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You continually roll a fair $10$ sided dice. What is the expected number of rolls until the lowest common multiple of all numbers that have appeared is greater than $2000$?

The primes in the numbers $1$ to $10$ are $2,3,5,7$. The lowest common multiple of these numbers is $210$. However, different numbers could come in different frequencies (one $9$ is worth two $3$s). How can you deal with this?

I am happy for solutions with approximate answers. The true value by simluation is around $18.8$.

Trajan
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2 Answers2

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The least common multiple of all the numbers $1-10$ is $2520$. Since half of this number is less than $2000$, it follows that you need to achieve $2520$ as an lcm, which means you need to see rolls of $7, 8, 9,$ and one of either $5$ or $10$. Other rolls are irrelevant to getting $2520$ as lcm.

Let $T$ be the number of rolls it takes to see a $7, 8, 9,$ and either a $5$ or $10$. Then $T$ can be written as $$T = X_1 + X_2 + X_3 + X_4,$$ where $X_1, X_2, X_3, X_4$ are independent geometric r.v.'s representing the time to get the first, second, third, and fourth of the four necessary prime factors of $2520$. $X_1$ always has a parameter of $\frac{5}{10}$, but the parameters of $X_2, X_3, X_4$ depend on whether the $5$ or $10$ has been rolled yet after the first, second, third prime factors of $2520$ have been collected.
Let $A_i$, $i = 1, 2, 3, 4$, be the event that the $5$ or $10$ is the $i$th to be achieved out of the $4$ types of roll necessary.
Then we can use conditional expectations to calculate $$\Bbb{E}[T] = \sum_{i=1}^4 \Bbb{E}[T | A_i] \Bbb{P}(A_i).$$ The probabilities of the $A_i$ are easy to find: \begin{align*} \Bbb{P}(A_1) &= \frac{2}{5}, \\ \Bbb{P}(A_2) &= \frac{3}{5} \times \frac{2}{4} = \frac{3}{10}, \\ \Bbb{P}(A_3) &= \frac{3}{5} \times \frac{2}{4} \times \frac{2}{3} = \frac{1}{5}, \\ \Bbb{P}(A_4) &= \frac{3}{5} \times \frac{2}{4} \times \frac{1}{3} = \frac{1}{10}. \ \end{align*}

Conditioned on the $A_i$, we also find \begin{align*} \Bbb{E}[T | A_1] &= \sum_{j=1}^4 \Bbb{E}[X_j | A_1] = \frac{10}{5} + \frac{10}{3} + \frac{10}{2} + \frac{10}{1} = \frac{61}{3}, \\ \Bbb{E}[T | A_2] &= \sum_{j=1}^4 \Bbb{E}[X_j | A_2] = \frac{10}{5} + \frac{10}{4} + \frac{10}{2} + \frac{10}{1} = \frac{39}{2}, \\ \Bbb{E}[T | A_3] &= \sum_{j=1}^4 \Bbb{E}[X_j | A_3] = \frac{10}{5} + \frac{10}{4} + \frac{10}{3} + \frac{10}{1} = \frac{107}{6}, \\ \Bbb{E}[T | A_4] &= \sum_{j=1}^4 \Bbb{E}[X_j | A_4] = \frac{10}{5} + \frac{10}{4} + \frac{10}{3} + \frac{10}{2} = \frac{77}{6}, \\ \end{align*}

and putting it all together, \begin{align*} \Bbb{E}[T] &= \sum_{i=1}^4 \Bbb{E}[T | A_i] \Bbb{P}(A_i) \\ &= \frac{2}{5} \times \frac{61}{3} + \frac{3}{10} \times \frac{39}{2} + \frac{1}{5} \times \frac{107}{6} + \frac{1}{10} \times \frac{77}{6} \\ &= \frac{113}{6} \approx 18.83333... \ \end{align*}

So, it should take a little less than $19$ rolls, on average, to see an lcm of $2520$.

