Calculate the probability with a finite arithmetic progression












1












$begingroup$


We have a finite arithmetic progression $ a_n $, where $ n geq 3 $ and its $rneq 0 $.
We draw three different numbers. We have to calculate the probability, that
these three numbers in the order of drawing will create another arithmetic progression.



My proposition :



$ Omega={n!over (n-3)!}$



$ mathbf{A}= {n choose 3} cdot 2$



But I think that my way of thinking about way of counting $ mathbf{A}$ is incorrect.



Any suggestions how can I count it? Thanks in advice, Kuba!






probability problem-solving arithmetic-progressions







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Is the progression finite?
    $endgroup$
    – EuxhenH
    Jan 7 at 18:05










  • $begingroup$
    I totally forgot about it :/. Yes, the progression is finite.
    $endgroup$
    – Kukoz
    Jan 7 at 18:15
















1












$begingroup$


We have a finite arithmetic progression $ a_n $, where $ n geq 3 $ and its $rneq 0 $.
We draw three different numbers. We have to calculate the probability, that
these three numbers in the order of drawing will create another arithmetic progression.



My proposition :



$ Omega={n!over (n-3)!}$



$ mathbf{A}= {n choose 3} cdot 2$



But I think that my way of thinking about way of counting $ mathbf{A}$ is incorrect.



Any suggestions how can I count it? Thanks in advice, Kuba!






probability problem-solving arithmetic-progressions







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Is the progression finite?
    $endgroup$
    – EuxhenH
    Jan 7 at 18:05










  • $begingroup$
    I totally forgot about it :/. Yes, the progression is finite.
    $endgroup$
    – Kukoz
    Jan 7 at 18:15














1












1








1





$begingroup$


We have a finite arithmetic progression $ a_n $, where $ n geq 3 $ and its $rneq 0 $.
We draw three different numbers. We have to calculate the probability, that
these three numbers in the order of drawing will create another arithmetic progression.



My proposition :



$ Omega={n!over (n-3)!}$



$ mathbf{A}= {n choose 3} cdot 2$



But I think that my way of thinking about way of counting $ mathbf{A}$ is incorrect.



Any suggestions how can I count it? Thanks in advice, Kuba!






probability problem-solving arithmetic-progressions







share|cite|improve this question











$endgroup$




We have a finite arithmetic progression $ a_n $, where $ n geq 3 $ and its $rneq 0 $.
We draw three different numbers. We have to calculate the probability, that
these three numbers in the order of drawing will create another arithmetic progression.



My proposition :



$ Omega={n!over (n-3)!}$



$ mathbf{A}= {n choose 3} cdot 2$



But I think that my way of thinking about way of counting $ mathbf{A}$ is incorrect.



Any suggestions how can I count it? Thanks in advice, Kuba!






probability problem-solving arithmetic-progressions




probability problem-solving arithmetic-progressions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 7 at 19:14







Kukoz

















asked Jan 7 at 17:28









KukozKukoz

409




409








  • 2




    $begingroup$
    Is the progression finite?
    $endgroup$
    – EuxhenH
    Jan 7 at 18:05










  • $begingroup$
    I totally forgot about it :/. Yes, the progression is finite.
    $endgroup$
    – Kukoz
    Jan 7 at 18:15














  • 2




    $begingroup$
    Is the progression finite?
    $endgroup$
    – EuxhenH
    Jan 7 at 18:05










  • $begingroup$
    I totally forgot about it :/. Yes, the progression is finite.
    $endgroup$
    – Kukoz
    Jan 7 at 18:15








2




2




$begingroup$
Is the progression finite?
$endgroup$
– EuxhenH
Jan 7 at 18:05




$begingroup$
Is the progression finite?
$endgroup$
– EuxhenH
Jan 7 at 18:05












$begingroup$
I totally forgot about it :/. Yes, the progression is finite.
$endgroup$
– Kukoz
Jan 7 at 18:15




$begingroup$
I totally forgot about it :/. Yes, the progression is finite.
$endgroup$
– Kukoz
Jan 7 at 18:15










2 Answers
2






active

oldest

votes


















1












$begingroup$

$Omega = n(n-1)(n-2)$



We're interested in triples whose elements differ by $r$, $2r$ ... up to $lceil n/3rceil r$

Notice, that if a triple $a_i,a_k,a_l$ is OK, then so is $a_l,a_k,a_i$.

