Probability question for the math heads

I'm trying to figure out the odds of the PowerBall drawing.

you have 55 numbers, from which to pick five.

Then, the powerball has 44 numbers from which to pick one.

How should I figure this out?

I'm thinking, the probability of getting the first five is 50!/55!. Then, the probability of getting the PowerBall correct is 1/44, so just multiply them and get 50!/(55!*44) leaving us with 1/18367858080.

Is this right?

You are going to your friends house. You know your friend has two children. When you arrive at the house and knock on the door a young boy answers the door. What are the odds that the other child is a boy as well?

Let the wailing and gnashing of teeth begin.

:evilgrin:

Its a bit of a gordian knot when you understand the problem.

:-)

mathematician (15th century) that I think you're talking about.

I still don't see how it isn't 50/50, ruling out hermaphrodites. It's an either or situation. A flip of the coin. No matter how many times it comes up heads, the odds each time that it will come up heads again is the same.

The odds of any given child being male or female is 50/50. But the odds of a woman having 2 boys is not the same. We know that the odds are as follows:

boy/boy
boy/girl
girl/boy
girl/girl

Since a boy answered the door already we can eliminate g/g leaving us with 3 possibilities. Of those only one renders us a second boy. So the odds are 1/3 that the other child is a boy.

It's the coin toss problem. The existence of one male child does not have a bearing on what the other child will be. The only odds that matter are the odds that the process of pregnancy will create a male child. It is no more likely to be a girl just because the first child is a boy. As numbers approach infinity, then, yes, the odds will bear your argument out. But, we're not talking about a results set. We're talking about an individual result. Each pregnancy is a separate event that is disconnected from all other pregnancies. The question is what are the odds that next individual child you'll see will be male. Those odds aren't affected by what happened before.

Now, if you're going to open the door to both children at the same time,the odds that both will be boys will not be 50/50 for the reasons you stated. But, if the question is what are the odds that specific child will be a boy, the odds are 50/50.

Real world solution is its about 1/3 chance the other child is a boy. Statistics world its 50/50. This one has driven much greater minds batty.

The way the question is phrased, it is asking what the odds are the next child will be a boy. The odds that the next child will be a boy is not 1/3. If you're standing there, and you've seen one child and don't know what the next one will be, you have 50/50 chance of getting it right. You're talking about a discrete individual event that is not influenced by previous similar events. Therefore the odds are the odds for the individual event. If you're going to place a bet on what the next child will be after you've seen the first one, and guess a boy, you have a 50% chance of being right, not a 33% chance.

You will find that the distribution of b/b, b/g, g/b, g/g families is equal. Thus giving us a real world examination of the matter giving us a 1/3 chance of the next child being a boy (after eliminating the g/g option).

Its global vs local. Real vs Discrete.

The question wasn't asking for the results set. It was asking about an individual event. The birth of one child. It isn't asking "What are the odds someone has two boys". It is asking "What are the odds that that particular child is a boy". Those odds are 50/50.

I've flipped ten coins. All ten of them have come up heads. What are the odds that the 11th coin flip will also come up heads? 50/50. Because the 11th coin flip is NOT affacted by any of the other coin flips. The birth of the second child is not affected by the birth of a previous child. If the quesion is "What are the odds the next child will be a boy", it is 50/50. If the question is "What are the odds both children will be boys", it is not 50/50. If you're standing there, and you have no idea what sex that child is, and someone tells you to place a bet, you will have a 50/50 chance of being right because there are only two possible outcomes. If someone asks you to bet on the odds that BOTH children are boys, then the odds aren't 50/50. I think you're mistaking the two.

What are the odds that you will have picked a family with 2 boys?

You aren't asking an either or question then. Because there are more than two outcomes.

It's all in the question asked. What are the odds a specific child is a boy or a girl? What are the odds two children will be boys? Two entirely different questions with completely different answers.

"Given that the first child you met is a boy, what is the probability that the other child is a boy".

Which is not the same as "what is the probability of two boys".

It's the Monty Hall problem - should you switch doors or not? And yes, you should, because you stand a 2/3 chance of winning that way.

I'm familiar with that, and the math does bear that out. If you switch doors after the first choice, it increases your odds.

The question is "Given that the first child is a boy, what are the odds the second child was born a boy". The problem, for you, is that the given part of that sentence has no affect on whether or not the other child was born a boy. Given that my first ten coin tosses came up heads, what are the odds my 11th coin toss will come up heads? The first ten coin tosses have nothing to do with the 11th. They don't influence the coin to fall on the tails

In the Monty Hall problem, opening the first door completely changes the problem becauase now you've gone from 3 possibilities to 2 possibilites. You're not doing that with the child problem because whether or not a child is born male or female has nothing to do with whether their siblings were born male or female. In the Monty Hall problem, there is one prize that can go in only one of three places. If it didn't go in door 1, we then know that the problem is one prize in one of two places. That's the differnce. That's why it's not the Monty Hall problem.

