Mean number of attempts for 3 events to happenwaiting for TT and H?Probability game of independent eventsAn intuitive solution to this problem (Using probability tree)Bernoulli trials are run until 3 successful trials are completed. If probability of a success is 0.3 what is the expected number of failures?Probability of sequential successesProbability of multiple independent events occurring after repeated attemptsAssume $E[X^2]=20$. Find the mean and variance of $X$ if the mean and variance are equal.Total expected timeProbability - three-headed dragon and three knightsWhat's the probability of guessing a secret code if the attempts are limited and you stop at the first success?Conditional Probability about succeeding in nth attempt given the failure in n-1 attempts

Removing files under particular conditions (number of files, file age)

Is a bound state a stationary state?

If a character has darkvision, can they see through an area of nonmagical darkness filled with lightly obscuring gas?

How can Trident be so inexpensive? Will it orbit Triton or just do a (slow) flyby?

How should I respond when I lied about my education and the company finds out through background check?

On a tidally locked planet, would time be quantized?

Loading commands from file

How to explain what's wrong with this application of the chain rule?

Travelling outside the UK without a passport

Creepy dinosaur pc game identification

Is it improper etiquette to ask your opponent what his/her rating is before the game?

Closed-form expression for certain product

Does a 'pending' US visa application constitute a denial?

A social experiment. What is the worst that can happen?

Why do we read the Megillah by night and by day?

It grows, but water kills it

Lowest total scrabble score

The IT department bottlenecks progress. How should I handle this?

What is this called? Old film camera viewer?

Should I stop contributing to retirement accounts?

Why Shazam when there is already Superman?

Pre-mixing cryogenic fuels and using only one fuel tank

Problem with TransformedDistribution

What was the exact wording from Ivanhoe of this advice on how to free yourself from slavery?



Mean number of attempts for 3 events to happen


waiting for TT and H?Probability game of independent eventsAn intuitive solution to this problem (Using probability tree)Bernoulli trials are run until 3 successful trials are completed. If probability of a success is 0.3 what is the expected number of failures?Probability of sequential successesProbability of multiple independent events occurring after repeated attemptsAssume $E[X^2]=20$. Find the mean and variance of $X$ if the mean and variance are equal.Total expected timeProbability - three-headed dragon and three knightsWhat's the probability of guessing a secret code if the attempts are limited and you stop at the first success?Conditional Probability about succeeding in nth attempt given the failure in n-1 attempts













1












$begingroup$


For example, 3 consecutive fixed tasks to complete, if one is failed, a new attempt is made from the beginning.

I'm saying an "attempt" is an attempt to complete all 3, a "try" is to try and complete a task.



Task A must be completed before B is tried, A and B must be completed before C is tried.
Each task has the same probability $p$ of being successful.



I have calculated:



1st task failed = $p$

1st task completed, 2nd task failed = $p(1-p)$

All 3 completed = $(1-p)^2$



and I've calculated the mean number of tries:$$p^2-3p+3$$



I tried to figure out the mean number of attempts for an "all three complete" but can't and I've been told the answer is, where w is the probability of all 3 completed: $$w + 2w(1-w) + 3w(1-w)^2 + ... + nw(1-w)^(n-1)$$
which reduces to $frac1w$, that is, $$frac1(1-p)^2$$



Could someone explain the maths and maybe intuition behind this please?










share|cite|improve this question









$endgroup$
















    1












    $begingroup$


    For example, 3 consecutive fixed tasks to complete, if one is failed, a new attempt is made from the beginning.

    I'm saying an "attempt" is an attempt to complete all 3, a "try" is to try and complete a task.



    Task A must be completed before B is tried, A and B must be completed before C is tried.
    Each task has the same probability $p$ of being successful.



    I have calculated:



    1st task failed = $p$

    1st task completed, 2nd task failed = $p(1-p)$

    All 3 completed = $(1-p)^2$



    and I've calculated the mean number of tries:$$p^2-3p+3$$



    I tried to figure out the mean number of attempts for an "all three complete" but can't and I've been told the answer is, where w is the probability of all 3 completed: $$w + 2w(1-w) + 3w(1-w)^2 + ... + nw(1-w)^(n-1)$$
    which reduces to $frac1w$, that is, $$frac1(1-p)^2$$



    Could someone explain the maths and maybe intuition behind this please?










    share|cite|improve this question









    $endgroup$














      1












      1








      1





      $begingroup$


      For example, 3 consecutive fixed tasks to complete, if one is failed, a new attempt is made from the beginning.

      I'm saying an "attempt" is an attempt to complete all 3, a "try" is to try and complete a task.



      Task A must be completed before B is tried, A and B must be completed before C is tried.
      Each task has the same probability $p$ of being successful.



