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Posers and Puzzles

Posers and Puzzles

  1. Standard member wolfgang59
    Infidel
    17 Feb '11 20:09
    I have 5 pairs of black socks with the days of the week printed on them (Monday-Friday). Unfortunately they all look identical when rolled up in my drawer.

    Being superstitious I think it will be a lucky day if I get the correct pair of socks for that day. (If I get a wrong pair I wear them any way).

    In the pub Thursday night I tell you I have been unlucky all week (ie I have never drawn the correct socks).

    What are the chances that the last pair are Friday's socks?
  2. Subscriber roma45
    st johnstone
    17 Feb '11 21:12
    Originally posted by wolfgang59
    I have 5 pairs of black socks with the days of the week printed on them (Monday-Friday). Unfortunately they all look identical when rolled up in my drawer.

    Being superstitious I think it will be a lucky day if I get the correct pair of socks for that day. (If I get a wrong pair I wear them any way).

    In the pub Thursday night I tell you I have been u ...[text shortened]... never drawn the correct socks).

    What are the chances that the last pair are Friday's socks?
    5 to 1
  3. 17 Feb '11 21:24
    Originally posted by roma45
    5 to 1
    Are all the socks separate, so on Monday you could wear Tuesday and Thursday's, or are they correctly paired in the drawer? It is quite a lot harder if they are separate!
  4. Standard member wolfgang59
    Infidel
    17 Feb '11 22:08
    My socks are all balled in perfect pairs.
  5. 18 Feb '11 12:05
    Originally posted by roma45
    5 to 1
    I agree, 5 to 1.

    Imagine he gets out his socks on a Sunday evening, and lays them on a table, mondays first, then tuesdays, etc.

    Then picking Mondays socks first, then Tuesdays, etc., is the same as picking Fridays, then Thursdays, etc.

    So there is a 1-in-5 chance of him having an unlucky friday; the rest is just noise.
  6. 18 Feb '11 21:06
    Originally posted by Swlabr
    I agree, 5 to 1.

    Imagine he gets out his socks on a Sunday evening, and lays them on a table, mondays first, then tuesdays, etc.

    Then picking Mondays socks first, then Tuesdays, etc., is the same as picking Fridays, then Thursdays, etc.

    So there is a 1-in-5 chance of him having an unlucky friday; the rest is just noise.
    So there is a 1-in-5 chance of him having an unlucky friday; the rest is just noise.

    But I thought this problem boils down to epistemic probability. If so, it would seem wrong to hold that any information you may learn about the preceding days is automatically just noise. If, for example in the extreme, you learned on Thursday night that he had been lucky every day thus far during the week, it would seem ridiculous to then give only 1-in-5 credence to the proposition that the remaining pair is the Friday pair.

    I may well be misconstruing the problem, but I do not think the answer is 1-in-5. I get an answer that is less than 20%.
  7. 18 Feb '11 22:13 / 1 edit
    Originally posted by LemonJello
    [b]So there is a 1-in-5 chance of him having an unlucky friday; the rest is just noise.

    But I thought this problem boils down to epistemic probability. If so, it would seem wrong to hold that any information you may learn about the preceding days is automatically just noise. If, for example in the extreme, you learned on Thursday night that he ha ...[text shortened]... ng the problem, but I do not think the answer is 1-in-5. I get an answer that is less than 20%.[/b]
    LemonJello is right, because of the extra information you're given in the problem, you can narrow down the number of possible outcomes which changes your odds. Factoring this information in (at least the way I did it) requires a bit of messy math to avoid double counting.

    53 possible ways to get no socks correct after Thursday and of these 53 only 9 end with the correct socks on Friday. 9/53 = 16.98% which is slightly less than 1/5.
  8. Standard member forkedknight
    Defend the Universe
    18 Feb '11 22:23
    Originally posted by Arctic Penguin
    LemonJello is right, because of the extra information you're given in the problem, you can narrow down the number of possible outcomes which changes your odds. Factoring this information in (at least the way I did it) requires a bit of messy math to avoid double counting.