If you know nothing about two siblings, and the odds of having a boy or girl are fifty percent, here are your possibilities, all of them equally possessed of a 1/4 chance of occuring:

B, G
G, B
B, B
G, G

So out of eight options, you have a fifty/fifty shot of being right, since there are four chances for one of the siblings to be a girl, and four chances for one of the siblings to be a boy.

But if we know one is a boy, things are different. Since there is still a fifty percent chance of having a boy or girl, these possibilities are all equally possible--but each has a third chance of being the case rather than a fourth chance. It's impossible to still have each possessed of a 1/4 chance of occuring, because the chance of having G, G is now zero--it is not a possiblity. We have only three possiblities:

B, G
G, B
B, B

We know one is a boy. There is only one solitary chance for the other child in a combination of two to be a boy if that is the case. The only way to have two boys is to have two boys--you have a boy first, and a boy second. There are TWO ways to have one boy and one girl. You have the boy first, the girl second, OR the girl first, then the boy second. See how that works? 50% chance of having a boy or a girl in all cases, but in a sequence of two there are two ways to have a girl and a boy and only one way to have two boys. For your theory of 50% to be correct, there would have to be an equal number of ways to have two boys as there is to have a boy and a girl. And that isn't the case.

Show me the mathematical proof. If you're right, you should be able to mathematically prove your contention. You won't be able to do that, because you're arguing that the birth of the first child affects the birth of the second. It's simply not the case. Your argument only works in a mathematical or logical sense if you're treating the two children as one set. If the children are individuals, then the gender of the first child has nothing to do with the gender of the second. Asking me "Knowing that the first child is a boy, what are the odds the second child is a boy" is asking me what the odds of that particular child being a boy are. And the odds are 50/50. It is either a boy or a girl. Ask me "What are the odds both children are boys", and you're treating them as a set. And the odds that that set turns out to be matched *aren't* 50/50.

Each time there is a child behind a door, I have a 50/50 chance of guessing right.

Gah, I'm having keyboard problems, so if there are typos that's why :)

You're thinking that, you've thrown a heads, what are the odds of throwing another? In that case you're right. Here are the possiblities for your second throw:

T
H

They are equally likely. Now, let's say you've thrown twice but don't know the results. All these chances are equally likely:

HT
TH
HH
TT

Now someone tells you one of the throws is heads. That leaves you with

HT
TH
HH

Let's eliminate one heads from each, because we know one exists, to show the possibilities of that other throw:

T
T
H

Now tell me, if we admit that the above:

HT
TH
HH
TT

are all equally likely to occur, and then someone tells us we have a heads which narrows our unknown not-heads throw to:

T
T
H

These are still all equally likely to occur, because there is a fifty percent chance of throwing each, and if that's the case, each has a one third chance of occuring. If each has a one third chance of occuring, since two of the possibilities are tails, then (1/3) + (1/3) = (2/3) and there must be only a one third chance that the other throw is heads. That's why you can throw a heads and a tails in sequence two separate ways, and two heads one way. There's still a fifty percent chance of throwing heads or tails each time, but you have more ways of throwing a tails and a heads than throwing two heads.

My answer was correct. :) The question was asking for the odds of that particular child. And the odds that that particular child was a boy was 50/50.

Here's the way the question should have been asked if they were asking this from a set perspective: "Given that a family has two children, what are the odds they will both turn out to be boys?" In this situation, we're looking at the two boys as part of a set.

Look at it like this.

I flip a coin. It comes up H. What are the possibilities that the second flip is H? It is 50%. That is the same as asking if I have a boy, what are the odds the second child will be a boy? 50%.

In both cases, the two "things" are independent of one another, not part of a set. In your cases, the coins or children are considered part of a set and you are asking "given what I know about this set,and given what we know about the probability distribution of these possible sets, what does that tell me about the second member of the set?"

The question doesn't ask



Note the word SECOND in your post. The question actually is asked in a way to make the sequence of births unknown.

It asks:

There are two children. One of which, born EITHER first OR second is a boy. What are the odds the other is a boy?

There are important differences. Let's take your example. Two kids are born. What are the possibilities?

BG
GB
BB
GG

We are told the firstborn was a boy. That leaves:

BG
BB

Eliminating the brith we know that gives us:

xG
xB

or

G
B

Hooray! You are right. 1 in 2 shot of the unknown sibling being a boy.

