# Number of Rolls of Die Until Sum Exceeds 6

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 $$\frac{6}{3.5}$$?

• (Dice is plural. You have one die.) For a sum of 7 or more, the simple formula would give exactly 2 expected rolls. But that case always requires at least 2 rolls, and sometimes takes more, so the expected value for that case is definitely more than 2. Commented Nov 15, 2023 at 22:01
• @aschepler (That might have been true long ago, but by now "dice" as singular is so prevalent it can no longer be seen as incorrect.) Commented Nov 15, 2023 at 22:18
• I just go for the pluras dices. And the singular douse. Commented Nov 15, 2023 at 22:23
• @Arthur: I agree with you about informal usage of the word "dice", but if you are writing up mathematics it is good idea to be a little more careful as you will irritate a lot of readers by not making a useful distinction. Commented Nov 15, 2023 at 22:37
• Note that when the expected value of $X$ is $E[X]$, it doesn't mean that the expected value of $1/X$ is $1/E[X]$. For instance, consider a die that rolls $1000$ with probability $1/1000$, and zero otherwise. Its expected value is $1$. How many rolls do you expect to need to exceed $5$? $10$? $100$? Commented Nov 16, 2023 at 1:44

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.

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 $$E_7$$.

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

• I was too lazy to compute by hand so I asked chatgpt to do it. Entering the recurrence and the initial condition + asking to perform the 7 computations manually lead it to the answer of $\frac{1543}{648}\approx 2.38$, which of course has to be triple-checked. Let me know if you get the same result! Commented Nov 15, 2023 at 23:00
• @ALG that cannot be the answer. The probability of less than 2 rolls is 0, and the probability of 2 rolls is 7/12. Since you have a probability of 5/12 of taking at least 3 rolls, it has to be bigger than what you have.
– Paul
Commented Nov 15, 2023 at 23:10
• Yes, it did some blatant errors with the fractions. I ran a program now and got $\dfrac{117649}{46656}\approx2.52$. Commented Nov 15, 2023 at 23:28
• It should not come as a surprise that the expected number of rolls is $\left(\frac76\right)^6$ Commented Nov 15, 2023 at 23:30
• But if we wanted to find the expected rolls to get a sum of 7, would it simply be 2?
– Anon
Commented Nov 15, 2023 at 23:47