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Combinatorics puzzle. Test yourself.

ACK Posts: 428Subscriber
edited December 2014 in Low Content Forum
1) If I flip two coins and one of them is heads, what is the probability that they are both heads?

2) Mrs Smith has two children, one of which is a boy. What is the probability she has two boys?

3) Mrs Smith has two children, one of which is a boy born on a Tuesday. What is the probability she has two boys?
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Comments

  • RecreationalRogerRecreationalRoger Posts: 789Subscriber
    50% for all 3
  • AesahAesah Posts: 1,048Pro
    edited December 2014
    Somewhat ambiguous wording, but typically the proper interpretation would be an if and only if.

    1) So here it would mean if we say "ok you flip two coins, look at them, and tell me nothing if they are both tails otherwise say something" then you can quickly see it's 1/3 looking at the 4 total combos being (TT/HT/TH/HH). However I'll write out the parallel solution since the next one is a bit more confusing, you can compare it to this one.

    -Each coin has 2 possibilities
    -There are 4 total combos
    -3/4 of the total combos are considered (i.e., qualify for "one of them is heads")
    -Only one of the three is valid (i.e., qualify for "both heads")



    3) Extracting the same logic, then we say "ok let's take a stranger with 2 children, and you tell me something if and only if one of the children is a boy born on Tuesday, otherwise you say nothing", then it's 13/27 because:

    -Each children has 14 possibilities (7 days times 2 sexes)
    -There are 14x14 = 196 total combos
    -27/196 are considered (i.e., qualify for "at least 1 is a boy born on Tuesday"). This is found by 1-(13/14)^2, since (13/14)^2 don't qualify.
    -13 of the 27 are valid (i.e., qualify for "both are boys"). We have the first boy being ThursdayBoy times 7 combos for the second, then vice versa and subtract the one double counted combo where both are ThursdayBoy.



    2) is the obviously the same as question 1) if we're assuming each child can only be a boy or a girl.
  • ACK Posts: 428Subscriber
    Haha yeah
    .
    1/2, 1/2 and 13/27

    Pretty unintuitive how the Tuesday changes it from .33 to .48
  • ACK Posts: 428Subscriber
    I don't think the wording is ambiguous. "IF x is true THEN what is the probably of y".
  • beauregard Posts: 1,592Subscriber
    Aesah said:

    3) Extracting the same logic, then we say "ok let's take a stranger with 2 children, and you tell me something if and only if one of the children is a boy born on Tuesday, otherwise you say nothing", then it's 13/27 because:

    -Each children has 14 possibilities (7 days times 2 sexes)
    -There are 14x14 = 196 total combos
    -27/196 are considered (i.e., qualify for "at least 1 is a boy born on Tuesday"). This is found by 1-(13/14)^2, since (13/14)^2 don't qualify.
    -13 of the 27 are valid (i.e., qualify for "both are boys"). We have the first boy being ThursdayBoy times 7 combos for the second, then vice versa and subtract the one double counted combo where both are ThursdayBoy.
    What?
    the Tuesday info has no bearing on the calculations.
    Mrs Smith either has a boy and a girl, or two boys.

    if we didn't know the sex of one, then she could have B/G, G/G, B/B or G/B
    but once we know the sex of one - she can only have B/G or B/B
    unless we're counting hermaphrodites...

  • AesahAesah Posts: 1,048Pro
    edited December 2014
    of course it is relevant, that is like saying it's not relevant what cards you have in poker you either win or you don't, 50/50.

    Like you said yourself, B/G and G/B are different entities. The reason it's slightly under 50% in this case is because the scenario of *both* boys being born on Tuesday is only counts 1 time.
  • beauregard Posts: 1,592Subscriber
    ACK said:
    What is the probability she has two boys?
    the question has no days of the week in it.
    it's like saying your Villain is wearing a blue shirt, what is the probability he's got AA?
    one thing has nothing to do with the other!
  • AesahAesah Posts: 1,048Pro
    edited December 2014
    Not quite sure what your thought process is right now, can you start by saying what you think the answer to 3) is then?

    Anyway the question is intentionally layered to not be trivial, but if you think of a simpler version of the same problem, it may be more obvious. Let's say that Mrs Smith has 1 child. I will ask Mrs Smith the gender/birthday, and I will relay the information to you if and only if it's a boy who is born on any day *other* than Christmas. Otherwise I won't tell you anything.

    I come back and tell you nothing. What are the chances her child is a boy?

    Spoiler:
    Here it should be apparent without doing any math that it's very very unlikely to be a boy and not anywhere close to 50/50. Anyway the correct answer here is 1/366 (not counting leap years/etc.)
  • beauregard Posts: 1,592Subscriber
    lol
    you're blowing my mind, here.

    if we didn't know the sex of one child (or side of a coin) - then there's more combinations to consider. but once we know one, it seems that it no longer should be considered in the possibilities.

    Mrs Smith has 2 kids. what are the odds that BOTH are boys - is a very different question than Mrs Smith has 2 kids, one is a boy - what are the odds that the other is a boy.

    The other can only be a boy or girl. 50/50. The sex of the first doesn't influence the sex/gender of the other. The date of birth also doesn't effect the calculations as the question doesn't ask for probability of when the other might be a boy.

    same holds true with coins.
    I've got 8,455 coins in which 8,454 are heads - what's the probability that the 8,455th is heads? That last coin's probability is independent of the other coins. Again, only 2 possibilities - so 50/50.