    [hidden]53 possible ways to get no socks correct after Thursday and of these ...[text shortened]... ith the correct socks on Friday. [b]9/53 = 16.98%
    which is slightly less than 1/5.[/hidden][/b]
    Not only is there extra information, but since you have to read all the socks when you unroll them to know if it's going to be a lucky day, you will know with 100% certainly whether or not Friday will be an unlucky day on Thursday.

    If you haven't already unrolled Friday's socks, then Friday will be a lucky day.

    If you're asking the probability of Friday being an unlucky day given an unlucky Mon-Thurs, then I think the ocrrect probability is 0.2 (1 in 5)
  9. 18 Feb '11 22:37 / 5 edits
    Originally posted by forkedknight
    Not only is there extra information, but since you have to read all the socks when you unroll them to know if it's going to be a lucky day, you will know with 100% certainly whether or not Friday will be an unlucky day on Thursday.

    If you haven't already unrolled Friday's socks, then Friday will be a lucky day.

    If you're asking the probability of ...[text shortened]... unlucky day given an unlucky Mon-Thurs, then I think the ocrrect probability is 0.2 (1 in 5)
    While wolfgang will know if he's worn the Friday socks by Thursday night and could be 100% certain about Friday's outcome with the information he has, we in the pub don't have enough information to be certain. All we can do is narrow down the number of scenarios and come up with a probability, which is why it isn't quite 20% for us but slightly lower as calculated above.
  10. Subscriber roma45
    st johnstone
    18 Feb '11 23:29
    Originally posted by Arctic Penguin
    While wolfgang will know if he's worn the Friday socks by Thursday night and could be 100% certain about Friday's outcome with the information he has, we in the pub don't have enough information to be certain. All we can do is narrow down the number of scenarios and come up with a probability, which is why it isn't quite 20% for us but slightly lower as calculated above.
    he has only 5 to choose from, we already know that he has been wrong all week, but it is still a 1 in 5 chance of being right. the only fact we need is the number of choices and hes been wrong on the other days. if he was correcy on every day then it would be a certainty he would be right on a friday as well. as it is its a 1 in 5 chance
  11. 18 Feb '11 23:52 / 12 edits
    Originally posted by roma45
    he has only 5 to choose from, we already know that he has been wrong all week, but it is still a 1 in 5 chance of being right. the only fact we need is the number of choices and hes been wrong on the other days. if he was correcy on every day then it would be a certainty he would be right on a friday as well. as it is its a 1 in 5 chance
    It's true that if you had correct pairs for the first four days, you'd know the fifth day was also correct, but that isn't the problem given. Here we're told a bit of information that on the surface seems not to affect the odds, but when you look deeper you see that it does (though to be fair, it only reduces the odds by a small amount).

    I'll demonstrate any easy to visualize brute force method on a very similar problem. Consider wolfgang labels four kinds of socks for each day in his four day work week from Monday-Thursday. Wednesday night in the pub you hear that wolfgang picked the wrong sock pair for the first three nights and are asked about the odds that wolfgang is left with the correct sock pair for Thursday. 1 in 4 chance you say? Let's see.

    I'll use M for Monday, T for Tuesday, W for Wednesday, R for Thursday. There are 4! = 24 permutations of sock pairs for the four day week.

    MTWR
    MTRW
    MWTR
    MWRT
    MRTW
    MRWT
    TMWR
    TMRW
    TWMR
    TWRM
    TRMW
    TRWM
    WMTR
    WMRT
    WTMR
    WTRM
    WRMT
    WRTM
    RMTW
    RMWT
    RTMW
    RTWM
    RWMT
    RWTM

    Now go through and cross off all the ones that have a match on M, T, or W (i.e. M in the 1st slot OR T in the 2nd slot OR W in the 3rd slot). The information given to us rules out these permutations, leaving only the permutations that have been wrong all three days so far.

    TMRW
    TWMR
    TWRM
    TRMW
    WMTR
    WMRT
    WRMT
    WRTM
    RMTW
    RWMT
    RWTM

    We're left with only 11 permutations that are possible. This represents our remaining pool of all possible outcomes and the denominator in our probability. Now let's count how many of these 11 possible outcomes have a match on the last day meaning an R in the 4th slot.

    TWMR
    WMTR

    Only 2 out of 11 or 18%, much less than the 25% you'd predict without being told that wolfgang picked wrong on Monday, Tuesday and Wednesday.