My idea was to "stick together" those desired elements that form each triple and calculate their quantity. To be precise: there are $2(n-2)$ triples whose elements differ by r, $2(n-4)$ triples whose elements differ by $2r$ .... and $2(n-2lceil n/3rceil)$ triples whose elements differ by $lceil n/3rceil r$. Then:
$P(A) = frac{2(n-2+n-4...+n-2 lceil n/3rceil)}{n(n-1)(n-2)}$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Not exactly we have different possibilities For example when the rest of $ {n 3} $ is $0$, your idea is good, -when the rest of $ {n 3} $ is $ 1 $ $mathbf{A}= 2[n-2+n-4...+n-2 lceil n/3rceil]+1$, when the rest of $ {n 3} $ is $ 2 $ $mathbf{A}= 2[n-2+n-4...+n-2 lceil n/3rceil]+2$, so this is completely good answer :) Cheers
    $endgroup$
    – Kukoz
    Jan 10 at 14:41








  • 1




    $begingroup$
    No. The ceiling function isn't the floor one. en.wikipedia.org/wiki/Floor_and_ceiling_functions <- look at the plot of the "ceiling function" on the right
    $endgroup$
    – The Cat
    Jan 10 at 16:28






  • 1




    $begingroup$
    $lceil 0.5rceil $=1
    $endgroup$
    – The Cat
    Jan 10 at 16:29








  • 1




    $begingroup$
    I think you just mistook those two together.
    $endgroup$
    – The Cat
    Jan 10 at 16:30






  • 1




    $begingroup$
    Yeah, now everything is good! Big thanks for you, and shoutout!
    $endgroup$
    – Kukoz
    Jan 10 at 16:33



















0












$begingroup$

Let the sequence be $a_1,a_2,a_3,...a_n$. Then $|Omega | = {nchoose 3} = {n(n-1)(n-2)over 6}$



and each good 3-sequence is uniquely defined by it's middle term and the difference $d$. So if middle term is $a_m$ then the difference can be $d,2d,...,(m-1)d$.



So if we have $n=2k$, then we have $$1+2+...+(k-1)+(k-1)+...+2+1 = k(k-1)$$
good 3-sequences. So the probability is $$P = {6k(k-1)over 4k(2k-1)(k-1)} = {3over 2(n-1)}$$



If $n=2k+1$ then we have $$1+2+...+(k-1)+k+(k-1)+...+2+1 = k(k-1)+k=k^2$$
good 3-sequences. So the probability is $$P = {6k^2over 2(2k+1)k(2k-1)} = {3(n-1)over 2n(n-2)}$$






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    $Omega = n(n-1)(n-2)$



    We're interested in triples whose elements differ by $r$, $2r$ ... up to $lceil n/3rceil r$

    Notice, that if a triple $a_i,a_k,a_l$ is OK, then so is $a_l,a_k,a_i$.

    My idea was to "stick together" those desired elements that form each triple and calculate their quantity. To be precise: there are $2(n-2)$ triples whose elements differ by r, $2(n-4)$ triples whose elements differ by $2r$ .... and $2(n-2lceil n/3rceil)$ triples whose elements differ by $lceil n/3rceil r$. Then:
    $P(A) = frac{2(n-2+n-4...+n-2 lceil n/3rceil)}{n(n-1)(n-2)}$






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Not exactly we have different possibilities For example when the rest of $ {n 3} $ is $0$, your idea is good, -when the rest of $ {n 3} $ is $ 1 $ $mathbf{A}= 2[n-2+n-4...+n-2 lceil n/3rceil]+1$, when the rest of $ {n 3} $ is $ 2 $ $mathbf{A}= 2[n-2+n-4...+n-2 lceil n/3rceil]+2$, so this is completely good answer :) Cheers
      $endgroup$
      – Kukoz
      Jan 10 at 14:41








    • 1




      $begingroup$
      No. The ceiling function isn't the floor one. en.wikipedia.org/wiki/Floor_and_ceiling_functions <- look at the plot of the "ceiling function" on the right
      $endgroup$
      – The Cat
      Jan 10 at 16:28






