{ "export": { "source": "https://urth.darkrealm.vip/thread/urth/6656", "note": "Exported from an unofficial mirror of a public mailing list. Copyright in each message remains with the person who wrote it. Email addresses are obscured as `user at host`, the form the original archive published. This file was rebuilt from parsed fields rather than retained headers, so it is faithful in content but not byte-exact.", "addresses": "obscured as 'user at host'" }, "archive": "urth", "thread": { "id": 6656, "subject": "math probability question for seven American nights", "messages": 9, "voices": 4, "first": "2014-09-07T12:34:02+00:00", "last": "2014-09-07T20:49:47+00:00" }, "messages": [ { "message_id": "", "from": { "name": "Marc Aramini", "address": "marcaramini at gmail.com" }, "date": "2014-09-07T12:34:02+00:00", "subject": "(urth) math probability question for seven American nights", "in_reply_to": null, "reply_link": null, "blocks": [ { "kind": "text", "depth": 0, "text": "I had a quick question but trying to articulate it to look it up\nindependently is hard.\n\nIs there a statistically most probable day for Nadan to eat the special one\nof the six eggs on any given day given that the first day is 1/6 and then\nfrom there the probability that the egg is there begins to be something\nlike 1/5 on the next day ( but the chance that it isn't there should be\nfactored in somehow, but I wasn't sure if it was 1/5 - 1/6 ... Then\n1/4-1/5-1/6 etc. ) Or is the system set up in such a way that the chance of\ngetting the egg is always 1/6 regardless on any given morning assuming no\neggs are stolen or disappear?" } ] }, { "message_id": "<540C5B6C.8050407@bindweed.com>", "from": { "name": "Gerry Quinn", "address": "gerry at bindweed.com" }, "date": "2014-09-07T13:19:40+00:00", "subject": "(urth) math probability question for seven American nights", "in_reply_to": "", "reply_link": "header", "blocks": [ { "kind": "quote", "depth": 1, "text": "On 07/09/2014 13:34, Marc Aramini wrote:\nI had a quick question but trying to articulate it to look it up \nindependently is hard.\n\nIs there a statistically most probable day for Nadan to eat the \nspecial one of the six eggs on any given day given that the first day \nis 1/6 and then from there the probability that the egg is there \nbegins to be something like 1/5 on the next day ( but the chance that \nit isn't there should be factored in somehow, but I wasn't sure if it \nwas 1/5 - 1/6 ... Then 1/4-1/5-1/6 etc. ) Or is the system set up in \nsuch a way that the chance of getting the egg is always 1/6 \nregardless on any given morning assuming no eggs are stolen or disappear?" }, { "kind": "text", "depth": 0, "text": "I don't have the book to hand, but if he is given six eggs, one special, \nand eats one every day at random, the chance of getting the special egg \non any day is indeed 1/6, as you reckoned in your other post.\n\nThe easiest way to see it is to imagine he decided at random the order \nto eat them in advance, and laid all six in a row, each marked with its \nday for eating. Clearly the chance is the same for each day!\n\n- Gerry Quinn" } ] }, { "message_id": "<540C5D11.5080106@bindweed.com>", "from": { "name": "Gerry Quinn", "address": "gerry at bindweed.com" }, "date": "2014-09-07T13:26:41+00:00", "subject": "(urth) math probability question for seven American nights -\n CORRECTION", "in_reply_to": "<540C5B6C.8050407@bindweed.com>", "reply_link": "header", "blocks": [ { "kind": "quote", "depth": 1, "text": "On 07/09/2014 14:19, Gerry Quinn wrote:\n\nOn 07/09/2014 13:34, Marc Aramini wrote:" }, { "kind": "quote", "depth": 2, "text": "I had a quick question but trying to articulate it to look it up \nindependently is hard.\n\nIs there a statistically most probable day for Nadan to eat the \nspecial one of the six eggs on any given day given that the first day \nis 1/6 and then from there the probability that the egg is there \nbegins to be something like 1/5 on the next day ( but the chance \nthat it isn't there should be factored in somehow, but I wasn't sure \nif it was 1/5 - 1/6 ... Then 1/4-1/5-1/6 etc. ) Or is the system set \nup in such a way that the chance of getting the egg is always 1/6 \nregardless on any given morning assuming no eggs are stolen or \ndisappear?" }, { "kind": "quote", "depth": 1, "text": "I don't have the book to hand, but if he is given six eggs, one \nspecial, and eats one every day at random, the chance of getting the \nspecial egg on any day is indeed 1/6, as you reckoned in your other post.