin Billy.
So even if my first kid is a girl, I can still continue and have a second boy, which I then name Billy. That makes sense.
What if instead of naming, I go by age? Is age somehow relevant?
"My younger kid is a boy, what's the probability that my older kid is a girl?"
Is it still 2/3?
Is this equivalent to:
"The first coin that I flipped was heads, what's the probability that the second was tails?"
Assuming that I flipped the coins in sequence, instead of flipping both and choosing one (either if HH) that landed on heads to be the "first".
More may-be-difficult-to-process thoughts:
[spoiler]If I flip the first coin and it lands tails, I can still flip the second and have it land heads. Then I name the second coin Billy.
So even if my first kid is a girl, I can still continue and have a second boy, which I then name Billy. That makes sense.
What if instead of naming, I go by age? Is age somehow relevant?
"My younger kid is a boy, what's the probability that my older kid is a girl?"
Is it still 2/3?
Is this equivalent to:
"The first coin that I flipped was heads, what's the probability that the second was tails?"
Assuming that I flipped the coins in sequence, instead of flipping both and choosing one (either if HH) that landed on heads to be the "first".
50% is the answer to both. In both, you have established an ORDERED pair of entities (ordered according to age/place in a sequence) ALONG WITH designating an independly arising characteristic (from a set of two binary characteristics--such as boy/girl or H/T) to the "first" member of the ordered pair. The "second" therefore will have either one of the binary characteristics with a 50% chance.
Edit: The difference between these and that Andy/Billy question is that even though those children are ordered in some single way like age, size, etc., not enough information about a single ordering is known (by he/she who is trying to answer that question) for the probability to be changed to 50%.
I roll 2 dice in a row;
1) One of them is 6, the probability of the other being 6 is - 1/11
2) The first one is 6, the probability of the other being 6 is - 1/6
I flip 2 coins in a row;
1) One of them is heads, the probability of the other being tails is - 2/3
2) The first one is heads, the probability of the other being tails is - 1/2
I have 2 kids;
1) Billy is a boy, the probability of my other kid being a girl is - 2/3?
2) My younger kid is a boy, the probability of my older kid being a girl is - 1/2?
With the dice and coins, it's obvious that when you say "the first one", you're referring to a specific coin.
With the kids, it's also obvious that when you say "my younger kid", you're referring to a specific kid.
BUT...When you say "Billy" - common sense indicates that you are referring to a specific kid, since that's what we give people names for.
Looking at the above examples, apparently "Billy" is not specific, or as turquoisepower puts it, not part of an "ordered pair of entities". How is that possible?
EDIT: Perhaps it's a bit clearer if you compare these two instead:
I have 2 kids;
1) One of my kids is a boy (and his name is Billy for all it matters), the probability of my other kid being a girl is - 2/3
2) My first (i.e. younger) kid is a boy, the probability of my older kid being a girl is - 1/2
In that case, yeah. The answer is 2/3.
But god DAMN, putting a name on someone really confuses me into thinking it's case 2.
Always remember what the original sample space is: {BB, BG, GB, GG}. As a commonly used sample space, it is not generally spoken of in terms of "Billy" vs. "Not Billy"; you don't always know if Billy is the name of either child in a given pair of two. BUT you could always designate a reason for why one letter is on the left and the other is on the right.
You could ALWAYS, if needed, say that the reason entails the left letter's representing the oldest child and the right letter's representing the other one, regardless of whether anybody is named Billy or not.
Saying "I have two children; I have a boy named Billy (or whatever his name is)" really only means that some boy is within a group of two children and can thus be found among one of the sample space elements that have at least one boy: BB, BG, or GB. 3 possibilities are therefore in view.
On the other hand, saying "I have two children; I have a boy (named whatever), AND the oldest child is a boy" means that you look at the sample space in terms of age order and reduce it to BG and BB. 2 possibilities are therefore in view.