    • 1




      $begingroup$
      $lceil 0.5rceil $=1
      $endgroup$
      – The Cat
      Jan 10 at 16:29








    • 1




      $begingroup$
      I think you just mistook those two together.
      $endgroup$
      – The Cat
      Jan 10 at 16:30






    • 1




      $begingroup$
      Yeah, now everything is good! Big thanks for you, and shoutout!
      $endgroup$
      – Kukoz
      Jan 10 at 16:33
















    1












    $begingroup$

    $Omega = n(n-1)(n-2)$



    We're interested in triples whose elements differ by $r$, $2r$ ... up to $lceil n/3rceil r$

    Notice, that if a triple $a_i,a_k,a_l$ is OK, then so is $a_l,a_k,a_i$.

    My idea was to "stick together" those desired elements that form each triple and calculate their quantity. To be precise: there are $2(n-2)$ triples whose elements differ by r, $2(n-4)$ triples whose elements differ by $2r$ .... and $2(n-2lceil n/3rceil)$ triples whose elements differ by $lceil n/3rceil r$. Then:
    $P(A) = frac{2(n-2+n-4...+n-2 lceil n/3rceil)}{n(n-1)(n-2)}$






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Not exactly we have different possibilities For example when the rest of $ {n 3} $ is $0$, your idea is good, -when the rest of $ {n 3} $ is $ 1 $ $mathbf{A}= 2[n-2+n-4...+n-2 lceil n/3rceil]+1$, when the rest of $ {n 3} $ is $ 2 $ $mathbf{A}= 2[n-2+n-4...+n-2 lceil n/3rceil]+2$, so this is completely good answer :) Cheers
      $endgroup$
      – Kukoz
      Jan 10 at 14:41








    • 1




      $begingroup$
      No. The ceiling function isn't the floor one. en.wikipedia.org/wiki/Floor_and_ceiling_functions <- look at the plot of the "ceiling function" on the right
      $endgroup$
      – The Cat
      Jan 10 at 16:28






    • 1




      $begingroup$
      $lceil 0.5rceil $=1
      $endgroup$
      – The Cat
      Jan 10 at 16:29








    • 1




      $begingroup$
      I think you just mistook those two together.
      $endgroup$
      – The Cat
      Jan 10 at 16:30






    • 1




      $begingroup$
      Yeah, now everything is good! Big thanks for you, and shoutout!
      $endgroup$
      – Kukoz
      Jan 10 at 16:33














    1












    1








    1





    $begingroup$

    $Omega = n(n-1)(n-2)$



    We're interested in triples whose elements differ by $r$, $2r$ ... up to $lceil n/3rceil r$

    Notice, that if a triple $a_i,a_k,a_l$ is OK, then so is $a_l,a_k,a_i$.

    My idea was to "stick together" those desired elements that form each triple and calculate their quantity. To be precise: there are $2(n-2)$ triples whose elements differ by r, $2(n-4)$ triples whose elements differ by $2r$ .... and $2(n-2lceil n/3rceil)$ triples whose elements differ by $lceil n/3rceil r$. Then:
    $P(A) = frac{2(n-2+n-4...+n-2 lceil n/3rceil)}{n(n-1)(n-2)}$






    share|cite|improve this answer











    $endgroup$



    $Omega = n(n-1)(n-2)$



    We're interested in triples whose elements differ by $r$, $2r$ ... up to $lceil n/3rceil r$

    Notice, that if a triple $a_i,a_k,a_l$ is OK, then so is $a_l,a_k,a_i$.

    My idea was to "stick together" those desired elements that form each triple and calculate their quantity. To be precise: there are $2(n-2)$ triples whose elements differ by r, $2(n-4)$ triples whose elements differ by $2r$ .... and $2(n-2lceil n/3rceil)$ triples whose elements differ by $lceil n/3rceil r$. Then:
    $P(A) = frac{2(n-2+n-4...+n-2 lceil n/3rceil)}{n(n-1)(n-2)}$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 10 at 18:33

























    answered Jan 7 at 19:35









    The CatThe Cat

    25112




    25112








    • 1




      $begingroup$
      Not exactly we have different possibilities For example when the rest of $ {n 3} $ is $0$, your idea is good, -when the rest of $ {n 3} $ is $ 1 $ $mathbf{A}= 2[n-2+n-4...+n-2 lceil n/3rceil]+1$, when the rest of $ {n 3} $ is $ 2 $ $mathbf{A}= 2[n-2+n-4...+n-2 lceil n/3rceil]+2$, so this is completely good answer :) Cheers
      $endgroup$
      – Kukoz
      Jan 10 at 14:41








