Statistics & Probability Quiz: Find Expected Value Of A Game
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Find Expected Value Of A GameQuestion 1 of 20

A simple wager works like this: you pay $1 to flip a fair coin. If it lands heads, you receive $3; if it lands tails, you receive $0. (Net payoff = winnings minus $1.) What is the expected value (long-run average net gain/loss) for the player on one play?

$2.00
$0.50
$1.50
-\0.50$
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Statistics & Probability Quiz

Statistics & Probability Quiz: Find Expected Value Of A Game

Practice Find Expected Value Of A Game in Statistics & Probability with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Find Expected Value Of A Game, giving you a quick way to practice the rules, question types, and explanations that matter most for Statistics & Probability.

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Question 1

A simple wager works like this: you pay $1 to flip a fair coin. If it lands heads, you receive $3; if it lands tails, you receive $0. (Net payoff = winnings minus $1.) What is the expected value (long-run average net gain/loss) for the player on one play?

  1. $2.00
  2. $0.50 (correct answer)
  3. $1.50
  4. -\0.50$
Explanation: Expected value (EV) for this coin flip wager is the average net payoff over many flips, guiding long-run expectations. Net payoffs are 2forheads(probability1/2)and−2 for heads (probability 1/2) and -1 for tails (1/2), giving EV = (2)(1/2) + (-1)(1/2) = 0.50. Over repeated plays, you'd expect to gain $0.50 per flip on average, tying EV to prolonged results. This positive EV makes the game favorable for the player. A common misconception is equating EV to the median outcome, but here EV is positive while half the time you lose $1. This distinction clarifies EV's role in probability.

Question 2

Raffle: 1,000 tickets at $2 each; prizes: one $800, three $100. Find expected profit per ticket.

  1. $0.90
  2. -$0.90 (correct answer)
  3. -$2.00
  4. $1.10
Explanation: Total prize money is 800 + 3(100) = 1,100. Expected prize per ticket is 1,100 / 1,000 = $1.10. Since each ticket costs 2,expectedprofitis1.10−2=−2, expected profit is 1.10 - 2 = -0.90. The tempting mistake is stopping at the $1.10 expected prize and forgetting to subtract the ticket price.

Question 3

Each play costs $2. A play pays $10 with probability .25, else $0. What is the expected net profit for 5 plays?

  1. $12.50
  2. $10.00
  3. $2.50 (correct answer)
  4. $0.50
Explanation: Each play has expected payout 10 × 0.25 = 2.50. After paying 2 to play, expected net per play is 0.50. Over 5 plays, that is 5 × 0.50 = 2.50. The tempting 12.50 is total expected payout, not profit, because it ignores the 10 total cost.

Question 4

Game costs $5. Roll two fair dice; win $6 on sum 7, $1 on even sum, else $0. Find expected net gain.

  1. -$4.00
  2. $3.50
  3. -$3.50 (correct answer)
  4. $1.50
Explanation: Two dice have 36 equally likely outcomes. Sum 7 has 6 outcomes, so its expected payout is 6/36 × $6 = $1. Even sum occurs 18 out of 36, adding 18/36 × $1 = $0.50. So expected payout is $1.50; subtract the 5costtoget−5 cost to get -3.50. The tempting $1.50 is only the payout before paying for the game.

Question 5

Lottery ticket costs $4. Payouts: $1000 with probability .001, $20 with probability .01, otherwise $0. Find expected net gain.

  1. -$3.80
  2. $2.80
  3. $1.20
  4. -$2.80 (correct answer)
Explanation: Expected payout is 1000 times .001 = $1, plus 20 times .01 = $0.20, for a total of $1.20. Subtract the $4 ticket price: $1.20 - 4=−4 = -2.80. The tempting $1.20 is just the expected payout, not the net gain after paying for the ticket.

Question 6

A game pays $30 with probability .15 and $5 with probability .4, else $0. What entry fee makes expected profit zero?

  1. $6.50 (correct answer)
  2. $35.00
  3. $4.50
  4. $2.00
Explanation: Multiply each payout by its probability: 30 times 0.15 is $4.50, and 5 times 0.40 is $2.00. The $0 case adds nothing, so the expected payout is $4.50 + $2.00 = $6.50. A fee of $6.50 makes expected profit zero. The tempting $4.50 is wrong because it ignores the $5 payout.

Question 7

A student pays $3 to play a game once. A bag contains 5 red marbles and 5 blue marbles. You randomly draw 1 marble. If it is red, the player receives $8. If it is blue, the player receives $0. What is the expected value (long-run average net gain/loss) for the player for one play, including the $3 cost?

  1. $1 (correct answer)
  2. -\4$
  3. -\1$
  4. $4
Explanation: The expected value (EV) is the long-term average net gain or loss from playing the game repeatedly. The expected payout is (5/10)×8+(5/10)×8 + (5/10)×0 = $4. Subtract the $3 cost: $4 - $3 = $1. Over many plays, the player averages $1 gain per game. In long-run terms, 100 plays might net about $100 profit. A common misconception is not accounting for the cost, leading to an EV of $4 instead of $1. Additionally, people often mistake EV for the most likely single outcome, but here both $8 and $0 are equally likely, not $1.

