Connecting noodles probability question











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I don't know how to solve this. You have 100 noodles in your soup bowl. Being blindfolded, you are told to take two ends of some noodles (each end of any noodle has the same probability of being chosen) in your bowl and connect them. You continue until there are no free ends. The number of loops formed by the noodles this way is stochastic. Calculate the expected number of circles.










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    I don't know how to solve this. You have 100 noodles in your soup bowl. Being blindfolded, you are told to take two ends of some noodles (each end of any noodle has the same probability of being chosen) in your bowl and connect them. You continue until there are no free ends. The number of loops formed by the noodles this way is stochastic. Calculate the expected number of circles.










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      up vote
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      I don't know how to solve this. You have 100 noodles in your soup bowl. Being blindfolded, you are told to take two ends of some noodles (each end of any noodle has the same probability of being chosen) in your bowl and connect them. You continue until there are no free ends. The number of loops formed by the noodles this way is stochastic. Calculate the expected number of circles.










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      I don't know how to solve this. You have 100 noodles in your soup bowl. Being blindfolded, you are told to take two ends of some noodles (each end of any noodle has the same probability of being chosen) in your bowl and connect them. You continue until there are no free ends. The number of loops formed by the noodles this way is stochastic. Calculate the expected number of circles.







      probability






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      asked Sep 1 '15 at 16:49









      Jojo

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      435517






















          2 Answers
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          up vote
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          accepted










          Let's do it in a more general case.
          Let the expected number of loops from $n$ noodles be $E(n)$.



          Obviously, $E(1)=1$.



          For $n>1$, if you pick up two ends, the possibility for those two ends belongs to the same noodle will be $frac{1}{2n-1}$. Then you have one loop now and there are $n-1$ noodles to keep on going. Otherwise you get no loop and still have $n-1$ noodles to go(The two you pick before are now connected as one). Hence $E(n)=E(n-1)+frac{1}{2n-1}$



          Inductively you may expect that $E(n) = 1+frac{1}{3}+frac{1}{5}+...+frac{1}{2n-1}$.






          share|cite|improve this answer





















          • So roughly spoken we have $E(75)approx pi$.
            – principal-ideal-domain
            Sep 1 '15 at 20:47










          • If you are familiar with harmonic numbers you can express $E(n)$ in such terms as $E(n)=H_{2n}-frac{1}{2}H_n$.
            – principal-ideal-domain
            Sep 2 '15 at 20:45




















          up vote
          2
          down vote













          The accepted answer is correct, but I think should be more carefully detailed.



          $E(1) = 1$



          Pick up two ends. They either belong to the Same Noodle ($S$) or they don't ($bar{S}$). The $S$ event can be used to "compute expectations by conditioning" in the same spirit of this question. So:



          $$E(n) = E(n-1 | S)p(S) + E(n-1| bar{S}) p(bar{S})$$



          Now $p(S) = frac{1}{2n-1}$ and $p(bar{S}) = 1-frac{1}{2n-1}$. Now the trick is that $E(n-1|S) = E(n-1)+1$ (i.e. if you picked the two ends from the same noodle you've added one loop to the bowl) and $E(n-1|bar{S}) = E(n-1)$ (i.e if you've picked two ends from different noodles you didn't increase the number of loops).



          $$E(n) = E(n-1 | S)p(S) + E(n-1| bar{S}) p(bar{S}) = \
          = left[E(n-1)+1right]frac{1}{2n-1} + E(n-1)left[1-frac{1}{2n-1}right] = \
          = E(n-1) + frac{1}{2n-1}$$






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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            4
            down vote



            accepted










            Let's do it in a more general case.
            Let the expected number of loops from $n$ noodles be $E(n)$.



            Obviously, $E(1)=1$.



            For $n>1$, if you pick up two ends, the possibility for those two ends belongs to the same noodle will be $frac{1}{2n-1}$. Then you have one loop now and there are $n-1$ noodles to keep on going. Otherwise you get no loop and still have $n-1$ noodles to go(The two you pick before are now connected as one). Hence $E(n)=E(n-1)+frac{1}{2n-1}$



            Inductively you may expect that $E(n) = 1+frac{1}{3}+frac{1}{5}+...+frac{1}{2n-1}$.






            share|cite|improve this answer





















            • So roughly spoken we have $E(75)approx pi$.
              – principal-ideal-domain
              Sep 1 '15 at 20:47










            • If you are familiar with harmonic numbers you can express $E(n)$ in such terms as $E(n)=H_{2n}-frac{1}{2}H_n$.
              – principal-ideal-domain
              Sep 2 '15 at 20:45

















            up vote
            4
            down vote



            accepted










            Let's do it in a more general case.
            Let the expected number of loops from $n$ noodles be $E(n)$.



