Delicious LiveJournal Links for 7-9-2010
- Jul. 9th, 2010 at 1:31 AM
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"I have two children. One is a boy born on a Tuesday. What is the probability I have two boys?" Pete Birks provides the first convincing explanation I have seen as to why the correct answer is 13/27 and not 1/3.(tags: mathematics)
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Daleks in your home!(tags: doctorwho)
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seawasp : (no subject) [+27] - (Anonymous) : (no subject) [+0]
gareth_rees : (no subject) [+3]
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Comments
Do read through the debate between Pete Birks (who has convinced me) and Pete Doubleday (who takes your side) at the link. Pete Birks puts the first half of the case well: Then the days of the week argument takes the same technique to a slightly different place.
His use of "about" shows that he is not doing the same thing.
In the original case, we **KNOW** that one of the children is a boy. Whether the boy is older or younger is utterly irrelevant, because we know two facts:
1) You *already have* two children (so no variation of probability applies to the number of children)
2) We *already know* that one of the children is a boy (thus no variation of probability applies to whether *at least* one of the children is a boy.
Thus, the ONLY variable is the sex of the other child. And, barring very odd genetic variations, there are only two options -- boy and girl -- and these are generally considered to be equally likely (50/50).
Therefore, there is a 1/2 probability that you have two children.
In the example above, both black and green balls are randomized. To be equivalent to the ORIGINAL scenario, you must proceed as follows;
You first give one ball at random to all your 1,000 guests. Then everyone who has a black ball is asked to raise their hands.
At that point, anyone who has a green ball is asked to leave. This leaves you with ("about") 500 people in the room *known* to have one black ball. You then pass out to these five hundred one more ball each, at random. The number who will then raise their hands to say they have now got TWO black balls will be ("about") 250, or, as I said, 1/2.
THIS is the equivalent of the original case.
So, I flip two coins randomly and hide them from you. You ask me, "did at least one of the coins land heads?" I answer "yes". What is the probability that both coins landed heads?
(For a mathematician, this is a straightforward application of Bayes' theorem, but if you're not intimate with this theorem, you might need to think about it a bit.)
Probability that both landed heads BEFORE you told me, 1/4.
In the case of the original question, it's analagous to the first case. We've already been told that you have one son. The probability the other child is a son -- and thus that you have two boys -- is then 1/2.
This is correct: P(A and B) = P(A) × P(B) if A and B are independent events.
Probability that both coins landed heads, AFTER you have told me one of them did, is 1/2.
Have you tried this experiment yourself? It's easy to do, and I hope the actual results will be more convincing than some guy on the Internet. If you're right, then after you've got at least one head 60 times, about 30 of those times you'll have got two heads.
Consider the following sequence of flips:
H T T H H T H T T H T T T T T H H H
If I guess what you're envisioning, it would be better written as
H T
T H
H T
H T
T H
T T
T T
T H
H T
In which all but two of these have at least one head but none have two heads
In my view, you only bother to check probability when you HAVE a head, so the sequence goes
H T T H H T H T T H T T T T T H H H
H: Get head next? T No
Tail, ignore, Head: Get head next? H Yes!
Tail, ignore, Head: Get head next? T No
Tail, ignore, Head: Get head next? T no
Tail, ignore, tail, ignore, tail, ignore, tail, ignore, Head: Get head next? H Yes!
Head: waiting for next flip.
Here, I am not calculating probability or counting until I already HAVE a head (just as in the original), and I have to ONLY calculate the next flip's probability (which, assuming a fair coin, is 1/2) of being a head. After that is done, I start anew, waiting for a head to show up.
If you tell me: I have two kids
And you tell me: One is a son
It doesn't matter to me that there are different orders in which you could obtain the kids. It doesn't matter that there are many other potential sequences in which you could not have had a son. All that matters is that you now have one child whose sex is indeterminate, and the chance that THAT child is male is 1/2 barring any oddities in the gene pool.
Yes, I have a particular experiment in mind and I am trying to convey as clearly as I can. If I can successfully convey it to you then we'll be in business! Remember that I described it like this:
I flip two coins randomly and hide them from you. You ask me, "did at least one of the coins land heads?" I answer "yes".
So to carry out one trial you need to:
1. Flip two coins
2. If they are both tails, this can't be an instance of the experiment (because I will answer "no" to the question in this case). So go back to step 1 and try again.
3. If both coins are heads, write down "YES"
4. If one coin is a head and the other a tail, write down "NO".
After carrying out sufficiently many trials, you'll be able to answer my original question, What is the probability that both coins landed heads? by looking at the number of times you've written YES and NO.
If you tell me: I have two kids
Let's wait until we've sorted out the coins before moving on to the boys.
There is literally no logical way for this probability to be other than 1/2. The first object doesn't count. It's fixed, no probabilities can be assigned to it (other than 1).
Your approach assumes a sequence of pairs, mine assumes a single instance in which I am presented with a single set of facts: there are two things. Thing one is in state 1. Thing two may be (with equal probability) in state 1 or state 2. What is the probability that both are in State 1? Well, we KNOW the first Thing is state 1, so we can ignore it. Thus we have only to consider Thing two, which is stated to have a 50/50 chance of being state 1 or state 2. Thus the probability of both being in State 1 is (1) [the certainty that the first is in State 1] * 0.50 [the probability that Thing 2 is in State 1], for a total of 0.5 or 1/2 that both are State 1.