      I have calculated:



      1st task failed = $p$

      1st task completed, 2nd task failed = $p(1-p)$

      All 3 completed = $(1-p)^2$



      and I've calculated the mean number of tries:$$p^2-3p+3$$



      I tried to figure out the mean number of attempts for an "all three complete" but can't and I've been told the answer is, where w is the probability of all 3 completed: $$w + 2w(1-w) + 3w(1-w)^2 + ... + nw(1-w)^(n-1)$$
      which reduces to $frac1w$, that is, $$frac1(1-p)^2$$



      Could someone explain the maths and maybe intuition behind this please?










      share|cite|improve this question









      $endgroup$




      For example, 3 consecutive fixed tasks to complete, if one is failed, a new attempt is made from the beginning.

      I'm saying an "attempt" is an attempt to complete all 3, a "try" is to try and complete a task.



      Task A must be completed before B is tried, A and B must be completed before C is tried.
      Each task has the same probability $p$ of being successful.



      I have calculated:



      1st task failed = $p$

      1st task completed, 2nd task failed = $p(1-p)$

      All 3 completed = $(1-p)^2$



      and I've calculated the mean number of tries:$$p^2-3p+3$$



      I tried to figure out the mean number of attempts for an "all three complete" but can't and I've been told the answer is, where w is the probability of all 3 completed: $$w + 2w(1-w) + 3w(1-w)^2 + ... + nw(1-w)^(n-1)$$
      which reduces to $frac1w$, that is, $$frac1(1-p)^2$$



      Could someone explain the maths and maybe intuition behind this please?







      probability






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Mar 15 at 17:52









      WilliamWilliam

      104




      104




















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          You can think of this problem as a Markov chain with 4 states - A, B, C, and D, where D is an absorbing state, i.e. you are done when you get there.



          • From state A, you go back to state A with probability (w.p.) 1-p and you go to state B w.p. p

          • From state B, you go back to state A w.p. 1-p and you go to state C w.p. p

          • From state C, you go back to state A w.p. 1-p and you go to state D (done!) w.p. p

          Now, let $T_A, T_B, T_C$ denote the expected times to transit from states A, B, C to the end of the game (state D).



          Based on the state transitions described above, you can write the following equations:



          $$T_A = (1-p)(1+T_A) + p(1+T_B)$$
          $$T_B = (1-p)(1+T_A) + p(1+T_C)$$
          $$T_C = (1-p)(1+T_A) + p(1)$$



          You can solve for $T_A$ from the above equations.



          $T_A = 1 over p + 1 over p^2 + 1 over p^3$ if I did the algebra right.






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            Thanks Aditya Dua, that is really helpful. Unfortunately I've realised the analogy I used to form my question is wrong, which is possibly where some of my confusion lies. I'll try and apply your method to my problem! Could someone advise me if it's better to edit the question so it is correct (regrettably rendering Aditya Dua's answer redundant) or to somehow delete the question and repost it? I don't want to leave it up as is since it might confuse people looking at it.
            $endgroup$
            – William
            Mar 16 at 12:56










          • $begingroup$
            Even though it's not what you wanted, the question in itself is fine (and so is the answer, I believe). You can leave this up there and maybe post a new question which describes the model pertinent to the problem you are trying to solve. My 2c.
            $endgroup$
            – Aditya Dua
            Mar 18 at 21:57










          Your Answer





          StackExchange.ifUsing("editor", function ()
          return StackExchange.using("mathjaxEditing", function ()
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          );
          );
          , "mathjax-editing");

          StackExchange.ready(function()
          var channelOptions =
          tags: "".split(" "),
          id: "69"
          ;
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function()
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled)
          StackExchange.using("snippets", function()
          createEditor();
          );

          else
          createEditor();

          );

          function createEditor()
          StackExchange.prepareEditor(
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader:
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          ,
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          );



          );













          draft saved

          draft discarded


















          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3149605%2fmean-number-of-attempts-for-3-events-to-happen%23new-answer', 'question_page');

          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          You can think of this problem as a Markov chain with 4 states - A, B, C, and D, where D is an absorbing state, i.e. you are done when you get there.



          • From state A, you go back to state A with probability (w.p.) 1-p and you go to state B w.p. p

          • From state B, you go back to state A w.p. 1-p and you go to state C w.p. p

          • From state C, you go back to state A w.p. 1-p and you go to state D (done!) w.p. p

          Now, let $T_A, T_B, T_C$ denote the expected times to transit from states A, B, C to the end of the game (state D).



          Based on the state transitions described above, you can write the following equations:



          $$T_A = (1-p)(1+T_A) + p(1+T_B)$$
          $$T_B = (1-p)(1+T_A) + p(1+T_C)$$
          $$T_C = (1-p)(1+T_A) + p(1)$$



          You can solve for $T_A$ from the above equations.