            Obviously, $E(1)=1$.



            For $n>1$, if you pick up two ends, the possibility for those two ends belongs to the same noodle will be $frac{1}{2n-1}$. Then you have one loop now and there are $n-1$ noodles to keep on going. Otherwise you get no loop and still have $n-1$ noodles to go(The two you pick before are now connected as one). Hence $E(n)=E(n-1)+frac{1}{2n-1}$



            Inductively you may expect that $E(n) = 1+frac{1}{3}+frac{1}{5}+...+frac{1}{2n-1}$.






            share|cite|improve this answer





















            • So roughly spoken we have $E(75)approx pi$.
              – principal-ideal-domain
              Sep 1 '15 at 20:47










            • If you are familiar with harmonic numbers you can express $E(n)$ in such terms as $E(n)=H_{2n}-frac{1}{2}H_n$.
              – principal-ideal-domain
              Sep 2 '15 at 20:45















            up vote
            4
            down vote



            accepted







            up vote
            4
            down vote



            accepted






            Let's do it in a more general case.
            Let the expected number of loops from $n$ noodles be $E(n)$.



            Obviously, $E(1)=1$.



            For $n>1$, if you pick up two ends, the possibility for those two ends belongs to the same noodle will be $frac{1}{2n-1}$. Then you have one loop now and there are $n-1$ noodles to keep on going. Otherwise you get no loop and still have $n-1$ noodles to go(The two you pick before are now connected as one). Hence $E(n)=E(n-1)+frac{1}{2n-1}$



            Inductively you may expect that $E(n) = 1+frac{1}{3}+frac{1}{5}+...+frac{1}{2n-1}$.






            share|cite|improve this answer












            Let's do it in a more general case.
            Let the expected number of loops from $n$ noodles be $E(n)$.



            Obviously, $E(1)=1$.



            For $n>1$, if you pick up two ends, the possibility for those two ends belongs to the same noodle will be $frac{1}{2n-1}$. Then you have one loop now and there are $n-1$ noodles to keep on going. Otherwise you get no loop and still have $n-1$ noodles to go(The two you pick before are now connected as one). Hence $E(n)=E(n-1)+frac{1}{2n-1}$



            Inductively you may expect that $E(n) = 1+frac{1}{3}+frac{1}{5}+...+frac{1}{2n-1}$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Sep 1 '15 at 17:30









            Asydot

            628410




            628410












            • So roughly spoken we have $E(75)approx pi$.
              – principal-ideal-domain
              Sep 1 '15 at 20:47










            • If you are familiar with harmonic numbers you can express $E(n)$ in such terms as $E(n)=H_{2n}-frac{1}{2}H_n$.
              – principal-ideal-domain
              Sep 2 '15 at 20:45




















            • So roughly spoken we have $E(75)approx pi$.
              – principal-ideal-domain
              Sep 1 '15 at 20:47










            • If you are familiar with harmonic numbers you can express $E(n)$ in such terms as $E(n)=H_{2n}-frac{1}{2}H_n$.
              – principal-ideal-domain
              Sep 2 '15 at 20:45


















            So roughly spoken we have $E(75)approx pi$.
            – principal-ideal-domain
            Sep 1 '15 at 20:47




            So roughly spoken we have $E(75)approx pi$.
            – principal-ideal-domain
            Sep 1 '15 at 20:47












            If you are familiar with harmonic numbers you can express $E(n)$ in such terms as $E(n)=H_{2n}-frac{1}{2}H_n$.
            – principal-ideal-domain
            Sep 2 '15 at 20:45






            If you are familiar with harmonic numbers you can express $E(n)$ in such terms as $E(n)=H_{2n}-frac{1}{2}H_n$.
            – principal-ideal-domain
            Sep 2 '15 at 20:45












            up vote
            2
            down vote













            The accepted answer is correct, but I think should be more carefully detailed.



