Well sure, the assumptions are:
If any of these assumptions are not valid, then of course the whole thing is going to be off.
As I said before, I agree with many people here that the wording is very important. The math is trivial.
What might be interesting for some people, in order to grasp this intuitively, is what causes the sway towards girls.
The wording, if done correctly, selects all households that contain at least one boy. It gives you all the households that have two boys, and the ones that have a boy and girl, regardless of which one was the the boy or the girl. It’s the “at least one” wording that gives you twice as many households where the other one is a girl.
If the wording had been “sibling number 1” is a boy, it would be completely different. Choose “sibling number 1” however you want to, but you are only selecting half of boy/girl households this time.
The easiest way to undestand this is to draw up every possible household and carefully examine the text to see which ones are selected.
But the point is - no there isn’t an order.
This objection would be true if it was always the same one sibling answering the door. (Firstborn? Most energetic? Friendliest? One who ordered pizza?) But it could be either sibling. You are assigning a distinguishing criteria that is randomly selected and does not really exist until the door is opened.
If you flip two coins, and the quarter answers the door first every time, then yes thier Heads means the odds are 50-50 the nickel is Heads or Tails. But if it could also be the nickel that answers the door. So the door gets answered, nickel or quarter (We’re foreign and don’t know one US coin from another), we see then only that it’s Heads ansering the door. What’s the other coin - Head or Tails?
The key here is that firstborn does not affect gender of second-born; nor does flipping a quarter affect the ooutcome of flipping a nickel. In case we’re hung up on birth order problem - call it Bob and Charles both had one child, and the girl answering the door could be Bob’s or Chuckie’s - we can’t tell whose child. What’s the gender of the other child? The choices are that Bg answered and the other could be Cb or Cg, or that Cg answered and the other could be Bg or Bb. The only case eliminated is Bb-Cb.
Choices are Bg-Cg, Bb-Cg, and Bg-Cb. all equally likely.
Of course there is an order. You are still confusing combinatorics with probabilities. If there is no order, or other distinguishing mechanism, you cannot divide the possibilities into four. There are then three possibilities and they do not have the same probability.
Without order there is no such thing and BG and GB as distinct possibilities. There is {B, G} with probability one half. If you decide that there is no mechanism to distinguish BG from GB when the door is answered, and thus only disqualify BB, you are required to update the probabilities, not claim that somehow magically your indistinguishable classes are now suddenly distinguishable and now have equal probabilities of one third.
Sure, and? It doesn’t matter when the distinguishing criterion comes into existence. All that matters is that there is a criterion. Any criterion that distinguishes the two is equivalent. It could be firstborn and secondborn. It could be taller and shorter. It could be the one named Alex and the one not named Alex. It could be the one who answers the door and the one who doesn’t. All give the same effect.
People keep saying the same things, and I can tell they’re saying the same things, yet I think I’m still failing to understand. Like many math things, understanding seems to come and go in my brain.
Let’s simplify to a single coin, flipped twice. It is a fair coin so
heads = 50%
tails = 50%
Before any flips we know that the probability for a pair of flips is:
heads and heads = 25%
heads and tails = 50%
tails and tails = 25%
We flip the coin once, and it lands on heads. The chance of landing on heads or tails on the next flip is 50/50, but the probabilities of the pair of flips (the last flip (heads), and the next flip (unknown)) is:
heads and heads = 50%
heads and tails = 50%
tails and tails = 0%
If I, behaving honestly, flip the coin twice without showing you, and record the results, then as far as you know, the probability of the pair of flips is still 25/50/25. I tell you one of the results was heads, so that updates the probabilities of the pair to be
heads and heads = 33% or is it 50%
heads and tails = 66% or is it 50%
tails and tails = 0%
This is where I get confused. The flips are independent, nobody denies that the chance of the coin landing on heads or tails for an additional flip is 50/50. It is also seems that the chance for the pair of flips is as written above 33/66. Going backwards from the chances of the pair being heads/heads or heads/tails is 33/66, it seems that the chance of the next flip being tails is 66%.
How does that all work? What am I misunderstanding?
I look at the results of both flips, so I have all of the information, and I tell you one of them was heads. If they were both tails, I would distract you, and re-flip until I get at least one head. That is the essence of the Monty Hall problem, right? I have information you do not, which changes the odds, because I rig telling you the results.
If I flip once, and it is heads, I tell you that, and then I flip again, and tell you the result, I do not have any information that you don’t have.
No. The difference is not order, it is distinctness.
There is a distinguishing mechanism, just that if it is still hidden to us - so we must allow that either case is possible and include that in the calculations.
A: If we know “there is at least one girl” and we know “a girl answered the door” those are two limiting criteria, as apposed to
B: “we don’t know beforehand what the genders are” and “a girl answers the door”.
In case A there was the odds 2 in 6 that a boy answers the door. The second revelation may change the calculation.
In case B there was a 50-50 that a boy or girl answers. When a girl answers the door, we only know that one thing still.
Distinctness: If we consider Bob’s and Charlie’s respective single children, it does not matter which is older, or if they were born simultaneously. What matters is Bob’s child is not Charlie’s child (we hope) and we don’t know which is which, even after the door is opened. They are two distinct entities. Therefore, there are 4 distinct possibilities equally likely, only one of which is eliminated when the door is opened and we are greeted by a child. (A nickel is not a quarter, either.)
