Brain Teasers
My Three Sons
Probability
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.Probability
Mrs. Mathemagican has three children. If AT LEAST one of her children is a boy, what is the probability that she has three sons?
Answer
The probability of her having three sons is 1/7 .Given that she has at least one boy, then there are seven equally likely outcomes for her three children: BBB, BBG, BGB, GBB, GBG, GGB, and BGG. The only one of the 8 total possible outcomes that you can rule out is GGG. Hence, the probability of three boys is 1/7.
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I think I spent more time reviewing this teaser than I have on any other. The answer "feels" wrong, but after an extensive amount of research, it turns out to be correct. And for those of you who don't believe so, BBG really is a different possibility than GBB or BGB.
Good puzzle and yes I agree totally with the answer. Many of these condidtional probability teasers have evoked a lot of discussion on these boards.
This teaser didn't "feel" right for me either, but after doing some homework...
I just can't imagine a couple going into a doctors office with their baby boy and 1 on the way and stating to the doc "We are plannig on two more siblings for our little Billy to play with, what are the odds he'll have 2 brothers?" And the doctor replies, "1/7 of a chance" Hee hee hee, just doesn't look right which makes it a Great Teaser!!
I just can't imagine a couple going into a doctors office with their baby boy and 1 on the way and stating to the doc "We are plannig on two more siblings for our little Billy to play with, what are the odds he'll have 2 brothers?" And the doctor replies, "1/7 of a chance" Hee hee hee, just doesn't look right which makes it a Great Teaser!!
Its like the problem "I flip two coins. At least one is tails. What is the probability the other is tails?" and the answer is 1/3. I frazzled every brain cell over that one...
No, it WOULD be 1/3. BBG, BGB and GBB all mean 1 girl and 2 boys. The kids arent in any order!
Sorry, pink, but you are very much incorrect. It's already been explained correctly, so if you still don't believe the answer you'll have to search the web or take a probability lesson. Look at it this way: I have two children. What are the odds that they are both boys? It should be obvious that the odds are 1/4 (1/2 * 1/2). If you use your logic, however, your options would be BB, BG, and GG, making the odds 1/3. You HAVE to include GB in this, because it truly is a different possibility than BG. If you don't include it, then you are saying that the first child being a girl and the second child being a boy is not possible, which of course is not true.
I agree with pink. Plus they could be triplets.
I don't know how many times I can say it. The answer is absolutely correct. Try flipping coins or something to prove it to yourself. Also, it doesn't make any difference whether they are triplets or not. Three sons is three sons no matter how you slice it.
the answer is wrong. order is not a factor. it says she already HAS three children and at least one of them is a boy. that leaves two other children. they can be both boys, both girls, or one of each. order does NOT matter.
The answer is absolutely correct. We are only given the fact that one of the children is a boy but we don't know which one. So we can only eliminate one of the eight possibilities and that leaves us with a probability of 1/7. The reason why we should consider GBB, BGB and BBG differently is because each one of them is as likely as, say, GGG. If we took them together as one, then we won't be dealing with equally likely elements and cannot conclude that the probability is 1/3.
Though I agree with the answer, basically, I'd like to throw a wrench in the works and say that statistically, boys and girls aren't exactly 50% each. Just trying to cause trouble
I must agree with cvlspaz. I have three coins. One has two heads. When I flip the three coins, I am always guaranteed at least one head. The remaining two coins have four possible outcomes: both heads, both tails, one head and one tail, one tail and one head. I can flip all three coins at the same time (as in triplets) or individually in any order. The outcome is the same. The answer is 1 in 4.
Cmnme, your analogy is not the same as the teaser. In your example you KNOW which coin will show up heads. In the teaser above you don't know, and this results in more possibilities. In general you figure out probabilities by dividing the number of outcomes that satisfy the given conditions by the TOTAL
(continued - sorry) TOTAL number of possible outcomes. In the teaser there are eight possible outcomes: BBB, BBG, BGB, GBB, GGB, GGG, GBG, & BGG. You can only rule one of these out: Since you know that there is at least one boy, then GGG is ruled out. Therefore there are 7 possible outcomes. The only one that satisfies the given conditions in the quesion (three boys) is BBB. ONE out of the possible SEVEN. I said it before and I'll say it again: The answer given is ABSOLUTELY CORRECT. If you think it's wrong, may I suggest going to a mathematics web site and studying up on probability calculations.