\n\nThe easiest way to see it is to imagine he decided at random the order \nto eat them in advance, and laid all six in a row, each marked with \nits day for eating. Clearly the chance is the same for each day!\n\n- Gerry Quinn" }, { "kind": "text", "depth": 0, "text": "Actually, that is the chance before the experiment, If the first egg is \nto be eaten on Sunday, then before he eats it he knows that the chance \nof eating the special egg on any given day (say Tuesday) is 1/6.\n\nBut that presupposes he has no clue as to when he has eaten the special \negg. If he can tell immediately, then after he eats Sunday's egg. he \nknows that either it was the special one, or it wasn't. Depending on \nwhich, the chance that he will eat the special egg on Tuesday becomes \neither zero, or 1/5.\n\nIf he has some information that might help him decide which egg is \nspecial, but cannot be certain, the answer is somewhere in between. I \ncannot remember the story well, but I suspect this is most likely the \nsituation!\n\n- Gerry Quinn" } ] }, { "message_id": "", "from": { "name": "Marc Aramini", "address": "marcaramini at gmail.com" }, "date": "2014-09-07T14:13:33+00:00", "subject": "(urth) math probability question for seven American nights -\n\tCORRECTION", "in_reply_to": "<540C5D11.5080106@bindweed.com>", "reply_link": "header", "blocks": [ { "kind": "text", "depth": 0, "text": "Thanks.\n\n(Almost every Middle Eastern name in the text is from the Adventures of\nHajji Baba, and interestingly enough, the detective sent to look for Nadan\nat the start of the book, Hassan Kerbalai, is the name of Hajji Baba's\nfather. I may have to read at least some of that book beyond the chapters\ndealing with Nadan and Mirza and Osman Aga (a writer in Seven American\nNights and Baba's master in the older text, who sends him to purchase\nlamb-skins to start the novel). Baba means father or grandfather, and\nwhen (in Wolfe's story) Nadan sees the old American man who has the writing\nmachine at the musuem he calls him grandfather. Pretty sure Hajji Baba\ntries to defraud Nadan in Morier's book)\n\nThe play Visit to a Small Planet is by (EuGENE) Gore Vidal and the cat in\nthe play is named Rosemary - the second play, Mary Rose, with the two\ndisappearances of the titular character for periods of time, also seems to\nplay on Rosemary's name besides the allusions to Nadan's disappearance.\n\nEven though The Adventures of Hajji Baba was very critical of the Middle\nEastern/Persian characters, it was popular there as kind of a satire that\nrevealed their societal shortcomings. The same is true of Visit to a Small\nPlanet and its take on McCarthyism for America. Ironically these Americans\nsee it as a kind of testament to their old greatness." }, { "kind": "quote", "depth": 1, "text": "On Sun, Sep 7, 2014 at 6:26 AM, Gerry Quinn wrote:\n\nOn 07/09/2014 14:19, Gerry Quinn wrote:" }, { "kind": "quote", "depth": 2, "text": "On 07/09/2014 13:34, Marc Aramini wrote:" }, { "kind": "quote", "depth": 3, "text": "I had a quick question but trying to articulate it to look it up\nindependently is hard.\n\nIs there a statistically most probable day for Nadan to eat the special\none of the six eggs on any given day given that the first day is 1/6 and\nthen from there the probability that the egg is there begins to be\nsomething like 1/5 on the next day ( but the chance that it isn't there\nshould be factored in somehow, but I wasn't sure if it was 1/5 - 1/6 ...\nThen 1/4-1/5-1/6 etc. ) Or is the system set up in such a way that the\nchance of getting the egg is always 1/6 regardless on any given morning\nassuming no eggs are stolen or disappear?" }, { "kind": "quote", "depth": 2, "text": "I don't have the book to hand, but if he is given six eggs, one special,\nand eats one every day at random, the chance of getting the special egg on\nany day is indeed 1/6, as you reckoned in your other post.