Problem 6.1 Easy Free 5 variations

Alternating Until First Success

Two players alternate flipping their own coins, hoping to be first to flip heads. Annie’s coin shows heads with probability $1/3$ and Brittany’s with probability $2/5$ . Annie flips first, then Brittany, and they keep alternating until someone gets heads; that person wins. Find the probability that Annie wins.

Work it out, then compare with the answer.

SolutionShow the worked solution

Let $p$ be the probability that Annie wins. Condition on the first round. Annie wins at once with probability $\tfrac13$ . With probability $\tfrac23\cdot\tfrac35=\tfrac25$ both players flip tails. The flips are independent, so the game then starts afresh with Annie to flip, and her winning probability is again $p$ . In every other case Brittany wins. Hence

p=\frac13+\frac25\,p \quad\Longrightarrow\quad \frac35\,p=\frac13 \quad\Longrightarrow\quad p=\frac59 .

\boxed{\mathbb{P}(\text{Annie wins})=\tfrac59}

Generalisation. Suppose the first player’s coin shows heads with probability $a$ and the second player’s with probability $b$ . The same argument gives $\mathbb{P}(\text{first player wins})=\dfrac{a}{1-(1-a)(1-b)}=\dfrac{a}{a+b-ab}$ .

Variations

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

Variation 6.1a

You and an opponent will play the following game: you each take turns flipping a fair coin, with you going first. If a player flips heads, the other player must pay them $30; if a player flips tails, the coin simply passes to the other player, and play continues. The game ends as soon as someone flips heads. Before the game starts, you are offered the chance to pay a fee for the right to be the one who flips first (instead of it being assigned by some other means). What is the largest fee you should be willing to pay for this right?

Work it out, then compare with the answer.

Solution

By the generalisation with $a=b=\tfrac12$ , the first flipper wins with probability $\dfrac{1/2}{1-1/4}=\tfrac23$ .

Going first therefore has expected value

30\cdot\tfrac23-30\cdot\tfrac13=10 .

If the order is assigned by some neutral means, such as a fair toss, the game is symmetric and has expected value $0$ to you. The right to go first is therefore worth $\$10$ .

If the only alternative were being forced to go second, worth $-\$10$ , the right would instead be worth $\$20$ .

\boxed{\$10}

Variation 6.1b

Abby and Ben alternately flip coins, with Abby going first, continuing until someone flips heads; that person wins. Abby’s coin shows heads with probability $p$ , while Ben’s coin is fair. Find the value of $p$ for which Abby and Ben are equally likely to win.

Work it out, then compare with the answer.

Solution

Take $a=p$ and $b=\tfrac12$ . Then

\mathbb{P}(\text{Abby wins})=\frac{p}{1-\frac{1-p}{2}}=\frac{2p}{1+p}.

Setting this equal to $\tfrac12$ gives $4p=1+p$ , so $p=\tfrac13$ .

\boxed{p=\tfrac13}

Variation 6.1c

Three riflemen, $A$ , $B$ , and $C$ , take turns shooting at a target in the cyclic order $A, B, C, A, B, C, \dots$ , stopping as soon as someone hits it. Each shot is independent and hits with probability $1/2$ . Find the probability that $A$ is the one who hits the target.

Work it out, then compare with the answer.

Solution

Let $q$ be the probability that $A$ hits.

For $B$ to hit, $A$ must first miss, which has probability $\tfrac12$ . After that, $B$ is in exactly $A$ ’s original position. So $\mathbb{P}(B)=\tfrac12 q$ . By the same reasoning, $\mathbb{P}(C)=\tfrac14 q$ .

Someone hits with probability $1$ , so

q\left(1+\tfrac12+\tfrac14\right)=1 \quad\Longrightarrow\quad q=\tfrac47 .

\boxed{\tfrac47}

Variation 6.1d

Two players alternately roll a fair six-sided die, keeping a running total of all rolls made by either player. Whoever’s roll first brings the running total to a multiple of $7$ wins. Before play begins, you may choose whether to roll first or second. Assuming you choose optimally, what is your probability of winning?

Work it out, then compare with the answer.

Solution

The first roll. It gives a total between $1$ and $6$ , which is never a multiple of $7$ , so the first roll cannot win.

Every later roll. Suppose the current total $S$ is not a multiple of $7$ . The six possible new totals $S+1,\dots,S+6$ cover every residue mod $7$ except that of $S$ . That residue is nonzero, so exactly one face makes the total a multiple of $7$ . Each later roll therefore wins with probability $\tfrac16$ , independently of the past.

Choosing to go second. You then take the first roll that can win. Your winning probability $w$ satisfies

w=\tfrac16+\left(\tfrac56\right)^2 w \quad\Longrightarrow\quad w=\frac{1/6}{11/36}=\frac{6}{11},

whereas going first gives only $\tfrac5{11}$ . So you choose to go second.

\boxed{\tfrac6{11}}

Variation 6.1e

Alice and Bob alternately flip a fair coin, with Alice going first, each taking one flip per turn before passing the coin to the other. The game ends as soon as one player has, over the course of their own flips, obtained at least one heads and at least one tails (in either order); that player wins. Find the probability that Alice wins.

Work it out, then compare with the answer.

Solution

First flips. No one can win on their first flip. That flip only fixes the face the player still needs.

Later flips. From each player’s second flip onwards, every flip wins with probability $\tfrac12$ , independently.

Alice’s chance. Alice makes the first such flip, so her winning probability $P$ satisfies

P=\tfrac12+\tfrac14 P \quad\Longrightarrow\quad P=\tfrac23 .

\boxed{\tfrac23}

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