On getting suckered: Often when it comes to seeing whether you have a winning or a losing bet, instead of looking at it from the bettor's standpoint, look at the whole payout system from the HOUSE standpoint. The HOUSE only had to give out money 1 time out of 36, on a 6-6 roll. Yet the betters are going to bet 11 times out of 36, any time they see a die with a 6 on it. House only has to pay 1 in 36, yet they collect 11 in 36. Obviously a $8 to $1 payout means the bettor is a loser.
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Re: Jamie and billy:
There is no such thing as a "universal" probability when it comes to how many boys or girls that father has. The father KNOWS has EXACTLY what he has.
The term "probabililty" comes into play for the PLAYER, YOU, trying to GUESS, because you don't know.
Through all your lines of reasoning in this thread, I get the subtle impression that your working concept of the probability is some "quantity" that exists separate from a POV.
If the dad says "I have a boy named Jamie" and you heard it, then from YOUR POV, the odds of him having at least one girl is now 2/3.
If I was standing right there next to you, but I am just hard of hearing, then from MY POV, the odds of him having at least one girl remains at 3/4.
Your answer of 2/3 is valid.
And my answer of 3/4 is valid.
Because probability is based on information.
I dont' know of a "general" way to tell you how to recognize when information you're given is useful vs irrelevant. I think turquoisepower summarized it about as well as you can, and its still far from straightforward. If it were straightforward, there wouldn't be so many people getting the answer wrong.
Quote from izzetmage »
EDIT: Perhaps it's a bit clearer if you compare these two instead:
I have 2 kids;
1) One of my kids is a boy (and his name is Billy for all it matters), the probability of my other kid being a girl is - 2/3
2) My first (i.e. younger) kid is a boy, the probability of my older kid being a girl is - 1/2
In that case, yeah. The answer is 2/3.
But god DAMN, putting a name on someone really confuses me into thinking it's case 2.
Though having any given child is an event with independent probability, the fact that we are limiting the problem to families with two children (or a pair of independent events) makes the ordering relevant information. The solution space starts out as all families with 2 children, whether they had one of them in 1931, or 1965, or 2012, and whether they had them 9 months apart or 50 years apart. Its just families with 2 children, and we know the distributions of male female in 2 child families based on the assumptions we make about those families.
We know nothing about actual age distributions in this population at all, some could be 80, 100000 or 1000000 years old, because we haven't specified whether these are special, long lived people. We haven't designated a distribution for names either... There could be 10000 jamies, or just one.
Because the way the problem is set up, ONLY pairs of children, our starting point is:
boy then boy
boy then girl
girl then boy
girl then girl
With each time of 2 kid family being equally likely.
(A) What are the chances that a family has at least 1 girl?
3/4
boy then boy boy then girl girl then boy girl then girl
(B) If you learn that a family has at least one boy, what are the chances that it has at least 1 girl?
2/3
boy then boy boy then girl girl then boy girl then girl
(C) If you learn that the FIRST kid is a boy, what are the chances that the family has at least 1 girl?
1/2
25% boy then boy 25% boy then girl 25% girl then boy
25% girl then girl
Which category does "If you learn that a family has at least one boy, his name is Jamie. Jamie is 19."* fall into all this? IT falls under (A).
Why is the information "His name is Jamie" irrelevant? Because we know absolutely nothing about the distribution of names.
Why is the information about "Jamie is 19" irrelevant? Because we know absolutely nothing about the distribution of ages of children.
* and obviously trying to argue that semantically, you're saying that the family has EXACTLY one boy, named Jamie, then the probability problem is just 100% chance of a girl, and its just a stupid trick question of the "read my mind" variety.
Here's another way to think about the girl/boy problem.
Suppose I'm meeting two families. The Andersons have two kids, and the Bartons have one kid, and the wife is pregnant. The Bartons do not know the gender of the fetus.
If I ask the Andersons, "Do you have at least one boy?", the answer they give me conveys information about both of their kids. If I ask the Bartons the same, it only conveys information about their one kid, and not about the mystery fetus. As a result, the Andersons have a 1/3 chance of having two boys if they say "yes", whereas the Bartons have a 1/2 chance of having two boys.