\n\nThe easiest way to see it is to imagine he decided at random the order to\neat them in advance, and laid all six in a row, each marked with its day\nfor eating. Clearly the chance is the same for each day!\n\n- Gerry Quinn" }, { "kind": "quote", "depth": 1, "text": "Actually, that is the chance before the experiment, If the first egg is\nto be eaten on Sunday, then before he eats it he knows that the chance of\neating the special egg on any given day (say Tuesday) is 1/6.\n\nBut that presupposes he has no clue as to when he has eaten the special\negg. If he can tell immediately, then after he eats Sunday's egg. he knows\nthat either it was the special one, or it wasn't. Depending on which, the\nchance that he will eat the special egg on Tuesday becomes either zero, or\n1/5.\n\nIf he has some information that might help him decide which egg is\nspecial, but cannot be certain, the answer is somewhere in between. I\ncannot remember the story well, but I suspect this is most likely the\nsituation!\n\n- Gerry Quinn\n\n\n\n\n\n_______________________________________________" } ] }, { "message_id": "", "from": { "name": "Gwern Branwen", "address": "gwern at gwern.net" }, "date": "2014-09-07T14:16:58+00:00", "subject": "(urth) math probability question for seven American nights", "in_reply_to": "<540C5B6C.8050407@bindweed.com>", "reply_link": "header", "blocks": [ { "kind": "quote", "depth": 1, "text": "On Sun, Sep 7, 2014 at 9:19 AM, Gerry Quinn wrote:\nI don't have the book to hand, but if he is given six eggs, one special, and\neats one every day at random, the chance of getting the special egg on any\nday is indeed 1/6, as you reckoned in your other post.\n\nThe easiest way to see it is to imagine he decided at random the order to\neat them in advance, and laid all six in a row, each marked with its day for\neating. Clearly the chance is the same for each day!" }, { "kind": "text", "depth": 0, "text": "Exactly. On the other hand, if he has some source of information about\nthe egg, then conditional on his new information each day, he may be\nable to update his beliefs as he goes along and do better than\nguessing just 1/6th each day.\n\nAs it happens, there's some literary precedent here: Graham Green's\nnovel https://en.wikipedia.org/wiki/Doctor_Fischer_of_Geneva_or_The_bomb_party\n\nAnd if one doesn't know the egg is there, then it becomes an\ninteresting statistical problem:\nhttp://gwern.net/docs/statistics/1994-falk" }, { "kind": "sig", "depth": 0, "text": "gwern" } ] }, { "message_id": "", "from": { "name": "Dimitar Nikolov", "address": "dimitar.g.nikolov at gmail.com" }, "date": "2014-09-07T20:06:13+00:00", "subject": "(urth) math probability question for seven American nights -\n\tCORRECTION", "in_reply_to": "<540C5D11.5080106@bindweed.com>", "reply_link": "header", "blocks": [ { "kind": "text", "depth": 0, "text": "It does depend what question exactly you're asking. Indeed, the probability\nstarts at 1/6, then assuming you know that wasn't the special egg, the\nprobability for the next day is 1/5 and so on.\n\nYou could however, ask the question, on which day is he most likely to eat\nthe egg?\n\nThen, you start with 1/6 for the first day once again. But the probability\nthe egg is eaten on the second day, is 1/6 times the probability it wasn't\neaten on the first day (=5/6), so .~.139. The probability it was eaten on\nthe third day is 1/6 times the probability it wasn't eaten on either\nprevious day, and so on. This is given by the geometric distribution that\nyou can read about on Wikipedia and results in the following probabilities:\n\neaten on day 1: 0.167\neaten on day 2: 0.139\neaten on day 3: 0.116\neaten on day 4: 0.096\neaten on day 5: 0.080\neaten on day 6: 0.067\n\nSo, in this phrasing of the question, which I think was the original\nquestion, he was most likely to eat the egg on the first day. Perhaps\nsomeone else can verify or challenge this, as I am not familiar with this\nparticular story.\n\nBest,\nDimitar" }, { "kind": "quote", "depth": 1, "text": "On Sun, Sep 7, 2014 at 9:26 AM, Gerry Quinn wrote:\n\nOn 07/09/2014 14:19, Gerry Quinn wrote:" }, { "kind": "quote", "depth": 2, "text": "On 07/09/2014 13:34, Marc Aramini wrote:" }, { "kind": "quote", "depth": 3, "text": "I had a quick question but trying to articulate it to look it up\nindependently is hard.