Lost In Translation (Google Translate Can't Do Math Notation ...)

Discussion in 'Computers and The Internet' started by AceK, Jun 13, 2016.

  1. AceK

    AceK Scientia Potentia Est

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    I took some text I wrote (part of a conversation i started on /r/math), translated it to Finnish, then took the Finnish and translated that back to English; and this is what I got. Tell me which one is most clear to you (if you can't read Finnish, skip the Finnish, I can't read Finnish and don't know how good it is but I know the final English re-translation is absolutely destroyed basically.) Google translate can't do subscript/superscript and other math notation and such (and also screwed up some punctuation), so that is kinda messed up from the beginning but it's still understandable at least in the beginning.



    Original English text I wrote as part of conversation on /r/math
    ___________________________________________________________________________________


    Your equation appears to give the product of a number of probabilities Which, to me, seems like it'll give you the probability of an event of length i rather than the probability of an event of length k in i iterations.

    Basically, that's what I was trying to write. Just simply the theoretical probability of some arbitrarily long sequence of events occurring. In the simplest terms you might say the probability of producing the favorable outcome for each event in the series and making it to the end of the series without failing. Dice are often used as examples and a very simple one would be, "what is the probability of rolling a die 100 times and getting all 6s." I wasn't trying to figure the probability of this sequence occurring in some arbitrary number of trials The upper limit i is the length of the sequence. k is just the index on the general term that goes to the limit i The length of the "event", is defined as the limit i, which is really a sequence of favorable events. k is used as a subscript of n and the second product series for E ... this is because we may not be dealing with die throws where every event in the sequence has the same number of favorable outcomes; it could be some arbitrary sequence of events with different numbers of favorable outcomes, that is each n may be a different n, thus a different probability for each independent event in the series which has to be accounted for if this were to be the case. The two expressions on the left (the ones with the big Pi) are just equivalent, I probably could have written 1 over the reciprocal of P(Ek), but that would be superfluous, as that is basically what 1/n is, and make things look more complex than necessary, which is counterproductive. The fraction on the left with the summation in the numerator, that's definitely more complex than needed, and it isn't equivalent to the first two expressions if the probability of a favorable outcome for each event in the series isn't the same (in the denominator of the fractions in the numerator of the entire expression, i believe that n should be nkk since each "n" in the sequence {sn} where sn is the general term may not be the same. The expression on the right though doesn't handle this properly even if you changed this, because it is a fraction and the denominator isn't part of the summation, it only works equivalently when the probability for each event in the sequence is the same* That math shouldn't be there since it is dis-proven that it is equivalent, though I would like to somehow make it equivalent. ... I fixed part of it I think, it shouldn't be in there, but I will edit my OP, maybe someone will have something to say besides it not being equivalent.

    The probability of randomly producing a sequence of favorable outcomes becomes lower and lower the longer the sequence is since you have more chances to fail at producing the next favorable outcome. *What I can't think of how to deal with right now is the fact that in the limit of infinity the probability of any event should be 1, i.e certain if it's possible at all, i.e non-zero probability. I don't know how to deal with this, because of what infinity is, a limit. You can always add another term and its still infinity, and you will never reach infinity either at least not in the real world. But that would I believe only apply if the sequence of favorable outcomes is a finite number, if our sequence is an infinite sequence then the probability I believe would tend to zero as i tends toward infinity (unless the probability for every event in the sequence was certain) but I don't think would actually be 0 (unless at least one of the events in the sequence is impossible), just infinitesimal though at least right now it looks like it would converge toward 0, and I believe the limit of the expression as i tends toward infinite would be zero... you might need some hardcore set theory to transcend past infinity and I'm not gonna say anymore about stuff like that because I'd have no clue what I was talking about.

    What you are talking about above though in your comment, I will have to come back to later to think about more when I'm a bit more rested , I started commenting about it and realized that I've been awake too long that I probably shouldn't be really be doing math. Your function looks pretty simple but I tried it and got strange results, e.g. negative probabilities, e.g. -1.02 x 10^21 iirc or something similar.

    Consider rolling 3 sixes in a row with only 11 dice rolls. The probability of this happening is ~0.04%.

    Don't know if I entered things wrong but I did try some quick and dirty math and got 0.0046...= ~0.46%. This looks a lot more similar to your ~0.04%, just an order of magnitude off. Idk if it's coincidence, but its a lot closer than than the number mentioned above, off by about 20 orders of magnitude and the significand did not begin with 4. What expression did you use to come up with this ~0.04%.