Now let's take the real question. Two children are born. What are the possibilities?

BG
GB
BB
GG

We are told that EITHER the first OR second is a boy. That gives us:

BG
GB
BB

Which is not at all the same as your example. Now if we eliminate the sibling whose gender we know, we are left with:

xG
Gx
xB

or

G
G
B

Do you understand the difference there?

I'm sorry I wasn't clear. It's late and my space bar is stuck, so I'm able to type as fluidly as I usually can.

I'm not misreading the question, though. The odds that the other boy will turn out to be a boy is still 50/50 given everything I've argued in this thread. As I said earlier, by identifying one child already, as the question does, you've gone from the complete set of possibilities to the sub-set of possibilities in which on child is a boy. If you look at that sub-set, you have a 50% chance of having that other child be a boy. You're not talking about a set probability anymore because you've fixed one member of the set. So, we're back to the coin flip. Since one member of the set is fixed, we collapse back to a situation where the second member of the set is not determined by the first member of the set.

Of course one birth does not affect the gender of the next one (let us assume that, anyway);

HOWEVER, since the set has already been created, and one knows that set contains exactly two elements, each element of which can have one and only one of two states, and since b/g is different than g/b, once you check the state of one element, the probably of the state of the other element is no longer 1/2.

b/b
b/g
g/b
g/g

Each one of these combos has 1/4 chance of being the set on hand. But remember that b/g and g/b are different. So, we should see in the population that 1/2 of all two child families have a boy and a girl, though only 1/2 of those will have an eldest girl, and 1/2 will have an eldest boy; 1/4 should have two boys, and 1/4 should have two girls.

If the first element you check is "boy", that eliminates g/g, leaving the following three combos possible: g/b, b/g, b/b

b/b is 1/3 of that.

In the coin flipping thing, you can ignore the previous flips, because you are looking at every flip as an individual; in this one, you cannot ignore the previous "flip", because you are looking at it as a set of two.

Assuming you're the one standing at the door not knowing what the sex of the children are, because that is how it is asked. If you make a guess with either speficic child, you have a 50/50 chance of being right both times. Yes, the children are already born. But, if a boy answers the door, and you do not know what the other child is, you still have a 50/50 chance of being correct. It's either going to be a boy or a girl. Two outcomes. You have 1 out of 2 odds.

How can the odds be a third if their are only two possible answers? It's either a boy or a girl. Previous guesses and outcomes have no bearing. If the question is "What are the odds you're going to go to your friends house, and both children will be boys", then you're asking a different question with a different set of odds. I think people are confusing the two.

Take two coin tosses in a row. What are the possibilities?

H/H
H/T
T/H
T/T

Now, each of these are all equally possible, correct? The odds of tossing H/H are the same as tossing H/T, T/H and T/T. If that's the case, then you have a 1/4 chance of getting any of these possibilities, right? In that case, if you were to guess whether any toss of a combination of two tosses was heads or tails, you would have a 50% chance as you say of getting it right. Because there are eight tosses, and four possibilities for both the first and second toss--two of which are tails, and two of which are heads. Two out of four is 50%. So your chance of guessing one toss in isolation correctly is always 50%. Okay, now what if I tell you that one toss is heads?

Then there is zero chance for T/T to be true. T/T becomes impossible. So there aren't four possibilities, there are three because we know one toss was heads. Those possibilities are:

H/H
H/T
T/H

Two of those chances include a tails. Only one includes another heads. Therefore if I ask you to guess whether a toss is heads or tails and you KNOW one toss is heads, since the possibility of T/T no longer exists and there isn't any chance of NOTHING happening--it has to be one of those combinations--the probability of another heads MUST be reduced from 50% since there is only one way to toss two heads--H, H. You can't wholly elminate one possible combination and have nothing at all happen to your probability--that makes no sense. Since we know one toss is heads, there are only three possibilities for the other toss. There is only one way to toss two heads--H, H. There are two ways to toss one heads and one tails--H, T and T, H. There are no ways to toss no heads in this scenario. Therefore the odds of the other toss in a combination of two being tails when you know one toss was heads is actually two thirds because there are two ways to do it. The odds of two heads having been tossed when one toss must be heads can only be one third, because there are three possible combinations and only one way to toss two heads--H,H. Make sense?

Did any one of those tosses affect the odds of any other? No.

The question did not ask "What are the odds the family had two boys". The question asked "Knowing that one child is a boy, what are the odds the other child is also a boy". The fact that one child is already a boy has no bearing on whether the second child was a boy. When that mother had the second baby, she still had a 50/50 chance of having a boy.

