Probability of rolling first 3 with a fair die before 10th roll and after 4th roll.












2












$begingroup$


My question



Lets say we have a fair 6-sided die that we roll continuously. We roll this same dice 10 times in a row. What is the probability that we get a 3 after our 4th roll but before our 10th roll?



In other words, given the events



$$R_1,R_2,R_3,R_4,R_5,R_6,R_7,R_8,R_9,R_{10}$$



What is the probability that



$$R_5,R_6,R_7,R_8,R_9$$



We get our first 3?



My attempt



$R_1,R_2,R_3,R_4$ cannot be 3s. This means that they have probabilities $frac{5}{6}$.



The first 3 appearing somewhere here $R_5,R_6,R_7,R_8,R_9$ is probably something like $(5 text{ choose } 1) * 6^4$ (I have no reasoning, just intuition).



I have no idea how to handle the last part of the problem with "before $R_{10}$"...






probability combinatorics







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    My question



    Lets say we have a fair 6-sided die that we roll continuously. We roll this same dice 10 times in a row. What is the probability that we get a 3 after our 4th roll but before our 10th roll?



    In other words, given the events



    $$R_1,R_2,R_3,R_4,R_5,R_6,R_7,R_8,R_9,R_{10}$$



    What is the probability that



    $$R_5,R_6,R_7,R_8,R_9$$



    We get our first 3?



    My attempt



    $R_1,R_2,R_3,R_4$ cannot be 3s. This means that they have probabilities $frac{5}{6}$.



    The first 3 appearing somewhere here $R_5,R_6,R_7,R_8,R_9$ is probably something like $(5 text{ choose } 1) * 6^4$ (I have no reasoning, just intuition).



    I have no idea how to handle the last part of the problem with "before $R_{10}$"...






    probability combinatorics







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      My question



      Lets say we have a fair 6-sided die that we roll continuously. We roll this same dice 10 times in a row. What is the probability that we get a 3 after our 4th roll but before our 10th roll?



      In other words, given the events



      $$R_1,R_2,R_3,R_4,R_5,R_6,R_7,R_8,R_9,R_{10}$$



      What is the probability that



      $$R_5,R_6,R_7,R_8,R_9$$



      We get our first 3?



      My attempt



      $R_1,R_2,R_3,R_4$ cannot be 3s. This means that they have probabilities $frac{5}{6}$.



      The first 3 appearing somewhere here $R_5,R_6,R_7,R_8,R_9$ is probably something like $(5 text{ choose } 1) * 6^4$ (I have no reasoning, just intuition).



      I have no idea how to handle the last part of the problem with "before $R_{10}$"...






      probability combinatorics







      share|cite|improve this question











      $endgroup$




      My question



      Lets say we have a fair 6-sided die that we roll continuously. We roll this same dice 10 times in a row. What is the probability that we get a 3 after our 4th roll but before our 10th roll?



      In other words, given the events



      $$R_1,R_2,R_3,R_4,R_5,R_6,R_7,R_8,R_9,R_{10}$$



      What is the probability that



      $$R_5,R_6,R_7,R_8,R_9$$



      We get our first 3?



      My attempt



      $R_1,R_2,R_3,R_4$ cannot be 3s. This means that they have probabilities $frac{5}{6}$.



      The first 3 appearing somewhere here $R_5,R_6,R_7,R_8,R_9$ is probably something like $(5 text{ choose } 1) * 6^4$ (I have no reasoning, just intuition).



      I have no idea how to handle the last part of the problem with "before $R_{10}$"...






      probability combinatorics




      probability combinatorics






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 24 at 10:59









      N. F. Taussig

      44.6k103357




      44.6k103357










      asked Jan 24 at 2:46









      AlanSTACKAlanSTACK

      1,75921525




      1,75921525






















          1 Answer
          1






          active

          oldest

          votes


















          3












          $begingroup$

          I am making the assumption that you're going to roll the dice 10 times, even if you manage to roll a 3 at a desired location (e.g. if the first 3 appears at 7, you continue rolling to the 10th dice.)