Interestingly enough, even though Jamie and Andy can be ordered alphabetically, the fact is that if you DON'T say that "Jamie is a boy; what is the probability that his only sibling, Andy, is a girl?" (2/3) and you instead say, "the two children have different names, and the one who is alphabetically second is a boy," it is THEN that the probability that the other one is a girl is 50%.
I think this is because if Jamie, and Andy are KNOWN to be the names and they happen to be boys, then the ONE order possible for Boy - Boy is unquestionably Andy - Jamie, whereas if the names are not known, then a pair of boys can be corresponded to TWO possible orders--to the order represented by "Child A - Child B" and to the one represented by "Child B - Child A."
Yep, probability for the boy/girl question is 2/3.
The con game works to the advantage of the house.
Instead of:
1) Red 6, green non-6 1/6
2) Red 6, green 6 5/6
it's
1) Red 6, green non-6 5/11
2) Red 6, green 6 1/11
3) Nonred 6, green 6 5/11
If the game only starts when the red lands on 6 then it favors the player. But if the game starts when either die lands on 6 (i.e. when the red lands on 6, OR when the red lands on non-6 and the green lands on 6 - even though you have no chance of winning in the latter) then it favors the house.
The premise of the problem is that there is a 50% probability that any given child is a boy, and 50% probability of a girl, therefore the distribution of 2 child families is:
25% boy boy
25% boy girl
25% girl boy
25% girl girl
The father saying "Jamie is a boy does not say Jamie is the first child. It only says "at least one child is a boy, either the first or second, I'm not going to tell you which. However i can give you any number of useless details about the boy in order to conince you that you know him. Wanna know his name? I'll tell you regardless: it's one of an infinite number of possible names we could have chosen, and it's Jamie, btw. Wanna know which grandfather he takes after? He doesn't like mustard or cilantro. But sorry, I can't tell you whether he's the first child or second child. But I do have at least one boy. Jamie is his name. "
When the father makes the statement "Jamie is a boy" you are not specifying which child is a boy, because the name of the boy is an UTTERLY USELESS piece of information.
From a problem solving standpoint, the father saying "Jamie is a boy" is the equivalent of saying "I have a boy. His name is ~mumble~".
If I don't happen to catch the name, does it really affect the probability of the other kid being a girl? Was "~mumble~" information that helps me?
It's the same as the father saying "My child who likes pancakes more, who has blue eyes btw, here look at his picture, his favorite number is 4, and he is wearing a green shirt today, is a boy"
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Comparing it to the two color dice problem:
If I roll two dice, and YOU DONT KNOW THE COLORS.
I volunteer this information: "A beige dice shows 6. What does the other dice show?"
Do you think the probability of the other dice being a 6 is 1/6? It's not. It's 1/11. Because "beige" tells you nothing. Color information never tells us anything about the numbers unless we are going to consistently report the information for that color.
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(this actually reminds me of a really silly card trick, using red and black) but it also makes a cool CON too. I am not promoting gambling and betting is illegal. This is all hypothetical. I dont know if Ive stumbled on this variation on the problem new or if somebody already has thought of and presented this before. This MAY END UP CONFUSING a few people, so don't attack this until you've concluded that the OP's solution is 1/11, or 1:10. This might trip up some more people (or it might not) but here we go:
I set up a game, the red die, green die game. I have a perfectly balanced 6 sided red die, and a perfectly balanced 6 sided green die.
Whenever I roll just one 6, if it is red I will TRUTHFULLY announce: "The Red die shows 6! Want to bet $1 on whether i rolled 6-6? It pays off $8 on a $1 bet." and even show you the die. Whenever I roll just one 6, if it is green, I will TRUTHFULLY announce: "The green die shows 6! Want to bet $1 on whether i rolled 6-6? It pays off $8 on a $1 bet." and even show you the die.
Whenever I roll two 6s, I get to randomly choose whether to say red or green (I choose to flip a coin to decide), and EITHER TRUTHFULLY announce that "The red die shows 6! Want to bet $1 on whether i rolled 6-6? It pays off $8 on a $1 bet." & show you the red die, OR TRUTHFULLY announce that "The green die shows 6! Want to bet $1 on whether i rolled 6-6? It pays off $8 on a $1 bet." and show you the green die.