    • 1




      $begingroup$
      No. The ceiling function isn't the floor one. en.wikipedia.org/wiki/Floor_and_ceiling_functions <- look at the plot of the "ceiling function" on the right
      $endgroup$
      – The Cat
      Jan 10 at 16:28






    • 1




      $begingroup$
      $lceil 0.5rceil $=1
      $endgroup$
      – The Cat
      Jan 10 at 16:29








    • 1




      $begingroup$
      I think you just mistook those two together.
      $endgroup$
      – The Cat
      Jan 10 at 16:30






    • 1




      $begingroup$
      Yeah, now everything is good! Big thanks for you, and shoutout!
      $endgroup$
      – Kukoz
      Jan 10 at 16:33














    • 1




      $begingroup$
      Not exactly we have different possibilities For example when the rest of $ {n 3} $ is $0$, your idea is good, -when the rest of $ {n 3} $ is $ 1 $ $mathbf{A}= 2[n-2+n-4...+n-2 lceil n/3rceil]+1$, when the rest of $ {n 3} $ is $ 2 $ $mathbf{A}= 2[n-2+n-4...+n-2 lceil n/3rceil]+2$, so this is completely good answer :) Cheers
      $endgroup$
      – Kukoz
      Jan 10 at 14:41








    • 1




      $begingroup$
      No. The ceiling function isn't the floor one. en.wikipedia.org/wiki/Floor_and_ceiling_functions <- look at the plot of the "ceiling function" on the right
      $endgroup$
      – The Cat
      Jan 10 at 16:28






    • 1




      $begingroup$
      $lceil 0.5rceil $=1
      $endgroup$
      – The Cat
      Jan 10 at 16:29








    • 1




      $begingroup$
      I think you just mistook those two together.
      $endgroup$
      – The Cat
      Jan 10 at 16:30






    • 1




      $begingroup$
      Yeah, now everything is good! Big thanks for you, and shoutout!
      $endgroup$
      – Kukoz
      Jan 10 at 16:33








    1




    1




    $begingroup$
    Not exactly we have different possibilities For example when the rest of $ {n 3} $ is $0$, your idea is good, -when the rest of $ {n 3} $ is $ 1 $ $mathbf{A}= 2[n-2+n-4...+n-2 lceil n/3rceil]+1$, when the rest of $ {n 3} $ is $ 2 $ $mathbf{A}= 2[n-2+n-4...+n-2 lceil n/3rceil]+2$, so this is completely good answer :) Cheers
    $endgroup$
    – Kukoz
    Jan 10 at 14:41






    $begingroup$
    Not exactly we have different possibilities For example when the rest of $ {n 3} $ is $0$, your idea is good, -when the rest of $ {n 3} $ is $ 1 $ $mathbf{A}= 2[n-2+n-4...+n-2 lceil n/3rceil]+1$, when the rest of $ {n 3} $ is $ 2 $ $mathbf{A}= 2[n-2+n-4...+n-2 lceil n/3rceil]+2$, so this is completely good answer :) Cheers
    $endgroup$
    – Kukoz
    Jan 10 at 14:41






    1




    1




    $begingroup$
    No. The ceiling function isn't the floor one. en.wikipedia.org/wiki/Floor_and_ceiling_functions <- look at the plot of the "ceiling function" on the right
    $endgroup$
    – The Cat
    Jan 10 at 16:28




    $begingroup$
    No. The ceiling function isn't the floor one. en.wikipedia.org/wiki/Floor_and_ceiling_functions <- look at the plot of the "ceiling function" on the right
    $endgroup$
    – The Cat
    Jan 10 at 16:28




    1




    1




    $begingroup$
    $lceil 0.5rceil $=1
    $endgroup$
    – The Cat
    Jan 10 at 16:29






    $begingroup$
    $lceil 0.5rceil $=1
    $endgroup$
    – The Cat
    Jan 10 at 16:29






    1




    1




    $begingroup$
    I think you just mistook those two together.
    $endgroup$
    – The Cat
    Jan 10 at 16:30




