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Mean number of attempts for 3 events to happen

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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?

probability

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asked Mar 15 at 17:52

WilliamWilliam

104

$endgroup$

add a comment | 

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?

probability

share|cite|improve this question

asked Mar 15 at 17:52

WilliamWilliam

104

$endgroup$

add a comment | 

$begingroup$

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?

probability

share|cite|improve this question

asked Mar 15 at 17:52

WilliamWilliam

104

$endgroup$

share|cite|improve this question

asked Mar 15 at 17:52

WilliamWilliam

104

share|cite|improve this question

asked Mar 15 at 17:52

WilliamWilliam

104

asked Mar 15 at 17:52

WilliamWilliam

104

asked Mar 15 at 17:52

WilliamWilliam

104

1 Answer
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oldest

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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

edited Mar 15 at 18:40

answered Mar 15 at 18:32

Aditya DuaAditya Dua

1,15418

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

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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

edited Mar 15 at 18:40

answered Mar 15 at 18:32

Aditya DuaAditya Dua

1,15418

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

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

edited Mar 15 at 18:40

answered Mar 15 at 18:32

Aditya DuaAditya Dua

1,15418

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

$begingroup$

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

edited Mar 15 at 18:40

answered Mar 15 at 18:32

Aditya DuaAditya Dua

1,15418

$endgroup$

share|cite|improve this answer

edited Mar 15 at 18:40

answered Mar 15 at 18:32

Aditya DuaAditya Dua

1,15418

answered Mar 15 at 18:32

Aditya DuaAditya Dua

1,15418

answered Mar 15 at 18:32

Aditya DuaAditya Dua

1,15418