Let's play:
(1) so I roll a few times and then truthfully announce "The red die shows 6! Want to bet $1 on whether i rolled 6-6? It pays off $8 on a $1 bet." I pull back the curtain and show you the red die is clearly 6.
(2) I roll a few more times and then truthfully announce "The green die shows 6! Want to bet $1 on whether i rolled 6-6? It pays off $8 on a $1 bet." I pull back the curtain and show you the green die is clearly 6.
No tricks here.
Should you take these bets?
Is it a sucker bet?
Yes it's a suckers bet. The odds are still 1/11, yet I only pay you off at $8 on the $1 bet.
Why is it still 1/11 ? Because I'm not giving you usable information when I tell you the color there. Just the illusion of useful information.
However, the fact that the person who posed the question asked for the gender of the OTHER CHILD is not! The OTHER CHILD cannot refer to the child who was mentioned!
Jamie is a boy. What can my OTHER CHILD be? A boy or a girl.
Jamie is a boy. What can JAMIE be? A boy only, obviously.
"JAMIE" and my "OTHER CHILD" are two different entities! You can replace "Jamie" with "younger son" or whatever, but as soon as you reference your "other child", you're reducing the possibilities!
Look at the dice problem. "A beige die shows 6. What does the other die show?" Let's say the other die is purple. Here are the correct probabilities:
Beige 6, Purple not 6 - 5/6
Beige 6, Purple 6 - 1/6
Beige not 6, Purple 6 - 0
Beige not 6, Purple not 6 - 0
And not:
Beige 6, Purple not 6 - 5/11
Beige 6, Purple 6 - 1/11
Beige not 6, Purple 6 - 5/11
Beige not 6, Purple not 6 - 0
Lastly, your game isn't a sucker's bet. When I play the game, let's say you rolled a red 6 and announced it. There are only 2 possibilities:
1) You rolled a red 6, and a green 6 - 1/6
2) You rolled a red 6, and a green 1, 2, 3, 4 or 5 - 5/6
The same holds, with colors reversed, for you rolling & announcing a green 6.
The game would only be a sucker bet if you said "I rolled a 6, but I'm not going to say whether it's the red or green die that came up 6. I'll take you on 7-to-1 that the green die is 6."
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Big Johnny.
Boy boy
Boy girl
Girl boy
Girl girlyou don't know the other die is purple. How do you know it's not beige? Certainly you acknowledge that if they're both beige, that the probability is 1/11. If I am trying to beat you at this game, then whenever a single die comes up 6, I will say "A X colored die came up 6." because that apparently offering useless color or name information is sufficient to con you into thinking that the other color has 1 in 6 change of being a 6.
It's would be like me saying: "At least one die shows a 6. How likely is the other die a 6... But wait! This die showing 6 was made in Indonesia... Oh don't answer yet, I want to give you more useful information! This die was carved by a guy named Jamie! He's a boy! Now, tell me what is the chance that the die not made by Jamie, not made in Indonesia, shows 6?"
thank you. You've been suckered.
You see in your case (1), whenever I roll a red 6 AND a green 6, only half the time I will say "red 6", and the other half of the time I will say "green 6". Yeah there are 5 losers for green 6, and 5 losers for red 6, but red 6 and green 6 half to SPLIT the WINNER TWO WAYS!
It's an awesome sucker bet.
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Here let's break It down by rolling the dice 360 times:
50 chances out of 360 I will roll a red 6 and green non-6. I announce "I rolled a red 6" (Loser gimme $1)
50 chances out of 360 I will roll a green 6 and a red non-6. I announce "I rolled a green 6" (Loser gimme $1)
10 chance out of 360 I will roll a green 6 and a red 6, and 50% of these cases I choose to randomly announce red 6, and 50% of the time I will choose to announce a green 6. Therefore:
5 times out of 360 I announce "I rolled a red 6" (Winner collect $8)
5 times I announce "I rolled a green 6" (Winner collect $8)
You lose. After we play this 360 times, I have $50+$50 of your money, and you have $40+$40 of mine. I do believe that nets me $20. Thanks for playin' you've been suckered! Welcome to Reno! Please, pleeeeeeaaaase come again!