Each time you toss that coin, it's either going to be heads or tails. Knowing the first coin is heads tells me nothing about the second coin toss. You are not asking the same question. Each time you flip the coin, the odds are 50/50.

Knowing that the first child is a boy does not affect the odds for the second child. You and others are conflating set theory with number theory.

You have a coin. There is a fifty percent chance you will toss heads or tails no matter what. You toss the coin twice. You are told one of your tosses was a heads. You could have:

H, T
T, H
H, H

Now, let's eliminate one heads from each, since you know that one toss is heads. Here are the possiblities for the other toss:

T
T
H

You're confusing this with knowing that your first toss was heads and then asking what the second would be. In that case your possibilities are:

T
H

There you would have your fifty percent probability. See what the difference is? The order isn't known, and there are two ways to throw one heads and one tail, and only one way to throw two heads.

Knowing that the first child is a boy does not limit the possibilities in any way with respect of the second child. The question in this thread is not "Given this probability distribution and given that one child is a boy, what are the odds the second child is a boy?" In THAT case, you're right, because you've placed constraints on the question that make the gender of the first child meaningful. You're dealing with sets and probability distributions. But, the question did not set up that situation. The question told us that one specific child was a boy. That means that we're not dealing with probability distributions any more. We're dealing with discrete events. Since we're dealing with discrete events and not sets, we're back to the 50/50 proposition.

Here's a link explaining what I'm trying to say. As I'm not a mathematician, it may be that I'm just not being clear :)\
http://www.cut-the-knot.com/bears.shtml

Ignore the white/black part and concentrate on the sex.

Let me try again.

You have a boy. What are the chances of having a boy or a girl?

Boy - 1/2
Girl - 1/2

You have two children. What are the odds of having:

B, G (1/2) * (1/2) = (1/4)
G, B (1/2) * (1/2) = (1/4)
B, B (1/2) * (1/2) = (1/4)
G, G (1/2) * (1/2) = (1/4)

We agree this far, correct? You can see that there is an equal chance of having any of those possibilities. Now we are told that one of the children is a boy. These two kids were already born in sequence--we aren't told that the first was a boy, and asked what could the second be? We are told either the first or second child is a boy. That leaves us with:

B, G (1/2) * (1/2) = (1/4)
G, B (1/2) * (1/2) = (1/4)
B, B (1/2) * (1/2) = (1/4)

Now, is there still a (1/4) chance of there being two girls? No, we are told two girls did not happen. But unless the rules of math suddenly don't exist in your universe, there are still equal chances of having had a boy first and a girl second, a girl first and a boy second, or a boy first and a boy second. At all times the chance of having a boy or girl remains the same. We must admit that there is an equal 1/3 chance of each possibility having occured, for we know two girls did not occur, and we know that there is no (1/4) chance that NOTHING occured to take that possibility's place. So your odds of guessing one of these three and being correct are:

B, G = (1/3)
G, B = (1/3)
B, B = (1/3)

Let's eliminate the one boy we know we have from each:

G
G
B

All equally likely to occur, therefore two thirds chance the other sibling is a girl, one third chance it is a boy. Now this is as good as I can explain, so if you have an issue, point out exactly where it is. The odds of having a boy or girl remain 50% throughout my example.

You seem to be saying that knowing that the first child is a boy somehow limits the possibilities of what the second one will be. And it simply doesn't under the conditions of the question. You are NOT saying that ONE of the children is a boy. You are saying that the FIRST child is a boy. You are then dealing with a subset of the range of possibilities--you have already taken yourself out of the setprobablity distribution. You are no longer dealing with a true setquestion.

IF the firstchildis known to be a boy,then the actual possibilitiesare

boy-boy
boy-girl.

50% chance,then.

You have already eliminated

girl-boy
girl-girl

Because you told me that the firstone wasa boy. Therefor,we are not dealing with the full range of sets,and so therefor,the probability has nothing to do with the sets.

That's what I'm trying to get at. Thank you :)

I've got the ideas in my head, it's just a matter of presenting them, and I'm not always good at doing that.

Can you demonstrate why B,G is different from G,B?

I'm going to introduce and index to clarify something.

