Problem 4.27 Easy Free 3 variations

Dice Sum Divisible by Six

You roll ten fair six-sided dice and sum the values shown. Find the probability that the sum is divisible by $6$ .

Work it out, then compare with the answer.

SolutionShow the worked solution

Key idea. Condition on every die except the last. A single fair die covers every residue modulo $6$ exactly once.

Let $S$ be the sum of the first nine dice and $D$ the value of the tenth. $D$ is independent of $S$ and uniform on $\{1,\dots,6\}$ . Fix $S=s$ . As $D$ runs through $1,\dots,6$ , the sum $s+D$ runs through six consecutive integers. Exactly one of these is divisible by $6$ . Hence

\mathbb{P}(6 \mid S+D \mid S=s)=\tfrac16 \quad\text{for every } s .

Averaging over $s$ gives the same value unconditionally:

\boxed{\tfrac16}

Generalisation. Suppose a sum of independent integer-valued terms contains at least one term that is uniform modulo $k$ . Then the sum is uniform modulo $k$ . In particular, the sum of $n\ge1$ fair $k$ -sided dice is divisible by $k$ with probability $1/k$ .

Variations

The same idea with a twist. Each solution gives only the changed step.

Variation 4.27a

Among $n$ coins laid in front of you, one is fair and the other $n-1$ each show heads independently with probability $0<\lambda<1$ . All $n$ coins are flipped. Find the probability that the total number of heads is even.

Work it out, then compare with the answer.

Solution

Condition on the $n-1$ biased coins. Let $p$ be the probability that they show an even number of heads. The fair coin is independent of them.

The total is even in two ways: the biased coins give an even count and the fair coin shows tails, or they give an odd count and the fair coin shows heads. So

\mathbb{P}(\text{even})=\tfrac12 p+\tfrac12(1-p)=\tfrac12 .

Alternative. For independent coins with head probabilities $p_i$ ,

\mathbb{P}(\text{even})-\mathbb{P}(\text{odd})=\prod_i(1-2p_i).

The fair coin contributes a factor $1-2\cdot\tfrac12=0$ , so the two probabilities are equal.

\boxed{\tfrac12}

Variation 4.27b

Alice has two six-sided dice with faces labelled $1$ through $6$ . One die is fair; the other is loaded so that each face $k$ comes up with probability proportional to $k$ . Alice rolls both dice and adds the results. Find the probability that the sum is odd.

Work it out, then compare with the answer.

Solution

Condition on the loaded die’s value $L=\ell$ . The fair die is independent of $L$ and is odd with probability $\tfrac12$ . So $\ell$ plus the fair die is odd with probability $\tfrac12$ , whatever $\ell$ is.

The loaded die’s distribution therefore plays no role. As a check, $\mathbb{P}(L \text{ odd})=\frac{1+3+5}{21}=\frac37$ , and

\tfrac12\cdot\tfrac37+\tfrac12\cdot\tfrac47=\tfrac12 .

\boxed{\tfrac12}

Variation 4.27c

You flip two fair coins. For each head obtained, you then roll one fair six-sided die. Find the probability that the sum of all the die rolls (zero if no dice are rolled) is odd.

Work it out, then compare with the answer.

Solution

Let $N\sim\mathrm{Bin}(2,\tfrac12)$ be the number of dice rolled.

If $N=0$ , the sum is $0$ , which is even.

If $N\ge1$ , condition on all dice but the last. The last die is odd with probability $\tfrac12$ , independently of the others, so the sum is odd with probability $\tfrac12$ .

Hence

\mathbb{P}(\text{odd})=\mathbb{P}(N\ge1)\cdot\tfrac12=\tfrac34\cdot\tfrac12 .

\boxed{\tfrac38}

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