Basic question: Calculating expected value [closed]

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I'm sorry for posting such a basic question, but me and my friend were discussing a problem and we just cant find a common ground. The problem is:

A guy is playing a game. On n^th day the game ends with the probability of 1/(100-n). So on the first day probability that the game ends is 1/99, on the second, if it hasn't ended on the first, probability of ending would be 1/98, and so on until 99^th day when the game ends with probability 1 if it hasn't so far.

What is the expected value of duration of the game?

probability expected-value

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asked Jan 28 at 21:25

bearbear

61

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closed as off-topic by Did, Lee David Chung Lin, mrtaurho, Leucippus, Claude Leibovici Jan 29 at 9:28

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, Lee David Chung Lin, mrtaurho, Leucippus, Claude Leibovici

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What are the solutions you and your friend suggested to each other?
  $endgroup$
  – Did
  Jan 28 at 21:46
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  One was 50, like the accepted answer, and the other was 63 (i don't really know the reasoning behind that number though)
  $endgroup$
  – bear
  Jan 28 at 21:50
  

  
  
  
  

add a comment | 

0

$begingroup$

I'm sorry for posting such a basic question, but me and my friend were discussing a problem and we just cant find a common ground. The problem is:

A guy is playing a game. On n^th day the game ends with the probability of 1/(100-n). So on the first day probability that the game ends is 1/99, on the second, if it hasn't ended on the first, probability of ending would be 1/98, and so on until 99^th day when the game ends with probability 1 if it hasn't so far.

What is the expected value of duration of the game?

probability expected-value

share|cite|improve this question

asked Jan 28 at 21:25

bearbear

61

$endgroup$

closed as off-topic by Did, Lee David Chung Lin, mrtaurho, Leucippus, Claude Leibovici Jan 29 at 9:28

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, Lee David Chung Lin, mrtaurho, Leucippus, Claude Leibovici

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What are the solutions you and your friend suggested to each other?
  $endgroup$
  – Did
  Jan 28 at 21:46
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  One was 50, like the accepted answer, and the other was 63 (i don't really know the reasoning behind that number though)
  $endgroup$
  – bear
  Jan 28 at 21:50
  

  
  
  
  

add a comment | 

0

0

0

$begingroup$

I'm sorry for posting such a basic question, but me and my friend were discussing a problem and we just cant find a common ground. The problem is:

A guy is playing a game. On n^th day the game ends with the probability of 1/(100-n). So on the first day probability that the game ends is 1/99, on the second, if it hasn't ended on the first, probability of ending would be 1/98, and so on until 99^th day when the game ends with probability 1 if it hasn't so far.

What is the expected value of duration of the game?

probability expected-value

share|cite|improve this question

asked Jan 28 at 21:25

bearbear

61

$endgroup$

I'm sorry for posting such a basic question, but me and my friend were discussing a problem and we just cant find a common ground. The problem is:

A guy is playing a game. On n^th day the game ends with the probability of 1/(100-n). So on the first day probability that the game ends is 1/99, on the second, if it hasn't ended on the first, probability of ending would be 1/98, and so on until 99^th day when the game ends with probability 1 if it hasn't so far.

What is the expected value of duration of the game?

probability expected-value

probability expected-value

share|cite|improve this question

asked Jan 28 at 21:25

bearbear

61

share|cite|improve this question

asked Jan 28 at 21:25

bearbear

61

share|cite|improve this question

share|cite|improve this question

asked Jan 28 at 21:25

bearbear

61

asked Jan 28 at 21:25

bearbear

61

asked Jan 28 at 21:25

bearbear

61

61

closed as off-topic by Did, Lee David Chung Lin, mrtaurho, Leucippus, Claude Leibovici Jan 29 at 9:28

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, Lee David Chung Lin, mrtaurho, Leucippus, Claude Leibovici

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by Did, Lee David Chung Lin, mrtaurho, Leucippus, Claude Leibovici Jan 29 at 9:28

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, Lee David Chung Lin, mrtaurho, Leucippus, Claude Leibovici

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What are the solutions you and your friend suggested to each other?
  $endgroup$
  – Did
  Jan 28 at 21:46
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  One was 50, like the accepted answer, and the other was 63 (i don't really know the reasoning behind that number though)
  $endgroup$
  – bear
  Jan 28 at 21:50
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What are the solutions you and your friend suggested to each other?
  $endgroup$
  – Did
  Jan 28 at 21:46
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  One was 50, like the accepted answer, and the other was 63 (i don't really know the reasoning behind that number though)
  $endgroup$
  – bear
  Jan 28 at 21:50
  