You seem to be imposing an order to these pairs, let's call it age. The number I put after a child will show the order of his/her birth.
So in mixed pairs of kids we have
G1,B2 for older sister with younger brother
B1,G2 for older brother with younger sister
B1,B2 no explanation needed
G1,G2 " "

Now, we come to the door and there is a boy there. G1,G2 is no longer a possibility, so you say that there are 3 possibilities

He is B2 and has sister G1, he is B1 and has sister G2, or that he has a brother. This is where I believe you are wrong. You seem distinguish between sisters based on order/age, but you are ignoring one major fact. THE BOY AT THE DOOR COULD BE B1 OR B2 IN A BROTHER BROTHER PAIR (caps for emphasis, not to yell) Thus the true possibilities are as follows

We met B1 he has sister G2
We met B2 he has sister G1
We met B1 he has brother B2
We met B2 he has brother B1

Now it is a 50/50 split

If you consider age to be invalid, then you cannot consider there to be two combinations of a brother sister pair because that idea slips order/age into it. Without age there are simply these combinations
2 Brothers
1 brother and 1 sister
2 sisters.
Since we met a boy, 2 sisters is out, leaving us two options
either the boy is one of 2 brothers or he has a sister, 50/50 again

We met B1 he has brother B2
We met B2 he has brother B1

These come from the same possible set, (B1 B2). Whereas:

We met B1 he has sister G2
We met B2 he has sister G1

Come from these two separate sets, (B1 G2) and (G1 B2). You're getting involved in different boys coming to the door when that would only matter if there were a possible set (B2 B1), which there isn't. See the difference?

edit: To remove possible snark. Sorry about that.

See my post #58.

Here, you are treating the children as aggregate instead of individual.

Here are the sets:

bb
bg
gg.

You just told me that one child is b, so the possibilities are then limited to

bb
bg

Therefor, we are back to a 50/50 situation.

Remember, we where told that the FIRST child was a boy, so gb is eliminated right form the start.

Where do you see order of birth established?

B,G from G,B

The set has two positions: c1 and c2, so the possibilities are:

c1 c2
b b
b g
g b
g g

You have told me that EITHER c1 is b OR c2 is b. It has to be one or the other, since one child has been identified. If the child at the door is c2, then the sets are

c1 c2
g b
b b

So, we are looking at the probablity that c1 is a bpoy or a girls, since we know what c2. And its 50/50%

If the child identified is c1, then we the sets:

c1 c2
b b
b g

So, we are looking at the probablity that c2 is a boy or a girl, since we know that c1 is a boy. Again, 50/50.

Essentially, we have taken ourselves out of the full range of the set probablity by fixing one member of the set.

So lets put together all you have. The second option on your first example:

g b
b b

Is exactly the same as the first option on your second:

b b
b g

Same two kids in each option, b b. Whereas the other two are completely different sets of kids. Your example would only be true if the question were asked twice, like this:

There are two children.

b b
b g
g b
g g

One case says the second is a boy.

g b
b b

The other says the first is a boy.

b g
b b

If one says we have no idea which is a boy?

b g
g b
b b

And we're back where we started. Since the question doesn't indicate that, and it your modification changes the possibilities back to the simplistic "you've thrown heads, what are the odds you throw tails?" how can you insert that in? The actual question is "You have two throws, you know one is heads, what are the odds the other is heads?"

The boy cannot be both a c1 and a c2.

So, my original point remains. We KNOW that we either have

c1 c2
b X

or

c1 c2
x b

The two situations are mutually exclusive. Since the two situatiosn are mutally exclusive, they cannot effect each other. once you have locked in a specific person as a boy, then you have eliminated a rage of possibilities, and you aren;t dealing with your original set. To do so means dealing with a physical and logical impossibility.

:evilgrin:

:evilgrin:

And there are mathematicians that still disagree on this one.

the chances that the next child to come from the women's womb will be a male is 50% (assuming all things equal in the birth process); the chance of the ALREADY BORN child being male is 1/3.

If you flipped a coin twice, and it was heads one time, the chance that it was heads the other time is 1/3; the chance that they were both heads is 1/4. (h/h, h/t, t/h, t/t).

If you flipped a coin once and it was heads, the chance that it will be heads on the next flip is still just 1/2, like it always is.


Depending on how you want to look at it.

it is asking what the odds of the sex of the next child. There are only two possible outcomes (again, for sake of discussion, we're ruling out the other, less likely outcomes). Even if there are 5 children, and the first four you've seen are boys, if you're going to bet that the next one will be a boy, you still have a 50/50 chance of being right. If the question is "5 babies will be born. What are the odds all five will be boys?" then it isn't 50/50 because there are way more than two different outcomes. But, facing each individual outcome, the odds stay 50/50 every time.

Yes, any given child is a 50/50 chance. But as the child is already born it is part of the real world and falls within the range of current population. And current population demographics shows that the odds of the next child being a boy is about 1/3.

Global vs Local. The problem with this is that the question is worded badly. It deliberately does not convey the necissary basis upon which you want the odds related. But it doesn't seem to be missing such instruction because it is parsed in normal day to day language. So it doesn't trigger any alarms.