          $T_A = 1 over p + 1 over p^2 + 1 over p^3$ if I did the algebra right.






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            Thanks Aditya Dua, that is really helpful. Unfortunately I've realised the analogy I used to form my question is wrong, which is possibly where some of my confusion lies. I'll try and apply your method to my problem! Could someone advise me if it's better to edit the question so it is correct (regrettably rendering Aditya Dua's answer redundant) or to somehow delete the question and repost it? I don't want to leave it up as is since it might confuse people looking at it.
            $endgroup$
            – William
            Mar 16 at 12:56










          • $begingroup$
            Even though it's not what you wanted, the question in itself is fine (and so is the answer, I believe). You can leave this up there and maybe post a new question which describes the model pertinent to the problem you are trying to solve. My 2c.
            $endgroup$
            – Aditya Dua
            Mar 18 at 21:57















          1












          $begingroup$

          You can think of this problem as a Markov chain with 4 states - A, B, C, and D, where D is an absorbing state, i.e. you are done when you get there.



          • From state A, you go back to state A with probability (w.p.) 1-p and you go to state B w.p. p

          • From state B, you go back to state A w.p. 1-p and you go to state C w.p. p

          • From state C, you go back to state A w.p. 1-p and you go to state D (done!) w.p. p

          Now, let $T_A, T_B, T_C$ denote the expected times to transit from states A, B, C to the end of the game (state D).



          Based on the state transitions described above, you can write the following equations:



          $$T_A = (1-p)(1+T_A) + p(1+T_B)$$
          $$T_B = (1-p)(1+T_A) + p(1+T_C)$$
          $$T_C = (1-p)(1+T_A) + p(1)$$



          You can solve for $T_A$ from the above equations.



          $T_A = 1 over p + 1 over p^2 + 1 over p^3$ if I did the algebra right.






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            Thanks Aditya Dua, that is really helpful. Unfortunately I've realised the analogy I used to form my question is wrong, which is possibly where some of my confusion lies. I'll try and apply your method to my problem! Could someone advise me if it's better to edit the question so it is correct (regrettably rendering Aditya Dua's answer redundant) or to somehow delete the question and repost it? I don't want to leave it up as is since it might confuse people looking at it.
            $endgroup$
            – William
            Mar 16 at 12:56










          • $begingroup$
            Even though it's not what you wanted, the question in itself is fine (and so is the answer, I believe). You can leave this up there and maybe post a new question which describes the model pertinent to the problem you are trying to solve. My 2c.
            $endgroup$
            – Aditya Dua
            Mar 18 at 21:57













          1












          1








          1





          $begingroup$

          You can think of this problem as a Markov chain with 4 states - A, B, C, and D, where D is an absorbing state, i.e. you are done when you get there.



          • From state A, you go back to state A with probability (w.p.) 1-p and you go to state B w.p. p

          • From state B, you go back to state A w.p. 1-p and you go to state C w.p. p

          • From state C, you go back to state A w.p. 1-p and you go to state D (done!) w.p. p

          Now, let $T_A, T_B, T_C$ denote the expected times to transit from states A, B, C to the end of the game (state D).



          Based on the state transitions described above, you can write the following equations:



          $$T_A = (1-p)(1+T_A) + p(1+T_B)$$
          $$T_B = (1-p)(1+T_A) + p(1+T_C)$$
          $$T_C = (1-p)(1+T_A) + p(1)$$



          You can solve for $T_A$ from the above equations.



          $T_A = 1 over p + 1 over p^2 + 1 over p^3$ if I did the algebra right.






          share|cite|improve this answer











          $endgroup$



          You can think of this problem as a Markov chain with 4 states - A, B, C, and D, where D is an absorbing state, i.e. you are done when you get there.



          • From state A, you go back to state A with probability (w.p.) 1-p and you go to state B w.p. p

          • From state B, you go back to state A w.p. 1-p and you go to state C w.p. p

          • From state C, you go back to state A w.p. 1-p and you go to state D (done!) w.p. p

          Now, let $T_A, T_B, T_C$ denote the expected times to transit from states A, B, C to the end of the game (state D).



          Based on the state transitions described above, you can write the following equations:



          $$T_A = (1-p)(1+T_A) + p(1+T_B)$$
          $$T_B = (1-p)(1+T_A) + p(1+T_C)$$
          $$T_C = (1-p)(1+T_A) + p(1)$$



          You can solve for $T_A$ from the above equations.



