That problem, AS WORDED THERE, is ABSOLUTELY AMBIGUOUSLY WORDED, and the interpretation making 2/3 the answer is by far the most far fetched interpretation:
A man has two children. (crystal clear)
one is a boy (not clear what is meant by that phrase in normal english usage. When you say "one is a boy"...
(a) ...it could mean that you know both genders, and are saying that EXACTLY one is a boy, in which case the answer is 100% that the other is a girl, since you said ONLY or EXACTLY one is a boy, the other must be a girl.
(b) ...it could also mean you looked at one of the children and have identified that person as a boy.
...if it means what you claim it to mean
What are the odds that the other is a girl? (depends on how you interpreted "one is a boy")
(a) ...it could mean that EXACTLY one is a boy, in which case the answer is 100% that the other is a girl, since you said ONLY or EXACTLY one is a boy, the other must be a girl.
(b) ... it could also mean you looked at one of the children and have identified that person as a boy. In that case, the probability of the other being a girl is 50%.
(c) ...if it means that at least one of the two is a boy, which let me tell you, is probably the LEAST natural way to interpret the phrase "one is a boy", you could argue that the answer is 2/3, but that is probably the THIRD BEST ANSWER for such a terribly worded problem.
The answer of course is 2/3 or 66.66%
No, the WORDING OF THE QUESTIONS IS MOST DEFINITELY AMBIGUOUS.
The "AT LEAST" phrase was absolutely critical, and quite precise in clarifying the meaning of the first version of the problem by the OP. The use of the phrase "one is a boy" in your version of the problem, borders on the professor just being deliberately misleading and ambiguous.
Trust me when I say that if the professor worded it as you posted it in bold, then the correct answer is 0 or 50%, not 67%.
Actually it was even more delphic than that. His exact statement was
"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. Well that's two thirds." (continues with lecture)
For those of you who were wondering, this was my final understanding of the problem after months--which unfortunately the rest of my engineering class seemed to get instantly.
The odds of having a boy 50% odds of having a girl 50%
The possible outcomes are
Child 1: Child 2:
================
Boy Boy
Boy Girl
Girl Boy
Girl Girl
The statement that "one" of the children is a boy immediately eliminates the girl-girl possibility.
However, the other three possible outcomes remain.
Thus, the final probablity space looks like this:
Child 1: Child 2:
================
Boy Boy
Boy Girl
Girl Boy
Because the odds of having a boy and a girl are 50-50, each of these outcomes is equally likely.
In this scenario with two children, two of the possible outcomes has a girl, but only one has a boy. This leaves the final odds of the "other" child being a girl at 2/3 --two out of three possible remaining outcomes.
Actually it was even more delphic than that. His exact statement was
"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. Well that's two thirds." (continues with lecture)
That's a NIIIICE wording of that problem. Its less ambiguous than the way you phrased your problem, but more likely to fool a person trying to solve it...
"Billy just turned 12 the other day" is a clever way of saying "I have AT LEAST one boy".
He is saying "I have AT LEAST one boy" because
(1) He has EXACTLY ONE BOY (2 ways to have that: boy first girl second, girl first boy second)
OR
(2) He has EXACTLY TWO BOYS (1 way to have that: boy then boy)
- HOWEVER, his version of the question is a bit of a cheat. Because the announcer of the information, who already knows the outcome (two boys or not two boys), is picking and choosing the information given to the questioner (as opposed to answering a yes/no question posed by the questioner), it becomes hazy.
We have some fact (gender of the children), and the announcer simply chooses to volunteer selective information about them. I don't believe that logically or philosophically, this is 100% the same thing as ASKING the announcer "Do you have at least one boy", and having his answer "yes". Your professor's illustration serves to make the problem seem more "MIND BOGGLING" and counterintuitive, but it is cheating on the ambiguity. Here's an illustration:
PHRASING ONE: I have 30 kids. You ask "Are at least 29 of them boys?". I obey rules and answer truthfully "yes". If a person is equally likely to have a boy or a girl on a given birth, what are the chances that the other child is a girl?
I argue that it's 30 in 31, because I gave up the information to a question that was formulated by a person without prior knowledge of the answer. There are 30 times as many families with EXACTLY ONE GIRL in 30 children, as there are families with THIRTY BOYS in 30 children... but the volunteering of information
PHRASING TWO: I have 30 kids. I just tell you "At least 29 of them are boys". If a person is equally likely to have a boy or a girl on a given birth, what are the chances that the other child is a girl?
I argue that you might not necessarily be giving me ANY good information, when you VOLUNTEER that kind of carefully parsed information, since a person who has 29 boys might have a different probability of making that kind of commentary compared to a person with 30 boys. I feel the wording brings the father's psychology into play.
Quote from Ral Zarek »
You and I sit down at a table. You hand me a pair of mathematically perfect six-sided dice. I take the dice and roll them behind a screen so that I know what is rolled but you do not. You are then allowed to ask me any yes-or-no question about the dice, and I am required to answer only "yes" or "no" and I must answer completely truthfully. You ask me, "did you roll at least one 6?". I reply, "yes".
It's obviously 1/6. There are two events (two D6), one of which we already know the outcome. The other die has 1/6 chance to have a 6.