    I wasn't sure how to treat the dice rolls as events, but with such small numbers trying a few different things made no significant difference in the result I got. In the expression I wrote in the OP, its value should be the probability of an event, and that event is the favorable outcome of all events in the sequence {Sn} where Sn could be arbitrary, or a random function, it could be anything. So I consider the sequence of events occurring (in sequence) to be an event itself, an abstraction of all of the outcomes of the sequence, and there is a probability that this sequence will occur; there is also a probability against it occurring. So when considering the probability of that ENTIRE expression occurring after some number of iterations I'm not sure how to treat it. If you have a sequence of some finite number of terms, then it seems to me that you would have to find the probability of this sequence occurring within some certain number of terms or iterations, as as you say that number has to be larger than the number of terms in the sequence else the probability is zero. I just am not sure how to deal with the sequence of events as an event given some number of "trials" for each term, especially in the case that we don't necessarily know the number of terms in the sequence.

    Finnish translated from original English
    ___________________________________________________________

    Yhtälö näyttää antavan tuotetta useita todennäköisyyksien Joka minulle, tuntuu se tulee antaa sinulle todennäköisyys tapahtuman pituudeltaan i sijaan todennäköisyys tapahtuman, jonka pituus on k i toistojen.

    Pohjimmiltaan se mitä yritin kirjoittaa. Vain yksinkertaisesti teoreettinen todennäköisyys joidenkin mielivaltaisesti pitkä sarja tapahtumia. Yksinkertaisesti esitettynä, voitaisiin sanoa todennäköisyys tuottaa myönteiseen tulokseen jokaisen tapahtuman sarjassa ja tehdä sen loppuun sarjan ilman ei ole. Dice käytetään usein esimerkkeinä ja hyvin yksinkertainen olisi, "mikä on todennäköisyys liikkuvan muotin 100 kertaa ja saada kaikki 6s." En yrittänyt selvittää todennäköisyys tämän järjestyksessä esiintyy joissakin mielivaltaisen määrän kokeita yläraja i on sekvenssin pituus. k on vain indeksi yleinen termi, joka menee rajan i pituus "tapahtuma", on määritelty raja-i, joka on todella sekvenssi suotuisa tapahtumia. k käytetään alaindeksi n ja toisen tuotesarja E ... tämä on, koska emme saa käsitellä die heittää, jossa jokainen tapahtuma järjestyksessä on sama määrä työn tuloksia; se voisi olla joitakin mielivaltaisia ​​tapahtumasarjaa joissa on eri määrä suotuisia tuloksia, jotka on kukin n voi olla erilainen n, mikä on eri todennäköisyys kunkin itsenäisen tapahtuma sarja, joka on otettava huomioon, jos näin olevan näin. Kahden lausekkeen vasemmalla (ne joilla on suuret Pi) ovat aivan vastaavia, en luultavasti olisi voinut kirjoittaa 1 yli vastavuoroinen P (Ek), mutta se olisi tarpeetonta, koska se on periaatteessa mitä 1 / n on, ja tehdä asiat näyttävät monimutkaisempi kuin on tarpeen, mikä on haitallista. Fraktio vasemmalla kanssa summattu osoittajassa, joka on varmasti monimutkaisempi kuin tarvitaan, ja se ei vastaa ensimmäistä kahta ilmaisua, jos todennäköisyys myönteisen tuloksen kunkin tapahtuman sarjassa ei ole sama (vuonna nimittäjä jakeiden osoittajassa koko ilmaisun, uskon, että n pitäisi NKK koska jokainen "n" järjestyksessä {sn} jossa sn on yleisnimitys voi olla sama. ilmaisu oikealla vaikka doesnt 't hoitaa tämän oikein, vaikka olet muuttanut tätä, koska se on vain murto-ja nimittäjä ei ole osa summattu, se toimii vain vastaavasti, kun todennäköisyys kunkin tapahtuman järjestysnumero on sama * tätä matematiikkaa pitäisi olla siellä koska se on dIS-todistettu, että se vastaa, vaikka haluaisin jotenkin tehdä se vastaa. ... Korjasin osa sitä mielestäni ei pitäisi olla siellä, mutta aion muokata OP, ehkä joku on jotain sanottavaa lisäksi se ei ole vastaavaa.