The question is only worded badly if you meant "What are the odds both children will be boys". They are two sepearate, entirely different questions. The question, as it was worded, was "What are the odds the child will be a boy". That is an either or question. There isn't a third option. If there are only two outcomes, boy or girl, then the odds are 50/50.

If you ask the question "What are the odds the child will be a boy?" the only correct answer is 50/50. Every time.

If you ask the question "What are the odds both children will be boys" you are asking a very different question.

It depends on the question you're asking. If you ask the odds for a coin toss/ boy girl outcome, each time it is 50/50. If you are asking for the odds of a group outcome, you're asking an entirely different question.

If the question is "Do previous outcomes affect future results?" the answer is no. They do not. So, given that you know the first result was a boy, does it affect the odds that the next one will be a boy or a girl? And the answer is that it does not.

My second son had the same odds of being a girl as my firt born son. Being born second in no way changed the odds for him. I still had a 50/50 chance of having a boy or a girl with my second pregnancy.

I think you're right 50/50

The arguments for 66/33 seem to hinge on an ordering of the outcomes, which I don't see as justified by this problem.

My question is basically, why does it matter for a brother sister combo, whether the sister is older or younger, thus yielding 2 possible bro/sis combos? Somehow order is creeping into this problem which is why I suspect the gamblers fallacy at work. Alternately, I'd suggest if meeting the brother at the door suggests a 2 combinations, one where the brother is older than the sister one where he is younger, do we not have the same situation with the pair of brothers where it is possible that the first boy you meet is the older brother and then another possibility is that he is the younger brother, thus bringing 2 boy boy pairs and evening the odds?

It's conditional probability.

If Ms. Jones has two children, and we know that at least one of them is a boy (because he answered the door), there is a 1/3 chance that they are both boys.

The sample space, S = {(b,b),(b,g),(g,b),(g,g)}, and all outcomes are equally likely. This notation means that in (b,g), the older child is a boy and the younger child is a girl.

So, we know Ms. Jones has at least one son. Let B denote the event that both children are boys, and A the event that at least one of them is a boy.

P(B|A) = P(BA) / P(A) = P({b,b)}) / P({b,b),(b,g),(g,b)}) = 1/4 / 3/4 = 1/3

P(B|A) means the probability of B given A, and P(BA) means the probability of event B intersect event A (B and A).

Taken directly from my probability book (Sheldon Ross):

The answer to "what is the probability that both children are boys, given that at least one of them is a boy" is 1/3 - there is no dispute there. But that doesn't take into account which boy came to the door.

That's part of the problem. If you knew the boy who answered was the older or younger sibling, it would change the problem space. It's intentionally not specified. Looking back as Az's problem, he doesn't specify if it was the older or younger sibling.

for a boy to answer the door from a family of two boys.

But I don't see how that matters.

We're not saying, "Given a family with two children, what's the probability of a boy answering the door." We're saying, at this point, we know that the family has two children, and at least one of them is a boy. What is the probability that they are both boys?

You have three cards:

Blue/Blue
Blue/Red
Red/Red

You pull one card and look at one side; it is blue. What are the chances that the other side is also red? 2/3, not 1/2. UGH. Silly probability.

Either way, your chance of winning powerball is slim to fucking none, dude :)

I mean...right?

:evilgrin:

I don't understand.

Local thinking: The odds are that any given child a woman has is either a boy or a girl. Even split 50/50.

Global thinking: A woman with 2 children can have the following offspring combinations.

boy/boy
girl/boy
boy/girl
girl/girl

Since we already know one of them is a boy we can eliminte g/g leaving 3 options. Thus the chance the other child is a boy is 1 in 3 or 1/3.

If we crank it open to actual population statistics twins and beyond screw the stats out of synch even more.

<--confused

This question actually through the self proffessed smartest woman in the world for a while. The problem is its a badly worded question. Its asking for a statistical answer but does not give the basis in which it wants that based. What are the odds that a woman will have a boy? 50/50. What are the odds that a woman will have 2 boys? 1/3.

The chance for having a boy will always be 51%. The chance for having a girl will always be 49%. The odds for having a certain combination of genders in a row are different. We'll simplify the chance for having each gender in isolation to a 1/2 chance just to keep figures round.

There are only four possibilities here: boy/girl, boy/boy, girl/boy, and girl/girl. The odds of girl/girl are 0, since we know one child is a boy.

Ways to have two boys: 1
Ways to have one boy and one girl: 2
Ways to have two girls: 0 (since we know there is one boy)

So there are now only three possiblities, since there is no chance of there being two girls and there are two children. Two of the three possibilities include a girl. One of the three includes a boy. Therefore, there is a 1/3 chance of it being a boy. Make sense?