  
  
  
  

$begingroup$
What are the solutions you and your friend suggested to each other?
$endgroup$
– Did
Jan 28 at 21:46

$begingroup$
What are the solutions you and your friend suggested to each other?
$endgroup$
– Did
Jan 28 at 21:46

$begingroup$
One was 50, like the accepted answer, and the other was 63 (i don't really know the reasoning behind that number though)
$endgroup$
– bear
Jan 28 at 21:50

$begingroup$
One was 50, like the accepted answer, and the other was 63 (i don't really know the reasoning behind that number though)
$endgroup$
– bear
Jan 28 at 21:50

add a comment | 

2 Answers
2

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oldest

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1

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Probability game ends on day n is $$frac{98}{99} frac{97}{98} ldots frac{100-n}{100-n+1} frac{1}{100-n} = frac{1}{99}$$. So expected value of what day the game ends is
$$ 1frac{1}{99}+2frac{1}{99}+...+99frac{1}{99} = frac{4950}{99} = 50.
$$

Intuitively, as each day is equally likely to be finished on, the expected day is in the dead middle of the possible outcomes 1 through 99 (think about rolling a 6-sided die, you expect to obtain the average of the values which is 3.5).

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edited Jan 28 at 22:13

karakfa

2,025811

answered Jan 28 at 21:42

Cathal Ó CléirighCathal Ó Cléirigh

363

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Very nice, couldn't agree more. Thank you!
  $endgroup$
  – bear
  Jan 28 at 21:46
  

  
  
  
  

add a comment | 

0

$begingroup$

The probabilities are (starting from $1$),

$$frac1{99},frac{98}{99}frac1{98}=frac1{99},frac{98}{99}frac{97}{98}frac1{97}=frac1{99},cdots$$ and so on until day $99$. As all outcomes are equal, the expectation is the midpoint of $1$ and $99$, i.e. $50$.

share|cite|improve this answer

answered Jan 28 at 21:53

Yves DaoustYves Daoust

131k676229

$endgroup$

add a comment | 

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

1

$begingroup$

Probability game ends on day n is $$frac{98}{99} frac{97}{98} ldots frac{100-n}{100-n+1} frac{1}{100-n} = frac{1}{99}$$. So expected value of what day the game ends is
$$ 1frac{1}{99}+2frac{1}{99}+...+99frac{1}{99} = frac{4950}{99} = 50.
$$

Intuitively, as each day is equally likely to be finished on, the expected day is in the dead middle of the possible outcomes 1 through 99 (think about rolling a 6-sided die, you expect to obtain the average of the values which is 3.5).

share|cite|improve this answer

edited Jan 28 at 22:13

karakfa

2,025811

answered Jan 28 at 21:42

Cathal Ó CléirighCathal Ó Cléirigh

363

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Very nice, couldn't agree more. Thank you!
  $endgroup$
  – bear
  Jan 28 at 21:46
  

  
  
  
  

add a comment | 

1

$begingroup$

Probability game ends on day n is $$frac{98}{99} frac{97}{98} ldots frac{100-n}{100-n+1} frac{1}{100-n} = frac{1}{99}$$. So expected value of what day the game ends is
$$ 1frac{1}{99}+2frac{1}{99}+...+99frac{1}{99} = frac{4950}{99} = 50.
$$

Intuitively, as each day is equally likely to be finished on, the expected day is in the dead middle of the possible outcomes 1 through 99 (think about rolling a 6-sided die, you expect to obtain the average of the values which is 3.5).

share|cite|improve this answer

edited Jan 28 at 22:13

karakfa

2,025811

answered Jan 28 at 21:42

Cathal Ó CléirighCathal Ó Cléirigh

363

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Very nice, couldn't agree more. Thank you!
  $endgroup$
  – bear
  Jan 28 at 21:46
  

  
  
  
  

add a comment | 

1

1

1

$begingroup$

Probability game ends on day n is $$frac{98}{99} frac{97}{98} ldots frac{100-n}{100-n+1} frac{1}{100-n} = frac{1}{99}$$. So expected value of what day the game ends is
$$ 1frac{1}{99}+2frac{1}{99}+...+99frac{1}{99} = frac{4950}{99} = 50.
$$

