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Roll a 6-sided fair die until a 6 appears. Let X = the number of 1's that are rolled. Find Var(X).

Calculating error percentage for rolling diceGenerating a rating between 1 to 5 accounting for number of users who have ratedVariance of sum of linear combinationIs a random variable considered “identical” if it can choose a value from a sample space containing values of differing probabilities?Central Limit Theorem (Normal Approximation to Binomial)Polar form representation of $aX+bY+cZ$ ($X$, $Y$, and $Z$ are complex Gaussian random variable)Conditional Expectation of dice rollsA fair six-sided die is rolled repeatedlyWhat is the probability of NOT rolling 1 or 8 on an 8 sided fair die N times in a row?Let $ X $ Be the Number of Faces that Never Showed Up in $ n $ Dice Rolls - What's $ mathbbE left[ X right] $?

0

$begingroup$

Let X = the number of 1's that are rolled. Find E[X] and Var(X).

I can't seem to calculate Var(X).

I've calculated E[X] = 1.

I let R = the number of non-6 rolls, and I let Y = the number of rolls . R + 1 = Y. E[R + 1] = E[Y]. E[R] = E[Y] - 1. E[Y] = 6 because Y ~ Geom(1/6), so E[R] = 5.

R = $R_1 + R_2 + R_3 + R_4 + R_5$, where $R_i$ = the number of times face i has been rolled. All the $R_i$'s are identically distributed, so E[Y] = E[$R_1$] = 1/5E[R] = 1.

statistics random-variables expectation

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asked Mar 11 '14 at 0:55

user2526276user2526276

1

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: en.wikipedia.org/wiki/Law_of_total_variance If you still don't get it, I (or somebody)'ll answer.
  $endgroup$
  – elexhobby
  Mar 11 '14 at 1:46
  

  
  
  
  

add a comment | 

0

$begingroup$

Let X = the number of 1's that are rolled. Find E[X] and Var(X).

I can't seem to calculate Var(X).

I've calculated E[X] = 1.

I let R = the number of non-6 rolls, and I let Y = the number of rolls . R + 1 = Y. E[R + 1] = E[Y]. E[R] = E[Y] - 1. E[Y] = 6 because Y ~ Geom(1/6), so E[R] = 5.

R = $R_1 + R_2 + R_3 + R_4 + R_5$, where $R_i$ = the number of times face i has been rolled. All the $R_i$'s are identically distributed, so E[Y] = E[$R_1$] = 1/5E[R] = 1.

statistics random-variables expectation

share|cite|improve this question

asked Mar 11 '14 at 0:55

user2526276user2526276

1

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: en.wikipedia.org/wiki/Law_of_total_variance If you still don't get it, I (or somebody)'ll answer.
  $endgroup$
  – elexhobby
  Mar 11 '14 at 1:46
  

  
  
  
  

add a comment | 

$begingroup$

Let X = the number of 1's that are rolled. Find E[X] and Var(X).

I can't seem to calculate Var(X).

I've calculated E[X] = 1.

I let R = the number of non-6 rolls, and I let Y = the number of rolls . R + 1 = Y. E[R + 1] = E[Y]. E[R] = E[Y] - 1. E[Y] = 6 because Y ~ Geom(1/6), so E[R] = 5.

R = $R_1 + R_2 + R_3 + R_4 + R_5$, where $R_i$ = the number of times face i has been rolled. All the $R_i$'s are identically distributed, so E[Y] = E[$R_1$] = 1/5E[R] = 1.

statistics random-variables expectation

share|cite|improve this question

asked Mar 11 '14 at 0:55

user2526276user2526276

1

$endgroup$

share|cite|improve this question

asked Mar 11 '14 at 0:55

user2526276user2526276

1

share|cite|improve this question

asked Mar 11 '14 at 0:55

user2526276user2526276

1

asked Mar 11 '14 at 0:55

user2526276user2526276

1

asked Mar 11 '14 at 0:55

user2526276user2526276

1

1 Answer
1

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0

$begingroup$

For roll $i$, let $X_i=1$ if roll $i$ of the die is 1 and 0 otherwise. Hence,
$$R_1 = sumlimits_i=1^R X_i$$