          To deal with $R_5, R_6, R_7, R_8, R_9$, observe that we can model this with the binomial distribution. Let X denote the number of 3's in the rolls $R_5, R_6, R_7, R_8, R_9$.



          $$
          X sim BinBig(n=5, p=frac{1}{6}Big)
          $$



          (Where we have 5 tries and the probability of rolling a $3$ on any of these tries is $frac{1}{6}$)



          We want the probability:



          $$
          Pr(X > 0) = 1- Pr(X=0) = 1 - Big(frac{5}{6}Big)^5
          $$



          So, multiplying this with the probability that you arrived at for not rolling a 3 in the first 4 rolls, we obtain the final result:
          $$
          P(text{No 3's in the first four rolls} cap text{at least one 3 in rolls 5 to 9} ) = Big(frac{5}{6}Big)^4 Big( 1 - Big(frac{5}{6}Big)^5 Big)
          $$



          Which I believe works out to be $approx0.288$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            How does "but before our 10th roll?" factor into your answer?
            $endgroup$
            – AlanSTACK
            Jan 24 at 3:11






          • 1




            $begingroup$
            I'm assuming that you do not care about the 10th roll - whether or not the 10th roll is a 3 or otherwise, it's irrelevant. Only what happens in the first 9 rolls is relevant.
            $endgroup$
            – Sean Lee
            Jan 24 at 3:13










          • $begingroup$
            Does that part of the question not change the probability?
            $endgroup$
            – AlanSTACK
            Jan 24 at 3:15






          • 2




            $begingroup$
            To put it another way, success consists in getting a 3 in at least one of the trials $R_5,R_6,R_7,R_8,R_9$ after not getting a 3 in any of the trials $R_1,R_2,R_3,R_4$. So the solution doesn't require information about the outcomes of the tenth or later rolls.
            $endgroup$
            – hardmath
            Jan 24 at 3:19






          • 1




            $begingroup$
            @hardmath Thank you, that cleared it up for me.
            $endgroup$
            – AlanSTACK
            Jan 24 at 3:25











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          1 Answer
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          1 Answer
          1






          active

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          active

          oldest

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          active

          oldest

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          3












          $begingroup$

          I am making the assumption that you're going to roll the dice 10 times, even if you manage to roll a 3 at a desired location (e.g. if the first 3 appears at 7, you continue rolling to the 10th dice.)



          To deal with $R_5, R_6, R_7, R_8, R_9$, observe that we can model this with the binomial distribution. Let X denote the number of 3's in the rolls $R_5, R_6, R_7, R_8, R_9$.



          $$
          X sim BinBig(n=5, p=frac{1}{6}Big)
          $$



          (Where we have 5 tries and the probability of rolling a $3$ on any of these tries is $frac{1}{6}$)



          We want the probability:



          $$
          Pr(X > 0) = 1- Pr(X=0) = 1 - Big(frac{5}{6}Big)^5
          $$



          So, multiplying this with the probability that you arrived at for not rolling a 3 in the first 4 rolls, we obtain the final result:
          $$
          P(text{No 3's in the first four rolls} cap text{at least one 3 in rolls 5 to 9} ) = Big(frac{5}{6}Big)^4 Big( 1 - Big(frac{5}{6}Big)^5 Big)
          $$



          Which I believe works out to be $approx0.288$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            How does "but before our 10th roll?" factor into your answer?
            $endgroup$
            – AlanSTACK
            Jan 24 at 3:11






          • 1




            $begingroup$
            I'm assuming that you do not care about the 10th roll - whether or not the 10th roll is a 3 or otherwise, it's irrelevant. Only what happens in the first 9 rolls is relevant.
            $endgroup$
            – Sean Lee
            Jan 24 at 3:13










          • $begingroup$
            Does that part of the question not change the probability?
            $endgroup$
            – AlanSTACK
            Jan 24 at 3:15






          • 2




            $begingroup$
            To put it another way, success consists in getting a 3 in at least one of the trials $R_5,R_6,R_7,R_8,R_9$ after not getting a 3 in any of the trials $R_1,R_2,R_3,R_4$. So the solution doesn't require information about the outcomes of the tenth or later rolls.
            $endgroup$
            – hardmath
            Jan 24 at 3:19






          • 1




            $begingroup$
            @hardmath Thank you, that cleared it up for me.
            $endgroup$
            – AlanSTACK
            Jan 24 at 3:25
















          3












          $begingroup$

          I am making the assumption that you're going to roll the dice 10 times, even if you manage to roll a 3 at a desired location (e.g. if the first 3 appears at 7, you continue rolling to the 10th dice.)