          $T_A = 1 over p + 1 over p^2 + 1 over p^3$ if I did the algebra right.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Mar 15 at 18:40

























          answered Mar 15 at 18:32









          Aditya DuaAditya Dua

          1,15418




          1,15418











          • $begingroup$
            Thanks Aditya Dua, that is really helpful. Unfortunately I've realised the analogy I used to form my question is wrong, which is possibly where some of my confusion lies. I'll try and apply your method to my problem! Could someone advise me if it's better to edit the question so it is correct (regrettably rendering Aditya Dua's answer redundant) or to somehow delete the question and repost it? I don't want to leave it up as is since it might confuse people looking at it.
            $endgroup$
            – William
            Mar 16 at 12:56










          • $begingroup$
            Even though it's not what you wanted, the question in itself is fine (and so is the answer, I believe). You can leave this up there and maybe post a new question which describes the model pertinent to the problem you are trying to solve. My 2c.
            $endgroup$
            – Aditya Dua
            Mar 18 at 21:57
















          • $begingroup$
            Thanks Aditya Dua, that is really helpful. Unfortunately I've realised the analogy I used to form my question is wrong, which is possibly where some of my confusion lies. I'll try and apply your method to my problem! Could someone advise me if it's better to edit the question so it is correct (regrettably rendering Aditya Dua's answer redundant) or to somehow delete the question and repost it? I don't want to leave it up as is since it might confuse people looking at it.
            $endgroup$
            – William
            Mar 16 at 12:56










          • $begingroup$
            Even though it's not what you wanted, the question in itself is fine (and so is the answer, I believe). You can leave this up there and maybe post a new question which describes the model pertinent to the problem you are trying to solve. My 2c.
            $endgroup$
            – Aditya Dua
            Mar 18 at 21:57















          $begingroup$
          Thanks Aditya Dua, that is really helpful. Unfortunately I've realised the analogy I used to form my question is wrong, which is possibly where some of my confusion lies. I'll try and apply your method to my problem! Could someone advise me if it's better to edit the question so it is correct (regrettably rendering Aditya Dua's answer redundant) or to somehow delete the question and repost it? I don't want to leave it up as is since it might confuse people looking at it.
          $endgroup$
          – William
          Mar 16 at 12:56




          $begingroup$
          Thanks Aditya Dua, that is really helpful. Unfortunately I've realised the analogy I used to form my question is wrong, which is possibly where some of my confusion lies. I'll try and apply your method to my problem! Could someone advise me if it's better to edit the question so it is correct (regrettably rendering Aditya Dua's answer redundant) or to somehow delete the question and repost it? I don't want to leave it up as is since it might confuse people looking at it.
          $endgroup$
          – William
          Mar 16 at 12:56












          $begingroup$
          Even though it's not what you wanted, the question in itself is fine (and so is the answer, I believe). You can leave this up there and maybe post a new question which describes the model pertinent to the problem you are trying to solve. My 2c.
          $endgroup$
          – Aditya Dua
          Mar 18 at 21:57




          $begingroup$
          Even though it's not what you wanted, the question in itself is fine (and so is the answer, I believe). You can leave this up there and maybe post a new question which describes the model pertinent to the problem you are trying to solve. My 2c.
          $endgroup$
          – Aditya Dua
          Mar 18 at 21:57

















          draft saved

          draft discarded
















































          Thanks for contributing an answer to Mathematics Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid


          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.

          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3149605%2fmean-number-of-attempts-for-3-events-to-happen%23new-answer', 'question_page');

          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          Popular posts from this blog

          How should I support this large drywall patch? Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?How do I cover large gaps in drywall?How do I keep drywall around a patch from crumbling?Can I glue a second layer of drywall?How to patch long strip on drywall?Large drywall patch: how to avoid bulging seams?Drywall Mesh Patch vs. Bulge? To remove or not to remove?How to fix this drywall job?Prep drywall before backsplashWhat's the best way to fix this horrible drywall patch job?Drywall patching using 3M Patch Plus Primer

          random experiment with two different functions on unit interval Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Random variable and probability space notionsRandom Walk with EdgesFinding functions where the increase over a random interval is Poisson distributedNumber of days until dayCan an observed event in fact be of zero probability?Unit random processmodels of coins and uniform distributionHow to get the number of successes given $n$ trials , probability $P$ and a random variable $X$Absorbing Markov chain in a computer. Is “almost every” turned into always convergence in computer executions?Stopped random walk is not uniformly integrable

          Lowndes Grove History Architecture References Navigation menu32°48′6″N 79°57′58″W / 32.80167°N 79.96611°W / 32.80167; -79.9661132°48′6″N 79°57′58″W / 32.80167°N 79.96611°W / 32.80167; -79.9661178002500"National Register Information System"Historic houses of South Carolina"Lowndes Grove""+32° 48' 6.00", −79° 57' 58.00""Lowndes Grove, Charleston County (260 St. Margaret St., Charleston)""Lowndes Grove"The Charleston ExpositionIt Happened in South Carolina"Lowndes Grove (House), Saint Margaret Street & Sixth Avenue, Charleston, Charleston County, SC(Photographs)"Plantations of the Carolina Low Countrye