    Q3 - didn't say anything about "I'll tell you something if..." like your X-mas example.
    It just said one boy was born on a Tuesday. That info is as helpful as saying one is a boy who likes to play with dolls. Maybe the question was worded incorrectly - but the probabilities seem to all be the same.

    What's the probability that I'm wrong against Don Ding?
    Spoiler:
    100% :lol:
  • ACK Posts: 428Subscriber
    I think it is quite unintuitive if you don't do any maths at all but the maths doesn't lie.
  • AJD804AJD804 Posts: 184Subscriber
    The problem with Q3 is that you are saying after the fact that she cant have two boys born on the same day of the week. Now this must be some sort of Statistics thing, because most of us do not understand why that cant be. Can we get an explanation as to why that is? If you already did, then try again in layman's terms because I couldnt figure it out in the previous posts
  • AesahAesah Posts: 1,048Pro
    edited December 2014
    bearu I'm still not sure I understand where you're coming from- you're saying the answer to 3) is 50/50?

    What do you think the answer to 1) is?

    Note that the problem is not quite the same as "Mrs Smith has 2 kids, one is a boy - what are the odds that the other is a boy.", which you're right the answer would be 50/50. Because in that scenario, there is no differentiation between G/B or B/G combos whereas there is in the OP's problem (which, as the very first thing I posted in this thread, I believe is slightly ambiguous but the most correct interpretation)
  • AJD804AJD804 Posts: 184Subscriber
    Why does both boys being born on tuesday only count as one instance when one born on tueday and the other born on wednesday count as two? It is still two separate outcomes.
  • AesahAesah Posts: 1,048Pro
    AJD, because there are 2 combos... basically

    TuesdayBoy/WednesdayBoy
    WednesdayBoy/TuesdayBoy

    whereas there's only 1 of

    Tuesdayboy/Tuesdayboy
  • OminousCowOminousCow Posts: 702Subscriber
    edited December 2014
    AJD804 said:
    Why does both boys being born on tuesday only count as one instance when one born on tueday and the other born on wednesday count as two? It is still two separate outcomes.
    There is only one way for both boys to be born on Tuesday. I'll use this notation: Day for boy 1 / Day for boy 2.

    Both born on Tuesday:
    Tues/Tues

    There are two ways for one boy to be born on Wednesday and one on Tuesday:

    Wed/Tues
    Tues/Wed

    Remember that each child is not interchangeable. For ease let's say the first born is always blond and the second born is always has brown hair.

    Blond born on Tuesday, Brown born on Wednesday is not the same as Blond-Wednesday, Brown-Tuesday, hence two options not one.
  • AesahAesah Posts: 1,048Pro
    Basically it's like, WITHOUT card removal- that is, you draw a card and shuffle it back into the deck, then draw another card. There's a higher chance you'll have AK (2/169 chance) than AA (1/169 chance)
  • AJD804AJD804 Posts: 184Subscriber
    I understand now. I was looking at it from a poker perspective( I know silly me for looking at it that way on a poker forum). I understand where you guys are coming from now
  • BartBart Posts: 6,078AdministratorLeadPro
    edited December 2014
    These problems are a variation of what I call the "Lets Make a Deal" scenerio.

    Let's say there is 1 car behind one of three doors A, B and C. We choose what's behind door B. Then we are told that door A does not contain the prize. Will changing our decision to door C increase the odds that we win?

    What about someone at random who does not know the results of anything choosing door A and its empty. Does changing our selection to door C make a difference?
  • beauregard Posts: 1,592Subscriber
    Aesah said:
    bearu I'm still not sure I understand where you're coming from- you're saying the answer to 3) is 50/50?
    absolutely.
    Q3 never mentioned anything about "you tell me something if and only if one of the children is a boy born on Tuesday, otherwise you say nothing"
    so there are no circumstances provided that would affect the result
    Aesah said:
    What do you think the answer to 1) is?
    I'm with Roger - they're all 50/50.

    2 coins can be H/H, H/T, T/H & T/T
    with one being H - then you're left with H/H or H/T... those are your only options!... 50/50
    since the value of one is already known (the first H) - the result is only affected by one of two possibilities from the other coin... T/H and T/T are never possibilities because we've already said that coin #1 was and is only H
    Bart said:
    These problems are a variation of what I call the "Lets Make a Deal" scenerio.

    Let's say there is 1 car behind one of three doors A, B and C. We choose what's behind door B. Then we are told that door A does not contain the prize. Will changing our decision to door C increase the odds that we win?
    Before any info revealed - we have a 33% chance of being correct
    After info provided - our odds increase to 50/50
    technically - changing our choice does not give us better odds
    some would argue that by switching, we've gained the equity given away with the door opened - so we only have a 33% chance of winning if we stay but a 66% chance of winning if we switch. But I don't buy that logic.
    when a door is opened and two doors are left - we can "choose" to stay with our original selection or choose to switch. either way, we've got the same probability because the problem has now changed to choosing one of two doors. the newly opened door does not provide any additional info as to whether either selection is better.
  • AbeLimonAbeLimon Posts: 860Member
    So no matter what i answer someone will agree that im correct? this thread rulez.
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