          To deal with $R_5, R_6, R_7, R_8, R_9$, observe that we can model this with the binomial distribution. Let X denote the number of 3's in the rolls $R_5, R_6, R_7, R_8, R_9$.



          $$
          X sim BinBig(n=5, p=frac{1}{6}Big)
          $$



          (Where we have 5 tries and the probability of rolling a $3$ on any of these tries is $frac{1}{6}$)



          We want the probability:



          $$
          Pr(X > 0) = 1- Pr(X=0) = 1 - Big(frac{5}{6}Big)^5
          $$



          So, multiplying this with the probability that you arrived at for not rolling a 3 in the first 4 rolls, we obtain the final result:
          $$
          P(text{No 3's in the first four rolls} cap text{at least one 3 in rolls 5 to 9} ) = Big(frac{5}{6}Big)^4 Big( 1 - Big(frac{5}{6}Big)^5 Big)
          $$



          Which I believe works out to be $approx0.288$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            How does "but before our 10th roll?" factor into your answer?
            $endgroup$
            – AlanSTACK
            Jan 24 at 3:11






          • 1




            $begingroup$
            I'm assuming that you do not care about the 10th roll - whether or not the 10th roll is a 3 or otherwise, it's irrelevant. Only what happens in the first 9 rolls is relevant.
            $endgroup$
            – Sean Lee
            Jan 24 at 3:13










          • $begingroup$
            Does that part of the question not change the probability?
            $endgroup$
            – AlanSTACK
            Jan 24 at 3:15






          • 2




            $begingroup$
            To put it another way, success consists in getting a 3 in at least one of the trials $R_5,R_6,R_7,R_8,R_9$ after not getting a 3 in any of the trials $R_1,R_2,R_3,R_4$. So the solution doesn't require information about the outcomes of the tenth or later rolls.
            $endgroup$
            – hardmath
            Jan 24 at 3:19






          • 1




            $begingroup$
            @hardmath Thank you, that cleared it up for me.
            $endgroup$
            – AlanSTACK
            Jan 24 at 3:25














          3












          3








          3





          $begingroup$

          I am making the assumption that you're going to roll the dice 10 times, even if you manage to roll a 3 at a desired location (e.g. if the first 3 appears at 7, you continue rolling to the 10th dice.)



          To deal with $R_5, R_6, R_7, R_8, R_9$, observe that we can model this with the binomial distribution. Let X denote the number of 3's in the rolls $R_5, R_6, R_7, R_8, R_9$.



          $$
          X sim BinBig(n=5, p=frac{1}{6}Big)
          $$



          (Where we have 5 tries and the probability of rolling a $3$ on any of these tries is $frac{1}{6}$)



          We want the probability:



          $$
          Pr(X > 0) = 1- Pr(X=0) = 1 - Big(frac{5}{6}Big)^5
          $$



          So, multiplying this with the probability that you arrived at for not rolling a 3 in the first 4 rolls, we obtain the final result:
          $$
          P(text{No 3's in the first four rolls} cap text{at least one 3 in rolls 5 to 9} ) = Big(frac{5}{6}Big)^4 Big( 1 - Big(frac{5}{6}Big)^5 Big)
          $$



          Which I believe works out to be $approx0.288$






          share|cite|improve this answer









          $endgroup$



          I am making the assumption that you're going to roll the dice 10 times, even if you manage to roll a 3 at a desired location (e.g. if the first 3 appears at 7, you continue rolling to the 10th dice.)



          To deal with $R_5, R_6, R_7, R_8, R_9$, observe that we can model this with the binomial distribution. Let X denote the number of 3's in the rolls $R_5, R_6, R_7, R_8, R_9$.



          $$
          X sim BinBig(n=5, p=frac{1}{6}Big)
          $$



          (Where we have 5 tries and the probability of rolling a $3$ on any of these tries is $frac{1}{6}$)



          We want the probability:



          $$
          Pr(X > 0) = 1- Pr(X=0) = 1 - Big(frac{5}{6}Big)^5
          $$



          So, multiplying this with the probability that you arrived at for not rolling a 3 in the first 4 rolls, we obtain the final result:
          $$
          P(text{No 3's in the first four rolls} cap text{at least one 3 in rolls 5 to 9} ) = Big(frac{5}{6}Big)^4 Big( 1 - Big(frac{5}{6}Big)^5 Big)
          $$



          Which I believe works out to be $approx0.288$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 24 at 3:03









