Dice counting problem

Question

Six standard 6-sided dice are rolled, and the resulting numbers are multiplied together. What is the probability that the product is divisible by 125?

I know that at least 3 of the dice have to be 5's and the rest can be anything. I had gotten 4320 possibilities.

Answer 1

Your observation that for the product to be divisible by $125$ requires that at least three of the dice to be rolled as a five is correct and needs no correction. Counting the number of favorable cases however has a mistake.

Your attempt probably looked something like this:

• Choose three out of the six positions to be guaranteed to be occupied by $5$'s. This can be done in $\binom{6}{3}$ ways.

• Choose what the remaining dice show. At $6$ options for each die, this can be done in $6^3$ ways.

Multiplying gives $\binom{6}{3}\cdot 6^3 = 4320$.

This is however incorrect as it overcounts cases with more than three fives. For example the sequence $5~5~5~5~1~1$ was counted once when the three initially selected $5$'s were the first three, again when the three initially selected $5$'s were the first, third, and fourth, again when it was... etc...

To correct the count, let us instead count the number of ways of getting exactly three fives, add this to the number of ways of getting exactly four fives, etc...

To get exactly $k$ fives do the following:

• Pick which $k$ dice showed $5$. This can be done in $\binom{6}{k}$ ways

• Pick what the results of the remaining $6-k$ dice are. As none can be a five, they each only have five remaining options giving a total of $5^{6-k}$ ways to complete this step.

Adding from $k=3$ to $k=6$ we get a grand total number of outcomes:

$$\binom{6}{3}5^3 + \binom{6}{4}5^2+\binom{6}{5}5^1+\binom{6}{6}5^0 = 2906$$

These are taken out of the $6^6$ equally likely sequences of six dice tosses for a probability then of $\dfrac{2906}{6^6}$.

Answer 2

Because $5$ is the only multiple of $5$ on a $6$-sided die, the product is divisible by $125 = 5^3$ if and only if at least three of the dice show a $5$. There are $6^6$ equally likely outcomes for the six dice to be rolled, and so we count the number of successful outcomes (where at least three dice show a $5$) using PIE.

There are $\binom{6}{3} \cdot 6^3$ ways to choose three dice that show $5$s and have the other three dice show any numbers.

There are $\binom{6}{4}\cdot 6^2$ ways to choose four dice that show $5$s and have the remaining two show any numbers. We counted each such outcome $\binom{4}{3} = 4$ times earlier, so we want to subtract it $3$ times.

There are $\binom{6}{5}\cdot 6^1$ ways to choose five dice that show $5$s and have the remaining die show any number. We counted each such outcome $\binom{5}{3} - 3\binom{5}{4} = -5$ times earlier, so we want to add it $6$ times.

There is $1$ way to have all six dice show $5$s . We counted this outcome $\binom{6}{3} - 3\binom{6}{4} + 6\binom{6}{5} = 11$ times earlier, so we want to subtract it $10$ times.

Thus, the number of distinct outcomes with at least three $5$s is $\binom{6}{3}\cdot 6^3 - 3\cdot\binom{6}{4}\cdot 6^2 + 6\cdot\binom{6}{5}\cdot 6^1 - 10 \cdot 1 = 2906$, and the probability of getting such an outcome is $$\frac{2906}{6^6} = \frac{1453}{3 \cdot 6^5} = \boxed{\frac{1453}{23328}}.$$