Transition matrix generated by throwing 5 dice












0












$begingroup$



You throw five dice and set aside those dice that show sixes. Throw the
remaining dice and then again set aside the sixes. Continue until you
get all sixes. Exhibit the transition matrix for the associated Markov
chain, where $X_n$ is the number of sixes after $n$ throws.




I figured that the matrix should have $6$ states, that is state space $S={0,1,2,3,4,5}.$ Just by thinking about the first entry, $P_{00}$ we can think "starting with $0$ sixes, what is the probability that on the first throw, we throw $0$ sixes?". To do that, we simply need to throw each die with probability $5/6$, that is $P_{00}=(5/6)^5approx 0.4018$, which is correct.



Same reasoning for $P_{01}$ gives $P_{01}=(1/6)^1cdot (5/6)^4approx 0.08037.$ This is incorrect, for the correct answer here one should apparently multiply the answer with 5. Why? Why did we not need to do that in $P_{00}?$



The general matrix element is given by



$$P_{ij}={5-ichoose j-i}left(frac{1}{6}right)^{j-i}left(frac{5}{6}right)^{5-j}, quad forall quad 0leq i,jleq 5. tag{1}$$



I'm trying to build this formula step by step, intuitivly, but I fail to understand the binomial. Can someone give me an explanation on how to construct $(1)?$






probability probability-theory







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$endgroup$








  • 1




    $begingroup$
    The $P_{01}$ that you’ve computed is the probability that a specific die shows a six. That one die could be any one of the five dice, though, so you’ve undercounted.
    $endgroup$
    – amd
    Jan 3 at 21:35


















0












$begingroup$



You throw five dice and set aside those dice that show sixes. Throw the
remaining dice and then again set aside the sixes. Continue until you
get all sixes. Exhibit the transition matrix for the associated Markov
chain, where $X_n$ is the number of sixes after $n$ throws.




I figured that the matrix should have $6$ states, that is state space $S={0,1,2,3,4,5}.$ Just by thinking about the first entry, $P_{00}$ we can think "starting with $0$ sixes, what is the probability that on the first throw, we throw $0$ sixes?". To do that, we simply need to throw each die with probability $5/6$, that is $P_{00}=(5/6)^5approx 0.4018$, which is correct.



Same reasoning for $P_{01}$ gives $P_{01}=(1/6)^1cdot (5/6)^4approx 0.08037.$ This is incorrect, for the correct answer here one should apparently multiply the answer with 5. Why? Why did we not need to do that in $P_{00}?$



The general matrix element is given by



$$P_{ij}={5-ichoose j-i}left(frac{1}{6}right)^{j-i}left(frac{5}{6}right)^{5-j}, quad forall quad 0leq i,jleq 5. tag{1}$$



I'm trying to build this formula step by step, intuitivly, but I fail to understand the binomial. Can someone give me an explanation on how to construct $(1)?$






probability probability-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    The $P_{01}$ that you’ve computed is the probability that a specific die shows a six. That one die could be any one of the five dice, though, so you’ve undercounted.
    $endgroup$
    – amd
    Jan 3 at 21:35
















0












0








0





$begingroup$



You throw five dice and set aside those dice that show sixes. Throw the
remaining dice and then again set aside the sixes. Continue until you
get all sixes. Exhibit the transition matrix for the associated Markov
chain, where $X_n$ is the number of sixes after $n$ throws.




I figured that the matrix should have $6$ states, that is state space $S={0,1,2,3,4,5}.$ Just by thinking about the first entry, $P_{00}$ we can think "starting with $0$ sixes, what is the probability that on the first throw, we throw $0$ sixes?". To do that, we simply need to throw each die with probability $5/6$, that is $P_{00}=(5/6)^5approx 0.4018$, which is correct.



Same reasoning for $P_{01}$ gives $P_{01}=(1/6)^1cdot (5/6)^4approx 0.08037.$ This is incorrect, for the correct answer here one should apparently multiply the answer with 5. Why? Why did we not need to do that in $P_{00}?$



The general matrix element is given by



$$P_{ij}={5-ichoose j-i}left(frac{1}{6}right)^{j-i}left(frac{5}{6}right)^{5-j}, quad forall quad 0leq i,jleq 5. tag{1}$$



I'm trying to build this formula step by step, intuitivly, but I fail to understand the binomial. Can someone give me an explanation on how to construct $(1)?$






probability probability-theory







share|cite|improve this question











$endgroup$





You throw five dice and set aside those dice that show sixes. Throw the
remaining dice and then again set aside the sixes. Continue until you
get all sixes. Exhibit the transition matrix for the associated Markov
chain, where $X_n$ is the number of sixes after $n$ throws.




I figured that the matrix should have $6$ states, that is state space $S={0,1,2,3,4,5}.$ Just by thinking about the first entry, $P_{00}$ we can think "starting with $0$ sixes, what is the probability that on the first throw, we throw $0$ sixes?". To do that, we simply need to throw each die with probability $5/6$, that is $P_{00}=(5/6)^5approx 0.4018$, which is correct.