This as opposed to the classic combinatorics problems - such as, two pairs of socks are unpacked and placed in a drawer, red and blue. We blindly pick a sock at random. Here - we do not care if we pick the left hand red sock or the right hand red sock, since the socks are identical and indeed have no L-R distinction. Thus, order/distinctness between the two socks of the same pair does not matter, only the colour difference between the two pairs.
The key is “I tell you the first flip” vs. “I telllyou one of two flips.”
If I tell you the first flip, whatever it is, the odds for the other are 50-50. If I tell you the second flip only, what wer the odds fo the first flip? 50-50. If I tell you the first flip was H, still 50-50 for the second. Same if I reveal the second flip. If I don’t tell you which coin was H, what are the odds?
It’s all about collapsing the odds. By revealing information, you limit the possible outcomes that can happen and still have the revealed information.
Consider flipping 4 coins, then telling you at least one was H. wat are the odds they are all H? (1 in 15)
Now, flip one coin. I tell you it’s H. If I flip 3 more times, what are the odds the next 3 are H? (1 in 8)
I flip a coin 4 times. I tell you the last one was H. Waht are the odds of 4H? (1 in 8)
In the second or third scenario, a pertinent detail was nailed down before the remaining odds were calculated. In the first scenario, all but TTTT are possible. In the others, Hxxx or xxxH are the only scenarios. Knowing a specific, distinct (by sequence), probability is determined changes the odds.
OK, yes, that makes lots of sense, but is a very slippery distinction. I grasp it and then it slips away.
I think, when I’m understanding it, the big distinction is that telling me the first flip, the second coin state is 50/50, because the second coin state is based on one flip (independent events).
Telling me “one of two” the remaining coin state is based on two flips, not one. I can rule out some of the pairs, but not all of them, so there are non-50/50 odds for the remaining coin state.
So with the children thing. I see one child, it is a girl. Will the next child I see be a boy or girl? That is 50/50 (assuming seeing a child is random with respect to binary gender). But if you tell me I will be seeing two children soon, and one of them is a girl, then I’m not dealing with a single fertilization event, but two fertilization events, with some outcomes ruled out.
If I flip two coins and tell you at least one is heads. Then the chance is 2/3 that the other one is tails. That is because there are three distinct and equally probably cases where this is true, and in two of those the other one is tails.
At no point is the probability of a single coin flip anything other than 50/50. It’s all in the wording and the selection of possibilities.
What I am really saying is:
“The first coin was heads, the second one was heads or both were heads”. That is what the wording “at least one is heads means”. The key point is the sneaky wording that selects both the first one and the second one as heads.
The other sneaky wording is that it asks what the OTHER coin is. There are two possibilities here. First one is tails or the second one is tails. The only possibility where it isn’t true is if both at heads. (Also if both are tails, but that one was not selected in the first place)
Yeah, that part I understand, which is why I’m trying to strip away the word problem aspect of the probabilities. As discussed up thread, many of these gotcha type word problems rest on vagaries of language that can be interpreted multiple different ways. Or if not vagaries, then on tricking people to misread the problem, and therefore get a wrong answer.
…Depending on the process by which you reached that statement. If your process is "I’ll pick one at random, look at it, and then say either “At least one is heads” or “at least one is tails”, then it’s 1/2. If your process is “I’ll look at both of the coins, and if at least one is heads, I’ll say that, or say nothing if they’re both tails”, then it’s 2/3.
I think the second way of interpreting it is implied by the wording. I don’t really understand what you mean by the first way.
And this is exactly how misleading problems are designed. Where the word seem to imply one thing, but actually say another. Or are open to multiple interpretations and all are logically grammatically correct, but lead to different mathematical answers.
Saying nothing was not part of the question. The person says something - either one is heads or one is tails. Yes, we are making the assumption - the person picks one of the two at random, and reveals only that one, regardless of what it is.
Predetermining to mention only tails is a different question.
Yes, it’s a different question but one that could be easily understood by someone presented with this problem. The comment by Chronos is a very clear example of hidden assumptions and unstated attributes. It’s issues like this that make creating clear and unambiguous word problems difficult.
Let me just say that I find it amusing that @Danger_Man thought that the wording clearly implied my second method, and @md-2000 thought that the wording clearly implied my first method. Which illustrates that neither is clear. The question-writer doubtless intended one or the other, but lots of luck determining which one they meant.
True, there are plenty of assumptions in this and any word question. Nowhere are you stating the speech matches what you see on a coin, for example; I could see a heads and say tails, to get picky. If you are saying “I will predeterminedly pick either the first or second toss”, then announce that result, then yes it is 50-50 for the other. If you are saying “I will pick one of the tosses at random and announce that result” (i.e. I don’t know if I picked #1 or #2) then it’s 2/3.
There are multiple ways to interpret a word question unless the wording is very precise.
The essence is - if you know whether this is #1 or #2 when you announce it, then you know the other is 50-50. If this coin you announce could be either toss, then it’s 2/3. If you know but the person you ask doesn’t then you know more information than them and can figure out better odds.
If you have the situation where you say nothing, does the person guessing odds know That “if there is no tails, he will say nothing”? That also changes the odds for some circumstances.