Problem 6.2 Easy Free 3 variations

One Total Before Another

You roll a pair of fair six-sided dice repeatedly. Find the probability that a sum of $5$ appears before a sum of $2$ appears.

Work it out, then compare with the answer.

SolutionShow the worked solution

Key idea. Rolls showing neither total decide nothing, so only the first roll that shows a $5$ or a $2$ matters.

Of the $36$ equally likely ordered outcomes, four give a sum of $5$ : $(1,4),(2,3),(3,2),(4,1)$ . Only $(1,1)$ gives a sum of $2$ . So on each roll

\mathbb{P}(\text{sum }5)=\tfrac{4}{36},\qquad \mathbb{P}(\text{sum }2)=\tfrac{1}{36},\qquad \mathbb{P}(\text{neither})=\tfrac{31}{36}.

The rolls are independent. A sum of $5$ comes first exactly when the first $n-1$ rolls show neither total and roll $n$ shows a $5$ , for some $n\ge 1$ . Summing over $n$ ,

\mathbb{P}(5\text{ before }2)=\sum_{n\ge1}\Big(\tfrac{31}{36}\Big)^{n-1}\tfrac{4}{36} =\frac{4/36}{1-31/36}=\frac{4/36}{4/36+1/36}=\frac45 .

The same computation shows that the probability of neither total ever appearing is $\lim_n (31/36)^n=0$ , so the race is decided with probability $1$ .

In general, for repeated independent trials, $\mathbb{P}(A\text{ before }B)=\mathbb{P}(A)/(\mathbb{P}(A)+\mathbb{P}(B))$ .

\boxed{\tfrac45}

Variations

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

Variation 6.2a

Dave and Mike each roll a fair six-sided die simultaneously, repeating the following procedure: if the two values match, Mike wins outright; if Mike’s value is strictly greater than Dave’s, Dave wins outright; if Dave’s value is strictly greater than Mike’s, both dice are re-rolled and the procedure repeats. Find the probability that Mike eventually wins.

Work it out, then compare with the answer.

Solution

One round of two dice has three possible results:

the values match, with probability $\tfrac{6}{36}=\tfrac16$ , and Mike wins;
Mike’s value is higher, with probability $\tfrac{15}{36}=\tfrac5{12}$ , and Dave wins;
Dave’s value is higher, with the remaining probability $\tfrac5{12}$ , and the dice are re-rolled.

Only the round that ends the game matters. Applying the rule from the main problem,

\mathbb{P}(\text{Mike wins})=\frac{1/6}{1/6+5/12}=\frac{2/12}{7/12}=\frac27 .

\boxed{\tfrac27}

Variation 6.2b

Bill and Bob simultaneously and repeatedly flip coins that each land heads with probability $1/3$ , independently of each other and of past flips. Find the probability that Bill’s first head occurs on a strictly earlier flip than Bob’s first head (so a round in which both flip heads for the first time simultaneously does not count as a win for Bill).

Work it out, then compare with the answer.

Solution

Treat each pair of simultaneous flips as one round. Each coin lands heads with probability $\tfrac13$ , independently, so a round has four possible results:

Bill heads and Bob tails, with probability $\tfrac13\cdot\tfrac23=\tfrac29$ , and Bill wins;
Bob heads and Bill tails, with probability $\tfrac29$ , and Bill loses;
both heads, with probability $\tfrac19$ , and Bill loses;
both tails, with probability $\tfrac49$ , and play continues.

Only the first round that is not both tails decides the outcome. Hence

\mathbb{P}(\text{Bill first})=\frac{2/9}{2/9+2/9+1/9}=\frac25 .

\boxed{\tfrac25}

Variation 6.2c

You repeatedly roll two fair dice until you get either a double six or exactly one six (one die showing $6$ , the other not); any other outcome is rerolled. If the game stops on a double six you receive $100; if it stops on exactly one six you lose $x. Find the largest value of $x$ for which this game has non-negative expected value.

Work it out, then compare with the answer.

Solution

On each roll:

a double six has probability $\tfrac1{36}$ ;
exactly one six has probability $\tfrac{10}{36}$ , since the six can be on either die and the other die shows one of $5$ values.

Every other outcome is rerolled. So the game ends on a double six with probability

\frac{1/36}{11/36}=\frac1{11},

and on exactly one six with probability $\tfrac{10}{11}$ .

The expected value of the game is

\frac{100}{11}-\frac{10x}{11}.

This is non-negative exactly when $x\le 10$ .

\boxed{x=10}

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