How to calculate how lucky you are in a contest. Probability.












0












$begingroup$


You are on a beach full of rocks looking for treasure. The odds of finding treasure under any rock is $1/273$. You have looked under $5000$ rocks and found only $2$ pieces of treasure. How unlucky were you? What are the odds of looking under $5000$ rocks and finding only $2$ pieces of treasure (when the odds of looking under $1$ rock and finding treasure is $1/273$)? I assume you have to add the probability of finding $0$, $1$ and $2$ together?



I found this website, but the numbers aren't precise enough for me. I would like to convert the odds to something more manageable. I wanted to be able to say, this is equivalent to getting heads X number times in a row for coin tosses. And with my sample data it just turns into "$0.9999999$" on that site, so its not useful to me.



I think I need "Cumulative probability: $P(X ≤ x)$".



If someone could help explain it to me.



Ideally I would also like to know how much outside of the norm my sample set is. In other words, I question the validity of the 1/273 odds and want to know the best way to test it.






probability probability-distributions







share|cite|improve this question











$endgroup$












  • $begingroup$
    Regarding the question toward the end of your post: If you toss a coin $10$ times, you will always have $le 10$ heads.
    $endgroup$
    – angryavian
    Dec 31 '18 at 1:39










  • $begingroup$
    "How unlucky were you?" -- Unless you have a means of quantifying this, no one is going to be able to help you on that. (Unless it's a more colloquial statement answered by the others.)
    $endgroup$
    – Eevee Trainer
    Dec 31 '18 at 1:41










  • $begingroup$
    As for the fact that the calculators seem to be producing $100%$ or something of the sort, that's an artifact of how small the odds of this happening are, and the fact that the average program/calculator/whatever cannot readily calculate something to such a precision, so it just rounds as close as it can get. You will have to calculate it manually using the binomial distribution formulas if you want a more precise answer.
    $endgroup$
    – Eevee Trainer
    Dec 31 '18 at 1:43










  • $begingroup$
    Just use the binomial distribution with $p=frac 1{273}$. We get $P(X=0)+P(X=1)+P(X=2)=.00000202369$ so you would indeed have to be very unlucky not to have more than $2$ successes. The expected number of wins is $frac {5000}{273}=18.315$
    $endgroup$
    – lulu
    Dec 31 '18 at 1:44
















0












$begingroup$


You are on a beach full of rocks looking for treasure. The odds of finding treasure under any rock is $1/273$. You have looked under $5000$ rocks and found only $2$ pieces of treasure. How unlucky were you? What are the odds of looking under $5000$ rocks and finding only $2$ pieces of treasure (when the odds of looking under $1$ rock and finding treasure is $1/273$)? I assume you have to add the probability of finding $0$, $1$ and $2$ together?



I found this website, but the numbers aren't precise enough for me. I would like to convert the odds to something more manageable. I wanted to be able to say, this is equivalent to getting heads X number times in a row for coin tosses. And with my sample data it just turns into "$0.9999999$" on that site, so its not useful to me.



I think I need "Cumulative probability: $P(X ≤ x)$".



If someone could help explain it to me.



Ideally I would also like to know how much outside of the norm my sample set is. In other words, I question the validity of the 1/273 odds and want to know the best way to test it.






probability probability-distributions







share|cite|improve this question











$endgroup$












  • $begingroup$
    Regarding the question toward the end of your post: If you toss a coin $10$ times, you will always have $le 10$ heads.
    $endgroup$
    – angryavian
    Dec 31 '18 at 1:39










  • $begingroup$
    "How unlucky were you?" -- Unless you have a means of quantifying this, no one is going to be able to help you on that. (Unless it's a more colloquial statement answered by the others.)
    $endgroup$
    – Eevee Trainer
    Dec 31 '18 at 1:41










  • $begingroup$
    As for the fact that the calculators seem to be producing $100%$ or something of the sort, that's an artifact of how small the odds of this happening are, and the fact that the average program/calculator/whatever cannot readily calculate something to such a precision, so it just rounds as close as it can get. You will have to calculate it manually using the binomial distribution formulas if you want a more precise answer.
    $endgroup$
    – Eevee Trainer
    Dec 31 '18 at 1:43










