Fair Game and Expected Value (Faires Spiel)

15 min · 2 drills

Introduction

A fair-game task describes a random experiment (dice, coins, a spinner, cards or an urn) and attaches an amount of money to some of its outcomes. On some outcomes you receive money, on the rest you have to pay an unknown amount $X$. You are asked for the $X$ that makes the game fair.

The real work is almost always in the probabilities of the individual outcomes. After that, the equation for $X$ takes one line. That is why the tasks often ask first for the probabilities and the probability of losing, and only at the end for $X$.

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The Expected Value of the Gain

Write $V$ for the player's gain in one round. It is a random variable: depending on the outcome it takes one of the values $g_1,g_2,\dots ,g_m$. An amount you receive is positive, an amount you pay is negative.

The expected value (German: Erwartungswert) of the gain weights each value by its probability:

$$E(V)==g_1\cdot \cdot P(V==g_1)++g_2\cdot \cdot P(V==g_2)++\dots ++g_m\cdot \cdot P(V==g_m)$$

$E(V)$ is the average gain per round when the game is played very many times. In a single round this value usually never occurs: you always receive or pay one of the amounts $g_i$.

Example: You roll a die once. On a 6 you receive €4, otherwise you pay €1.

$$E(V)==4\cdot \cdot \frac{1}{6}++(-1)\cdot \cdot \frac{5}{6}==\frac{4-5}{6}==-\frac{1}{6}\approx \approx -0.17\text{€}$$

In the long run the player loses about 17 cents per round. The sign tells you whom the game favours:

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The Condition for a Fair Game

A game is called fair (German: faires Spiel) when the expected value of the gain is zero:

$$\boxed{E(V)=0}$$

In the tasks one amount is unknown, usually the amount $X$ you pay when you lose. If there are gains $g_1,g_2,\dots$ with probabilities $p_1,p_2,\dots$ and the loss $X$ with probability $P(\text{loss})$, the condition reads

$$g_1{p}_1++g_2{p}_2++\dots --X\cdot \cdot P(\text{loss})==0$$

$X$ enters with a minus sign because you pay it. Solved for $X$:

$$\boxed{X=\frac{g_1{p}_1+g_2{p}_2+\dots }{P(\text{loss})}}$$

The numerator is what you win on average per round, the denominator is how often you pay. The rarer the loss, the larger $X$ may be.

When all outcomes are equally likely, the common denominator cancels. Then the counts of outcomes are enough: $n_1,n_2,\dots$ for the gains and $n_X$ for the loss.

$$X==\frac{g_1{n}_1+g_2{n}_2+\dots }{n_X}$$

Procedure

1. Fix the events: which outcomes bring which amount? Every outcome belongs to exactly one event.
2. Find the probabilities, for each event separately.
3. Check: the probabilities add up to $1$. The probability of losing is what is left, $P(\text{loss})=1-p_1-p_2-\dots$
4. Write the ansatz $E(V)=0$, with $-X$ for the loss.
5. Solve for $X$ and round to cents only at the very end.

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Probabilities of the Events

The outcomes of these games are equally likely, so the Laplace rule $P=\frac{\text{favourable outcomes}}{\text{possible outcomes}}$ applies. Count like this:

Two dice thrown "at the same time" are still two different dice. The result "a 2 and a 5" consists of the two outcomes $(2,5)$ and $(5,2)$. "In order 1, then 2", on the other hand, is a single outcome.

Counting example with four dice: how many of the $6^4=1296$ outcomes show exactly one pair of equal numbers while the other two dice show two further, different numbers? Choose the two dice of the pair ($\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6$ ways), the number of the pair ($6$) and the numbers of the other two dice ($5\cdot 4=20$):

$$6\cdot \cdot 6\cdot \cdot 20==720,P==\frac{720}{1296}==\frac{5}{9}$$

"All equal" or "exactly three equal" are counted the same way: positions first, then numbers.

For the counting rules themselves see the topic "Combinatorics", for "exactly $k$ times" in repeated trials the topic "Binomial Distribution".

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Examples

Example 1: Two Outcomes

A spinner has eight equal sectors, three of them gold. If it stops on gold you receive €10, otherwise you pay $X$ euros.

$P(\text{gold})=\frac{3}{8}$ and $P(\text{loss})=\frac{5}{8}$.

$$10\cdot \cdot \frac{3}{8}--X\cdot \cdot \frac{5}{8}==0|\cdot \cdot 8$$ $$30==5X$$ $$\boxed{X=6\text{€}}$$

With counts: three sectors bring €10 each, five sectors cost $X$ each, so $30=5X$.