Oh, and if you want to test this out with coins, you can't use a double-headed coin. What you have to do is flip three regular coins and keep track of how many times you get three heads. What you have to do, however, is not count any of the flips where you get three tails. Then divide the number of times you get three heads by the total number of flips (not counting the TTT flips). This should come out to around 1/7, if you flip enough times.
My answer is 3/12 or 1/4.
We know that there is at least 1 boy (BOY). That BOY may be 1st, 2nd ,or third.
If BOY is 1st, the FOUR possible outcomes are BOYBB BOYBG BBOYGB BOYGG.
If BOY is 2nd , the FOUR possible outcomes are BBOYB BBOYG GBOYB GBOYG
If BOY is 3rd , the FOUR possible outcomes are BBBOY BGBOY GBBOY GGBOY.
There are therefore 12 possible outcomes in all. 3 of the 12 are BBB.
We know that there is at least 1 boy (BOY). That BOY may be 1st, 2nd ,or third.
If BOY is 1st, the FOUR possible outcomes are BOYBB BOYBG BBOYGB BOYGG.
If BOY is 2nd , the FOUR possible outcomes are BBOYB BBOYG GBOYB GBOYG
If BOY is 3rd , the FOUR possible outcomes are BBBOY BGBOY GBBOY GGBOY.
There are therefore 12 possible outcomes in all. 3 of the 12 are BBB.
Oh good grief! I give up. The answer is whatever you want it to be, folks. The same goes for all of the other probability teasers on the site with "wrong" answers.
1/4 it is. Good math Scrambel
Don't Make me think about it any more!!! *head explodes*
My mother is a Doctor of maths and she agrees with what 'pink' says, and so do i. 1 in 3 should be the correct answer.
The teaser states the chance of it happening, no the order in which things happen.
The teaser states the chance of it happening, no the order in which things happen.
One problem with your reasoning, scramble:
You listed 12 possibilities:
If boy given is first, then:
BBB or BBG or BGB or BGG
If second, then:
BBB or BBG or GBB or GBG
If third, then:
BBB or BGB or GBB or GGB
Notice you listed BBB three times, BBG, BGB, and GBB two times, and BGG, GBG, and GGB only one time.
You then proceed to give equal probability to each of these "twelve" options. This is incorrect. For example, if the first child is a boy and the third child is a girl, then the only possibilities are BBG and BGG. Obviously, the probability of the second child being a boy is 1/2. That is, if BBG is listed once in a list of possibilities, then BGG would be listed once. So, scramble, why did you list BBG TWICE and BGG only ONCE in your list?
I agree with the 1/7 answer. There are seven equally likely possibilities with only one that will make the statement correct.
You listed 12 possibilities:
If boy given is first, then:
BBB or BBG or BGB or BGG
If second, then:
BBB or BBG or GBB or GBG
If third, then:
BBB or BGB or GBB or GGB
Notice you listed BBB three times, BBG, BGB, and GBB two times, and BGG, GBG, and GGB only one time.
You then proceed to give equal probability to each of these "twelve" options. This is incorrect. For example, if the first child is a boy and the third child is a girl, then the only possibilities are BBG and BGG. Obviously, the probability of the second child being a boy is 1/2. That is, if BBG is listed once in a list of possibilities, then BGG would be listed once. So, scramble, why did you list BBG TWICE and BGG only ONCE in your list?
I agree with the 1/7 answer. There are seven equally likely possibilities with only one that will make the statement correct.
If anyone does not believe it is 1/7 please, lets gamble for money. I'll bring three coins and you bring lots of cash, cause it's going to cost you. A very similar Teaser on this sight is called "Swindlers Dice" It shows how people who can't believ this is the right answer can get taken for Money.
This is the problem when not everybody reads it in the same way.