  • $begingroup$
    Just use the binomial distribution with $p=frac 1{273}$. We get $P(X=0)+P(X=1)+P(X=2)=.00000202369$ so you would indeed have to be very unlucky not to have more than $2$ successes. The expected number of wins is $frac {5000}{273}=18.315$
    $endgroup$
    – lulu
    Dec 31 '18 at 1:44














0












0








0





$begingroup$


You are on a beach full of rocks looking for treasure. The odds of finding treasure under any rock is $1/273$. You have looked under $5000$ rocks and found only $2$ pieces of treasure. How unlucky were you? What are the odds of looking under $5000$ rocks and finding only $2$ pieces of treasure (when the odds of looking under $1$ rock and finding treasure is $1/273$)? I assume you have to add the probability of finding $0$, $1$ and $2$ together?



I found this website, but the numbers aren't precise enough for me. I would like to convert the odds to something more manageable. I wanted to be able to say, this is equivalent to getting heads X number times in a row for coin tosses. And with my sample data it just turns into "$0.9999999$" on that site, so its not useful to me.



I think I need "Cumulative probability: $P(X ≤ x)$".



If someone could help explain it to me.



Ideally I would also like to know how much outside of the norm my sample set is. In other words, I question the validity of the 1/273 odds and want to know the best way to test it.






probability probability-distributions







share|cite|improve this question











$endgroup$




You are on a beach full of rocks looking for treasure. The odds of finding treasure under any rock is $1/273$. You have looked under $5000$ rocks and found only $2$ pieces of treasure. How unlucky were you? What are the odds of looking under $5000$ rocks and finding only $2$ pieces of treasure (when the odds of looking under $1$ rock and finding treasure is $1/273$)? I assume you have to add the probability of finding $0$, $1$ and $2$ together?



I found this website, but the numbers aren't precise enough for me. I would like to convert the odds to something more manageable. I wanted to be able to say, this is equivalent to getting heads X number times in a row for coin tosses. And with my sample data it just turns into "$0.9999999$" on that site, so its not useful to me.



I think I need "Cumulative probability: $P(X ≤ x)$".



If someone could help explain it to me.



Ideally I would also like to know how much outside of the norm my sample set is. In other words, I question the validity of the 1/273 odds and want to know the best way to test it.






probability probability-distributions




probability probability-distributions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 31 '18 at 2:53







Curtis

















asked Dec 31 '18 at 1:35









CurtisCurtis

1064




1064












  • $begingroup$
    Regarding the question toward the end of your post: If you toss a coin $10$ times, you will always have $le 10$ heads.
    $endgroup$
    – angryavian
    Dec 31 '18 at 1:39










  • $begingroup$
    "How unlucky were you?" -- Unless you have a means of quantifying this, no one is going to be able to help you on that. (Unless it's a more colloquial statement answered by the others.)
    $endgroup$
    – Eevee Trainer
    Dec 31 '18 at 1:41










  • $begingroup$
    As for the fact that the calculators seem to be producing $100%$ or something of the sort, that's an artifact of how small the odds of this happening are, and the fact that the average program/calculator/whatever cannot readily calculate something to such a precision, so it just rounds as close as it can get. You will have to calculate it manually using the binomial distribution formulas if you want a more precise answer.
    $endgroup$
    – Eevee Trainer
    Dec 31 '18 at 1:43










  • $begingroup$
    Just use the binomial distribution with $p=frac 1{273}$. We get $P(X=0)+P(X=1)+P(X=2)=.00000202369$ so you would indeed have to be very unlucky not to have more than $2$ successes. The expected number of wins is $frac {5000}{273}=18.315$
    $endgroup$
    – lulu
    Dec 31 '18 at 1:44


















  • $begingroup$
    Regarding the question toward the end of your post: If you toss a coin $10$ times, you will always have $le 10$ heads.
    $endgroup$
    – angryavian
    Dec 31 '18 at 1:39