When you say stuff like "boy boy" or "boy girl", have you ever thought what those refer to? Here are the clear, definite possibilities:
Jamie boy, Andy boy
Jamie boy, Andy girl
Jamie girl, Andy boy
Jamie girl, Andy girl
Eliminate all those with "Jamie girl", because it is established that Jamie is a boy.
2) So you're saying that the two dice might be beige? Then you MUST take away the die that you referred to as the beige 6, so there can be no ambiguity as to which one is "the other".
My point is not that irrelevant information can alter the probabilities. My point is, as soon as you use the words "THE OTHER", you have made the two dice/kids distinct. You cannot play around ambiguities.
3) EDIT: OK, it seems like a sucker bet now, if you're forced to say "the red is 6" every time you roll 6-6 instead of choosing randomly, I lose every single time on "the green is 6". Still not sold on the other 2 though.
You like sucker bets? How about this:
If you flip 3 coins, at least 2 land the same way:
HHH, HHT, HTH, THH, TTT, TTH, THT, HTT
So the last one is either gonna match them, or it isn't. 1/2 chance, right?
But look at the possibilities - only 2 out of 8 have all 3 coins match up (HHH and TTT). That's not 1/2, that's 1/4.
The tricky bit here is, which is supposed to be the third coin?
Similarly with the dice/kids...which is supposed to be "the other"?
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Big Johnny.
So, we have the following possible states:
HHH
HHT
HTH
THH
HTT
THT
TTH
TTT
That gives us 8 equiprobable states. We konw that the current state (result of three flipped coins) is one of the states involve *at least* two matching states. Which, gives us no information, since *every* possible state involves two matching sets.
It becomes a simple matter then to identify that HHH and TTT are the two states that "win". Odds -- 2/8 (1/4).
How you can understand *this* and not understand the Jami problem I don't konw, since they are mathematically exactly the same principle.
The main catch is: any of the coins can be the "other" coin. just as either of the kids, or either of the dice can be the "other" dice.
The boy/girl problem: Jamie has been identified, so when you say "the other", you are not referring to him. I can pick out Jamie the boy, and there is no way in hell that "The other kid is a boy, and Jamie is a girl".
Here's the question from TomCat26, requoted:
"I have two children. Billy (or some other male name) just turned 12 the other day. What are the odds that the other one is a girl?"
Tell me, can "the other one" refer to Billy?
Because if it can't, then "girl boy" needs to be struck off, because "girl" refers to Billy. And the answer is 1/2.
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Big Johnny.
Let's make the game exactly the same, and you will BET every time I show you the red die 6. You will always bet on red.
I roll red 6:
(a) 5/6 times, red 6, green is not 6. I show you red die 6 (you must bet $1 and LOSE)
(b) 1/6 time, red 6, green is 6, I am supposed to show you red die and let you bet $1 and pay you off for a winner... But HOLD UP! We still have to flip a coin to decide whether I'm going to show you the red 6 or the green 6! (So Only 50% of the time when I roll 6-6, I will show you the red die you must bet $1 and win $8) the other 50% of the time when I roll 6-6, I show you green and you don't bet.
Get it now?
Somebody betting green die 6 gets the same con job.
You see, I show the red die 5.5 times per 36 rolls and I show the green die 5.5 times per 36 rolls, but I PAY OUT ONE TIME PER 36 rolls. A green better wins .5 out of 5.5, a red better wins .5 out of 5.5, and a guy betting red and green both wins 1 time out of 11. Cant you see it?
Surely you see that we will show a red or green die 11 times out of 36.
Surely you see that I will show red as often as green, meaning I show each 5.5 out of 36.
Surely you see that red has to win as often as green.
Surely you see that there is only one payout per 6 rolls.
Yeah you see a red die 6, but I've essentially skimmed half your winnings off the top before you ever see that red die.