Whats throwing people off in this question is the part: "If AT LEAST one of her children is a boy..." which is making people think 'ok 1 boy is given, what are the odds of the other 2 being boys.' -which would be 1/4. Thats not what it means. It is simply saying that even if she doesn't have 3 boys, she's at least got 1, so you can eliminate 1 out of the 8 combo possibilites.
The other problem is that people are confusing the total tally of b's & g's with the combo possibilities. Yes, order is irrelevant, but the # of possibilites are not. To make this clearer we can think of rolling 2 dice. Each combo of die number has an equal chance of coming out. (eg: 1&5, 6&3, 2&2) The *sum* or tally of the numbers though tends to be more often towards to the middle, since there are more possible combo's that make 5,6 or 7 than 2 or 12. The sums are different than the combo possibilities.
Allways back up theory with empirical observation. And we can do that here: http://shazam.econ.ubc.ca/flip/table.cgi
Note the 'percentage of heads found' of the 3 coin flips. all heads was %13 & all tails was %12. The mean is .125 or 1/8. Eliminate the GGG and your left with 1/7.
Confusing question at first.
The other problem is that people are confusing the total tally of b's & g's with the combo possibilities. Yes, order is irrelevant, but the # of possibilites are not. To make this clearer we can think of rolling 2 dice. Each combo of die number has an equal chance of coming out. (eg: 1&5, 6&3, 2&2) The *sum* or tally of the numbers though tends to be more often towards to the middle, since there are more possible combo's that make 5,6 or 7 than 2 or 12. The sums are different than the combo possibilities.
Allways back up theory with empirical observation. And we can do that here: http://shazam.econ.ubc.ca/flip/table.cgi
Note the 'percentage of heads found' of the 3 coin flips. all heads was %13 & all tails was %12. The mean is .125 or 1/8. Eliminate the GGG and your left with 1/7.
Confusing question at first.
I find that it helps to consider the phrase "at least one is a boy" as equivalent to "not all of them are girls." That way, only one possibility is eliminated, no more, and seven remain - one of which is correct. 1/7.
This one was easy... I'm hopeless at maths, but this isn't a brain teaser at all.
i apologize for apparently not understanding probability (i am pretty good at just about every other math function, rule, etc) but i just dont get it... i understand what youre saying, but, what about like this - one definitely IS a boy... so take the other two children and subject them to the law of averages - 1/4 BOTH will be boys. for example, i have a red sock and a blue sock. now, my first try, i got a red, i KNOW tha ti got a red... now, with that knowledge in hand... there is a 1/4 chance i will get two more reds (obviously i return the sock), because the one red sock is already a decidd event, so i dont see how it factors into probability.
Sep 28, 2005
Answer should be 1/4.
B
|
_____
b g
/ \ / \
b g b g
Possibble Outcomes
Bbb
Bbg
Bgb
Bgg
B = known results
b = unknown result or man
g = unknown result of girl
Once you know one of the results, it can no longer be used in probability statements because it is 100% going to occur. The known B must be removed from the probability equation so that the true statement of probabilty can be determined.
B
|
_____
b g
/ \ / \
b g b g
Possibble Outcomes
Bbb
Bbg
Bgb
Bgg
B = known results
b = unknown result or man
g = unknown result of girl
Once you know one of the results, it can no longer be used in probability statements because it is 100% going to occur. The known B must be removed from the probability equation so that the true statement of probabilty can be determined.
i don't know whether the answer is 1/3 or 1/4.. but i totally disagree with 1/7.
but my own answer is 1/3.
there is four probabilities: BBB, BBG, BGG, GGG. Because you've already known that one of your children is a boy, so we can eliminate GGG. There is only 3 probabilities left. To have three sons the probability is 1/3.
why i totally disagree that the answer is 1/7? coz you don't say that the first child is a boy, or the second child is a boy, or the third child is a boy. because of that, there is only 4 probabilities.
but my own answer is 1/3.
there is four probabilities: BBB, BBG, BGG, GGG. Because you've already known that one of your children is a boy, so we can eliminate GGG. There is only 3 probabilities left. To have three sons the probability is 1/3.
why i totally disagree that the answer is 1/7? coz you don't say that the first child is a boy, or the second child is a boy, or the third child is a boy. because of that, there is only 4 probabilities.