The dice cannot roll 6 in the first $R$ rolls, so $X_i=1$ with probability 1/5 and 0 with probability 4/5. Hence, $$E[R_1|R] = sumlimits_i=1^R E[X_i|R] = frac15R$$

$$Var[R_1|R] = sumlimits_i=1^R Var[X_i|R] = frac15frac45R$$

By the law of total variance,
beginalign*
Var[R_1] &= Var[E[R_1|R]] + E[Var[R_1|R]]\
&= Var[fracR5] + E[frac425R]\
&= frac125Var[R] + frac425E[R]\
&= frac125frac1-1/61/36 + frac4255\
&= 2
endalign*

share|cite|improve this answer

answered Mar 11 '14 at 2:25

elexhobbyelexhobby

929515

$endgroup$

add a comment | 

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0

$begingroup$

For roll $i$, let $X_i=1$ if roll $i$ of the die is 1 and 0 otherwise. Hence,
$$R_1 = sumlimits_i=1^R X_i$$

The dice cannot roll 6 in the first $R$ rolls, so $X_i=1$ with probability 1/5 and 0 with probability 4/5. Hence, $$E[R_1|R] = sumlimits_i=1^R E[X_i|R] = frac15R$$

$$Var[R_1|R] = sumlimits_i=1^R Var[X_i|R] = frac15frac45R$$

By the law of total variance,
beginalign*
Var[R_1] &= Var[E[R_1|R]] + E[Var[R_1|R]]\
&= Var[fracR5] + E[frac425R]\
&= frac125Var[R] + frac425E[R]\
&= frac125frac1-1/61/36 + frac4255\
&= 2
endalign*

share|cite|improve this answer

answered Mar 11 '14 at 2:25

elexhobbyelexhobby

929515

$endgroup$

add a comment | 

0

$begingroup$

For roll $i$, let $X_i=1$ if roll $i$ of the die is 1 and 0 otherwise. Hence,
$$R_1 = sumlimits_i=1^R X_i$$

The dice cannot roll 6 in the first $R$ rolls, so $X_i=1$ with probability 1/5 and 0 with probability 4/5. Hence, $$E[R_1|R] = sumlimits_i=1^R E[X_i|R] = frac15R$$

$$Var[R_1|R] = sumlimits_i=1^R Var[X_i|R] = frac15frac45R$$

By the law of total variance,
beginalign*
Var[R_1] &= Var[E[R_1|R]] + E[Var[R_1|R]]\
&= Var[fracR5] + E[frac425R]\
&= frac125Var[R] + frac425E[R]\
&= frac125frac1-1/61/36 + frac4255\
&= 2
endalign*

share|cite|improve this answer

answered Mar 11 '14 at 2:25

elexhobbyelexhobby

929515

$endgroup$

add a comment | 

$begingroup$

For roll $i$, let $X_i=1$ if roll $i$ of the die is 1 and 0 otherwise. Hence,
$$R_1 = sumlimits_i=1^R X_i$$

The dice cannot roll 6 in the first $R$ rolls, so $X_i=1$ with probability 1/5 and 0 with probability 4/5. Hence, $$E[R_1|R] = sumlimits_i=1^R E[X_i|R] = frac15R$$

$$Var[R_1|R] = sumlimits_i=1^R Var[X_i|R] = frac15frac45R$$

By the law of total variance,
beginalign*
Var[R_1] &= Var[E[R_1|R]] + E[Var[R_1|R]]\
&= Var[fracR5] + E[frac425R]\
&= frac125Var[R] + frac425E[R]\
&= frac125frac1-1/61/36 + frac4255\
&= 2
endalign*

share|cite|improve this answer

answered Mar 11 '14 at 2:25

elexhobbyelexhobby

929515

$endgroup$

share|cite|improve this answer

answered Mar 11 '14 at 2:25

elexhobbyelexhobby

929515

answered Mar 11 '14 at 2:25

elexhobbyelexhobby

929515

answered Mar 11 '14 at 2:25

elexhobbyelexhobby

929515