# Rolling a fair 6 sided die k times

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

• 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

• 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.

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

• Case 3: Roll 3

• $$E(S_3) = (3.5\times 3)+3=13.5$$ points
• Case 4: Roll 4

• $$E(S_4) = (3.5\times 4)+4=18$$ points
• Case 5: Roll 5

• $$E(S_5) = (3.5\times 5)+5=22.5$$ points
• Case 6: Roll 6

• $$E(S_6) = (3.5\times 6)+6=27$$ 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

• I think this is a law of total expectation problem, $E[X]=E[E(X|K)]$. Where $K$ is the face of the first roll, and $X$ is the sum of the $K$ rolls – gd1035 Nov 16 '18 at 15:49
• I think your attempt is correct. That is how I would have tried it anyways. – Sauhard Sharma Nov 16 '18 at 16:22

The expected value of a sum of random number $$N$$ of iid random variables $$X_i$$ is $$E\left[\sum_{i=1}^N X_i\right]=E[N]E[X_i]$$ In your case you add $$E[N]$$, so the answer is $$E[N]E[X_i]+E[N]=3.5\cdot 3.5+3.5 =4.5\cdot 3.5 = 15.75$$
• Why do we add $E[N]$ to the summation? – Arthur Green Nov 16 '18 at 19:13
• Because you include the 1st (actually zeroth) roll which determines $N$ into the sum, and because of linearity of expectations it adds $E[N]$. – kludg Nov 16 '18 at 20:05