\n\nIs there a statistically most probable day for Nadan to eat the special\none of the six eggs on any given day given that the first day is 1/6 and\nthen from there the probability that the egg is there begins to be\nsomething like 1/5 on the next day ( but the chance that it isn't there\nshould be factored in somehow, but I wasn't sure if it was 1/5 - 1/6 ...\nThen 1/4-1/5-1/6 etc. ) Or is the system set up in such a way that the\nchance of getting the egg is always 1/6 regardless on any given morning\nassuming no eggs are stolen or disappear?" }, { "kind": "quote", "depth": 2, "text": "I don't have the book to hand, but if he is given six eggs, one special,\nand eats one every day at random, the chance of getting the special egg on\nany day is indeed 1/6, as you reckoned in your other post.\n\nThe easiest way to see it is to imagine he decided at random the order to\neat them in advance, and laid all six in a row, each marked with its day\nfor eating. Clearly the chance is the same for each day!\n\n- Gerry Quinn" }, { "kind": "quote", "depth": 1, "text": "Actually, that is the chance before the experiment, If the first egg is\nto be eaten on Sunday, then before he eats it he knows that the chance of\neating the special egg on any given day (say Tuesday) is 1/6.\n\nBut that presupposes he has no clue as to when he has eaten the special\negg. If he can tell immediately, then after he eats Sunday's egg. he knows\nthat either it was the special one, or it wasn't. Depending on which, the\nchance that he will eat the special egg on Tuesday becomes either zero, or\n1/5.\n\nIf he has some information that might help him decide which egg is\nspecial, but cannot be certain, the answer is somewhere in between. I\ncannot remember the story well, but I suspect this is most likely the\nsituation!\n\n- Gerry Quinn\n\n\n\n\n\n_______________________________________________" } ] }, { "message_id": "", "from": { "name": "Marc Aramini", "address": "marcaramini at gmail.com" }, "date": "2014-09-07T20:17:21+00:00", "subject": "(urth) math probability question for seven American nights -\n\tCORRECTION", "in_reply_to": "", "reply_link": "header", "blocks": [ { "kind": "text", "depth": 0, "text": "hmmm ... but all the probabilities add up to .665 then ... and if he eats\nall six, he's GOING to eat the egg on one of those days\n... sometimes I feel like probabiity is the most nebulous and abstract\naspect of lower math.\n\n\nOn Sun, Sep 7, 2014 at 1:06 PM, Dimitar Nikolov wrote:" }, { "kind": "quote", "depth": 2, "text": "On 07/09/2014 14:19, Gerry Quinn wrote:" }, { "kind": "quote", "depth": 3, "text": "On 07/09/2014 13:34, Marc Aramini wrote:" }, { "kind": "quote", "depth": 4, "text": "I had a quick question but trying to articulate it to look it up\nindependently is hard.\n\nIs there a statistically most probable day for Nadan to eat the special\none of the six eggs on any given day given that the first day is 1/6 and\nthen from there the probability that the egg is there begins to be\nsomething like 1/5 on the next day ( but the chance that it isn't there\nshould be factored in somehow, but I wasn't sure if it was 1/5 - 1/6 ...\nThen 1/4-1/5-1/6 etc. ) Or is the system set up in such a way that the\nchance of getting the egg is always 1/6 regardless on any given morning\nassuming no eggs are stolen or disappear?" }, { "kind": "quote", "depth": 3, "text": "I don't have the book to hand, but if he is given six eggs, one special,\nand eats one every day at random, the chance of getting the special egg on\nany day is indeed 1/6, as you reckoned in your other post.\n\nThe easiest way to see it is to imagine he decided at random the order\nto eat them in advance, and laid all six in a row, each marked with its day\nfor eating. Clearly the chance is the same for each day!\n\n- Gerry Quinn" }, { "kind": "quote", "depth": 2, "text": "Actually, that is the chance before the experiment, If the first egg is\nto be eaten on Sunday, then before he eats it he knows that the chance of\neating the special egg on any given day (say Tuesday) is 1/6.\n\nBut that presupposes he has no clue as to when he has eaten the special\negg. If he can tell immediately, then after he eats Sunday's egg. he knows\nthat either it was the special one, or it wasn't. Depending on which, the\nchance that he will eat the special egg on Tuesday becomes either zero, or\n1/5.