  • $begingroup$
    "How unlucky were you?" -- Unless you have a means of quantifying this, no one is going to be able to help you on that. (Unless it's a more colloquial statement answered by the others.)
    $endgroup$
    – Eevee Trainer
    Dec 31 '18 at 1:41










  • $begingroup$
    As for the fact that the calculators seem to be producing $100%$ or something of the sort, that's an artifact of how small the odds of this happening are, and the fact that the average program/calculator/whatever cannot readily calculate something to such a precision, so it just rounds as close as it can get. You will have to calculate it manually using the binomial distribution formulas if you want a more precise answer.
    $endgroup$
    – Eevee Trainer
    Dec 31 '18 at 1:43










  • $begingroup$
    Just use the binomial distribution with $p=frac 1{273}$. We get $P(X=0)+P(X=1)+P(X=2)=.00000202369$ so you would indeed have to be very unlucky not to have more than $2$ successes. The expected number of wins is $frac {5000}{273}=18.315$
    $endgroup$
    – lulu
    Dec 31 '18 at 1:44
















$begingroup$
Regarding the question toward the end of your post: If you toss a coin $10$ times, you will always have $le 10$ heads.
$endgroup$
– angryavian
Dec 31 '18 at 1:39




$begingroup$
Regarding the question toward the end of your post: If you toss a coin $10$ times, you will always have $le 10$ heads.
$endgroup$
– angryavian
Dec 31 '18 at 1:39












$begingroup$
"How unlucky were you?" -- Unless you have a means of quantifying this, no one is going to be able to help you on that. (Unless it's a more colloquial statement answered by the others.)
$endgroup$
– Eevee Trainer
Dec 31 '18 at 1:41




$begingroup$
"How unlucky were you?" -- Unless you have a means of quantifying this, no one is going to be able to help you on that. (Unless it's a more colloquial statement answered by the others.)
$endgroup$
– Eevee Trainer
Dec 31 '18 at 1:41












$begingroup$
As for the fact that the calculators seem to be producing $100%$ or something of the sort, that's an artifact of how small the odds of this happening are, and the fact that the average program/calculator/whatever cannot readily calculate something to such a precision, so it just rounds as close as it can get. You will have to calculate it manually using the binomial distribution formulas if you want a more precise answer.
$endgroup$
– Eevee Trainer
Dec 31 '18 at 1:43




$begingroup$
As for the fact that the calculators seem to be producing $100%$ or something of the sort, that's an artifact of how small the odds of this happening are, and the fact that the average program/calculator/whatever cannot readily calculate something to such a precision, so it just rounds as close as it can get. You will have to calculate it manually using the binomial distribution formulas if you want a more precise answer.
$endgroup$
– Eevee Trainer
Dec 31 '18 at 1:43












$begingroup$
Just use the binomial distribution with $p=frac 1{273}$. We get $P(X=0)+P(X=1)+P(X=2)=.00000202369$ so you would indeed have to be very unlucky not to have more than $2$ successes. The expected number of wins is $frac {5000}{273}=18.315$
$endgroup$
– lulu
Dec 31 '18 at 1:44




$begingroup$
Just use the binomial distribution with $p=frac 1{273}$. We get $P(X=0)+P(X=1)+P(X=2)=.00000202369$ so you would indeed have to be very unlucky not to have more than $2$ successes. The expected number of wins is $frac {5000}{273}=18.315$
$endgroup$
– lulu
Dec 31 '18 at 1:44










1 Answer
1






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oldest

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1












$begingroup$

"More precise" isn't necessarily the right viewpoint here, especially when you're having issues with a theoretically accurate calculation rounding to $0$ or $1$. Instead, we should look for an approximate model that handles this extreme well.



And on that note - there is such a model. The Poisson distribution arises as a limit of binomial distributions with large numbers of trials, low probability, and fixed mean. That mean becomes the distribution's sole parameter $lambda$, which will be $frac{5000}{273}approx 18.315$ in this case.



In a Poisson distribution with parameter $lambda$, the probability of getting exactly $k$ is $frac{lambda^k}{k!}e^{-k}$. Here, that gives (using Excel for the calculations, since I happen to have it open), probabilities of 1.11145E-08 for $0$, 2.03562E-07 for $1$, and 1.86412E-06 for $2$. The probability of being at most $2$ is their sum 2.07879E-06.