All those who disagree with the answer of 1/7 should take probability lessons with someone who has done a doctrate in probability.
I do not mean to offend anybody, but its true that probability is not an easy concept to understand. Even many Maths doctorates find it difficult to fully understand probability. My maths teachers at school were doctorates in Maths, but even they did not feel comfortable teaching probability, and so we specially hired a probability doctorate to teach us the Probability course.In Probability the order of the things IS important, and cant be ignored. Probability is not always governed by normal maths rules. So before you start thrashing others' correct answers please take probability lessons with someone whi is a doctorate in Probability. That will clarify your concepts and will make you understand that the answer to this teaser is indeed 1/7.
I do not mean to offend anybody, but its true that probability is not an easy concept to understand. Even many Maths doctorates find it difficult to fully understand probability. My maths teachers at school were doctorates in Maths, but even they did not feel comfortable teaching probability, and so we specially hired a probability doctorate to teach us the Probability course.In Probability the order of the things IS important, and cant be ignored. Probability is not always governed by normal maths rules. So before you start thrashing others' correct answers please take probability lessons with someone whi is a doctorate in Probability. That will clarify your concepts and will make you understand that the answer to this teaser is indeed 1/7.
Just got back to the site and have enjoyed reading comments.I didn't expect this teaser to be overly dificult or generate so much disagreement. I have taught probability and statistics for many years at universities, so I can state emphatically that the given answer is correct. What makes it a nice teaser is that the answer doesn't seem like the right answer, though it most definitely is.
Arguments from authority will not likely help those who misunderstand the teaser. Arguments should be able to hold their own weight and stand up to criticism.
"Mrs. Mathemagican has three children."
Immediately we should imagine the number of possibilities for these three from youngest to oldest:
BBB
GGG
BGG
GBB
BBG
GGB
BGB
GBG
There are eight equally possible outcomes for the gender of her children.
"If AT LEAST one of her children is a boy, what is the probability that she has three sons?"
Notice that the teaser does not tell us which child is a boy. It could be the youngest, oldest, or middle child. The information given does not tell us who is the boy. Since this is an event that has already happened, we can not assume that the first child is a boy and that we are calculating the odds of the next two being boys. That is a different scenario entirely. Therefore we can only safely eliminate the outcome where there are no boys (GGG).
If the teaser told us that the oldest child was a boy, then we could narrow our outcomes to the following:
BBB
BGG
BBG
BGB
That specific information gives us 1 in 4 outcomes where there are three boys, but this information was not given. We do not know which child is the boy. We do not know what the outcome is.
She could have had any one of 8 possible outcomes, but we now know that she did not have three girls and so we can eliminate one.
The teaser could have been written as follows:
"Mrs. Mathemagican has three children and they are not all three girls. What is the probability that she has three sons?"
"Mrs. Mathemagican has three children."
Immediately we should imagine the number of possibilities for these three from youngest to oldest:
BBB
GGG
BGG
GBB
BBG
GGB
BGB
GBG
There are eight equally possible outcomes for the gender of her children.
"If AT LEAST one of her children is a boy, what is the probability that she has three sons?"
Notice that the teaser does not tell us which child is a boy. It could be the youngest, oldest, or middle child. The information given does not tell us who is the boy. Since this is an event that has already happened, we can not assume that the first child is a boy and that we are calculating the odds of the next two being boys. That is a different scenario entirely. Therefore we can only safely eliminate the outcome where there are no boys (GGG).
If the teaser told us that the oldest child was a boy, then we could narrow our outcomes to the following:
BBB
BGG
BBG
BGB
That specific information gives us 1 in 4 outcomes where there are three boys, but this information was not given. We do not know which child is the boy. We do not know what the outcome is.
She could have had any one of 8 possible outcomes, but we now know that she did not have three girls and so we can eliminate one.
The teaser could have been written as follows:
"Mrs. Mathemagican has three children and they are not all three girls. What is the probability that she has three sons?"
My answer was "slightly more than 1/7".
The expected answer of 1/7 was fairly obvious since there is only one way at least one child is not a boy out of the 2*2*2 possibilities.
However, there are two factors this answers fails to take into consideration.