The most important of these difficulties is that I am not convinced yet that you understand what I mean when I ask my question (I flip two coins randomly and hide them from you. You ask me, "did at least one of the coins land heads?" I answer "yes". What is the probability that both coins landed heads?)
If we can't agree on what this means, or how a question like this can be turned into series of trials, then we've got no business attempting anything more complicated!
So, do you agree that the experimental procedure I outlined above will answer my question?
The first one no longer matters. All that matters is that I have one remaining unknown which can be A or B.
Any other interpretation implies that prior events influence future ones -- the Gambler's Fallacy in which "The next one's GOTTA come up heads! The last three have been tails!".
Also, actual experiments in this area are far too messy (and long-drawn-out if you want the statistics to be reasonably close to the expected). I've seen wayyyyyy too many completely improbable sequences of dice rolls to trust flipping coins (and that's assuming I am able to flip coins reliably AND randomly).
That isn't quite right. What Gareth and I have in mind is more like this:
I have two things which can be A or B (with equal probaility). I uncover them both, find that at least one is A, and tell you. You then have to bet on whether the second one is A.
Do you see the difference?
Let's take step one back, and consider a simpler problem: I roll one six-sided die, you ask me "is it showing an even number of spots?" I answer "yes." What is the probability that it landed showing six?
You're introducing a sequence.
If I have two kids, and the eldest is a boy, then the other kid is 50:50 boy:girl
If I have two kids, and one of them is a boy, I could have boy/boy, boy/girl or girl/boy. Each alternative is equally likely, and only one alternative has two boys.
Would you change your mind if the question was phrased as "my children are not both girls"? Note that this is logically equivalent to "at least one of my children is a boy"...
I don't see it makes any difference. You thus have one boy or you have two boys. One is established therefore to be a boy, leaving one that could be either, with a 50/50 chance of it being a boy.
Ball 1 Black / Ball 2 Green (at least one black ball)
Ball 1 Black / Ball 2 Black (at least one black ball)
Ball 1 Green / Ball 2 Green
Ball 1 Green / Ball 2 Black (at least one black ball)
That's four possible combinations, three of which include at least one black ball. Only one of those three combinations is 'two black balls'
In your example, they get one ball at a time, and only the people with a black ball first get a second ball. So there are only two possible combinations of people with two balls.
Ball 1 Black / Ball 2 Green
Ball 1 Black / Ball 2 Black
There are two different questions being asked.
You are asking "If my first child is a boy, what are the chances that my second child is a boy?"
But the question above is, "If I have two children, and at least one of them is a boy, what are the chances that both of them are boys?"
You are being asked to flip two coins together, and see if either one comes up heads.
You're saying, "No, I'm going to flip this one coin. If it comes up heads I'll flip it again, but if it comes up tails I'm stopping there."
You can tell peterbirks is starting with some extremely flawed assumptions because he says 'the obvious answer is one in three' WTF??? He doesn't understand probability. No actually, he is making additional assumptions which are over-complicating.
Put it like this - you have two coins. One is hidden. The one you can see is heads-up. What is the probability both are heads-up? Not 13/27, not 1/3. It's one in two, because it depends entirely on a 50/50 chance.
But if I have two coins which you can't see, and tell you that one of them is heads-up, the chance that the other is heads-up is 1/3 - because you cannot know which coin I am talking about. See the black/green snooker ball analogy.
set-sequence with linear sequencethis isn't a helpful description, I'll think of better words in a minute. Next paragraph still stands.If you are undecided whether I am onto something consider this - how likely do you really think it is that having one child born on Tuesday affects the chances that one of your children is a girl? In reality you know it does not. Therefore if someone thinks they have invented a mathematical proof that it does, I would be very sceptical of their maths.
Edited at 2010-07-09 06:06 am (UTC)
quarsan
'A person has two children; you know one is a boy' can mean
- you only have information about one child, and he is a boy
- you have information about two children, but you are only revealing that one is a boy
From the initial interpretation of the sentence in one of these two ways, all the rest flows. Nick is using the second interpretation, which is why the bowling ball analogy works. Using the first interpretation the maths is wrong.
This explains why intelligent people are coming to such wildly different conclusions. It's about English not maths basically.
The trouble comes, of course, when you take a problem for mathematicians and present it to non-mathematicians, people who aren't familiar with the conventions. 90% of any ensuing conversation involves getting the non-mathematician up to the point where they understand the problem (in its intended form). My exchange above with
I would beg anyone who is disagreeing with the numbers presented by Peter Birks, to please go away and *do the experiment* described by Gareth Rees in the thread here, to answer this question:
I flip two coins randomly and hide them from you. You ask me, "did at least one of the coins land heads?" I answer "yes". What is the probability that both coins landed heads?
The answer to this is 1/3. It really, really is. If you are convinced that it is not 1/3, then you are incorrect, *but* it will not be possible for us to convince you of this. The only way to convince yourself is to do an experiment. It will take about the same amount of time as replying to this message.
You can do this yourself, with paper, pen, and two actual coins:
1. repeat sixty times:
- flip both coins.
- write down HH, HT, TH, or TT on a fresh line.
2. now go down the column, crossing out any TT lines.
3. now go down the column, ticking HH lines and crossing HT and TH lines.
4. how many ticks and crosses do you have?
Many years ago I didn't understand probability very well. Despite a good maths degree, I used to get into arguments about Monte Hall on Usenet. Finally I did a Monte Hall experiment, and then I understood.