Number of Rolls of Die Until Sum Exceeds 6

Question

What is the expected number of rolls of a 6-sided die until the sum exceeds 6?

As the expected value of one roll is 3.5, why is the answer not just $\frac{6}{3.5}$?

Answer 1

You have to roll the 🎲 at least once. You'll get some $j\in\{1\ldots 6\}$, and in each such case you'll reduce the total sum you want to achieve by $j$. Each such case has equal probability of occuring $\frac 1 6$.

Therefore, the total expected value is 1 (the first roll) plus the average of the expected values for reaching the reduced total sum in each subcase.

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In symbols, if $E_n$ is the expected number of rolls for achieving sum $n$, the previous sentence reads $$E_n = 1+\frac 1 6 \sum_{j=1}^6E_{n-i}.$$ In the bottom of this recurrence, when $n\le 0$, then $E_n=0$ because you don't have to toss the 🎲 at all to achieve a non-positive sum (the sum is $0$ before any rolls). From here on, I'd pick a calculator/compute by hand $E_1,E_2,\ldots$ until $E_7$.

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As @DanielMathias mentioned in the comments, initially, the sequence seems to behave like $(7/6)^{n-1}$. From 8 on, though, it does not:

$$E_{1}=1, E_{2}=\frac{7}{6}, E_{3}=\frac{49}{36}, E_{4}=\frac{343}{216}, E_{5}=\frac{2401}{1296}, E_{6}=\frac{16807}{7776}, E_{7}=\frac{117649}{46656}, E_{8}=\frac{776887}{279936}, E_{9}=\frac{5111617}{1679616}, E_{10}=\frac{33495175}{10077696}, E_{11}=\frac{218463217}{60466176},$$ or in decimal: $$E_{1}=1.00,E_{2}=1.17,E_{3}=1.36,E_{4}=1.59,E_{5}=1.85,E_{6}=2.16,E_{7}=2.52,E_{8}=2.78,E_{9}=3.04,E_{10}=3.32,E_{11}=3.61,E_{12}=3.91,E_{13}=4.20,E_{14}=4.48,E_{15}=4.76,E_{16}=5.05,E_{17}=5.33,E_{18}=5.62,E_{19}=5.91,E_{20}=6.19.$$ If you want to play with some Haskell: https://ideone.com/GthHDs