I'm still not sold on the boy/girl problem, as stated this way:
"I have two children. Billy (or some other male name) just turned 12 the other day. What are the odds that the other one is a girl?"
If you're saying that the answer is 2/3, I want you to look at your workings and confirm that you did not include "Billy is a girl and the other one is a boy" as one of the possibilities.
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Big Johnny.
Lets use families with 10 kids. There are 1024 families on the island, each with equally likely combination and order of boy-girl distribution.
There is exactly one family with bbbbbbbbbb
There is exactly one family with bbbbbbbbbg
There is exactly one family with bbbbbbbbgb
There is exactly one family with bbbbbbbbgg
...
There is exactly one family with gggggggggg
If you ask a dad, do you have at least 9 boys? And dad say "yes, their names are jimmy, jonny, jacob, joseph, jodi, jack, jed, jeremy, and jackson, what are the chances my OTHER child is a girl?"
What's your answer?
There are 10 families on the island with nine boys and a girl, and 1 family with 10 boys.
Does the fact that the father named 9 boys AND used your "magic word" ( "OTHER" ), mean that the other child has a 50% chance of being a girl?
The answer is "NO, there is a 10/11 possibility of the OTHER kid being a girl. The fact that you learned all their names means nothing."
Here are all the possible states for a family:
Child 1: boy, Child 2: boy
Child 1: boy, Child 2: girl
Child 1: girl, Child 2: boy
Child 1: girl, Child 2: girl
The identifying information "Jamie is a boy" only eliminates the last state because we do not know which child is Jamie. Jamie could be child 1 OR child 2.
Thus, there are three possible (and equiprobable) states, two resulting in a "loss" and one in a "win" for the goal of two boys.
Where you are getting hung up, is you are assuming that the identifier "jamie is a boy" identifies one specific result (IE child 1), but it doesn't.
Now, the children are named. If Jamie is a boy, then it is still twice as likely that Andy is a girl than not. This is because there is one BB, but there are two permutations of boy-girl.
You can't necessarily assume that probabilities are equal between two TYPES of different things--e.g., Andy's being or girl and Andy's being a boy. You have to FIRST figure out how likely those types are to show up in the first place, which means treating the children as if they were first "dice in the womb."
If "at least" 9 kids are boys is general, and I can accept that the answer is 1 in 10. Like how Ral Zarek used "at least one die is 6" and the answer was 1/11.
Let's say you ask a dad, he says "My 9 youngest children are all male, what's the probability of my eldest child being female?" There are 2 families whose first 9 children were boys. They only differ in their last child.
bLatch: In two cases, Jamie is child 1. In one case, Jamie is Child 2. How do you reconcile this inconsistency?
My thought process right now:
I flip 2 coins, both hidden from you. If both land tails, I show them to you and start over.
If at least 1 lands heads, I cover the other one up. You guess that the other lands tails.
HH - 1/3, you lose
HT - 1/3, you win
TH - 1/3, you win
Let's say I name the revealed coin Billy. It doesn't change the chances.
Suppose I named the coin Billy before flipping it? And I only start the game if Billy lands on heads? The probability is 1/2.
Suppose I name my first kid Billy? And I only have a second kid if Billy is a boy? The probability of the second kid being a girl is 1/2?
Or do I have 2 kids, start over if they're both girls.
Then if at least 1 is a boy, I name him Billy and you guess that the other is a girl?
WTF? Who waits until their second child is born to name their first?
Is that settled? People name their kids when they're born. Not at the moment when the second is born, or when you tell a stranger that they just turned 12.
So the prof's question is poorly worded...right?
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Big Johnny.
But since it's not assumed that the first kid is named Billy, the fixed order of the kids is either Billy-Andy or Andy-Billy, and if Billy is a boy then you can say the order is Boy-Andy or Andy-Boy.
And if you fill in the sex of Andy as "Girl," you get Boy-Girl or Girl-Boy (=2).
If you fill in the sex of Andy as "Boy," you can only have Boy-Boy (=1).
So it is twice more likely that Andy is a girl.
(1) Non-relevant information posing as relevant information. I hope I can help with that.