Well, OK, we don't really have all of those digits - but it'll be pretty close. We can confidently say that if the probabilities were accurate, the likelihood of getting at most two successes would be approximately $2cdot 10^{-6}$. That's definitely a "reject the hypothesis" level.



How many consecutive head flips would it take to reach that level of unlikeliness? The probability of $n$ consecutive flips (with no extra tries) is $2^{-n}$. From the standard approximation $2^{10}approx 10^3$, we get $10^{-6}approx 2^{-20}$, and $2^{-19}approx 2cdot 10^{-6}$. It would take nineteen consecutive head flips to reach that level of improbability.



I tried a manual version of the binomial approximation as well - it came out about $2.5%$ different, a total probability of about $2.024cdot 10^{-6}$. It's technically more accurate, but is that worth switching from a simple exponential $e^{-lambda}$ to a high power $(1-p)^N$ of something very close to $1$?






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    $begingroup$

    "More precise" isn't necessarily the right viewpoint here, especially when you're having issues with a theoretically accurate calculation rounding to $0$ or $1$. Instead, we should look for an approximate model that handles this extreme well.



    And on that note - there is such a model. The Poisson distribution arises as a limit of binomial distributions with large numbers of trials, low probability, and fixed mean. That mean becomes the distribution's sole parameter $lambda$, which will be $frac{5000}{273}approx 18.315$ in this case.



    In a Poisson distribution with parameter $lambda$, the probability of getting exactly $k$ is $frac{lambda^k}{k!}e^{-k}$. Here, that gives (using Excel for the calculations, since I happen to have it open), probabilities of 1.11145E-08 for $0$, 2.03562E-07 for $1$, and 1.86412E-06 for $2$. The probability of being at most $2$ is their sum 2.07879E-06.



    Well, OK, we don't really have all of those digits - but it'll be pretty close. We can confidently say that if the probabilities were accurate, the likelihood of getting at most two successes would be approximately $2cdot 10^{-6}$. That's definitely a "reject the hypothesis" level.



    How many consecutive head flips would it take to reach that level of unlikeliness? The probability of $n$ consecutive flips (with no extra tries) is $2^{-n}$. From the standard approximation $2^{10}approx 10^3$, we get $10^{-6}approx 2^{-20}$, and $2^{-19}approx 2cdot 10^{-6}$. It would take nineteen consecutive head flips to reach that level of improbability.



    I tried a manual version of the binomial approximation as well - it came out about $2.5%$ different, a total probability of about $2.024cdot 10^{-6}$. It's technically more accurate, but is that worth switching from a simple exponential $e^{-lambda}$ to a high power $(1-p)^N$ of something very close to $1$?






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      "More precise" isn't necessarily the right viewpoint here, especially when you're having issues with a theoretically accurate calculation rounding to $0$ or $1$. Instead, we should look for an approximate model that handles this extreme well.



      And on that note - there is such a model. The Poisson distribution arises as a limit of binomial distributions with large numbers of trials, low probability, and fixed mean. That mean becomes the distribution's sole parameter $lambda$, which will be $frac{5000}{273}approx 18.315$ in this case.



      In a Poisson distribution with parameter $lambda$, the probability of getting exactly $k$ is $frac{lambda^k}{k!}e^{-k}$. Here, that gives (using Excel for the calculations, since I happen to have it open), probabilities of 1.11145E-08 for $0$, 2.03562E-07 for $1$, and 1.86412E-06 for $2$. The probability of being at most $2$ is their sum 2.07879E-06.



      Well, OK, we don't really have all of those digits - but it'll be pretty close. We can confidently say that if the probabilities were accurate, the likelihood of getting at most two successes would be approximately $2cdot 10^{-6}$. That's definitely a "reject the hypothesis" level.



      How many consecutive head flips would it take to reach that level of unlikeliness? The probability of $n$ consecutive flips (with no extra tries) is $2^{-n}$. From the standard approximation $2^{10}approx 10^3$, we get $10^{-6}approx 2^{-20}$, and $2^{-19}approx 2cdot 10^{-6}$. It would take nineteen consecutive head flips to reach that level of improbability.