This is a simple way of looking at it, but this method works with coins as well. Take three consecutive coin tosses. Let's say we want to know the chance of tossing three heads in a row. What are your possibilities? There is exactly one way to toss three heads in a row (HHH). There are three ways to toss two heads and one tail (HHT, HTH, THH) and three ways to toss two tails and one head (TTH, THT, HTT). There is exactly one way to toss three tails (TTT).

Ways to toss three heads: 1
Ways to toss two heads and one tail: 3
Ways to toss two tails and one head: 3
Ways to toss three tails: 1

So the chance of tossing three heads is 1/8. There are eight possibile combinations, and only one will yield three heads. The short way to this conclusion is using the formula (1/2)^3, the 1/2 representing the odds of tossing heads and the 3 representing the number of throws.

To simplify, suppose we have one family with b1 and b2, one with g1 and g2, one with b3 and g3, and one with b4 and g4, each equally likely to be selected.

You randomly select one of these families and one of the two siblings randomly answers the door. This gives eight ordered pairs, where the first entrant is the child who answered the door:

(b1,b2)
(b2,b1)
(g1,g2)
(g2,g1)
(b3,g3)
(g3,b3)
(b4,g4)
(g4,b4)

Each ordered pair is equally likely to occur. Now, since we know a boy answered the door, that only leaves (b1,b2),(b2,b1),(b3,g3), and (b4,g4) as possibilities, each still equally likely. Two of them have both boys. The probability is 1/2.

If I am wrong, I would like to see a clear explanation of why.

BTW I am familiar with the Monty Hall problem and this is different.

These two groups possibilities each come from the same sibling set in your example:

(b1, b2)
(b2, b1)

(g1, g2)
(g2, g1)

The possibilities for birth remain:

(b1, b2)
(g1, g2)
(g3, b3)
(b4, g4)

You're getting tied up with who is answering the door. It is true there are two ways the brothers can answer the door, and the heterogenous group can only answer in one way according to the rules of the question. But are you arguing that because there are two boys and therefore two ways they can answer the door, it is more likely that two boys be born in sequence than one girl and one boy in any order? Try flipping a coin twice in succession. Are two heads in a row more or less likely to occur than one heads and one tails in any order?

You can only have two boys one way: (b1, b2) and you can have one boy or one girl two ways (b3, g3) and (g4, b4). That's why it ends up being 1/3.

I'm "getting tied up" with who answers the door because that was how the question was phrased! I guess if you decide to answer another question you can decide what to and what not to "get tied up with."

Of course, if you flip a coin twice, you are twice as likely to get a head and a tail (in either order) than two heads. Now, could you address the specific error in my simple argument?

You're positing that there is a possibility the girl in a heterogenous pair could answer the door. I didn't think we were supposed to admit that possibility, because the question becomes unanswerable. There's no way to calculate the probability of it since it involves human behavior. With a pair of two boys a boy would be 100% likely to answer the door, whereas a pair with only one the probability would of course be less, but it's impossible to determine who in a heterogenous pair would more likely to answer the door, unless we assume that boys and girls have equal proclivity toward door-answering, equal position relative to the door at the time of the knock, equal door-answering ability, &c. My assumption has been that no matter what the situation, the boy answers the door because the question is unsolvable otherwise.

In other words I've read it as: someone flips a coin twice, and tells you once it came up heads--what is the probability the other time it was heads as well? Your reading would be more like: someone flips two coins under a box and reveals one at random and it is heads, and you have to say what's the probability the other is heads. Does that make sense?

and the answer is, of course, 1/2.

In AZ's problem, one boy is revealed at random.

BTW have you voted in my poll on this question yet? B-)

http://www.democraticunderground.com/discuss/duboard.php?az=view_all&address=105x4173709

Your sets are wrong, but that's beside the point. The only thing I can really say is that you're imposing an order onto the problem whereas I was assuming none. I guess my probability book's problem is stated more clearly, where there is not assumption of order, just knowledge of the fact that one of the two children is a boy.

merely incorporating the fact that one child answers the door into the construction of my sample space.

If there were no order, but we knew one of the children was a boy, of the three possible sets

{b,b}, {b,g}, {g,g}

for the sexes in a two-child family, it's clear to see that there's only a one third chance of both children being boys. But you're saying, let's take the possible combinations, and put them into ordered pairs

(b,g)
(g,b)
(b,b)
(b,b)
(g,g)
(g,g)

And only consider the pairs in which a boy comes first (read: answers the door). That excludes the ordered pair of (g,b), which is completely different from simply saying that we know at least one of the children is a boy.