    Todennäköisyys satunnaisesti tuottaa sekvenssin suotuisia madaltuu ja laske pidempi jono on, koska sinulla on enemmän mahdollisuuksia epäonnistua tuottamaan seuraavan myönteisen tuloksen. * Voin ajatella miten käsitellä juuri nyt on se, että raja ääretön todennäköisyys tapauksessa pitäisi olla 1, ts tiettyjä jos se on suinkin mahdollista, eli ei-nolla todennäköisyys. En tiedä, miten käsitellä tätä, koska mitä ääretön on, raja. Voit aina lisätä toisen aikavälillä ja sen edelleen ääretön, ja et koskaan saavuttaa äärettömyyttä joko ainakaan reaalimaailmassa. Mutta joka uskon soveltaa vain, jos mainittu sekvenssi työn tuloksia on äärellinen määrä, jos meidän sekvenssi on ääretön jono niin todennäköisyys uskon pyrkisi nollaksi i pyrkii kohti ääretöntä (ellei todennäköisyys jokaisen tapahtuman sekvenssi oli tietyt), mutta en usko, olisi todella olla 0 (ellei vähintään yksi tapahtumista järjestyksessä on mahdotonta), vain äärettömän joskin ainakin nyt näyttää siltä, ​​että se lähestyvät 0, ja uskon rajan lauseketta i taipumus kohti ääretöntä olisi nolla ... ehkä jotkut hardcore joukko-oppi ylittämään ohi äärettömään ja en aio sanoa enää noin tuollaista koska olin ei ole aavistustakaan, mitä puhuin.

    Mitä te puhutte yläpuolella vaikka kommenttisi, minun täytyy palata myöhemmin ajatella enemmän, kun olen vähän enemmän lepoa, aloin kommentoi siitä ja tajusin, että olen ollut hereillä liian pitkä, että en luultavasti ei pitäisi oikeastaan ​​olla tekemässä matematiikkaa. Your toiminto näyttää melko yksinkertainen, mutta yritin sitä ja sai outoja tuloksia, esim. negatiivinen todennäköisyydet, esim. -1,02 X 1021 IIRC tai jotain vastaavaa.

    Tarkastellaan liikkuvan 3 kuutosta peräkkäin vain 11 noppaa sämpylöitä. Todennäköisyys, että näin tapahtuu on ~ 0,04%.

    En tiedä, jos olen tullut asioita väärin, mutta tein kokeilla nopea ja likainen matematiikan ja sai 0,0046 ... = ~ 0,46%. Tämä näyttää paljon samankaltaisia ​​kuin ~ 0,04%, vain kertaluokkaa pois. Idk jos se on sattumaa, mutta sen paljon lähempänä kuin kuin määrä Edellä mainittujen pois noin 20 kertaluokkaa ja significand ei alkanut 4. Mitä ilme käytit keksiä tähän ~ 0,04%.

    En ollut varma, miten käsitellä noppaa rullat tapahtumina, mutta niin pieniä määriä yrittää muutamia eri asioita tehty mitään merkittävää eroa tuloksen sain. Ilmaisussa kirjoitin OP, sen arvon tulisi olla todennäköisyys tapahtuman, ja että tapahtuma on myönteinen tulos kaikkien tapahtumien järjestyksessä {Sn} jossa Sn voi olla mielivaltainen, tai satunnainen funktio, se voi olla mitä tahansa . Joten katson tapahtumaketju esiintyy (järjestyksessä) on tapahtuma itsessään, abstraktio kaikkien tulosten järjestyksessä, ja on todennäköistä, että tämä järjestys tapahtuu; on myös todennäköisyys sitä vastaan ​​esiintyy. Joten kun otetaan huomioon todennäköisyys, että koko lauseke esiintyy jonkin toistojen En ole varma, miten käsitellä sitä. Jos sinulla on sekvenssi joidenkin rajallinen määrä ehtoja, niin minusta tuntuu, että sinulla olisi löytää todennäköisyys tämän järjestyksessä esiintyvien joitakin tiettyjä termejä tai toistojen, niin kuin sanot, että määrä on oltava suurempi kuin termien lukumäärä sekvenssin muu todennäköisyys on nolla. En vain ole varma, miten käsitellä tapahtumaketju tapahtumana antanut joitakin määrä "kokeita" jokaiselle aikavälillä erityisesti siinä tapauksessa, että emme välttämättä tiedä termien lukumäärä järjestyksessä.



    AND THIS IS THAT SAME FINNISH TRANSLATED BACK TO ENGLISH
    _________________________________________________________________________________



    The equation seems to give the product a number of probabilities Which to me seems like it will give you the length of the probability of an event instead of i the probability of an event, with a length of k i iterations.

