Problem 21.1 Easy Free 8 variations

Reroll a Die Once

A fair six-sided die is rolled once. You may either accept the payout equal to the value shown, or discard it and roll the die a second time, in which case you are paid the value of the second roll instead. Assuming you play optimally, find the fair value (expected payout) of this game.

Work it out, then compare with the answer.

SolutionShow the worked solution

After the first roll, the decision is final. Rerolling gives the value of a fresh roll, whose expectation is

\mathbb{E}[X_2]=\frac{1+2+\dots+6}{6}=\frac72 .

Seeing the first roll $X_1=x$ , you therefore keep it if $x>\tfrac72$ and reroll if $x<\tfrac72$ . So you keep $4,5,6$ and reroll $1,2,3$ . The value of the game is

\mathbb{E}\!\left[\max\!\left(X_1,\tfrac72\right)\right] =\frac16\left(3\cdot\frac72+4+5+6\right) =\frac16\cdot\frac{51}{2}=\frac{17}{4}.

\boxed{\dfrac{17}{4}=4.25}

Key idea. With one decision left, the value is $\mathbb{E}[\max(\text{current payout},\ \text{continuation value})]$ .

Variations

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

Variation 21.1a

A fair $8$ -sided die is rolled once. You may keep the payout equal to the value shown, or instead roll a fair $12$ -sided die once and be paid its value. Find your expected payout under optimal play.

Work it out, then compare with the answer.

Solution

The continuation value is now the mean of a $12$ -sided die, $\frac{13}{2}$ . Keep the $8$ -sided roll only if it shows $7$ or $8$ ; otherwise roll the $12$ -sided die. The value is

\frac{7+8}{8}+\frac68\cdot\frac{13}{2}=\frac{15}{8}+\frac{39}{8}=\frac{27}{4}.

\boxed{\dfrac{27}{4}=6.75}

Variation 21.1b

A fair $6$ -sided die is rolled. If the face shown is odd, the payout equals the face value; if it is even, the payout equals twice the face value. You may keep this payout or reroll the die once and accept the payout (under the same odd/even rule) from the new roll. Find the expected payoff under optimal play.

Work it out, then compare with the answer.

Solution

The possible payouts are $1,4,3,8,5,12$ . Their mean is $\frac{33}{6}=\frac{11}{2}$ , which is the continuation value. Keep only the payouts $8$ and $12$ (faces $4$ and $6$ ) and reroll otherwise:

\frac{8+12}{6}+\frac46\cdot\frac{11}{2}=\frac{20}{6}+\frac{22}{6}=7.

\boxed{7}

Variation 21.1c

A number is drawn uniformly at random from $(0,1)$ . You may keep this number as your payout, or discard it and draw a new uniform number from $(0,1)$ once, in which case you are paid that new number. Find your expected payout under optimal play.

Work it out, then compare with the answer.

Solution

A fresh draw is worth $\frac12$ on average, so keep $U_1$ if and only if $U_1>\frac12$ . The value is

\mathbb{E}\!\left[\max\!\left(U_1,\tfrac12\right)\right]=\int_0^{1/2}\tfrac12\,du+\int_{1/2}^1 u\,du=\frac14+\frac38=\frac58.

\boxed{\dfrac58}

Variation 21.1d

Alice rolls a fair $6$ -sided die and observes the value shown. She may stop and receive that value as her payout, or pay $\$1$ to roll the die a second time and receive the value of that second roll instead. Assuming Alice plays optimally, find her expected payout.

Work it out, then compare with the answer.

Solution

Rolling again is worth $\frac72-1=\frac52$ net of the cost. Alice therefore keeps $3,4,5,6$ and rerolls on $1$ or $2$ . Her expected net payout is

\frac{3+4+5+6}{6}+\frac26\cdot\frac52=\frac{18}{6}+\frac56=\frac{23}{6}.

\boxed{\dfrac{23}{6}\approx 3.83}

Variation 21.1e

Ten cards numbered $1$ through $10$ lie face down. You draw one card at random and look at its value. You may take a fixed payout of $\$3.50$ , or instead take the face value of the card. Find the fair value of this game under optimal play.

Work it out, then compare with the answer.

Solution

Take the face value when it exceeds $3.5$ , which happens for cards $4$ to $10$ . Take the fixed $\$3.50$ for cards $1$ to $3$ . The value is

\frac{3}{10}\cdot\frac72+\frac{4+5+\dots+10}{10}=\frac{21}{20}+\frac{49}{10}=\frac{119}{20}.

\boxed{\$5.95}

Variation 21.1f

Two fair $6$ -sided dice are rolled, and you observe the product of the two values. You may keep this product as your payout, or pay $\$4$ to reroll both dice once and take the product of the new roll instead. Find your expected payout under optimal play.

Work it out, then compare with the answer.

Solution

By independence, $\mathbb{E}[XY]=\left(\frac72\right)^2=\frac{49}{4}$ , so rerolling is worth $\frac{49}{4}-4=\frac{33}{4}=8.25$ . You therefore reroll exactly when the product is at most $8$ .

Count the pairs with product at most $8$ by fixing the first die:

first die $1$ : $6$ pairs, with products summing to $21$ ;
first die $2$ : $4$ pairs, summing to $20$ ;
first die $3$ : $2$ pairs, summing to $9$ ;
first die $4$ : $2$ pairs, summing to $12$ ;
first die $5$ : $1$ pair, summing to $5$ ;
first die $6$ : $1$ pair, summing to $6$ .

That gives $16$ pairs whose products sum to $73$ . Over all $36$ pairs the products sum to $21^2=441$ , so the kept products sum to $441-73=368$ . The value is

\frac{368}{36}+\frac{16}{36}\cdot\frac{33}{4}=\frac{368+132}{36}=\frac{125}{9}.

\boxed{\dfrac{125}{9}\approx 13.89}

Variation 21.1g

Four fair coins are flipped, and you are paid $\$1$ for each head shown. You also have the option, before being paid, to pay $\$1$ to reflip all four coins once and be paid based on the new outcome instead. Suppose you follow the rule of rerolling whenever the first flip shows $0$ or $1$ heads. Find the fair value of this game under that rule.

Work it out, then compare with the answer.

Solution

Let $H\sim\mathrm{Bin}(4,\tfrac12)$ be the number of heads. Then $\mathbb{P}(H=k)=\binom4k/16$ .

On $H\le1$ (probability $\frac5{16}$ ) you reflip, worth $2-1=1$ on average.
Otherwise you keep $H$ , which contributes $\frac{2\cdot6+3\cdot4+4\cdot1}{16}=\frac{28}{16}$ .

The fair value is

\frac{28}{16}+\frac{5}{16}\cdot1=\frac{33}{16}.

\boxed{\dfrac{33}{16}\approx 2.06}

Variation 21.1h

Eight fair coins are flipped simultaneously. You may then choose, independently for each coin, whether to reflip it once more. After all chosen reflips, you are paid $\$1$ for every coin showing heads. Assuming optimal play, find the fair value of this game.

Work it out, then compare with the answer.

Solution

The payout is a sum over coins, and each reflip is independent of everything else. So it is optimal to decide for each coin separately.

A head is worth $1$ if kept and $\frac12$ if reflipped, so keep it.
A tail is worth $0$ if kept and $\frac12$ if reflipped, so reflip it.

Each coin then ends on heads unless it shows tails twice, so it ends on heads with probability $1-\frac14=\frac34$ . By linearity the value is

8\cdot\frac34=6.

\boxed{6}

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