    I'll leave the 5 day problem posed in the first post to you, but hopefully this example helps you visualize why it isn't correct to just divide 1 into the number of types of socks and get 1/5.
  12. Standard member wolfgang59
    Infidel
    19 Feb '11 02:46 / 1 edit
    Originally posted by Arctic Penguin
    It's true that if you had correct pairs for the first four days, you'd know the fifth day was also correct, but that isn't the problem given. Here we're told a bit of information that on the surface seems not to affect the odds, but when you look deeper you see that it does (though to be fair, it only reduces the odds by a small amount).

    I'll demon why it isn't correct to just divide 1 into the number of types of socks and get 1/5.
    Nice work AP. (and LJ)

    The 2 day problem gives a probability of zero (obviously)

    The 3 day problem (interestingly) gives a probability of 1/3.

    The 4 day problem (as AP has demonstrated) gives a probability of 2/11

    Anyone care to do the N day problem?
  13. 19 Feb '11 03:31
    Originally posted by Arctic Penguin
    LemonJello is right, because of the extra information you're given in the problem, you can narrow down the number of possible outcomes which changes your odds. Factoring this information in (at least the way I did it) requires a bit of messy math to avoid double counting.

    [hidden]53 possible ways to get no socks correct after Thursday and of these ...[text shortened]... ith the correct socks on Friday. [b]9/53 = 16.98%
    which is slightly less than 1/5.[/hidden][/b]
    Right.

    And in my mind it certainly makes intuitive sense that the given information should drag the probability below 20%. For example, if you considered a default stance of total ignorance concerning the status of wolfgang's luck in the first 4 days of the week, then you would expect his chances of being lucky on Friday to be 20% and you would also expect the chances that he has been lucky in at least one of the first 4 days of the week to be something >0 (whatever it is, it is certainly nonzero). But, now, if you are given just the information that he in fact has been unlucky all week, then that tells you have to give more probabilistic weight above the default level of ignorance to the notion that the Friday socks have already been consumed in making him unlucky on some earlier day in the week. Hence, we are now below the default of 20% regarding his being lucky on Friday.
  14. Subscriber roma45
    st johnstone
    19 Feb '11 13:29
    Originally posted by LemonJello
    Right.

    And in my mind it certainly makes intuitive sense that the given information should drag the probability below 20%. For example, if you considered a default stance of total ignorance concerning the status of wolfgang's luck in the first 4 days of the week, then you would expect his chances of being lucky on Friday to be 20% and you would also e ...[text shortened]... y in the week. Hence, we are now below the default of 20% regarding his being lucky on Friday.
    how can it possibly be less than 20%, previous days only matter if he was lucky, since we know that he was wrong on the 1st four days it has to be 1 in 5 chance. same as tossing a coin 100 times the 1st 99 turn up heads. whats the odds the last toss will be heads? its 50%. previous outcomes dont effect the future if it did we would all be millionaires
  15. 19 Feb '11 18:06 / 1 edit
    Originally posted by roma45
    how can it possibly be less than 20%, previous days only matter if he was lucky, since we know that he was wrong on the 1st four days it has to be 1 in 5 chance. same as tossing a coin 100 times the 1st 99 turn up heads. whats the odds the last toss will be heads? its 50%. previous outcomes dont effect the future if it did we would all be millionaires
    previous outcomes dont effect the future if it did we would all be millionaires

    Sometimes they don't, but sometimes they do. In this case, previous outcomes can influence our epistemic situation and the associated epistemic probability or credence level one should give to the proposition that the last pair remaining is the Friday pair. I have already given one extreme example in my first post to make clear that knowledge of previous outcomes can certainly influence the situation. Here is another one: you learn on Thursday that wolfgang considered himself unlucky on Monday when he drew and wore the Friday socks. Are you still going to hold that previous outcomes do not affect your epistemic situation and that you should still give 1-in-5 credence to the proposition that the Friday socks are the last pair remaining in wolfgang's drawer? Or are you construing probability differently here?

    As I hinted before, there are certainly many ways to construe probability. We may be construing it differently for this problem. But I think this problem is about epistemic probability, and it seems patently false that knowledge of previous outcomes cannot matter here.