    Basically, that's what I was trying to write. Just simply theoretical probability of some arbitrarily long series of events. Simply put, you could say the probability of producing a positive result in each event and make it to the end of the series without failing. Dice is often used as an example and should be very simple, "what is the probability of rolling the die 100 times and get all the 6s." I'm trying to figure out the probability of this sequence occurring in some arbitrary number of tests the upper limit i is the length of the sequence. the index k is a general term which goes length limit i "event" is defined as the border-i, which is very favorable sequence of events. k subscript is used and a second product line E ... this is because we do not deal with die-casts, where each event the order is the same amount of results of the work; it could be some arbitrary sequence of events with a different number of beneficial results, which are each n may be the difference, which is a different probability of each independent series of events which must be taken into account, if this is the case. on the left of the two expressions (those with large Pi) are quite similar, I probably would have been able to write one more than the reciprocal P (Ek), but it would be unnecessary, because it is basically what the 1 / n is, and to make things seem more complicated than it is necessary what is harmful. Fraction left with the summation in the numerator, which is certainly more complex than needed, and it does not correspond to the first two terms, if the probability of a positive outcome of each event series is not the same (in the denominator of the fractions in the numerator full of expression, I believe that should NKK because each "n" sequence {sn}, where sn is the generic name can be the same. the expression on the right even if doesn 't handle this correctly, even if you have changed this, because it is only a fraction, and the denominator is not part of the summation, it will only work, respectively, when the probability of each event sequence number this is the same * mathematics should be there because it is a dis-proven fact that it is responsible for, although I would like to somehow make it responsible. ... I fixed part of it, I think should not be there, but I'm going to edit the OP, maybe someone has something to say in addition to it is not equivalent.

    The probability of randomly generate the sequence of favorable lowered and the lower the longer the queue is, because you will have more chances to fail to produce a positive result to the next. * I can think of how to deal with right now is the fact that the threshold of infinite probability event should be 1, that is certain, if it is at all possible, that is, non-zero probability. I do not know how to deal with this, because what infinity is the limit. You can always add a second term, and continues to infinity, and you will never reach infinity, either, at least not in the real world. But that I believe apply only if the sequence results of the work is a finite number, if our sequence is an infinite sequence so the likelihood of faith would tend to zero i tends towards infinity (unless the probability of each event sequence was certain), but I do not think would actually be 0 (unless at least one of the events in the order it is impossible), only infinite but at least now it seems that it is approaching 0, and i think the limit clause i tends towards infinity is zero ... maybe some hardcore set theory to cross over to infinity and i'm not going to say anymore about things like that because I do not have a clue what I was talking about.

    What you are talking about even if the above comment, I have to come back later to think more when I'm a bit more rest, I began to comment about it and I realized that I've been awake too long, that I probably should not really be doing math. Your activity seems pretty simple, but I tried it and got strange results, eg. A negative probabilities, for example. -1.02 X 1021 IIRC or something similar.

    Consider a rolling three sixes in a row, only 11 dice rolls. The probability of this happening is ~ 0.04%.

    I do not know if I have become things wrong, but I did try the quick and dirty math and got 0.0046 ... = ~ 0.46%. This looks much similar to ~ 0.04%, only an order of magnitude off. Idk if it is a coincidence, but a lot closer than the amount of the above-mentioned off about 20 orders of magnitude and the significand is not started 4. What look did you come up with this ~ 0.04%.

    I was not sure how to handle the dice rolls events, but in such small quantities to try a few different things made no significant difference in the result I got. I wrote in the expression of OP, its value should be the probability of the event, and that event is a positive outcome for all events in sequence {Sn} where Sn can be arbitrary, or random function, it can be anything. So I look at the chain of events occurs (order) is an event in itself, an abstraction in order for all of the results, and it is likely that this sequence occurs; is also the probability of it occurring. So when you take into account the probability that the entire expression occurs in some iterations I'm not sure how to handle it. If you have a sequence of some of the limited number of conditions, so it seems to me that you would have to find the probability of this sequence occurring in some specific terms or iterations, as you say, that the amount must be greater than the number of terms in the sequence of other probability is zero. I'm just not sure how to deal with the chain of events as an event given some number of "tests" for each term, especially in the event that we do not necessarily know the number of terms in order.




    SEE HOW MUCH HAS BEEN CHANGED.

    I suspect the last translation to English is probably the worst, and doesn't make very much sense to me. I don't know about the Finnish translation, because I can't read Finnish, but is suspect it makes more sense than that last English translation, but is worse than the original text ;)
     
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