\n\nIf he has some information that might help him decide which egg is\nspecial, but cannot be certain, the answer is somewhere in between. I\ncannot remember the story well, but I suspect this is most likely the\nsituation!\n\n- Gerry Quinn\n\n\n\n\n\n_______________________________________________" } ] }, { "message_id": "<540CBE0D.4020705@bindweed.com>", "from": { "name": "Gerry Quinn", "address": "gerry at bindweed.com" }, "date": "2014-09-07T20:20:29+00:00", "subject": "(urth) math probability question for seven American nights -\n CORRECTION", "in_reply_to": "", "reply_link": "header", "blocks": [ { "kind": "quote", "depth": 1, "text": "On 07/09/2014 21:06, Dimitar Nikolov wrote:\nIt does depend what question exactly you're asking. Indeed, the \nprobability starts at 1/6, then assuming you know that wasn't the \nspecial egg, the probability for the next day is 1/5 and so on.\n\nYou could however, ask the question, on which day is he most likely to \neat the egg?\n\nThen, you start with 1/6 for the first day once again. But the \nprobability the egg is eaten on the second day, is 1/6 times the \nprobability it wasn't eaten on the first day (=5/6), so .~.139. The \nprobability it was eaten on the third day is 1/6 times the probability \nit wasn't eaten on either previous day, and so on. This is given by \nthe geometric distribution that you can read about on Wikipedia and \nresults in the following probabilities:\n\neaten on day 1: 0.167\neaten on day 2: 0.139\neaten on day 3: 0.116\neaten on day 4: 0.096\neaten on day 5: 0.080\neaten on day 6: 0.067" }, { "kind": "text", "depth": 0, "text": "Those probabilities don't add to 1! You're forgetting that tghere is \none fewer egg every day.\n\nYour figures work for a scenario in which he adds an ordinary egg every \nday, so he always has six, and chooses at random from them. Of course \nthis could go on forever, as the special egg might never be chosen. \nYou'll find that the limit of the sum of probabilities goes to 1.0 after \ninfinitely many days.\n\n- Gerry Quinn" } ] }, { "message_id": "", "from": { "name": "Dimitar Nikolov", "address": "dimitar.g.nikolov at gmail.com" }, "date": "2014-09-07T20:49:47+00:00", "subject": "(urth) math probability question for seven American nights -\n\tCORRECTION", "in_reply_to": "<540CBE0D.4020705@bindweed.com>", "reply_link": "header", "blocks": [ { "kind": "text", "depth": 0, "text": "You are right! This would only work for an infinite number of eggs. Sorry\nfor bringing my own confusion into this." }, { "kind": "quote", "depth": 1, "text": "On Sun, Sep 7, 2014 at 4:20 PM, Gerry Quinn wrote:\n\nOn 07/09/2014 21:06, Dimitar Nikolov wrote:" }, { "kind": "quote", "depth": 2, "text": "It does depend what question exactly you're asking. Indeed, the\nprobability starts at 1/6, then assuming you know that wasn't the special\negg, the probability for the next day is 1/5 and so on.\n\nYou could however, ask the question, on which day is he most likely to\neat the egg?\n\nThen, you start with 1/6 for the first day once again. But the\nprobability the egg is eaten on the second day, is 1/6 times the\nprobability it wasn't eaten on the first day (=5/6), so .~.139. The\nprobability it was eaten on the third day is 1/6 times the probability it\nwasn't eaten on either previous day, and so on. This is given by the\ngeometric distribution that you can read about on Wikipedia and results in\nthe following probabilities:\n\neaten on day 1: 0.167\neaten on day 2: 0.139\neaten on day 3: 0.116\neaten on day 4: 0.096\neaten on day 5: 0.080\neaten on day 6: 0.067" }, { "kind": "quote", "depth": 1, "text": "Those probabilities don't add to 1! You're forgetting that tghere is one\nfewer egg every day.\n\nYour figures work for a scenario in which he adds an ordinary egg every\nday, so he always has six, and chooses at random from them. Of course this\ncould go on forever, as the special egg might never be chosen. You'll find\nthat the limit of the sum of probabilities goes to 1.0 after infinitely\nmany days.\n\n\n- Gerry Quinn\n_______________________________________________" } ] } ] }