            $E(1) = 1$



            Pick up two ends. They either belong to the Same Noodle ($S$) or they don't ($bar{S}$). The $S$ event can be used to "compute expectations by conditioning" in the same spirit of this question. So:



            $$E(n) = E(n-1 | S)p(S) + E(n-1| bar{S}) p(bar{S})$$



            Now $p(S) = frac{1}{2n-1}$ and $p(bar{S}) = 1-frac{1}{2n-1}$. Now the trick is that $E(n-1|S) = E(n-1)+1$ (i.e. if you picked the two ends from the same noodle you've added one loop to the bowl) and $E(n-1|bar{S}) = E(n-1)$ (i.e if you've picked two ends from different noodles you didn't increase the number of loops).



            $$E(n) = E(n-1 | S)p(S) + E(n-1| bar{S}) p(bar{S}) = \
            = left[E(n-1)+1right]frac{1}{2n-1} + E(n-1)left[1-frac{1}{2n-1}right] = \
            = E(n-1) + frac{1}{2n-1}$$






            share|cite|improve this answer



























              up vote
              2
              down vote













              The accepted answer is correct, but I think should be more carefully detailed.



              $E(1) = 1$



              Pick up two ends. They either belong to the Same Noodle ($S$) or they don't ($bar{S}$). The $S$ event can be used to "compute expectations by conditioning" in the same spirit of this question. So:



              $$E(n) = E(n-1 | S)p(S) + E(n-1| bar{S}) p(bar{S})$$



              Now $p(S) = frac{1}{2n-1}$ and $p(bar{S}) = 1-frac{1}{2n-1}$. Now the trick is that $E(n-1|S) = E(n-1)+1$ (i.e. if you picked the two ends from the same noodle you've added one loop to the bowl) and $E(n-1|bar{S}) = E(n-1)$ (i.e if you've picked two ends from different noodles you didn't increase the number of loops).



              $$E(n) = E(n-1 | S)p(S) + E(n-1| bar{S}) p(bar{S}) = \
              = left[E(n-1)+1right]frac{1}{2n-1} + E(n-1)left[1-frac{1}{2n-1}right] = \
              = E(n-1) + frac{1}{2n-1}$$






              share|cite|improve this answer

























                up vote
                2
                down vote










                up vote
                2
                down vote









                The accepted answer is correct, but I think should be more carefully detailed.



                $E(1) = 1$



                Pick up two ends. They either belong to the Same Noodle ($S$) or they don't ($bar{S}$). The $S$ event can be used to "compute expectations by conditioning" in the same spirit of this question. So:



                $$E(n) = E(n-1 | S)p(S) + E(n-1| bar{S}) p(bar{S})$$



                Now $p(S) = frac{1}{2n-1}$ and $p(bar{S}) = 1-frac{1}{2n-1}$. Now the trick is that $E(n-1|S) = E(n-1)+1$ (i.e. if you picked the two ends from the same noodle you've added one loop to the bowl) and $E(n-1|bar{S}) = E(n-1)$ (i.e if you've picked two ends from different noodles you didn't increase the number of loops).



                $$E(n) = E(n-1 | S)p(S) + E(n-1| bar{S}) p(bar{S}) = \
                = left[E(n-1)+1right]frac{1}{2n-1} + E(n-1)left[1-frac{1}{2n-1}right] = \
                = E(n-1) + frac{1}{2n-1}$$






                share|cite|improve this answer














                The accepted answer is correct, but I think should be more carefully detailed.



                $E(1) = 1$



                Pick up two ends. They either belong to the Same Noodle ($S$) or they don't ($bar{S}$). The $S$ event can be used to "compute expectations by conditioning" in the same spirit of this question. So:



                $$E(n) = E(n-1 | S)p(S) + E(n-1| bar{S}) p(bar{S})$$



                Now $p(S) = frac{1}{2n-1}$ and $p(bar{S}) = 1-frac{1}{2n-1}$. Now the trick is that $E(n-1|S) = E(n-1)+1$ (i.e. if you picked the two ends from the same noodle you've added one loop to the bowl) and $E(n-1|bar{S}) = E(n-1)$ (i.e if you've picked two ends from different noodles you didn't increase the number of loops).



                $$E(n) = E(n-1 | S)p(S) + E(n-1| bar{S}) p(bar{S}) = \
                = left[E(n-1)+1right]frac{1}{2n-1} + E(n-1)left[1-frac{1}{2n-1}right] = \
                = E(n-1) + frac{1}{2n-1}$$







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                share|cite|improve this answer



                share|cite|improve this answer








                edited Oct 21 at 7:29









                miracle173

                7,31222247




                7,31222247










                answered Apr 15 at 15:14









                oneloop

                1213




                1213






























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