      I tried a manual version of the binomial approximation as well - it came out about $2.5%$ different, a total probability of about $2.024cdot 10^{-6}$. It's technically more accurate, but is that worth switching from a simple exponential $e^{-lambda}$ to a high power $(1-p)^N$ of something very close to $1$?






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        "More precise" isn't necessarily the right viewpoint here, especially when you're having issues with a theoretically accurate calculation rounding to $0$ or $1$. Instead, we should look for an approximate model that handles this extreme well.



        And on that note - there is such a model. The Poisson distribution arises as a limit of binomial distributions with large numbers of trials, low probability, and fixed mean. That mean becomes the distribution's sole parameter $lambda$, which will be $frac{5000}{273}approx 18.315$ in this case.



        In a Poisson distribution with parameter $lambda$, the probability of getting exactly $k$ is $frac{lambda^k}{k!}e^{-k}$. Here, that gives (using Excel for the calculations, since I happen to have it open), probabilities of 1.11145E-08 for $0$, 2.03562E-07 for $1$, and 1.86412E-06 for $2$. The probability of being at most $2$ is their sum 2.07879E-06.



        Well, OK, we don't really have all of those digits - but it'll be pretty close. We can confidently say that if the probabilities were accurate, the likelihood of getting at most two successes would be approximately $2cdot 10^{-6}$. That's definitely a "reject the hypothesis" level.



        How many consecutive head flips would it take to reach that level of unlikeliness? The probability of $n$ consecutive flips (with no extra tries) is $2^{-n}$. From the standard approximation $2^{10}approx 10^3$, we get $10^{-6}approx 2^{-20}$, and $2^{-19}approx 2cdot 10^{-6}$. It would take nineteen consecutive head flips to reach that level of improbability.



        I tried a manual version of the binomial approximation as well - it came out about $2.5%$ different, a total probability of about $2.024cdot 10^{-6}$. It's technically more accurate, but is that worth switching from a simple exponential $e^{-lambda}$ to a high power $(1-p)^N$ of something very close to $1$?






        share|cite|improve this answer









        $endgroup$



        "More precise" isn't necessarily the right viewpoint here, especially when you're having issues with a theoretically accurate calculation rounding to $0$ or $1$. Instead, we should look for an approximate model that handles this extreme well.



        And on that note - there is such a model. The Poisson distribution arises as a limit of binomial distributions with large numbers of trials, low probability, and fixed mean. That mean becomes the distribution's sole parameter $lambda$, which will be $frac{5000}{273}approx 18.315$ in this case.



        In a Poisson distribution with parameter $lambda$, the probability of getting exactly $k$ is $frac{lambda^k}{k!}e^{-k}$. Here, that gives (using Excel for the calculations, since I happen to have it open), probabilities of 1.11145E-08 for $0$, 2.03562E-07 for $1$, and 1.86412E-06 for $2$. The probability of being at most $2$ is their sum 2.07879E-06.



        Well, OK, we don't really have all of those digits - but it'll be pretty close. We can confidently say that if the probabilities were accurate, the likelihood of getting at most two successes would be approximately $2cdot 10^{-6}$. That's definitely a "reject the hypothesis" level.



        How many consecutive head flips would it take to reach that level of unlikeliness? The probability of $n$ consecutive flips (with no extra tries) is $2^{-n}$. From the standard approximation $2^{10}approx 10^3$, we get $10^{-6}approx 2^{-20}$, and $2^{-19}approx 2cdot 10^{-6}$. It would take nineteen consecutive head flips to reach that level of improbability.



        I tried a manual version of the binomial approximation as well - it came out about $2.5%$ different, a total probability of about $2.024cdot 10^{-6}$. It's technically more accurate, but is that worth switching from a simple exponential $e^{-lambda}$ to a high power $(1-p)^N$ of something very close to $1$?







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 31 '18 at 2:28









        jmerryjmerry

        17.1k11633




        17.1k11633






























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            Ida-Boy-Ed-Garten

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            Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
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