Intuitively, as each day is equally likely to be finished on, the expected day is in the dead middle of the possible outcomes 1 through 99 (think about rolling a 6-sided die, you expect to obtain the average of the values which is 3.5).

share|cite|improve this answer

edited Jan 28 at 22:13

karakfa

2,025811

answered Jan 28 at 21:42

Cathal Ó CléirighCathal Ó Cléirigh

363

$endgroup$

Probability game ends on day n is $$frac{98}{99} frac{97}{98} ldots frac{100-n}{100-n+1} frac{1}{100-n} = frac{1}{99}$$. So expected value of what day the game ends is
$$ 1frac{1}{99}+2frac{1}{99}+...+99frac{1}{99} = frac{4950}{99} = 50.
$$

Intuitively, as each day is equally likely to be finished on, the expected day is in the dead middle of the possible outcomes 1 through 99 (think about rolling a 6-sided die, you expect to obtain the average of the values which is 3.5).

share|cite|improve this answer

edited Jan 28 at 22:13

karakfa

2,025811

answered Jan 28 at 21:42

Cathal Ó CléirighCathal Ó Cléirigh

363

share|cite|improve this answer

share|cite|improve this answer

edited Jan 28 at 22:13

karakfa

2,025811

edited Jan 28 at 22:13

karakfa

2,025811

edited Jan 28 at 22:13

karakfa

2,025811

2,025811

answered Jan 28 at 21:42

Cathal Ó CléirighCathal Ó Cléirigh

363

answered Jan 28 at 21:42

Cathal Ó CléirighCathal Ó Cléirigh

363

answered Jan 28 at 21:42

Cathal Ó CléirighCathal Ó Cléirigh

363

363

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Very nice, couldn't agree more. Thank you!
  $endgroup$
  – bear
  Jan 28 at 21:46
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Very nice, couldn't agree more. Thank you!
  $endgroup$
  – bear
  Jan 28 at 21:46
  

  
  
  
  

$begingroup$
Very nice, couldn't agree more. Thank you!
$endgroup$
– bear
Jan 28 at 21:46

$begingroup$
Very nice, couldn't agree more. Thank you!
$endgroup$
– bear
Jan 28 at 21:46

add a comment | 

0

$begingroup$

The probabilities are (starting from $1$),

$$frac1{99},frac{98}{99}frac1{98}=frac1{99},frac{98}{99}frac{97}{98}frac1{97}=frac1{99},cdots$$ and so on until day $99$. As all outcomes are equal, the expectation is the midpoint of $1$ and $99$, i.e. $50$.

share|cite|improve this answer

answered Jan 28 at 21:53

Yves DaoustYves Daoust

131k676229

$endgroup$

add a comment | 

0

$begingroup$

The probabilities are (starting from $1$),

$$frac1{99},frac{98}{99}frac1{98}=frac1{99},frac{98}{99}frac{97}{98}frac1{97}=frac1{99},cdots$$ and so on until day $99$. As all outcomes are equal, the expectation is the midpoint of $1$ and $99$, i.e. $50$.

share|cite|improve this answer

answered Jan 28 at 21:53

Yves DaoustYves Daoust

131k676229

$endgroup$

add a comment | 

0

0

0

$begingroup$

The probabilities are (starting from $1$),

$$frac1{99},frac{98}{99}frac1{98}=frac1{99},frac{98}{99}frac{97}{98}frac1{97}=frac1{99},cdots$$ and so on until day $99$. As all outcomes are equal, the expectation is the midpoint of $1$ and $99$, i.e. $50$.

share|cite|improve this answer

answered Jan 28 at 21:53

Yves DaoustYves Daoust

131k676229

$endgroup$

The probabilities are (starting from $1$),

$$frac1{99},frac{98}{99}frac1{98}=frac1{99},frac{98}{99}frac{97}{98}frac1{97}=frac1{99},cdots$$ and so on until day $99$. As all outcomes are equal, the expectation is the midpoint of $1$ and $99$, i.e. $50$.

share|cite|improve this answer

answered Jan 28 at 21:53

Yves DaoustYves Daoust

131k676229

share|cite|improve this answer

share|cite|improve this answer

answered Jan 28 at 21:53

Yves DaoustYves Daoust

131k676229

answered Jan 28 at 21:53

Yves DaoustYves Daoust

131k676229

answered Jan 28 at 21:53

Yves DaoustYves Daoust

131k676229

131k676229

add a comment | 

add a comment | 

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