Example 2: Two Gains, One Loss

An urn holds 2 gold, 5 silver and 8 black balls. One ball is drawn. Gold pays you €30, silver €6, and on black you pay $X$ euros.

Of the 15 balls, 8 are black, so $P(\text{loss})=\frac{8}{15}\approx 0.5333$.

$$30\cdot \cdot \frac{2}{15}++6\cdot \cdot \frac{5}{15}--X\cdot \cdot \frac{8}{15}==0|\cdot \cdot 15$$ $$60++30==8X$$ $$\boxed{X=\frac{90}{8}=11.25\text{€}}$$

Example 3: "Exactly One" with Two Dice

Two dice are thrown. If both show a number greater than 4, you receive €9. If exactly one shows a number greater than 4, you receive €2. Otherwise you pay $X$ euros.

Each die shows more than 4 with $2$ of its $6$ numbers (5 or 6). Of the 36 outcomes:

• both greater than 4: $2\cdot 2=4$
• exactly one: first yes and second no, or the other way round, $2\cdot 4+4\cdot 2=16$
• neither: $4\cdot 4=16$

Check: $4+16+16=36$.

$$9\cdot \cdot 4++2\cdot \cdot 16--X\cdot \cdot 16==0$$ $$36++32==16X$$ $$\boxed{X=\frac{68}{16}=4.25\text{€}}$$

Counting only $2\cdot 4=8$ outcomes for "exactly one" forgets the second order, and the total no longer comes to 36.

Example 4: Events That Overlap

Two dice are thrown. If the sum is 7, you receive €5. If at least one 1 shows but the sum is not 7, you receive €1. Otherwise you pay $X$ euros.

The table shows all 36 outcomes: S for sum 7, E for at least one 1 without sum 7, a dot for a loss.

"At least one 1" has $36-25=11$ outcomes. Two of them, $(1,6)$ and $(6,1)$, have sum 7 and already belong to the €5 gain. That leaves $9$ outcomes for the €1 gain and $36-6-9=21$ for the loss.

$$5\cdot \cdot 6++1\cdot \cdot 9--X\cdot \cdot 21==0$$ $$39==21X$$ $$\boxed{X=\frac{39}{21}=\frac{13}{7}\approx 1.86\text{€}}$$

Example 5: Drawing Without Replacement

An urn holds 4 gold and 8 black balls. Three balls are drawn at once. If all three are gold, you receive €50. If exactly two are gold, you receive €5. Otherwise you pay $X$ euros.

There are $\left(\genfrac{}{}{0ex}{}{12}{3}\right)=220$ equally likely draws.

• three gold: $\left(\genfrac{}{}{0ex}{}{4}{3}\right)=4$
• exactly two gold and one black: $\left(\genfrac{}{}{0ex}{}{4}{2}\right)\cdot \left(\genfrac{}{}{0ex}{}{8}{1}\right)=6\cdot 8=48$
• loss: $220-4-48=168$, so $P(\text{loss})=\frac{168}{220}\approx 0.7636$

The tree gives the same result for "three gold": $\frac{4}{12}\cdot \frac{3}{11}\cdot \frac{2}{10}=\frac{1}{55}=\frac{4}{220}$.

$$50\cdot \cdot 4++5\cdot \cdot 48--X\cdot \cdot 168==0$$ $$200++240==168X$$ $$\boxed{X=\frac{440}{168}=\frac{55}{21}\approx 2.62\text{€}}$$

Example 6: Repetition and the Complement

A spinner is one quarter red. It is spun three times. If it stops on red all three times, you receive €40. If it stops on red at least once but not three times, you receive €2. If it never stops on red, you pay $X$ euros.

The spins are independent, so the probabilities multiply:

$$P(\text{red three times})=={\left(\frac{1}{4}\right)}^3==\frac{1}{64},P(\text{never red})=={\left(\frac{3}{4}\right)}^3==\frac{27}{64}$$

The middle case is the rest: $1-\frac{1}{64}-\frac{27}{64}=\frac{36}{64}$.

$$40\cdot \cdot \frac{1}{64}++2\cdot \cdot \frac{36}{64}--X\cdot \cdot \frac{27}{64}==0|\cdot \cdot 64$$ $$40++72==27X$$ $$\boxed{X=\frac{112}{27}\approx 4.15\text{€}}$$

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EPM Problem: Full Solution

Task: Two cards are drawn without replacement from a 52-card deck (13 of them hearts). If both cards are hearts, you receive €20. If exactly one card is a heart, you receive €2. In all other cases you have to pay $X$ euros.

a) Probability of two hearts? b) Probability of exactly one heart? c) Probability of losing? d) For which $X$ is the game fair?

