Csdrhrt

The question:

Suppose we roll a fair 6 sided die with the number [1,6] written on them. After the first die roll we roll the die $k$ times where $k$ is the number on the first die roll. The number of points you score is the sum of the face-values on all die rolls (including the first). What is the expected number of points you will score?

Attempt:

There are six different cases for the first roll

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  Case 1: Roll 1

  
  
  
  
  
  • 
    $P(D_1=1) =frac{1}{6}$
  
  

  • The expected number on the next roll is $E(D_2)=frac{1}{6}times(1+2+3+4+5+6)=3.5$

  
  

  • $E(S_1) = 3.5+1=4.5$ points

  
  
  
  


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  Case 2: Roll 2

  
  
  
  
  

  • We know what the expected value of one roll is, and since rolling the die is independent we can use the previous expected value for the next roll.

  
  

  • $E(S_2) = (3.5times 2)+2=9$ points

  
  
  
  


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  Case 3: Roll 3

  
  
  
  
  
  • 
    $E(S_3) = (3.5times 3)+3=13.5$ points
  
  
  
  


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  Case 4: Roll 4

  
  
  
  
  
  • 
    $E(S_4) = (3.5times 4)+4=18$ points
  
  
  
  


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  Case 5: Roll 5

  
  
  
  
  
  • 
    $E(S_5) = (3.5times 5)+5=22.5$ points
  
  
  
  


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  Case 6: Roll 6

  
  
  
  
  
  • 
    $E(S_6) = (3.5times 6)+6=27$ points
  
  
  
  

Therefore, the expected number of points scored is

$=frac{1}{6}times(4.5+9+13.5+18+22.5+27)=15.75$ points

The expected value of a sum of random number $N$ of iid random variables $X_i$ is
$$Eleft[sum_{i=1}^N X_iright]=E[N]E[X_i]$$
In your case you add $E[N]$, so the answer is
$$E[N]E[X_i]+E[N]=3.5cdot 3.5+3.5 =4.5cdot 3.5 = 15.75$$