First, between 51% to 51.5% of all children are male. This preference for male children is believed to have evolved in order to account for a higher mortality rate for males.
Second, the sex of a second child is not independent of the sex of the first child. If you have two children it is slightly more likely they are the same sex than different. This is because each male will have a different probability distribution for the sex of their children.
So assuming a single father, we know the father did not have three girls. Therefore it is slightly more likely than random that he will have three boys because knowing at least one is a boy increases the odds of each other child being a boy above the average.
Given that the information to calculate these probabilities is missing, all we can say is that the 1/7 is an underestimation of the actual probability.
The expected answer of 1/7 was fairly obvious since there is only one way at least one child is not a boy out of the 2*2*2 possibilities.
However, there are two factors this answers fails to take into consideration.
First, between 51% to 51.5% of all children are male. This preference for male children is believed to have evolved in order to account for a higher mortality rate for males.
Second, the sex of a second child is not independent of the sex of the first child. If you have two children it is slightly more likely they are the same sex than different. This is because each male will have a different probability distribution for the sex of their children.
So assuming a single father, we know the father did not have three girls. Therefore it is slightly more likely than random that he will have three boys because knowing at least one is a boy increases the odds of each other child being a boy above the average.
Given that the information to calculate these probabilities is missing, all we can say is that the 1/7 is an underestimation of the actual probability.
Oct 16, 2009
After analyzing all the previous comments, I landed on 1/7.
Those who are saying 1/4 are wrong because u r saying a specific child is a son. For example: first child is son, or second child is son. This is totally wrong assumption. See the explanation by "scienth".
Those who are saying 1/3 are also wrong because u r assuming probability of having 1 son is equal to the probablility of having 3 son. i.e. u r asuming P(1 son, 2 daughter) = P(2 son, 1 daughter) = P(3 son, 0 daughter) = P(0 son, 3 daughter). This is wrong assumption. Think this way. if you have a 3 children, what is the probability of having all three children as sons? (of course: (1/2) * (1/2) * (1/2))
Those who are saying 1/4 are wrong because u r saying a specific child is a son. For example: first child is son, or second child is son. This is totally wrong assumption. See the explanation by "scienth".
Those who are saying 1/3 are also wrong because u r assuming probability of having 1 son is equal to the probablility of having 3 son. i.e. u r asuming P(1 son, 2 daughter) = P(2 son, 1 daughter) = P(3 son, 0 daughter) = P(0 son, 3 daughter). This is wrong assumption. Think this way. if you have a 3 children, what is the probability of having all three children as sons? (of course: (1/2) * (1/2) * (1/2))
Mar 25, 2010
The ans is correct
@marrianna I think your logic about the doctor is wrong, if we create the sample space and assign each column the age of the child where the first column is the eldest and knowing the eldest (only child alive at this time) is a boy we have
b-g-g
b-b-g
b-g-b
b-b-b
So a 1 in 4 chance.
Or in other words, the children to come are independent of those already born so 0.5*0.5=0.25
The reason I think for the difference is that the question states at least ONE of the children is a boy, where as in you scenrio the eldest MUST be a boy.
@marrianna I think your logic about the doctor is wrong, if we create the sample space and assign each column the age of the child where the first column is the eldest and knowing the eldest (only child alive at this time) is a boy we have
b-g-g
b-b-g
b-g-b
b-b-b
So a 1 in 4 chance.
Or in other words, the children to come are independent of those already born so 0.5*0.5=0.25
The reason I think for the difference is that the question states at least ONE of the children is a boy, where as in you scenrio the eldest MUST be a boy.
This is a more complex variant of the old Boy or Girl paradox that dates back to Martin Gardner in the 1950s. The strictly correct answer is that the problem is underdetermined. Without additional information about how we came to be told that at least one child is a boy, the answer could be 1/7 or 1/4, or anything in between. If the lady tells us that she has 3 children and we ask, Is one of them a boy? and she says, Yes, then the answer is 1/7. On the other hand, if the lady simply volunteers the information, then in the event that she has at least one girl and at least one boy, if she might have chosen to tell us that she has at least one girl, then the answer is not 1/7 and depends on the probability distribution that describes her decision.
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