Solution: There are $\left(\genfrac{}{}{0ex}{}{52}{2}\right)=1326$ equally likely pairs of cards.

a) $P(\text{two hearts})=\frac{\left(\genfrac{}{}{0ex}{}{13}{2}\right)}{1326}=\frac{78}{1326}=\frac{1}{17}\approx 0.05882$

Check with the tree: $\frac{13}{52}\cdot \frac{12}{51}=\frac{1}{17}$.

b) $P(\text{exactly one heart})=\frac{13\cdot 39}{1326}=\frac{507}{1326}=\frac{13}{34}\approx 0.38235$

c) No heart: $\left(\genfrac{}{}{0ex}{}{39}{2}\right)=741$ pairs, so $P(\text{loss})=\frac{741}{1326}=\frac{19}{34}\approx 0.55882$. Check: $78+507+741=1326$.

d) With the counts:

$$20\cdot \cdot 78++2\cdot \cdot 507--X\cdot \cdot 741==0$$ $$1560++1014==741X$$ $$\boxed{X=\frac{2574}{741}=\frac{66}{19}\approx 3.47\text{€}}$$

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Typical EPM Questions

The task text almost always follows the same pattern: "Folgendes Spiel wird vereinbart: ... bekommt man ... In allen anderen Fällen muss man $X$ Euro zahlen. Bestimmen Sie $X$ so, dass das Spiel fair ist!"

• Probability of losing (German: Verlustwahrscheinlichkeit): the probability of having to pay $X$. It is the denominator of the formula for $X$ and is often a part of its own.
• Parts a) to d): first the probabilities of the winning events, then the probability of losing, finally $X$. Each part uses the results of the one before.
• Wording with "but not" or "no" (German: aber nicht, kein): it separates two events that would otherwise overlap, as in Example 4. If a part asks for the probability of the whole event without that restriction, the number differs from that of the winning event.
• Stating the result: probabilities as a fraction or to four or five decimal places, $X$ in euros rounded to cents.

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Common Mistakes

Mistake 1: $X$ with the wrong sign

$X$ is an amount you pay, so it carries a minus sign in the ansatz. Writing $+X$ gives a negative $X$. A negative result is the sign that a sign is wrong.

Mistake 2: Counting overlaps twice

If an outcome belongs to two winning events, it is counted twice and the probabilities add up to more than $1$. The check "sum $=1$" exposes this at once. The fix is to assign every outcome to exactly one event, as in the table of Example 4.

Mistake 3: Getting the order wrong

"Exactly one of two dice" has two orders (Example 3). "In order 1, 2, 3" has exactly one: ${\left(\frac{1}{6}\right)}^3$. The same three numbers in any order have $3!=6$ outcomes.

Mistake 4: Rounding too early

Working Example 4 with the rounded probabilities $0.17$, $0.25$ and $0.58$ gives

$$X==\frac{5\cdot 0.17+1\cdot 0.25}{0.58}\approx \approx 1.90\text{€}$$

instead of €1.86. Work with fractions or counts and round only the final result.

Mistake 5: Subtracting a stake on top

In these tasks "you receive €5" is already the gain of the round, and $X$ is paid only when you lose. There is no further stake to subtract from every gain. Only when the text explicitly names a stake (German: Einsatz) paid before every round is the gain of an outcome the payout minus that stake.

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Check Your Understanding

Your turn1 of 2

A spinner has five equal sectors. On two sectors you receive €6, on the other three you pay €3. What is true of this game?

AIt is fair.BIt favours the player, with $E(V)=0.60$ € per round.CIt is unfavourable to the player, with $E(V)=-0.60$ € per round.DIt favours the player, with $E(V)=3$ € per round.

Check

Your turn2 of 2

Two coins are tossed. Two heads pay you €6, exactly one head €2, and on two tails you pay $X$ euros. For which $X$ is the game fair?

A$X=8$ €B$X=2.50$ €C$X=10$ €D$X=6$ €

Check

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Formulas to Remember

Expected value of the gain:

$$E(V)==\sum _i{g}_i\cdot \cdot P(V==g_i)$$

Fair game:

$$E(V)==0$$

Fair loss amount:

$$X==\frac{g_1{p}_1+g_2{p}_2+\dots }{P(\text{loss})}$$

Probability of losing:

$$P(\text{loss})==1--p_1--p_2--\dots$$

At least one hit:

$$P(\text{at least one})==1--P(\text{none})$$