Same reasoning for $P_{01}$ gives $P_{01}=(1/6)^1cdot (5/6)^4approx 0.08037.$ This is incorrect, for the correct answer here one should apparently multiply the answer with 5. Why? Why did we not need to do that in $P_{00}?$



The general matrix element is given by



$$P_{ij}={5-ichoose j-i}left(frac{1}{6}right)^{j-i}left(frac{5}{6}right)^{5-j}, quad forall quad 0leq i,jleq 5. tag{1}$$



I'm trying to build this formula step by step, intuitivly, but I fail to understand the binomial. Can someone give me an explanation on how to construct $(1)?$






probability probability-theory




probability probability-theory






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share|cite|improve this question













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edited Jan 3 at 17:47







Parseval

















asked Jan 3 at 15:01









ParsevalParseval

2,9571719




2,9571719








  • 1




    $begingroup$
    The $P_{01}$ that you’ve computed is the probability that a specific die shows a six. That one die could be any one of the five dice, though, so you’ve undercounted.
    $endgroup$
    – amd
    Jan 3 at 21:35
















  • 1




    $begingroup$
    The $P_{01}$ that you’ve computed is the probability that a specific die shows a six. That one die could be any one of the five dice, though, so you’ve undercounted.
    $endgroup$
    – amd
    Jan 3 at 21:35










1




1




$begingroup$
The $P_{01}$ that you’ve computed is the probability that a specific die shows a six. That one die could be any one of the five dice, though, so you’ve undercounted.
$endgroup$
– amd
Jan 3 at 21:35






$begingroup$
The $P_{01}$ that you’ve computed is the probability that a specific die shows a six. That one die could be any one of the five dice, though, so you’ve undercounted.
$endgroup$
– amd
Jan 3 at 21:35












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$begingroup$

Look at the case for just two dice: then you have a transition matrix where $X_n in {0, 1, 2}$. What is the dimension of this matrix? Then if $X_n = 0$, what is the chance that $X_{n+1} = 0$? What kind of outcomes correspond to such an event? Similarly, what is the chance that $X_{n+1} = 2$ if $X_n = 0$? Finally, based on these previous results, what is the chance that $X_{n+1} = 1$ if $X_n = 0$?



Then do the same for $X_n = 1$: Is it possible that $X_{n+1} = 0$ if $X_n = 1$?



Finally, consider $X_n = 2$. Is this an absorbing state?



If you answer these correctly, then think about how this generalizes to more than two dice.






share|cite|improve this answer









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    $begingroup$

    Look at the case for just two dice: then you have a transition matrix where $X_n in {0, 1, 2}$. What is the dimension of this matrix? Then if $X_n = 0$, what is the chance that $X_{n+1} = 0$? What kind of outcomes correspond to such an event? Similarly, what is the chance that $X_{n+1} = 2$ if $X_n = 0$? Finally, based on these previous results, what is the chance that $X_{n+1} = 1$ if $X_n = 0$?



    Then do the same for $X_n = 1$: Is it possible that $X_{n+1} = 0$ if $X_n = 1$?



    Finally, consider $X_n = 2$. Is this an absorbing state?



    If you answer these correctly, then think about how this generalizes to more than two dice.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Look at the case for just two dice: then you have a transition matrix where $X_n in {0, 1, 2}$. What is the dimension of this matrix? Then if $X_n = 0$, what is the chance that $X_{n+1} = 0$? What kind of outcomes correspond to such an event? Similarly, what is the chance that $X_{n+1} = 2$ if $X_n = 0$? Finally, based on these previous results, what is the chance that $X_{n+1} = 1$ if $X_n = 0$?



      Then do the same for $X_n = 1$: Is it possible that $X_{n+1} = 0$ if $X_n = 1$?



      Finally, consider $X_n = 2$. Is this an absorbing state?



      If you answer these correctly, then think about how this generalizes to more than two dice.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Look at the case for just two dice: then you have a transition matrix where $X_n in {0, 1, 2}$. What is the dimension of this matrix? Then if $X_n = 0$, what is the chance that $X_{n+1} = 0$? What kind of outcomes correspond to such an event? Similarly, what is the chance that $X_{n+1} = 2$ if $X_n = 0$? Finally, based on these previous results, what is the chance that $X_{n+1} = 1$ if $X_n = 0$?



        Then do the same for $X_n = 1$: Is it possible that $X_{n+1} = 0$ if $X_n = 1$?



        Finally, consider $X_n = 2$. Is this an absorbing state?



        If you answer these correctly, then think about how this generalizes to more than two dice.






        share|cite|improve this answer









        $endgroup$



        Look at the case for just two dice: then you have a transition matrix where $X_n in {0, 1, 2}$. What is the dimension of this matrix? Then if $X_n = 0$, what is the chance that $X_{n+1} = 0$? What kind of outcomes correspond to such an event? Similarly, what is the chance that $X_{n+1} = 2$ if $X_n = 0$? Finally, based on these previous results, what is the chance that $X_{n+1} = 1$ if $X_n = 0$?



        Then do the same for $X_n = 1$: Is it possible that $X_{n+1} = 0$ if $X_n = 1$?



        Finally, consider $X_n = 2$. Is this an absorbing state?



        If you answer these correctly, then think about how this generalizes to more than two dice.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 4 at 1:05









        heropupheropup

        64.1k762102




        64.1k762102






























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