    $begingroup$
    I think you just mistook those two together.
    $endgroup$
    – The Cat
    Jan 10 at 16:30




    1




    1




    $begingroup$
    Yeah, now everything is good! Big thanks for you, and shoutout!
    $endgroup$
    – Kukoz
    Jan 10 at 16:33




    $begingroup$
    Yeah, now everything is good! Big thanks for you, and shoutout!
    $endgroup$
    – Kukoz
    Jan 10 at 16:33











    0












    $begingroup$

    Let the sequence be $a_1,a_2,a_3,...a_n$. Then $|Omega | = {nchoose 3} = {n(n-1)(n-2)over 6}$



    and each good 3-sequence is uniquely defined by it's middle term and the difference $d$. So if middle term is $a_m$ then the difference can be $d,2d,...,(m-1)d$.



    So if we have $n=2k$, then we have $$1+2+...+(k-1)+(k-1)+...+2+1 = k(k-1)$$
    good 3-sequences. So the probability is $$P = {6k(k-1)over 4k(2k-1)(k-1)} = {3over 2(n-1)}$$



    If $n=2k+1$ then we have $$1+2+...+(k-1)+k+(k-1)+...+2+1 = k(k-1)+k=k^2$$
    good 3-sequences. So the probability is $$P = {6k^2over 2(2k+1)k(2k-1)} = {3(n-1)over 2n(n-2)}$$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Let the sequence be $a_1,a_2,a_3,...a_n$. Then $|Omega | = {nchoose 3} = {n(n-1)(n-2)over 6}$



      and each good 3-sequence is uniquely defined by it's middle term and the difference $d$. So if middle term is $a_m$ then the difference can be $d,2d,...,(m-1)d$.



      So if we have $n=2k$, then we have $$1+2+...+(k-1)+(k-1)+...+2+1 = k(k-1)$$
      good 3-sequences. So the probability is $$P = {6k(k-1)over 4k(2k-1)(k-1)} = {3over 2(n-1)}$$



      If $n=2k+1$ then we have $$1+2+...+(k-1)+k+(k-1)+...+2+1 = k(k-1)+k=k^2$$
      good 3-sequences. So the probability is $$P = {6k^2over 2(2k+1)k(2k-1)} = {3(n-1)over 2n(n-2)}$$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Let the sequence be $a_1,a_2,a_3,...a_n$. Then $|Omega | = {nchoose 3} = {n(n-1)(n-2)over 6}$



        and each good 3-sequence is uniquely defined by it's middle term and the difference $d$. So if middle term is $a_m$ then the difference can be $d,2d,...,(m-1)d$.



        So if we have $n=2k$, then we have $$1+2+...+(k-1)+(k-1)+...+2+1 = k(k-1)$$
        good 3-sequences. So the probability is $$P = {6k(k-1)over 4k(2k-1)(k-1)} = {3over 2(n-1)}$$



        If $n=2k+1$ then we have $$1+2+...+(k-1)+k+(k-1)+...+2+1 = k(k-1)+k=k^2$$
        good 3-sequences. So the probability is $$P = {6k^2over 2(2k+1)k(2k-1)} = {3(n-1)over 2n(n-2)}$$






        share|cite|improve this answer









        $endgroup$



        Let the sequence be $a_1,a_2,a_3,...a_n$. Then $|Omega | = {nchoose 3} = {n(n-1)(n-2)over 6}$



        and each good 3-sequence is uniquely defined by it's middle term and the difference $d$. So if middle term is $a_m$ then the difference can be $d,2d,...,(m-1)d$.



        So if we have $n=2k$, then we have $$1+2+...+(k-1)+(k-1)+...+2+1 = k(k-1)$$
        good 3-sequences. So the probability is $$P = {6k(k-1)over 4k(2k-1)(k-1)} = {3over 2(n-1)}$$



        If $n=2k+1$ then we have $$1+2+...+(k-1)+k+(k-1)+...+2+1 = k(k-1)+k=k^2$$
        good 3-sequences. So the probability is $$P = {6k^2over 2(2k+1)k(2k-1)} = {3(n-1)over 2n(n-2)}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 10 at 17:36









        greedoidgreedoid

        42.9k1153105




        42.9k1153105






























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