Rivers McForge
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    You need either a $5$ or a $10$, not just a $5$, among your rolls – Ross Millikan Oct 12 '21 at 20:15
  • why cant you get three 2s instead of an 8? – Trajan Oct 12 '21 at 20:30
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    @Trajan, the lcm of $2,2,$ and $2$ is $2$, not $8$. – TonyK Oct 13 '21 at 11:31
  • @TonyK my mistake sorry – Trajan Oct 13 '21 at 11:32
  • I think @RossMillikan's comment suddenly makes the problem more difficult! – TonyK Oct 13 '21 at 11:36
  • @Trajan, perhaps you should unaccept it? – TonyK Oct 13 '21 at 11:37
  • @TonyK i have done so – Trajan Oct 13 '21 at 11:39
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    Because either a $5$ or a $10$ can supply the necessary factor $5$, instead of an abbreviated version of the coupon collector's problem this is an abbreviated version of the coupon collector's problem with unequal probabilities. Expected number of rolling a pair of dice to generate all possible sums may be relevant. – David K Oct 13 '21 at 11:39
  • This solution is incorrect and should be edited or deleted. As has been remarked, it is not true that you need to see a $5$, a $10$ would suffice. – lulu Oct 13 '21 at 13:54
  • +1 for the insight that you need to observe at least one each 7, 8, 9 and 5 (or 10), even if the answer as currently written contains a mistake due to neglecting the possibility of obtaining 5 as a factor by rolling 10. – Ilmari Karonen Oct 13 '21 at 19:58
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    OK, I think I fixed the analysis above. My updated answer of $\frac{113}{6} \approx 18.8333...$ agrees much more closely with the OP's average from simulations. – Rivers McForge Oct 14 '21 at 04:18
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    @RiversMcForge Can you clarify how to obtain the conditioning results? In my understanding, when you calculate E[T|A1] = 10/5+10/3+10/2+10, the 10/5 allows for the probability that a 7, 8, or 9 occurs before the first (5 or 10); as such, why isn't the correct expression E[T|A1] = 7/2+10/3+10/2+10, since the first term can't be a 7, 8, or 9? – Cyclopropane Sep 13 '23 at 21:15
  • @Cyclopropane The idea is that the expected waiting time prior to seeing one of ${ 5, 10 }$ never changes under any scenario, and seeing those only increases the expected waiting times after they've been seen. You can also check that your way doesn't work directly from the unconditional known value of $E[X_1] = 2$: your method would give $$E[X_1 | A_1] P(A_1) + E[X_1 | A_1^c] P(A_1^c) = \frac{7}{2} * \frac{2}{5} + \frac{8}{3} * \frac{3}{5} = 3,$$ which is wrong. – Rivers McForge Sep 16 '23 at 14:01
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(sketch of the method)

Best to proceed by states.

We remark that $$\text{lcm}(2,3,4,5,6,7,8,9,10)=2^3\times 3^2\times 5\times 7=2520$$ is so close to $2000$ that no divisor of it will suffice. It follows that you need to see the $8$, the $9$, the $7$ and either a $5$ or a $10$.

Introduce states $(a,b,c,d)$ according to whether or not the relevant prime power, $(2^3,3^2,5,7)$, has been fully seen. Here $a,b,c,d\in \{0,1\}$ and, $(1,1,0,0)$, say, means that you have seen $8,9$ but none of $5,7,10$. We remark that there are $16$ states which we can number according to the binary representations of the numbers $0,\cdots, 15$ in an obvious way. Since $14 =1110_2$ we'd have state $14$ be $(1,1,1,0)$, meaning that you have only to see the $7$. Similarly, the state $8=(1,0,0,0)$ would mean that you had seen the $8$ roll but sill needed the $9$, the $7$ and either the $5$ or the $10$.

Let $E_i$ be the expected number of tosses it will take given that you are starting in state $i$. Of course, the answer to the question is $E_0$.
We see that $E_{15}=0$ since state $15$ means you have already seen enough tosses.

$E_{14}=10$ since you expect it to take $10$ tosses to see the $7$.

$E_{13}=5$ since you expect it to take $5$ tosses to see either the $5$ or the $10$.

For $E_{12}=E_{(1,1,0,0)}$ we note that $$E_{12}=\frac 7{10}\times (E_{12}+1)+\frac 2{10}\times (E_{14}+1)+\frac 1{10}\times (E_{13}+1)\implies$$ $$\implies E_{12}=\frac {35}3$$

Continue in this manner. As you see, the computations are tedious but not difficult. They are also a bit error prone, so I suggest simulation to confirm the result.

lulu
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