I'm not saying your interpretation of the problem is wrong, it's just a different problem from the one I was assuming.

but I really feel I am addressing the problem as asked. There are different ways for a boy to come to the door, they are equally likely to occur, so they must be factored into the sample space and the computation of the probability.

Whereas in a pair of a boy and a girl, it can only be less. That would have an effect, would it not? If you admit a girl can answer and not only the boy, it adds a lot of problems. That's the stumbling block, and I think reading it that way makes the question unanswerable--at least in any precise way.

I was conveniently ignoring that in my post, though I did stop to think about it while trying to figure out how to answer alarcojon's post. It all still comes back to ordered versus non ordered. The simple solution, and the one for which I based my original argument, way up there, is that we just know that at least one of the children is a boy. I simply think now that the problem needs to be stated in a more clear way.

I, however, have to teach tomorrow, and am therefore off to bed.

If we have order, then we have to consider order for all possibilities. If we want to keep order out of it then the only possibilities are b,g g,g and b,b.

g,b is not a distinct possibility from b,g unless we want to introduce an order to this

Flip a coin twice in succession. Is it more likely to get two heads in a row, or one heads and one tails in any order?

Let's say you're going for two heads in a row. You need to get heads each time, and cannot get tails once. There is only 1/2 chance that you will succeed each time, and you must do it twice.

(1/2) * (1/2) = (1/4)

Okay so far? Now let's say you need to get one heads and one tails in any order. You need to get heads first and tails second, or heads second and tails first. The odds of throwing each remain as always 50%. So let's try for heads first, tails second. There is only 1/2 chance that you will succeed each time, and you must do so twice.

(1/2) * (1/2) = (1/4)

The same thing goes for tails first, heads second, because the odds of getting each remain the same. But note the order can change, whereas there is no second way to get two heads other than getting one heads and then another. Even if you were to label them by which came first, you can get (h1, h2), but (h2, h1) can never be, whereas both (h1, t2) and (t1, h2) are possible. Here are the odds for tails first, heads second--there is still only one chance you will succeed each time, and you must again do it twice:

(1/2) * (1/2) = (1/4)

So what is the chance of getting one heads and one tails in any order? We have seen two possible combinations that provide (1/4) chances of doing so. Adding these together gives us the total probability of success in getting one heads and one tails in any order as:

(1/4) + (1/4) = (2/4) = (1/2)

Whereas the probability for getting two heads in a row remains:

(1/4)

There is no second combination possible when attempting to roll two heads in a row. There are however two combinations possible when trying to roll one head and one tails in any order--heads then tails, or tails then heads. You can't do this with two heads in a row, it remains "heads then heads" in every instance. Show me with a coin two ways to throw heads twice in a row--it happens the same every time: one heads, then another. But there are two ways to throw one heads and one tails in any order: either heads then tails, or tails then heads.

Therefore it is twice as likely to get one heads and one tails in any order than it is to get two heads in a row. If that's the case, and if the possibility of two tails is completely removed by our knowledge that we have thrown at least one heads first or second, the above options--two heads or one heads and one tails in any order--are the only possiblities. It must either be a head then a head, a tails then a head, or a head then a tails. How can we discover the exact probability of each now that the possibility of throwing a tails and then a tails is removed? Well, we know we have two chances. One head and one tails in any order being twice as likely as two heads in a row, and added together they must equal 1 since there are no other possibilities and it is impossible that nothing happened. This leads to the equations:

y = 2x
x + y = 1

which, after plugging in the y, becomes:

x + (2x) = 1
3x = 1
x = 1/3

and therefore:

y = 2(1/3)
y = 2/3

So if you know either the first or second throw was heads, the chance that the other is also heads is only (1/3). The chance that there is one heads and one tails in some orderis (2/3). This is the most mathematical way to explain my answer.

your chances are really good - and the more tickets you buy - even if it is instead of paying the heating bill - the better your chances...

what they think the day after each drawing: "Suckers!"

I was actually doing two - for one thing, there's only 42 powerballs, not 44.

And I forgot to divide by 5! (that's factoral, not explamation point)

Thanks!

Now the numbers work. I was doing a permutation, not a combination.

I was doing a permutation, not a combination:dunce:

A combination is not ordered.

e.g. powerball choices are counted as combinations because you pick some numbers and it doesn't matter which one is first or second or last. It's just numbers. For a permuatation it does matter which is first.

Regarded as combinations (1 2 3) and (2 3 1) would be considered the same, but as permutations they're considered different.

1 in 146,107,962.00

(Silly...It's online at the Powerball site!)

http://www.powerball.com/powerball/pb_prizes.asp

in his name?

I'm trying to think a bit,here :evilgrin: