My OperaSample Space
Sunday, January 4, 2009 1:42:17 AM
This is a followup on my recent entries on Jeff Atwood's article about the other child. I will restate the original scenario here:
Jeff tries to say this is 66% (2/3). But it's not clear at all.
At first, I was 100% certain that it was 50% because one child has no effect on the other. I was then bothered by the fact that while you could pick any one of 4 girls in the sample space GB, BG and GG, two of the girls are in the same family. So TWO of the girls point to the same family. If you can only have ONE representative from each family, then each girl in a GG family can only represent the family 50% of the time that a GG family is encountered.
After this, I had to accept that two of the families had boys as the other sibling, so I admitted that I was wrong. Only thing is that I wasn't.
Let's say, hypothetically speaking, you met someone who told you they had two children, and one of them is a girl. What are the odds that person has a boy and a girl?
Jeff tries to say this is 66% (2/3). But it's not clear at all.
At first, I was 100% certain that it was 50% because one child has no effect on the other. I was then bothered by the fact that while you could pick any one of 4 girls in the sample space GB, BG and GG, two of the girls are in the same family. So TWO of the girls point to the same family. If you can only have ONE representative from each family, then each girl in a GG family can only represent the family 50% of the time that a GG family is encountered.
After this, I had to accept that two of the families had boys as the other sibling, so I admitted that I was wrong. Only thing is that I wasn't.
After being involved in some of the discussion on Jeff's blog, I now see why there was so much confusion, myself included.
I will set up a different scenario that is equivalent to Jeff's scenario.
Say I have two decks of cards. 52 cards in each deck. 26 black cards and 26 red cards.
I have two spots labelled "Card #1" and Card #2" which are each assigned a specific deck also called Deck #1 and Deck #2. I then pick one card from deck #1 and put it face down over the label "Card #1". I do the same thing for the other card choosing from the other deck. I use two decks to keep the odds of each card 50/50 black or red in exactly the same fashion that it's equally likely to get a boy or a girl for each child.
Now, I put you behind a curtain where you do not peek. I go and do something. You do not know what. After I'm done, I tell you that at least one of the cards is red.
What are the odds that the other card is black?
Most people rightfully say 50/50. The answer 2/3 can only happen under certain specific situations.
Where lies the confusion? How can it be 50/50 in some cases and 2/3 in others?
The question lies in the sample space. All of us assume that the sample space is about the cards. Let's examine those.
In the first case, some say that one card has no effect on the other, so the sample space for the other card is either RED or BLACK. This would mean 50/50.
In the second case, some say that the two cards can form FOUR different hands RR, RB, BR and BB. Since one of the cards is red, then it cannot be BB leaving the sample space {RR, RB, BR}. This would mean 2/3.
Let me tell you right now that while both results may occur, we need more information in order to tell us which one is correct. And we're not supplied this information.
How would we know what kind of information we would need to make that determination?
It just so happens that the way we obtain information is just as important as the cards themselves. What I did when you were behind the curtain is critically important.
For example, if I decide that I only want two red cards, I could have decided to only talk to you if it happened that both cards were red. In that event, there is ZERO chance that the other card is black.
If I decide to only talk to you when there are red cards, then the odds of the other card being black is 2/3, but I've been awfully deceitful by eliminating certain pairs that anyone would expect to occur under normal conditions.
Also, if I talk to you every time, but I only report on red cards FIRST if any, then the odds are 2/3 that the other card is black. But again, I've intentionally skewed the results without telling you.
Why do most people answer 50%? Because they quite rightly don't make assumptions that can't be backed up.
The sample space isn't only about the cards. It's also about what kind of information *I* tell you about. What is THAT sample space?
For each of BB, BR, RB and RR, what would *I* tell you? Until you know this, you cannot answer any further questions about the cards. For each hand, *I* could tell you a variety of things with different probabilities. You need to know that sample space along with their probabilities. You also need to know what the probability is that *I* will report to you at all.
Let's take an example. Say I'm equally likely to say there is at least one red card as I am to say there is at least one black card! I will report to you every time no matter the cards that come up.
Here, we have enough information to know the odds of the other card being black. We start with the odds of what *I* will tell you for each combination of BB, RB, BR and RR.
For BB, I will say "one black card" 100% of the time.
For RB, I will say "one black card" 50% of the time.
For RB, I will say "one red card" 50% of the time.
For BR, I will say "one black card" 50% of the time.
For BR, I will say "one red card" 50% of the time.
For RR, I will say "one red card" 100% of the time.
Now, we state the odds of each of BB, RB, BR and RR happening.
BB 25%
RB 25%
BR 25%
RR 25%
Note how we have TWO sample spaces?
Combine them:
BB, "one black card" 100% * 25% = 25%
RB, "one black card" 50% * 25% = 12.5%
RB, "one red card" 50% * 25% = 12.5%
BR, "one black card" 50% * 25% = 12.5%
BR, "one red card" 50% * 25% = 12.5%
RR, "one red card" 100% * 25% = 25%
That is the TRUE GLOBAL sample space. Without the odds of how you obtain information, the sample space is useless. If you remember ONE thing from all this is that you should always ask HOW the information was retrieved and what are ALL the different ways different information could have been given to you along with their probabilities. Without all this, it's a trick question.
Looking at our last table, we notice this is still not our final sample space. We are told more information. We are told that one card is red. So we only keep the entries where we are told that one card is red and ditch the others. We are left with:
RB, "one red card" 50% * 25% = 12.5%
BR, "one red card" 50% * 25% = 12.5%
RR, "one red card" 100% * 25% = 25%
To remap the odds, we simply divide the original odds by the remaining total. Let's do that.
Total = 12.5+12.5+25 = 50.
RB 12.5/50 = 25%
BR 12.5/50 = 25%
RR 25/50 = 50%
From this table, we see that there are two entries where the other card is black. So let's add up those percentages.
25%(RB) + 25%(BR) = 50%
So there is 50% chance that the other card is black.
Note that we are told that one card is red. The "telling of a card being red" is the permutation that we are interested in with respect to the cards and what we could have been told. It's not the cards alone.
Our global sample space is WHAT WE ARE TOLD in COMBINATION with the CARDS. Once we have the global sample space, we can cut down the possibilities.
I want to now look at another sample space for what we are told. Let's say I change the odds of being told that one card is red. Instead of even likelihood that we are told that, let's say that *I* HAVE to tell you about the red card if there is one. So if there is a red card in the hand, I'll tell you about it. If I do tell you there is a red card, what are the odds that the other card is black?
We start over with the odds of what *I* can tell you for each pair of cards.
For BB, I will say "one black card" 100% of the time (You would know both are black).
For RB, I will say "one red card" 100% of the time.
For BR, I will say "one red card" 100% of the time.
For RR, I will say "one red card" 100% of the time.
Now, we state the odds of each of BB, RB, BR and RR happening which are still the same.
BB 25%
RB 25%
BR 25%
RR 25%
Combine them:
BB, "one black card" 100%*25% = 25%
RB, "one red card" 100% * 25% = 25%
BR, "one red card" 100% * 25% = 25%
RR, "one red card" 100% * 25% = 25%
That is our global sample space.
Since we are told that one of them is red, let's eliminate the times where it states that you were told "one black card". Note that we don't eliminate cards. We eliminate the sample space ACCORDING TO WHAT WE WERE TOLD. We don't look at BB, RB, BR, RR at all just as we did above.
RB, 25%
BR, 25%
RR, 25%
Remap the odds.
Total = 25+25+25 = 75
RB, 25%/75 = 1/3
BR, 25%/75 = 1/3
RR, 25%/75 = 1/3
There are two entries where the other card is black. Add those up.
1/3(RB) + 1/3(BR) = 2/3
What can we say about all this? That the way you are told about the information is critical. You need to know this sample space just as much as the sample space that belongs to the target of the information.
Just a note about the above. I used BB, BR, RB and RR where the first letter is Card #1. But that was not actually justified. It's fine to use it, but it confuses the issue. There's simply no need for it at all. You're allowed to label the cards any which way you want. Instead of card 1 and card 2, you can label the first card to be the one your are first given information about. Just as long as you are consistent, you are perfectly justified to do this. In fact, this is how you should do it.
RR, BR, RB and BB will now be by order of introduction.
If you're equally likely to be told "red card" or "black card", we have this table instead:
BB, "one black card" 100% * 25% = 25%
RB, "one red card" 100% * 25% = 25%
BR, "one black card" 100% * 25% = 25%
RR, "one red card" 100% * 25% = 25%
If we are told "red card", then we only keep those.
RB, "one red card" 100% * 25% = 25%
RR, "one red card" 100% * 25% = 25%
Remap the odds.
Total: 25+25=50
RB, 25%/50 = 50%
RR, 25%/50 = 50%
Only one entry that has a black card, so the odds of that entry are 50%. Same as before.
And in the case *I* MUST tell you about the red card if there is one, we have a little more work to do.
BB, "one black card" 100%
RB, "one red card" 100%
RR, "one red card" 100%
BR, 0% (impossible)
BR is impossible since the red card must always be first.
Now I'm going to list all cards as if you can be told about any card first.
BB 25%
RB 25%
BR 25%
RR 25%
But in cases where you would normally would encounter BR, *I* have to tell you about the red card first. So I must convert this to RB so that the red one is told first.
BB 25%
RB 25%
RB 25%
RR 25%
We have two entries for RB, so we add their odds.
BB 25%
RB 50%
RR 25%
This makes sense since mixed cards are just as likely as same colored cards. Our table has the R in front if there is a R, so we're good to go. We can combine our two sample spaces.
BB, "one black card" 100%*25% = 25%
RB, "one red card" 100%*50% = 50%
RR, "one red card" 100%*25% = 25%
BR, 0% (impossible)
We're told "red card", so we keep those.
RB, 50%
RR, 25%
Remap their odds.
Total: 50+25 = 75
RB, 50%/75 = 2/3
RR, 25%/75 = 1/3
The entry that has a black card is 2/3 exactly as before.
So it doesn't matter which card came first or what girl was born first.
We are now ready to revisit Jeff's question.
Let's say, hypothetically speaking, you met someone who told you they had two children, and one of them is a girl. What are the odds that person has a boy and a girl?
We are told TWO pieces of information. That means we have TWO sample spaces. And we also have the target of that information with its own sample space. That makes THREE sample spaces.
Luckily, one of them is easy to reduce.
Odds of a parent of 1 child to tell you they have 1 child? 100%
Odds of a parent of 2 children to tell you they have 2 children? 100%
Odds of a parent of 3 children to tell you they have 3 children? 100%
...
Since we're told that the parent has two children, we keep that entry.
Odds of a parent of 2 children to tell you they have 2 children? 100%
The odds are 100%, so this won't affect any of the other sample spaces. Because it's 100%, we can leave it out for our calculations.
The other sample space is this (in bold):
you met someone who told you they had two children, and one of them is a girl.
Here, we're told by this someone that they have two children and we're also told that one of them is a girl.
Let's examine that sample space where we list these by order of whom we're told about.
For BB, Odds of being told one of them is a boy 100%
For BG, Odds of being told one of them is a boy ???
For BG, Odds of being told one of them is a girl ???
For GB, Odds of being told one of them is a boy ???
For GB, Odds of being told one of them is a girl ???
For GG, Odds of being told one of them is a girl 100%
We don't have enough information. So most people assume there is an even likelihood to be told about the boy as there is for the girl in mixed families. Isn't this the most probable scenario? One must take a leap of faith to assume that ALL families in the world would tell you about their girl ALL the time.
Still, we'll continue without the odds just for fun. I will use birth order since we have no idea what the sample space is for being told a boy or a girl. This is our third and final sample space.
BB, 25%
GB, 25%
BG, 25%
GG, 25%
Combine them.
BB, told one of them is a boy 100% * 25% = 25%
BG, told one of them is a boy A*25% = .25A
BG, told one of them is a girl B*25% = .25B
GB, told one of them is a boy C*25% = .25C
GB, told one of them is a girl D*25% = .25D
GG, told one of them is a girl 100%*25% = 25%
We're told about the girl. So we keep all those entries where this is the case.
BG, told one of them is a girl B*25% = .25B
GB, told one of them is a girl D*25% = .25D
GG, told one of them is a girl 100%*25% = 25%
Remap the odds.
Total: .25C + .25D + .25
BG, told one of them is a girl .25B/(.25B+.25D+.25)
GB, told one of them is a girl .25D/(.25B+.25D+.25)
GG, told one of them is a girl 25%/(.25B+.25D+.25)
The entries where there is a boy are the first two. Add them up.
(.25B+.25D)/(.25B+.25D+.25)
(B+D)/(B+D+1)
B and D amount to the odds of being told it's a girl when encountering mixed families. We can simplify since B and D are both for mixed families.
2A/(2A+1)
If ALL families in the world would ALWAYS report on their girl first if they have one, then A is 100% or 1.
(2*1)/(2*1+1) = 2/3
But if mixed families are just as likely to report on their boy as they are on their girl, then A is 50% or .5.
(2*.5)/(2*.5+1) = 1/2
The second scenario is how most people expect parents to act. In fact, there is absolutely no basis for the 2/3 scenario. Nothing in the problem statement says that all parents MUST report on the girl. It only says that this one specific family did so.
So Jeff is wrong to say:
Most people answer 50%.
Unfortunately, this isn't correct.
It is correct if humans aren't totally biased toward reporting on their daughters.
Sample space. Know it!
Vladasvladas # Sunday, January 4, 2009 8:56:22 AM
I should tell that the original question isn't incomplete. Even more - it has a redundant unneeded information! Just as I told in my first reply.
To prove this I'll ask another simple question: Does the odds depend on the first child particular sex TELLING (Even if you decide that it's 66%)? NO! It's the same if you're told that was a girl, or you're told that was a boy! It doesn't matter at all. So, if it doesn't matter, this is a wrong erroneous information they try to provide you with.
And the answer (and odds) cannot be dependent on such ridiculous info.
And this was what I tried to say in my second reply.
Vladasvladas # Sunday, January 4, 2009 9:12:38 AM
What if a person will tell you: "My first child sex is..., but I won't tell you
Vorlath # Sunday, January 4, 2009 6:20:42 PM
Vladasvladas # Sunday, January 4, 2009 6:43:40 PM
My main point is that if you change the word 'girl' with the word 'boy' in the Question, you must get the same 66% based on your logic. Are you still going to eliminate BB case in this case. Or you'll switch onto GG case
So, my question still is: Does the answer depend on what gender was told to you for the one of children?
Vorlath # Sunday, January 4, 2009 8:30:05 PM
We have first child A. Second child B. But there's another entry for what we are told. Call this C. So there are 8 eight possibilities.
C is what we are told.
We need to know the odds of being told it's a boy vs. the odds of being told it's a girl. We need to think of this just like another child. Or a third coin. Or a third card.
X = We're told it's a girl.
Y = We're told it's a boy.
But we need to know the odds of X and the odds of Y for each of GG, GB, BG and BB. If the parent MUST tell about their girl, then the odds are this:
For BB, odds of X is 0%
For BB, odds of Y is 100%
For GB, odds of X is 100%
For GB, odds of Y is 0%
For GG, odds of X is 100%
For GG, odds of Y is 0%
For BG, odds of X is 0% (impossible)
For BG, odds of Y is 0% (impossible)
Then we list what are the odds of pairs of children where the first one is the one we're first told about.
Odds of BB is 25%
Odds of GB is 50%
Odds of GG is 25%
Odds of BG is 0% (impossible)
BG is impossible since the parent must tell us about the girl first.
Then we combine the odds so that we have our final sample space of ABC. The eight possibilities are BBX, BBY, GBX, GBY, GGX, GGY, BGX, BGY.
Take BBY. We look up the odds of BB. It's 25%. Then we look up the odds of Y for BB. It's 100%. Multiply them. 25%*100% and it's 25%. Do this for each entry and we get this table:
BBX (0%)
BBY (25%)
GBX (50%)
GBY (0%)
GGX (25%)
GGY (0%)
BGX (0%)
BGY (0%)
But we know that Y is what we have in the current case. The parent told us it was a girl. So we need to only keep those entries. We don't look at ANY of the children at all. It's not dependent on them at all. It's 100% dependent on what we are told. Are we told X or Y? The confusion comes from the fact that what we're being told has the word girl in it. Forget that. Think in terms of another coin toss or another card. We got X so that is what we keep.
BBX (0%)
GBX (50%)
GGX (25%)
BGX (0%)
Remap the odds
BBX (0%)
GBX (67%)
GGX (33%)
BGX (0%)
So there is a 66% chance of a girl (A) and a boy (B) when told it's a girl (C). ABC=GBX.
However, if the odds of X and Y are different, then it changes the overall odds. If the parents are just as likely to talk about their daughter as they are about their son, then we get this:
For BB, odds of X is 0%
For BB, odds of Y is 100%
For GB, odds of X is 50%
For GB, odds of Y is 50%
For GG, odds of X is 100%
For GG, odds of Y is 0%
For BG, odds of X is 0% (impossible)
For BG, odds of Y is 0% (impossible)
Combining these odds with the BB,GB,GG,BG table, we get these odds:
BBX (0%)
BBY (25%)
GBX (25%)
GBY (25%)
GGX (25%)
GGY (0%)
BGX (0%)
BGY (0%)
We are told that X is what we have. So keep those entries where X exists.
BBX (0%)
GBX (25%)
GGX (25%)
BGX (0%)
Remap the odds.
BBX (0%)
GBX (50%)
GGX (50%)
BGX (0%)
In this case, the odds of being told a girl (A) and the other child being a boy (B) when told it's a girl (C) is 50% because ABC=GBX.
So the specific gender isn't what's important. It's rather about what are the odds that you can be told about that gender.
Another way to phrase it is this. It's not the specific heads or tails that matters, it's the odds of being TOLD it's head or tails that matters.
Vladasvladas # Sunday, January 4, 2009 9:12:15 PM
Should additional information be taken into account when it carries no information at all (in this particular case)? When a given clue (it is a boy, or it is a girl) leads to the same result (66%). Should we consider it as useful information and so, change our logic? Does it really have any relation with initial probability matters?
To be precise, what I want to know: If we change the question to be told opposite (the one of them is a boy), would it provide us with different result? If no, why we should consider such clue in our further logic at all?!!!
This is an extract from Jeff's blog comments:
Vorlath # Sunday, January 4, 2009 11:08:32 PM
In Jeff's problem, nowhere does it say that the parent MUST tell about the girl. That's why what you quoted is so pertinent.
Jeff thinks that just because you know one of them is a girl, that this is enough information to get to 66%. It is not. The only way you can get to 66% is if you are FORCED to inform us about the girl first if there is one. That eliminates the alternate universe scenario (by stating that you're not allowed to say "boy" in mixed families). But there is nothing that says this must be so. So the answer to Jeff's problem must be 50%.
What you quoted is quite correct in demonstrating how Jeff's scenario is contradictory. Unless you eliminate those alternate scenarios somehow, you can't say the odds are anything other than even between all those alternate universes.
BTW, what you quoted is exactly what I'm talking about when saying "what we are told". One universe, "we are told girl". The other universe, "we are told boy". That's what my X and Y represent. It represents those alternate universes.
Sean Connerspc476 # Monday, January 5, 2009 8:47:12 PM
But, the odds of the two cards being red is 25%, 50% for being different colors (25% for red-black, and 25% for black-red), and 25% for being black. Again, nothing under dispute here and pretty much standard probability.
So now we're looking at two cards with four possibilities: red-red, red-black, black-red and black-black. We reveal one of the cards, and it's black. If we think of these two cards as coming from a four card deck (with two reds and two blacks) we've just revealed one of the cards. Of the three cards remaining, one is black (33%) and two are red (66%).
So, if you know that a parent has two kids, you have a 1 in 4 chance of getting them right (although half the time you'll get the order wrong). But if you know that one of the kids is red, um, a girl, then it's more likely that the other one is a boy, because you have additional information about the situation (although you still don't know the order of the kids).
Vorlath # Tuesday, January 6, 2009 4:37:58 AM
You can't use one of the red cards since I specifically stated it can only be used for Card #1. So it's 50/50.
In the same manner, if you have a boy, you can't keep the girl you didn't have as an option for later on. Once you've had a boy, that's it. The other option disapears. You can't use it anymore.
BUT!!! If you changed the scenario to use a four card deck, then yeah, the second card has a 66% chance of being red.
As for children, your scenario could only be true (again by changing the rules) if a parent decides to adopt two kids out of four where two are boys and two are girls. Otherwise, your scenario cannot happen.
But you DO know the order. The first one you're told about is a girl. The second one is all that's left to determine. You can order whatever you like in any order you like. This first born stuff is there to steer you in the wrong direction.
Sean Connerspc476 # Tuesday, January 6, 2009 9:00:56 AM
In other words, out of 100 pairs, 75% will have at least one girl (and conversely, 75% will have at least one boy), and out of said 75 pairs, 50 will have a boy (which is 66% of 75) and 25 will have another girl (which is 33% of 75).
There. I think that nails that.
Vladasvladas # Tuesday, January 6, 2009 10:24:45 AM
Vorlath # Tuesday, January 6, 2009 4:45:52 PM
Do you understand that these odds can never change?
Two Boys 25%
Mixed 50%
Two Girls 25%
NEVER! It's impossible to change those odds. The only way we can get different odds is when used in COMBINATION with other odds. I showed you what they were. That is lacking from your scenario.
I'll use your own words.
How did you set out that BB is not possible? It's because there is 0% possibility of this happening if we're told it's a girl, right? So show me the computation you did to turn BB @ 25% into 0%! Show me the sample space that says BB is @ 0% and all others are at 100%. That's what I want to see.
Here is what I got:
When encountering a family of two boys, what are the odds that the parent discloses a boy? 100%
When encountering a family of two boys, what are the odds that the parent discloses a girl? 0%
When encountering a family of two girls, what are the odds that the parent discloses a boy? 0%
When encountering a family of two girls, what are the odds that the parent discloses a girl? 100%
When encountering a family of one boy and one girl, what are the odds that the parent discloses a boy? 50%
When encountering a family of one boy and one girl, what are the odds that the parent discloses a girl? 50%
You considered these things for the BB case. So we must consider them for the other cases as well.
The only way your case holds up is if it's impossible for parents to disclose a boy when they have a mixed family. There was no mention of this.
See, it's easy to get it wrong when we only apply one entry (BB and told girl = 0%) and forget to figure out the odds for the other options.
Continuing on my scenario, I'm left with these odds when told it's a girl.
For a family of two boys, odds of being told it's a girl? 0%
For a family of two girls, odds of being told it's a girl? 100%
For a mixed family, odds of being told it's a girl? 50%
Combine the odds:
(When told it's a girl)
Two boys 25% * 0% = 0%
Two girls 25% * 100% = 25%
Mixed 50% * 50% = 25%
Remap the odds to 100%
(When told it's a girl)
Two boys 0%
Two girls 50%
Mixed 50%
See, I've SHOWN how BB got to 0%. I don't simply discard it. My last table is the combination of two sample spaces. As it must be since it's impossible to change the odds of mixed, GG and BB.
I want to see your other sample space. I want to see exactly how you got BB to 0% along with what odds you give all the other entries (not only mixed and GG, but also the cases where you could have been told about a boy).
Vladasvladas # Tuesday, January 6, 2009 5:14:06 PM
Everyone is happy
Sean Connerspc476 # Tuesday, January 6, 2009 10:40:16 PM
25 are boy/boy (25%)
25 are boy/girl (25%)
25 are girl/boy (25%)
25 are girl/girl (25%)
No dispute. And with that, we can make the following guesses:
If we say they're both boys, we're right 25% of the time.
If we say they're both girls, we're right 25% of the time.
If we say there's one boy and one girl (regardless of order), we're right 50%.
So far, so good?
Now we're told there's at least one girl. That drops 25 pairs (or 25%) from consideration. We now know:
25 are boy/girl (25% of the original 100, 33% of the remaining 75)
25 are girl/boy (25% of the original 100, 33% of the remaining 75)
25 are girl/girl (25% of the orginal 100, 33% of the remaining 75)
So far so good? The boy/boy combination was dropped because given that one of the kids is a girl, there can't be a boy/boy combination. So, given that we're told that one of the kids is a girl, we can make the following guesses:
If we say the other one is a boy, we're right 66% of the time.
If we say the other one is a girl, we're right 33% of the time.
So, where did I blow it this time?
Vorlath # Wednesday, January 7, 2009 9:48:03 AM
So your sampling space here is saying the parents will first report on a girl 75% of the time and only first report on the boy 25% of the time. Why would that be? Why would the parents prefer to report on the girl by a 3 to 1 ratio?
Sean Connerspc476 # Wednesday, January 7, 2009 9:04:42 PM
HERE'S MY COMMENT YOU QUOTED, BUT GENDERS SWAPPED!
"Now we're told that there's at least one boy. That drops 25 pairs (or 25%) from consideration. We now know:
"25 are boy/girl (25% of the original 100, 33% of the remaining 75)
"25 are girl/boy (25% of the original 100, 33% of the remaining 75)
"25 are boy/boy (25% of the original 100, 33% of the remaining 75)"
(My guess, you thought I double posted---I didn't, if you read closely enough)
The rest of the math *is the same* although the genders are swapped.
While it might seem odd that there's a 75% of a girl *and* a 75% of a boy, it's not really:
AT LEAST ONE GIRL --------------------------- girl/girl girl/boy boy/girl boy/boy ------------------------- AT LEAST ONE BOYThe two ranges overlap.
So, again, where did I go wrong?
Sean Connerspc476 # Wednesday, January 7, 2009 9:12:46 PM
Vorlath # Wednesday, January 7, 2009 9:13:08 PM
I can go and fix it if you'd like.
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It does add up to 150%. That's what you're showing. If there is an intersection, then you must remove what you double counted. Since you double counted GB and BG, their odds must be cut in half for each scenario.
Sean Connerspc476 # Wednesday, January 7, 2009 9:21:02 PM
and in the other half of the scenarios:
girl/girl girl/boy boy/girl boy/boy -------------------------In both cases, *at the same time* there's a 75% that the stated gender is correct. Yes, there's overlap, but you can't combine the two like you seem to want to do. It just means that once you pick one set, you have to treat it separately than the other case (but it just so happens that the two cases are symmetrical).
Vorlath # Wednesday, January 7, 2009 9:31:50 PM
Combine their odds and you get:
XA 0%
XB 0%
XC 0%
YA 1/6
YB 1/6
YC 1/6
Remap the odds:
YA 1/3
YB 1/3
YC 1/3
So people usually just eliminate all the X's and remap the odds that are leftover.
But for Jeff's scenario. We can't do that because the odds aren't all 100% and 0%.
For BB, odds of "told it's a girl" is 0%
For GB, odds of "told it's a girl" is 50%
For BG, odds of "told it's a girl" is 50%
For GG, odds of "told it's a girl" is 100%
So you get:
BB 25% * 0% = 0%
GB 25% * 50% = 12.5%
BG 25% * 50% = 12.5%
GG 25% * 100% = 25.5%
Remap the odds to 100%:
BB 0%
GB 25%
BG 25%
GG 50%
So the odds of the other being a boy is 25% + 25% = 50%
You have no problem eliminating BB, but you refuse to also adjust the other possibilities.
OTOH, if you're asking the parents if they have at least one girl, then 75% of them will answer yes and 25% will answer no. In that case, you get:
For BB, odds of "told it's a girl" is 0%
For GB, odds of "told it's a girl" is 100%
For BG, odds of "told it's a girl" is 100%
For GG, odds of "told it's a girl" is 100%
See the difference?
You can't just reject ONE case because it's impossible without regard to the other possibilities. In this last case, they happen to be 100% for each of GB, BG and GG, but it would be incorrect to just assume they are without looking at the problem statement.
Vorlath # Wednesday, January 7, 2009 9:35:04 PM
Sean Connerspc476 # Thursday, January 8, 2009 12:34:34 AM
Picking at least one girl: 75%
Picking at least one boy: 75%
Picking a boy/girl or girl/boy: 50%
(odd how that adds up to 200%, isn't it? And if not, why not?)
Now, assume that before you are asked, Monty opens one of the doors and reveals two boys. He how asks you to pick a door with a girl and a boy behind it. Odds are slightly different now. Why? Because you have more information.
And, would the odds be any different if Monty opens one of the doors and reveals two girls, then asks you to pick a door with one of each?
How is this any different than meeting a parent who says they have two kids, one daughter, and the odds of their other kid being a boy?
Vorlath # Thursday, January 8, 2009 7:26:51 AM
IOW, in your scenario, it's impossible to pick BB. But in the parent's scenario, it's quite possible to pick BB.
Sean Connerspc476 # Friday, January 9, 2009 9:57:01 PM
http://boston.conman.org/2009/01/09.1
Vorlath # Friday, January 9, 2009 10:46:43 PM
I then read my first comment on Jeff's site and I was going "no no no you idiot." heh. My original thoughts (my first blog entry on the subject) were actually close, but ultimately flawed because of ONE point.
I'm firmly on the 1/2 side now though because it has the least amount of assumptions. And I went back to traditional probability theory that I was taught in grade 9 where we had to list the entire sample space. It always seems to work out better that way although it's super long. There are faster ways, but you can miss critical items.
Anonymous # Wednesday, November 4, 2009 10:27:57 AM
Vorlath # Wednesday, November 4, 2009 9:19:52 PM