(2) Emphasis on the concept of using the word "OTHER" somehow changing things. I cant help you with this because I still have no idea why you think saying "other child" is reasonable.
As for the issue of (1), and Jamie. The name doesnt matter.
The father has two boys, Jamie and Andy, he would have said "I have a boy named Andy" 50% of the time, instead of announcing "I have a boy named Jamie"
If the farmer is required to truthfully reveal that he has a boy, and he has two, then just like in the red dice / green dice sucker game, he is prefiltering his answer.
He says "I have a boy named Jamie" only half as often if he has two boys, as he does if he has a boy named Jamie and a girl.
So even if my first kid is a girl, I can still continue and have a second boy, which I then name Billy. That makes sense.
What if instead of naming, I go by age? Is age somehow relevant?
"My younger kid is a boy, what's the probability that my older kid is a girl?"
Is it still 2/3?
Is this equivalent to:
"The first coin that I flipped was heads, what's the probability that the second was tails?"
Assuming that I flipped the coins in sequence, instead of flipping both and choosing one (either if HH) that landed on heads to be the "first".
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Big Johnny.
50% is the answer to both. In both, you have established an ORDERED pair of entities (ordered according to age/place in a sequence) ALONG WITH designating an independly arising characteristic (from a set of two binary characteristics--such as boy/girl or H/T) to the "first" member of the ordered pair. The "second" therefore will have either one of the binary characteristics with a 50% chance.
Edit: The difference between these and that Andy/Billy question is that even though those children are ordered in some single way like age, size, etc., not enough information about a single ordering is known (by he/she who is trying to answer that question) for the probability to be changed to 50%.
It's not an inconsistency?
1) One of them is 6, the probability of the other being 6 is - 1/11
2) The first one is 6, the probability of the other being 6 is - 1/6
I flip 2 coins in a row;
1) One of them is heads, the probability of the other being tails is - 2/3
2) The first one is heads, the probability of the other being tails is - 1/2
I have 2 kids;
1) Billy is a boy, the probability of my other kid being a girl is - 2/3?
2) My younger kid is a boy, the probability of my older kid being a girl is - 1/2?
With the dice and coins, it's obvious that when you say "the first one", you're referring to a specific coin.
With the kids, it's also obvious that when you say "my younger kid", you're referring to a specific kid.
BUT...When you say "Billy" - common sense indicates that you are referring to a specific kid, since that's what we give people names for.
Looking at the above examples, apparently "Billy" is not specific, or as turquoisepower puts it, not part of an "ordered pair of entities". How is that possible?
EDIT: Perhaps it's a bit clearer if you compare these two instead:
I have 2 kids;
1) One of my kids is a boy (and his name is Billy for all it matters), the probability of my other kid being a girl is - 2/3
2) My first (i.e. younger) kid is a boy, the probability of my older kid being a girl is - 1/2
In that case, yeah. The answer is 2/3.
But god DAMN, putting a name on someone really confuses me into thinking it's case 2.
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Big Johnny.
You could ALWAYS, if needed, say that the reason entails the left letter's representing the oldest child and the right letter's representing the other one, regardless of whether anybody is named Billy or not.
Saying "I have two children; I have a boy named Billy (or whatever his name is)" really only means that some boy is within a group of two children and can thus be found among one of the sample space elements that have at least one boy: BB, BG, or GB. 3 possibilities are therefore in view.
On the other hand, saying "I have two children; I have a boy (named whatever), AND the oldest child is a boy" means that you look at the sample space in terms of age order and reduce it to BG and BB. 2 possibilities are therefore in view.
On getting suckered: Often when it comes to seeing whether you have a winning or a losing bet, instead of looking at it from the bettor's standpoint, look at the whole payout system from the HOUSE standpoint. The HOUSE only had to give out money 1 time out of 36, on a 6-6 roll. Yet the betters are going to bet 11 times out of 36, any time they see a die with a 6 on it. House only has to pay 1 in 36, yet they collect 11 in 36. Obviously a $8 to $1 payout means the bettor is a loser.
