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When throwing a die $3$ times, find the probability of the product of the three results being a multiple of $9$.

I tried drawing a table for the possible outcomes of the first TWO dice. Then I tried considering the cases for each when the third die is rolled - as you can imagine not very efficient. Perhaps I could use $x+y+z = 9k$ as a divisibility test, where $k$ is an integer. I'm pretty sure this would involve stars and bars, but I'm not sure how to implement it in this case. Would I have to solve for $k$?

I also worked out the boundaries but I'm not sure if that will help either: $9≤x+y+z≤216$.

Do I need to use stars and bars for this question?

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    $\begingroup$ The product is a multiple of $9$ if and only if at least two of the rolls are a multiple of $3$. Does that help? $\endgroup$ – diracdeltafunk Mar 7 at 4:53
  • $\begingroup$ Take two of the rolls as 3 or 6, then multiply all cases of the third roll. $\endgroup$ – Righter Mar 7 at 4:55
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    $\begingroup$ You can't use stars and bars as assignments using stars and bars aren't equiprobable $\endgroup$ – true blue anil Mar 7 at 5:03
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    $\begingroup$ Where did you get $x + y + z$? Nothing is added in this problem. The numbers on each roll of the die are multiplied. $\endgroup$ – David K Mar 7 at 5:53
  • $\begingroup$ @DavidK oh yes sorry, I was using the divisibility rule that if the sum of the digits is divisible by $9$, then the entire number is divisible by $9$. Hence $ 𝑥+𝑦+𝑧=9𝑘$ where $k$ is integral $\endgroup$ – user71207 Mar 7 at 6:32
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I do not agree with this solution.

I obtained the result $\frac{7}{27}$

In fact, the probability to have a product divisible by 9 means that, among 3 die's rolls, at least 2 must be #$3$ or #$6$, thus

$$\binom{3}{2}\left(\frac{1}{3}\right)^2\cdot\frac{2}{3}+\left(\frac{1}{3}\right)^3=\frac{7}{27}$$

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  • $\begingroup$ Thanks, this makes it look so easy. For some reason divisibility by $9$ meaning two threes or two sixes didn't come across to me $\endgroup$ – user71207 Mar 7 at 10:13
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You can solve it quickly by counting the cases when the product is not divisible by $9$.

  • Number of all throws: $\color{blue}{6^3}$
  • Number of throws containing neither $3$ nor $6$: $\color{blue}{4^3}$
  • Number of throws containing exactly one $3$: $\color{blue}{3\cdot 4^2}$
  • Number of throws containing exactly one $6$: $\color{blue}{3\cdot 4^2}$

Hence, the probability $P$ of a throw with a product divisible by $9$ is

$$P = \color{blue}{\frac{6^3-4^3 -3\cdot 4^2 - 3\cdot 4^2}{6^3}}= \frac{7}{27}$$

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I agree with tommik's solution. Here is a simulation in R to verify.

# Number of simulations
nsims <- 1e7

roll <- function() sample(1:6, size=1)

# Generate products
products <- purrr::map_dbl(1:nsims, ~roll()*roll()*roll())

# Get proportion divisible by 9
sum((products %% 9) == 0) / nsims
0.2592488

which is about $\frac{7}{27} = 0.259259259...$

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