So here you have it: 1/6. It's not difficult.
No, because Baye's Theorem strongly disagrees with you.
It's impossible to explain in a way that is intuitive. If you analyse the math behind it, you can understand it, but it still won't be intuitive.
Here's another way to see it. You enter a lotto... The numbers they used on the lotto tickets are 0s and 1s. The tickets have 20 binary digits on them,
e.g. A ticket might read:
01010 10011 00010 00111
They issue exactly 1048576 unique tickets (2^20 = 1048576) distributed in random order, and sell them for $1 apiece. winner will get $1 million.
After all tickets are sold in random order, the winning ticket is announced!
Winning ticket is 11111 11111 11111 11111 and the ticket holder will get $1 million!!
You did not buy a ticket, but are allowed to ask any ticket holder the question: "does your ticket have at least nineteen 1s?" and the ticket holder must answer truthfully. You also have a temporary multimillion dollar credit line of your own, and you have the option of using it to buy any ticket from any person for $100,000 and he must sell it to you if you offer that price. You may buy form as many people as you like.
So you ask a bunch of people until finally a ticket holder says "yes, my ticket has at least nineteen 1s"
What are the chances that the other digit is a 1? Is it really 50%? What are the chances that his ticket has 20 1s and is the winning ticket? Should you buy his ticket for $100,000? Should you buy tickets from each person who answers "yes" to your question?
ANSWER:
obviously there is only 1 winning ticket with twenty 1s on it and 20 tickets with exactly nineteen 1s on it. If he says he has at least nineteen 1s, he is far more likely to be holding one of the tickets with exactly 19 1s, not the only ticket with 20 1s.
And obviously if you pay $100,000 each to acquire all 21 tickets that have at least 19 1s on them, you will pay out $2.1 million to win $1 million, which is dumb.
-
You can do the same exercise with the lotto tickets using 2 digits. The tickets are
00
01
10
11
If the winning ticket is 11 and you ask a ticket holder if he has At least one 1 on his ticket, and he says "yes", then all you know is that he's not holding the 00 lotto ticket.
He could be holding the 01, the 10, or the 11. Most likely he is not holding the winning ticket with all 1s on it. There are 2 chances in 3 that he has a 0 on his ticket.
Anybody who says 1/6 should be ejected from this forum.
It's 1/11 and if you don't believe me, just roll two dice enough to simulate this.
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Anybody who says 1/6 should be ejected from this forum.
It's 1/11 and if you don't believe me, just roll two dice enough to simulate this.
I'd not be so harsh Power - as I stated a ways back - it's very easy to fall into the "grammar trap" of it and make a false assumption.
Even I KNOWING the right answer attempting to make a better wording forgot two key words that actually made it 1:11 not 1:5. (Note: Technically you should be typing those as 1:11 and 1:5 or 1/6 and 1/12 with my understanding of how your supposed to present odds 1 incidences of vs 11 w/o and 1 w/ vs 5 w/o rather than 1 in 6 and 1 in 12... 1/11 would be a technically incorrect answer from my understanding of the notation while 1/12 would be correct [1:11 != 1/11])
I'd not be so harsh Power - as I stated a ways back - it's very easy to fall into the "grammar trap" of it and make a false assumption.
Even I KNOWING the right answer attempting to make a better wording forgot two key words that actually made it 1:11 not 1:5. (Note: Technically you should be typing those as 1:11 and 1:5 or 1/6 and 1/12 with my understanding of how your supposed to present odds 1 incidences of vs 11 w/o and 1 w/ vs 5 w/o rather than 1 in 6 and 1 in 12... 1/11 would be a technically incorrect answer from my understanding of the notation while 1/12 would be correct [1:11 != 1/11])
That would be 1:10, or 1/11.
1 way to win, 10 ways to lose. 1 chance in 11 to win.
So payoff should be 11 to 1. You bet $1 eleven times, so you put up $11. You get paid $11 once in those 11 times. Breaks even.
It is NOT 1:11 and not 1/12.
-
In the example of the couple having two kids:
Some people find that their intuition tells them that learning that a couple has a boy shouldn't increase the chances that a couple has a girl... And in fact it doesn't increase the chance. It actually decreases the chance of the couple having a girl.
Because you have to understand that the starting point of having NO information the odds are 3/4 that a couple has least one girl. We start at 3/4 couples having a girl:
gg
gb
bg
bb
When you say you have AT LEAST one boy, you are only eliminating one possibility.
gg
gb
bg
bb
Instead Of looking at it as increasing the chances of a specific child being a girl from 1/2 to 2/3 (which is incorrect thinking).. Gaining the information about the existence of at least one boy of random age DECREASES the overall probability that the couple has at least one girl from 3/4 to 2/3!
You are being given less information than if the person tells you: "my first child was a boy, what are the chances that my second child is a girl?". Which looks like this:
gg
gb
bg
bb
And in which case the answer is 1/2. The chance of 1 girl in the family is 1/2.
Isn't 1/2 the same as 1:1 though? 1 chance of either in the two possibilities? (Assuming they're equal in incidence - b/g being slightly askew of equal supposedly probably isn't the best to use though, perfectly weighted coin w/ a perfect flip instead?)
Isn't 1/2 the same as 1:1 though? 1 chance of either in the two possibilities? (Assuming they're equal in incidence - b/g being slightly askew of equal supposedly probably isn't the best to use though, perfectly weighted coin w/ a perfect flip instead?)
What are you asking? Please clarify.
1/2 is 1:1, yes.
And b/g is fine since we've stipulated that this is theoretical equal probability of boy or girl.
Common answer: 1 out of 6 (on the assumption that one die is a six based on the wording of the question)
Answer separating probability and odds (ver. 1): 1 against 5 (based on the previous answer)
Mathematical answer: 1 out of 11 (on the idea of rolling one or more sixes against rolling two sixes)
Answer separating probability and odds (ver. 2): 1 against 10 (based on the previous answer)
Total probability answer: 1 out of 36 (the odds of rolling a twelve, or two sixes, at any given time)
Total odds answer: 1 against 35 (based on the previous answer)
Meta-answer: 1 out of 7 (based on the idea that only one of these answers is correct, and that we are to calculate the odds of getting the right answer based on the number of potential answers to the riddle)
Meta-answer odds: 1 against 6 (based on the previous answer)
Epistemological answer: 0 (which would be my answer if I were asked this)
I participate yearly in a survival challenge in which I am dropped off in the wilderness with 1 days rations, a knife, and no other supplies, and have to find my way home. This is typically several weeks hiking from civilization.
I have personally killed several bears in the course of this challenge, with the use of snares and other traps.
^Grammar issues again. There's a difference between:
1) AT LEAST (i.e., either) one of the die is 6, and
2) One of the die is 6.
Let's say there's one blue and one red die. Saying that EITHER the red or blue die is 6 is different from saying that the blue die is 6.
TomCat26: please show your professor pane 5 of this comic. Or my personal favorite, "Two people scratched your car. One of them is not me."
"One of them is a boy", followed by what are the phrase "what are the odds that the other is a girl" narrows down the sample space to BG or BB.
When you say "one of them", and follow up with "the other", you're making it clear that the probability of the "other" child being female DEPENDS on the first child being male. Thus all sample spaces with the "first" child being female must be excluded.
Let's phrase it another way, but maintaining the important details:
"Jamie is a boy. What are the odds that Andy is a girl?" - phrased like this, there are only 2 possible scenarios:
Jamie is a boy, Andy is a boy.
Jamie is a boy, Andy is a girl.
What your professor is doing is including "Jamie is a girl, Andy is a boy" with all the above possibilities - pure nonsense.
In order for such a case to be considered, the question should be phrased as "One is a boy. What are the odds that I have 1 girl and 1 boy?" - making it possible to include "Jamie girl, Andy boy" in the sample space.
Asking the question has no effect on probability of rolling the two dice. There are 36 combinations when you roll 2 dice, so the odds of getting any single combination (this one being 6,6) is 1:35.
Asking the question has no effect on probability of rolling the two dice. There are 36 combinations when you roll 2 dice, so the odds of getting any single combination (this one being 6,6) is 1:35.
Say what?
The dice are already rolled.
Gaining truthful information about the results reduced the solution space.
So if you ask "is there at least one 6", and he looks at them and says "No", are you saying that the odds that he rolled a 12 are still 1:35?
No, The probability is now 0%.
If you ask "is it a 12?", and he looks at them and says "yep", are you saying that the chances are still 1:35?
No, The probability is now 100%
Of course getting information changes the probability for you.
^Grammar issues again. There's a difference between:
1) AT LEAST (i.e., either) one of the die is 6, and
2) One of the die is 6.
Let's say there's one blue and one red die. Saying that EITHER the red or blue die is 6 is different from saying that the blue die is 6.
people keep talking about this like it's a "grammar issue" and some kind of "fine print shenanigans" but it's not. The wording is not intended to be a "gotcha", and it's no "trick".
It's actually a demonstration of the failure of people's natural intuition, and their mentally failing to understand how to construct the problem.
The wording is precise and there is no ambiguity when somebody says "I looked at the dice and at least one is a 6! How likely is i that they're both 6s?"
Quote from izzetmage »
TomCat26: please show your professor pane 5 of this comic. Or my personal favorite, "Two people scratched your car. One of them is not me."
"One of them is a boy", followed by what are the phrase "what are the odds that the other is a girl" narrows down the sample space to BG or BB.
no. It reduces the solution space to BG or BB or GB.
Or it reduces it to BG or GB (making the probability 100%) if you're reducing the original statement to mean "exactly one is a boy".
When you say "one of them", and follow up with "the other", you're making it clear that the probability of the "other" child being female DEPENDS on the first child being male. Thus all sample spaces with the "first" child being female must be excluded.
Let's phrase it another way, but maintaining the important details:
"Jamie is a boy. What are the odds that Andy is a girl?" - phrased like this, there are only 2 possible scenarios:
Jamie is a boy, Andy is a boy.
Jamie is a boy, Andy is a girl.
What your professor is doing is including "Jamie is a girl, Andy is a boy" with all the above possibilities - pure nonsense.
i think you're mistaken. You're getting confused about what information is provided by the statement "Jamie is a boy. What are the odds that Andy is a girl?"
Let's consider the actual 4 families in the village:
Family 1:
Boy boy
Family 2:
Boy girl
Family 3:
Girl Boy
Family 4:
Girl girl
(order is birth order)
Father says "Jamie is a boy. "Jamie is a boy. What are the odds that Andy is a girl? Could We be family 3?"
Correct Answers are 2/3 and yes.
Nothing about "Jamie is a boy. What are the odds that Andy is a girl?" eliminates family 3.
In order for such a case to be considered, the question should be phrased as "One is a boy. What are the odds that I have 1 girl and 1 boy?" - making it possible to include "Jamie girl, Andy boy" in the sample space.
i dont think so.
Telling you the name of a boy does not change the solution space.
"One is a boy. What are the odds that I have 1 girl and 1 boy? ... <pause>... By the way the one boy I mentioned's name is Jamie. So I'm asking you whether the kid not named Jamie is a girl."
How does adding "By the way the One boy I mentioned's name is Jamie. So I'm asking you whether the kid not named Jamie is a girl." change anything? It changes nothing.
Is exactly the same thing as asking "Jamie is a boy. What are the odds that Andy is a girl?"
Solution space is still:
Boy-boy
Boy-girl
Girl-boy
(order is birth order)
There is nothing special about saying "the other" child, and saying you have A boy and providing a name provides no other information than the fact that you have at least one boy. It tells you nothing about whether he was the first or second kid. Look at my lotto example.
Precisely, getting informaiton changes the probability for ME.
The question was asking for the odds Ral had rolled 2 6s, and not the odds that I would be correct when guessing he had rolled 2 6s.
The odds Ral rolled 2 6s is 1:35.
The odds that I would guess he rolled 2 6s and be correct is 1:11, now that I know he had rolled at least 1 6.
Now that sounds like you're just trying to tapdance your way out of your bad answer and claim you understood all along.
The question was asking for the odds Ral had rolled 2 6s, and not the odds that I would be correct when guessing he had rolled 2 6s
That verb "had rolled" refers to a completed event after the roll and after the information is given. Because the outcome is already established as a reality, any probability questions are based around limited information.
Post rolling, which is where the entire question lives, there is only a question of outcome and probability from one POV: yours. Ral already saw BOTH dice so he knows the outcome, so there really is no probability to calculate for him. You have extra information about the dice in their current state, and the probability from your limited information POV that it's two 6s is 1/11.
When you ask "what is the probability that Ral HAS ROLLED two 6s, after you know at least one of the dice showed 6?" that probability is NOT just 1/36.
This persuade double talk may persuade an extremely lenient teacher in school to get half credit on a problem, but the answer was wrong, and there is no valid way to rationalize it as right with respect to the OP.
That verb "had rolled" refers to a completed event after the roll and after the information is given. Because the outcome is already established as a reality, any probability questions are based around limited information.
Post rolling, which is where the entire question lives, there is only a question of outcome and probability from one POV: yours. Ral already saw BOTH dice so he knows the outcome, so there really is no probability to calculate for him. You have extra information about the dice in their current state, and the probability from your limited information POV that it's two 6s is 1/11.
When you ask "what is the probability that Ral HAS ROLLED two 6s, after you know at least one of the dice showed 6?" that probability is NOT just 1/36.
This persuade double talk may persuade an extremely lenient teacher in school to get half credit on a problem, but the answer was wrong, and there is no valid way to rationalize it as right with respect to the OP.
I gave it some further thought and realised the answer is actually 1:5 (or 1/6).
It doesn't matter which dice (the first or the second) that has the 6. The probability of the other dice is basically the answer.
By knowing the result of one dice, it has been set into stone. It is a past event that I know already.
I think i need to reread the whole thread.
EDIT: Add-on Notes:
1:35 = Odds of Ral getting 2 6s.
1:10 = Odds of Ral getting 2 6s, assuming we know 1 is a 6.
1:5 = Odds of Ral getting 2 6s and us guessing it correctly.
Rolling 2 dice there are 36 ways the dice can come up.
you have a 5/6 chance of not rolling a 6 on either dice. so there are 25 ways of not rolling a 6.
36-25 =11.
1/11 to get at least one 6.
----------------------------------------------------------
while rolling the dice are an independant event we are looking at the total outcome of rolling a 6 on at least 1 die.
You and I sit down at a table. You hand me a pair of mathematically perfect six-sided dice. I take the dice and roll them behind a screen so that I know what is rolled but you do not. You are then allowed to ask me any yes-or-no question about the dice, and I am required to answer only "yes" or "no" and I must answer completely truthfully. You ask me, "did you roll at least one 6?". I reply, "yes".
What are the odds that I rolled two 6's?
The odds against rolling two sixes in this case: ten to one. (The probability of rolling this is 1/11.)
I am a Mathematics major. Whoever says it is 1/6 is ... wrong. Basically, the poster called mystery45 shows an approach that my idea involves. First you want to reduce the possibilities down to the relevant ones. I.e., you only consider the possibilities that are implied by "at least one 6." The opposite of this is no sixes, which happens 5x5, or 25, times in a space of 36. Subtract "no sixes" from 36 to get 11, and out of these 11, there is only one double-six. The other ten are not it.
So the odds against are ten to one.
We don't care which die rolled 6. "At least one" = I don't care which die.
I can't believe you guys are reading so much into this.
Even though it is correct to say, "HEY, if one die is a six, then the other is either a one, a two, a three, a four, a five, or a six," these values do NOT all share the same probability of coming up. This is because there is only one double-six, whereas there are two permutations that are possible if we say, "one die is a six, and the other is specific different number X."
So ... following are the values of the probabilities of these: that the other die is a one, that it is a two, that it is a three, that it is a four, that it is a five, and that it is a six ---> 2/11, 2/11, 2/11, 2/11, 2/11, and--for a double-six--1/11 (ten to one, if you want it in terms of odds against).
Look at the 4 possibilities:
1) Jamie boy, Andy boy
2) Jamie boy, Andy girl
3) Jamie girl, Andy boy
4) Jamie girl, Andy girl
If Jamie is a boy, there's no way can they be family 3 (or 4). Jamie is a girl in those families.
The name matters because "the other" child cannot refer to Jamie, the boy, again. Look at it in the case of the two dice: if I say "the red die landed on 6, what's the probability of the other landing on 6 as well?" The answer will be completely different from asking "one die, either the blue or the red, landed on 6, what's the probability that the other landed on 6?"
The former case, 2 scenarios (including both success and failure) are possible from the given information:
1) The red die landed on 6, the blue die landed on 6
2) The red die landed on 6, the blue die didn't land on 6
The latter case:
1) The red die landed on 6, the blue die landed on 6
2) The red die landed on 6, the blue die didn't land on 6
3) The red die didn't land on 6, the blue die landed on 6
The following is me trying to explain the problem using another example:
"I have two children"
4 possibilities:
1) Younger boy, older boy
2) Younger boy, older girl
3) Younger girl, older boy
4) Younger girl, older girl
"My younger child is a boy. What's the probability that my older child is a girl?"
We can eliminate 2 possibilities, leaving 2:
1) Younger boy, older boy
2) Younger boy, older girl
It is obvious that when I say "older child", I can't possibly be referring to my younger child, which is the mistake you're making: "Girl boy" actually means "Younger girl, older boy", which is impossible given the above, truthful information.
Replace "younger child" with "Jamie" and "older child" with "Andy".
Actually it was even more delphic than that. His exact statement was
"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. Well that's two thirds." (continues with lecture)
The odds of having a boy 50% odds of having a girl 50%
The possible outcomes are
Child 1: Child 2:
================
Boy Boy
Boy Girl
Girl Boy
Girl Girl
The statement that "one" of the children is a boy immediately eliminates the girl-girl possibility.
However, the other three possible outcomes remain.
Thus, the final probablity space looks like this:
Child 1: Child 2:
================
Boy Boy
Boy Girl
Girl Boy
Because the odds of having a boy and a girl are 50-50, each of these outcomes is equally likely.
In this scenario with two children, two of the possible outcomes has a girl, but only one has a boy. This leaves the final odds of the "other" child being a girl at 2/3 --two out of three possible remaining outcomes.
"Billy just turned 12 the other day" is a clever way of saying "I have AT LEAST one boy".
He is saying "I have AT LEAST one boy" because
(1) He has EXACTLY ONE BOY (2 ways to have that: boy first girl second, girl first boy second)
OR
(2) He has EXACTLY TWO BOYS (1 way to have that: boy then boy)
- HOWEVER, his version of the question is a bit of a cheat. Because the announcer of the information, who already knows the outcome (two boys or not two boys), is picking and choosing the information given to the questioner (as opposed to answering a yes/no question posed by the questioner), it becomes hazy.
We have some fact (gender of the children), and the announcer simply chooses to volunteer selective information about them. I don't believe that logically or philosophically, this is 100% the same thing as ASKING the announcer "Do you have at least one boy", and having his answer "yes". Your professor's illustration serves to make the problem seem more "MIND BOGGLING" and counterintuitive, but it is cheating on the ambiguity. Here's an illustration:
PHRASING ONE: I have 30 kids. You ask "Are at least 29 of them boys?". I obey rules and answer truthfully "yes". If a person is equally likely to have a boy or a girl on a given birth, what are the chances that the other child is a girl?
I argue that it's 30 in 31, because I gave up the information to a question that was formulated by a person without prior knowledge of the answer. There are 30 times as many families with EXACTLY ONE GIRL in 30 children, as there are families with THIRTY BOYS in 30 children... but the volunteering of information
PHRASING TWO: I have 30 kids. I just tell you "At least 29 of them are boys". If a person is equally likely to have a boy or a girl on a given birth, what are the chances that the other child is a girl?
I argue that you might not necessarily be giving me ANY good information, when you VOLUNTEER that kind of carefully parsed information, since a person who has 29 boys might have a different probability of making that kind of commentary compared to a person with 30 boys. I feel the wording brings the father's psychology into play.
But that is wrong. We do not know the outcome of one of the two events.
You ask "did you roll AT LEAST one 6?".
If the answer is "yes", you still don't know the specific outcome of EITHER dice.
Do you know if the first event was a 6? Nope.
Do you know if the second event was a 6? Nope.
You know only aggregate information about the combination of the two events, which is that AT LEAST one of the two was a 6.
You got it wrong though.
Another way to think of it is this:
When a guy answers "yes" to the question "did you roll AT LEAST one 6?", WHY Does he give that answer?
He answers "yes" either because he rolled EXACTLY TWO SIXES, OR he answers "yes" because he rolled EXACTLY ONE SIX.
There is only 1 way to roll EXACTLY TWO SIXES, and there are 10 ways to roll EXACTLY ONE SIX.
When a guy tells you he rolled AT LEAST ONE SIX, then 10 times out of 11, he is telling you that because he rolled EXACTLY ONE SIX.
Make your Vegas wheel:
1-1
1-2
1-3
1-4
1-5
1-6
2-1
2-2
2-3
2-4
2-5
2-6
3-1
3-2
3-3
3-4
3-5
3-6
4-1
4-2
4-3
4-4
4-5
4-6
5-1
5-2
5-3
5-4
5-5
5-6
6-1
6-2
6-3
6-4
6-5
6-6
I spin it and I tell you I got AT LEAST ONE SIX.
10 times it means I got EXACTLY ONE SIX.
1 time it means I got EXACTLY TWO SIXES.
Warning for using mod text
No, because Baye's Theorem strongly disagrees with you.
It's impossible to explain in a way that is intuitive. If you analyse the math behind it, you can understand it, but it still won't be intuitive.
e.g. A ticket might read:
01010 10011 00010 00111
They issue exactly 1048576 unique tickets (2^20 = 1048576) distributed in random order, and sell them for $1 apiece. winner will get $1 million.
Ticket numbers:
00000 00000 00000 00000
00000 00000 00000 00001
00000 00000 00000 00010
00000 00000 00000 00011
.
. (count up in binary)
.
11111 11111 11111 11111
After all tickets are sold in random order, the winning ticket is announced!
Winning ticket is 11111 11111 11111 11111 and the ticket holder will get $1 million!!
You did not buy a ticket, but are allowed to ask any ticket holder the question: "does your ticket have at least nineteen 1s?" and the ticket holder must answer truthfully. You also have a temporary multimillion dollar credit line of your own, and you have the option of using it to buy any ticket from any person for $100,000 and he must sell it to you if you offer that price. You may buy form as many people as you like.
So you ask a bunch of people until finally a ticket holder says "yes, my ticket has at least nineteen 1s"
What are the chances that the other digit is a 1? Is it really 50%? What are the chances that his ticket has 20 1s and is the winning ticket? Should you buy his ticket for $100,000? Should you buy tickets from each person who answers "yes" to your question?
ANSWER:
obviously there is only 1 winning ticket with twenty 1s on it and 20 tickets with exactly nineteen 1s on it. If he says he has at least nineteen 1s, he is far more likely to be holding one of the tickets with exactly 19 1s, not the only ticket with 20 1s.
And obviously if you pay $100,000 each to acquire all 21 tickets that have at least 19 1s on them, you will pay out $2.1 million to win $1 million, which is dumb.
-
You can do the same exercise with the lotto tickets using 2 digits. The tickets are
00
01
10
11
If the winning ticket is 11 and you ask a ticket holder if he has At least one 1 on his ticket, and he says "yes", then all you know is that he's not holding the 00 lotto ticket.
He could be holding the 01, the 10, or the 11. Most likely he is not holding the winning ticket with all 1s on it. There are 2 chances in 3 that he has a 0 on his ticket.
Replace 1 with boy and 0 with girl.
It's 1/11 and if you don't believe me, just roll two dice enough to simulate this.
1. Have absolutely no idea what you're talking about.
2. Stubbornly argue for at least 1 page of posts.
3. Regurgitate things you read on Starcitygames or Channelfireball.
4. Hate all netdeckers
5. Claim your 2nd place FNM deck broke the format and Delver is not even that good.
6. Tell everybody that MBC is coming back every time a new black card is released.
I'd not be so harsh Power - as I stated a ways back - it's very easy to fall into the "grammar trap" of it and make a false assumption.
Even I KNOWING the right answer attempting to make a better wording forgot two key words that actually made it 1:11 not 1:5. (Note: Technically you should be typing those as 1:11 and 1:5 or 1/6 and 1/12 with my understanding of how your supposed to present odds 1 incidences of vs 11 w/o and 1 w/ vs 5 w/o rather than 1 in 6 and 1 in 12... 1/11 would be a technically incorrect answer from my understanding of the notation while 1/12 would be correct [1:11 != 1/11])
Re: People misusing the term Vanilla to describe a flying, unleash (sometimes trample) critter.
1 way to win, 10 ways to lose. 1 chance in 11 to win.
So payoff should be 11 to 1. You bet $1 eleven times, so you put up $11. You get paid $11 once in those 11 times. Breaks even.
It is NOT 1:11 and not 1/12.
-
In the example of the couple having two kids:
Some people find that their intuition tells them that learning that a couple has a boy shouldn't increase the chances that a couple has a girl... And in fact it doesn't increase the chance. It actually decreases the chance of the couple having a girl.
Because you have to understand that the starting point of having NO information the odds are 3/4 that a couple has least one girl. We start at 3/4 couples having a girl:
gg
gb
bg
bb
When you say you have AT LEAST one boy, you are only eliminating one possibility.
gggb
bg
bb
Instead Of looking at it as increasing the chances of a specific child being a girl from 1/2 to 2/3 (which is incorrect thinking).. Gaining the information about the existence of at least one boy of random age DECREASES the overall probability that the couple has at least one girl from 3/4 to 2/3!
You are being given less information than if the person tells you: "my first child was a boy, what are the chances that my second child is a girl?". Which looks like this:
gggb
bg
bb
And in which case the answer is 1/2. The chance of 1 girl in the family is 1/2.
Re: People misusing the term Vanilla to describe a flying, unleash (sometimes trample) critter.
1/2 is 1:1, yes.
And b/g is fine since we've stipulated that this is theoretical equal probability of boy or girl.
1/2 = 1:1
Re: People misusing the term Vanilla to describe a flying, unleash (sometimes trample) critter.
but if he's asking the probability of two dice rolling both sixes, my answer is 1/36
Answer separating probability and odds (ver. 1): 1 against 5 (based on the previous answer)
Mathematical answer: 1 out of 11 (on the idea of rolling one or more sixes against rolling two sixes)
Answer separating probability and odds (ver. 2): 1 against 10 (based on the previous answer)
Total probability answer: 1 out of 36 (the odds of rolling a twelve, or two sixes, at any given time)
Total odds answer: 1 against 35 (based on the previous answer)
Meta-answer: 1 out of 7 (based on the idea that only one of these answers is correct, and that we are to calculate the odds of getting the right answer based on the number of potential answers to the riddle)
Meta-answer odds: 1 against 6 (based on the previous answer)
Epistemological answer: 0 (which would be my answer if I were asked this)
How well do you do against one?
1) AT LEAST (i.e., either) one of the die is 6, and
2) One of the die is 6.
Let's say there's one blue and one red die. Saying that EITHER the red or blue die is 6 is different from saying that the blue die is 6.
TomCat26: please show your professor pane 5 of this comic. Or my personal favorite, "Two people scratched your car. One of them is not me."
"One of them is a boy", followed by what are the phrase "what are the odds that the other is a girl" narrows down the sample space to BG or BB.
When you say "one of them", and follow up with "the other", you're making it clear that the probability of the "other" child being female DEPENDS on the first child being male. Thus all sample spaces with the "first" child being female must be excluded.
Let's phrase it another way, but maintaining the important details:
"Jamie is a boy. What are the odds that Andy is a girl?" - phrased like this, there are only 2 possible scenarios:
Jamie is a boy, Andy is a boy.
Jamie is a boy, Andy is a girl.
What your professor is doing is including "Jamie is a girl, Andy is a boy" with all the above possibilities - pure nonsense.
In order for such a case to be considered, the question should be phrased as "One is a boy. What are the odds that I have 1 girl and 1 boy?" - making it possible to include "Jamie girl, Andy boy" in the sample space.
| Ad Nauseam
| Infect
Big Johnny.
Asking the question has no effect on probability of rolling the two dice. There are 36 combinations when you roll 2 dice, so the odds of getting any single combination (this one being 6,6) is 1:35.
The dice are already rolled.
Gaining truthful information about the results reduced the solution space.
So if you ask "is there at least one 6", and he looks at them and says "No", are you saying that the odds that he rolled a 12 are still 1:35?
No, The probability is now 0%.
If you ask "is it a 12?", and he looks at them and says "yep", are you saying that the chances are still 1:35?
No, The probability is now 100%
Of course getting information changes the probability for you.
people keep talking about this like it's a "grammar issue" and some kind of "fine print shenanigans" but it's not. The wording is not intended to be a "gotcha", and it's no "trick".
It's actually a demonstration of the failure of people's natural intuition, and their mentally failing to understand how to construct the problem.
The wording is precise and there is no ambiguity when somebody says "I looked at the dice and at least one is a 6! How likely is i that they're both 6s?"
no. It reduces the solution space to BG or BB or GB.
Or it reduces it to BG or GB (making the probability 100%) if you're reducing the original statement to mean "exactly one is a boy".
i think you're mistaken. You're getting confused about what information is provided by the statement "Jamie is a boy. What are the odds that Andy is a girl?"
Let's consider the actual 4 families in the village:
Family 1:
Boy boy
Family 2:
Boy girl
Family 3:
Girl Boy
Family 4:
Girl girl
(order is birth order)
Father says "Jamie is a boy. "Jamie is a boy. What are the odds that Andy is a girl? Could We be family 3?"
Correct Answers are 2/3 and yes.
Nothing about "Jamie is a boy. What are the odds that Andy is a girl?" eliminates family 3.
i dont think so.
Telling you the name of a boy does not change the solution space.
"One is a boy. What are the odds that I have 1 girl and 1 boy? ... <pause>... By the way the one boy I mentioned's name is Jamie. So I'm asking you whether the kid not named Jamie is a girl."
How does adding "By the way the One boy I mentioned's name is Jamie. So I'm asking you whether the kid not named Jamie is a girl." change anything? It changes nothing.
Is exactly the same thing as asking "Jamie is a boy. What are the odds that Andy is a girl?"
Solution space is still:
Boy-boy
Boy-girl
Girl-boy
(order is birth order)
There is nothing special about saying "the other" child, and saying you have A boy and providing a name provides no other information than the fact that you have at least one boy. It tells you nothing about whether he was the first or second kid. Look at my lotto example.
yep. I'm changing my answer to 1/11
rolls:
6-1
6-2
6-3
6-4
6-5
6-6
1-6
2-6
3-6
4-6
5-6
Precisely, getting informaiton changes the probability for ME.
The question was asking for the odds Ral had rolled 2 6s, and not the odds that I would be correct when guessing he had rolled 2 6s.
The odds Ral rolled 2 6s is 1:35.
The odds that I would guess he rolled 2 6s and be correct is 1:11, now that I know he had rolled at least 1 6.
That verb "had rolled" refers to a completed event after the roll and after the information is given. Because the outcome is already established as a reality, any probability questions are based around limited information.
Post rolling, which is where the entire question lives, there is only a question of outcome and probability from one POV: yours. Ral already saw BOTH dice so he knows the outcome, so there really is no probability to calculate for him. You have extra information about the dice in their current state, and the probability from your limited information POV that it's two 6s is 1/11.
When you ask "what is the probability that Ral HAS ROLLED two 6s, after you know at least one of the dice showed 6?" that probability is NOT just 1/36.
This persuade double talk may persuade an extremely lenient teacher in school to get half credit on a problem, but the answer was wrong, and there is no valid way to rationalize it as right with respect to the OP.
I gave it some further thought and realised the answer is actually 1:5 (or 1/6).
It doesn't matter which dice (the first or the second) that has the 6. The probability of the other dice is basically the answer.
By knowing the result of one dice, it has been set into stone. It is a past event that I know already.
I think i need to reread the whole thread.
EDIT: Add-on Notes:
1:35 = Odds of Ral getting 2 6s.
1:10 = Odds of Ral getting 2 6s, assuming we know 1 is a 6.
1:5 = Odds of Ral getting 2 6s and us guessing it correctly.
We don't care which die rolled 6. "At least one" = I don't care which die.
I can't believe you guys are reading so much into this.
Rolling 2 dice there are 36 ways the dice can come up.
you have a 5/6 chance of not rolling a 6 on either dice. so there are 25 ways of not rolling a 6.
36-25 =11.
1/11 to get at least one 6.
----------------------------------------------------------
while rolling the dice are an independant event we are looking at the total outcome of rolling a 6 on at least 1 die.
Thanks to Epic Graphics the best around.
Thanks to Nex3 for the avatar visit ye old sig and avatar forum
The odds against rolling two sixes in this case: ten to one. (The probability of rolling this is 1/11.)
I am a Mathematics major. Whoever says it is 1/6 is ... wrong. Basically, the poster called mystery45 shows an approach that my idea involves. First you want to reduce the possibilities down to the relevant ones. I.e., you only consider the possibilities that are implied by "at least one 6." The opposite of this is no sixes, which happens 5x5, or 25, times in a space of 36. Subtract "no sixes" from 36 to get 11, and out of these 11, there is only one double-six. The other ten are not it.
So the odds against are ten to one.
Even though it is correct to say, "HEY, if one die is a six, then the other is either a one, a two, a three, a four, a five, or a six," these values do NOT all share the same probability of coming up. This is because there is only one double-six, whereas there are two permutations that are possible if we say, "one die is a six, and the other is specific different number X."
So ... following are the values of the probabilities of these: that the other die is a one, that it is a two, that it is a three, that it is a four, that it is a five, and that it is a six ---> 2/11, 2/11, 2/11, 2/11, 2/11, and--for a double-six--1/11 (ten to one, if you want it in terms of odds against).
Look at the 4 possibilities:
1) Jamie boy, Andy boy
2) Jamie boy, Andy girl
3) Jamie girl, Andy boy
4) Jamie girl, Andy girl
If Jamie is a boy, there's no way can they be family 3 (or 4). Jamie is a girl in those families.
The name matters because "the other" child cannot refer to Jamie, the boy, again. Look at it in the case of the two dice: if I say "the red die landed on 6, what's the probability of the other landing on 6 as well?" The answer will be completely different from asking "one die, either the blue or the red, landed on 6, what's the probability that the other landed on 6?"
The former case, 2 scenarios (including both success and failure) are possible from the given information:
1) The red die landed on 6, the blue die landed on 6
2) The red die landed on 6, the blue die didn't land on 6
The latter case:
1) The red die landed on 6, the blue die landed on 6
2) The red die landed on 6, the blue die didn't land on 6
3) The red die didn't land on 6, the blue die landed on 6
The following is me trying to explain the problem using another example:
"I have two children"
4 possibilities:
1) Younger boy, older boy
2) Younger boy, older girl
3) Younger girl, older boy
4) Younger girl, older girl
"My younger child is a boy. What's the probability that my older child is a girl?"
We can eliminate 2 possibilities, leaving 2:
1) Younger boy, older boy
2) Younger boy, older girl
It is obvious that when I say "older child", I can't possibly be referring to my younger child, which is the mistake you're making: "Girl boy" actually means "Younger girl, older boy", which is impossible given the above, truthful information.
Replace "younger child" with "Jamie" and "older child" with "Andy".
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| Infect
Big Johnny.