          Sean LeeSean Lee

          483211




          483211












          • $begingroup$
            How does "but before our 10th roll?" factor into your answer?
            $endgroup$
            – AlanSTACK
            Jan 24 at 3:11






          • 1




            $begingroup$
            I'm assuming that you do not care about the 10th roll - whether or not the 10th roll is a 3 or otherwise, it's irrelevant. Only what happens in the first 9 rolls is relevant.
            $endgroup$
            – Sean Lee
            Jan 24 at 3:13










          • $begingroup$
            Does that part of the question not change the probability?
            $endgroup$
            – AlanSTACK
            Jan 24 at 3:15






          • 2




            $begingroup$
            To put it another way, success consists in getting a 3 in at least one of the trials $R_5,R_6,R_7,R_8,R_9$ after not getting a 3 in any of the trials $R_1,R_2,R_3,R_4$. So the solution doesn't require information about the outcomes of the tenth or later rolls.
            $endgroup$
            – hardmath
            Jan 24 at 3:19






          • 1




            $begingroup$
            @hardmath Thank you, that cleared it up for me.
            $endgroup$
            – AlanSTACK
            Jan 24 at 3:25


















          • $begingroup$
            How does "but before our 10th roll?" factor into your answer?
            $endgroup$
            – AlanSTACK
            Jan 24 at 3:11






          • 1




            $begingroup$
            I'm assuming that you do not care about the 10th roll - whether or not the 10th roll is a 3 or otherwise, it's irrelevant. Only what happens in the first 9 rolls is relevant.
            $endgroup$
            – Sean Lee
            Jan 24 at 3:13










          • $begingroup$
            Does that part of the question not change the probability?
            $endgroup$
            – AlanSTACK
            Jan 24 at 3:15






          • 2




            $begingroup$
            To put it another way, success consists in getting a 3 in at least one of the trials $R_5,R_6,R_7,R_8,R_9$ after not getting a 3 in any of the trials $R_1,R_2,R_3,R_4$. So the solution doesn't require information about the outcomes of the tenth or later rolls.
            $endgroup$
            – hardmath
            Jan 24 at 3:19






          • 1




            $begingroup$
            @hardmath Thank you, that cleared it up for me.
            $endgroup$
            – AlanSTACK
            Jan 24 at 3:25
















          $begingroup$
          How does "but before our 10th roll?" factor into your answer?
          $endgroup$
          – AlanSTACK
          Jan 24 at 3:11




          $begingroup$
          How does "but before our 10th roll?" factor into your answer?
          $endgroup$
          – AlanSTACK
          Jan 24 at 3:11




          1




          1




          $begingroup$
          I'm assuming that you do not care about the 10th roll - whether or not the 10th roll is a 3 or otherwise, it's irrelevant. Only what happens in the first 9 rolls is relevant.
          $endgroup$
          – Sean Lee
          Jan 24 at 3:13




          $begingroup$
          I'm assuming that you do not care about the 10th roll - whether or not the 10th roll is a 3 or otherwise, it's irrelevant. Only what happens in the first 9 rolls is relevant.
          $endgroup$
          – Sean Lee
          Jan 24 at 3:13












          $begingroup$
          Does that part of the question not change the probability?
          $endgroup$
          – AlanSTACK
          Jan 24 at 3:15




          $begingroup$
          Does that part of the question not change the probability?
          $endgroup$
          – AlanSTACK
          Jan 24 at 3:15




          2




          2




          $begingroup$
          To put it another way, success consists in getting a 3 in at least one of the trials $R_5,R_6,R_7,R_8,R_9$ after not getting a 3 in any of the trials $R_1,R_2,R_3,R_4$. So the solution doesn't require information about the outcomes of the tenth or later rolls.
          $endgroup$
          – hardmath
          Jan 24 at 3:19




          $begingroup$
          To put it another way, success consists in getting a 3 in at least one of the trials $R_5,R_6,R_7,R_8,R_9$ after not getting a 3 in any of the trials $R_1,R_2,R_3,R_4$. So the solution doesn't require information about the outcomes of the tenth or later rolls.
          $endgroup$
          – hardmath
          Jan 24 at 3:19




          1




          1




          $begingroup$
          @hardmath Thank you, that cleared it up for me.
          $endgroup$
          – AlanSTACK
          Jan 24 at 3:25




          $begingroup$
          @hardmath Thank you, that cleared it up for me.
          $endgroup$
          – AlanSTACK
          Jan 24 at 3:25


















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