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Re: Jamie and billy:
There is no such thing as a "universal" probability when it comes to how many boys or girls that father has. The father KNOWS has EXACTLY what he has.
The term "probabililty" comes into play for the PLAYER, YOU, trying to GUESS, because you don't know.
Through all your lines of reasoning in this thread, I get the subtle impression that your working concept of the probability is some "quantity" that exists separate from a POV.
If the dad says "I have a boy named Jamie" and you heard it, then from YOUR POV, the odds of him having at least one girl is now 2/3.
If I was standing right there next to you, but I am just hard of hearing, then from MY POV, the odds of him having at least one girl remains at 3/4.
Your answer of 2/3 is valid.
And my answer of 3/4 is valid.
Because probability is based on information.
I dont' know of a "general" way to tell you how to recognize when information you're given is useful vs irrelevant. I think turquoisepower summarized it about as well as you can, and its still far from straightforward. If it were straightforward, there wouldn't be so many people getting the answer wrong.
Though having any given child is an event with independent probability, the fact that we are limiting the problem to families with two children (or a pair of independent events) makes the ordering relevant information. The solution space starts out as all families with 2 children, whether they had one of them in 1931, or 1965, or 2012, and whether they had them 9 months apart or 50 years apart. Its just families with 2 children, and we know the distributions of male female in 2 child families based on the assumptions we make about those families.
We know nothing about actual age distributions in this population at all, some could be 80, 100000 or 1000000 years old, because we haven't specified whether these are special, long lived people. We haven't designated a distribution for names either... There could be 10000 jamies, or just one.
Because the way the problem is set up, ONLY pairs of children, our starting point is:
boy then boy
boy then girl
girl then boy
girl then girl
With each time of 2 kid family being equally likely.
(A) What are the chances that a family has at least 1 girl?
3/4
boy then boy
boy then girl
girl then boy
girl then girl
(B) If you learn that a family has at least one boy, what are the chances that it has at least 1 girl?
2/3
boy then boy
boy then girl
girl then boy
girl then girl(C) If you learn that the FIRST kid is a boy, what are the chances that the family has at least 1 girl?
1/2
25% boy then boy
25% boy then girl
25% girl then boy25% girl then girl
Which category does "If you learn that a family has at least one boy, his name is Jamie. Jamie is 19."* fall into all this? IT falls under (A).
Why is the information "His name is Jamie" irrelevant? Because we know absolutely nothing about the distribution of names.
Why is the information about "Jamie is 19" irrelevant? Because we know absolutely nothing about the distribution of ages of children.
* and obviously trying to argue that semantically, you're saying that the family has EXACTLY one boy, named Jamie, then the probability problem is just 100% chance of a girl, and its just a stupid trick question of the "read my mind" variety.
Suppose I'm meeting two families. The Andersons have two kids, and the Bartons have one kid, and the wife is pregnant. The Bartons do not know the gender of the fetus.
If I ask the Andersons, "Do you have at least one boy?", the answer they give me conveys information about both of their kids. If I ask the Bartons the same, it only conveys information about their one kid, and not about the mystery fetus. As a result, the Andersons have a 1/3 chance of having two boys if they say "yes", whereas the Bartons have a 1/2 chance of having two boys.
I think this is because if Jamie, and Andy are KNOWN to be the names and they happen to be boys, then the ONE order possible for Boy - Boy is unquestionably Andy - Jamie, whereas if the names are not known, then a pair of boys can be corresponded to TWO possible orders--to the order represented by "Child A - Child B" and to the one represented by "Child B - Child A."
The con game works to the advantage of the house.
Instead of:
1) Red 6, green non-6 1/6
2) Red 6, green 6 5/6
it's
1) Red 6, green non-6 5/11
2) Red 6, green 6 1/11
3) Nonred 6, green 6 5/11
If the game only starts when the red lands on 6 then it favors the player. But if the game starts when either die lands on 6 (i.e. when the red lands on 6, OR when the red lands on non-6 and the green lands on 6 - even though you have no chance of winning in the latter) then it favors the house.
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Big Johnny.