Question 8

A student buys a lottery-style ticket for $3. The player outcomes are: win $9 with probability $\tfrac{1}{6},win$3withprobability$13, win $3 with probability $\tfrac{1}{3}, and win $0 with probability $\tfrac{1}{2}$. What is the expected value (long-run average net gain/loss) of one ticket, including the $3 cost?

  1. $3.50
  2. -\3.50$
  3. -\0.50$ (correct answer)
  4. $0.50
Explanation: Expected value (EV) measures the long-run average net gain or loss per ticket over many purchases. Expected payout: (1/6)×9+(1/3)×9 + (1/3)×3 + (1/2)×$0 = $2.50. Net EV: $2.50 - 3=−3 = -0.50. This implies an average loss of $0.50 per ticket in the long term. Over 200 tickets, you'd expect to lose about $100. A misconception is omitting the cost, leading to $2.50 as EV. Another is confusing EV with the most probable outcome, which is 0,not−0, not -0.50.

Question 9

A teacher runs a simple wager game. A student pays $2 to roll a fair six-sided die once. If the roll is a 6, the player wins $10. Otherwise, the player wins $0. What is the expected value (long-run average net gain/loss) for the player for one play, including the $2 cost?

  1. -\\tfrac{1}{3}$ (correct answer)
  2. -\\tfrac{5}{3}$
  3. \\tfrac{5}{3}$
  4. \\tfrac{1}{3}$
Explanation: Expected value (EV) indicates the average net gain or loss in the long run over many plays. Expected payout: (1/6)×10+(5/6)×10 + (5/6)×0 = $10/6 ≈ $1.67. Net EV: $1.67 - 2=−2 = -1/3 ≈ -$0.33. This means a long-term average loss of about $0.33 per roll. For example, over 300 rolls, you'd expect to lose around $100. One misconception is ignoring the cost, yielding an EV of $1.67 instead. Another is believing EV means you'll lose $0.33 every time, but actual outcomes are $10 or $0, with EV as the average.

Question 10

In a classroom wager, you pay $3 to draw one marble from a bag of 10 marbles. If you draw a red marble (4 marbles), you receive $7. If you draw a blue marble (6 marbles), you receive $0. From the player's perspective, what is the expected value (long-run average net gain/loss) of one draw?​

  1. $4.00
  2. -\1.60$
  3. -\0.20$ (correct answer)
  4. $1.60
Explanation: The expected value (EV) for this marble draw measures the average net payoff over many repeated plays. Net payoffs are 4forred(probability4/10)and−4 for red (probability 4/10) and -3 for blue (probability 6/10), resulting in EV = (4)(4/10) + (-3)(6/10) = -0.20. In the long run, you'd anticipate losing $0.20 per draw on average, linking EV to overall trends rather than single results. This negative EV indicates the game benefits the house over time. A common misconception is averaging payoffs without probabilities, such as simply averaging $7 and $0, ignoring costs and chances. Grasping EV correctly reveals the game's inherent bias.

Question 11

A simple wager: you pay $2 to play. With probability $\tfrac{1}{5}youreceive$9;withprobability$45 you receive $9; with probability $\tfrac{4}{5} you receive $1. From the player's perspective, what is the expected value (long-run average net gain/loss) of 1 play?

  1. $1.40
  2. $-0.60
  3. $-1.40
  4. $0.60 (correct answer)
Explanation: Expected value in this wager is the average net gain or loss per play over the long run. Net gains are 7(probability1/5)and−7 (probability 1/5) and -1 (probability 4/5) after subtracting $2 cost, giving (1/5)×7 + (4/5)×(-1) = 0.60. This positive value means an expected gain of 60 cents per play in the long term, beneficial for the player. It connects to repeated plays by averaging outcomes to show overall trend. A common misconception is not subtracting the play cost, resulting in expected receipt of $2.60, which exaggerates the positivity. This concept is key for assessing game fairness.

Question 12

A school club sells a lottery-style ticket for $2. Exactly one of the following outcomes happens when you buy 1 ticket: with probability 0.100.10 you win $10; with probability 0.200.20 you win $3; with probability 0.700.70 you win $0. From the player's perspective, what is the expected value (long-run average net gain/loss) of buying 1 ticket?

  1. $0.00
  2. $-0.40 (correct answer)
  3. $-1.60
  4. $1.60
Explanation: The expected value in a game like this lottery represents the long-run average net gain or loss per ticket if you were to buy many tickets over time. To compute it, calculate the net gain for each outcome by subtracting the $2 cost from the prize, then multiply by the respective probabilities and sum them up. Here, the net gains are $8 (probability 0.10), 1(probability0.20),and−1 (probability 0.20), and -2 (probability 0.70), yielding an expected value of 0.10×8 + 0.20×1 + 0.70×(-2) = -0.40. This negative value indicates that, on average, you lose 40 cents per ticket in the long run, making the game unfavorable for the player. A common misconception is forgetting to subtract the ticket cost from each prize, which would incorrectly give an expected prize of $1.60 instead of the net expected value. Recognizing this distinction helps players understand the true financial implications of participating in such games.

Question 13

A prize wheel costs $1 per spin. Outcomes: win $7 with probability $\tfrac{1}{10},win$1withprobability$310, win $1 with probability $\tfrac{3}{10}, win $0 with probability $\tfrac{6}{10}$. From the player's perspective, what is the expected value (long-run average net gain/loss) of 1 spin?

  1. $-0.40
  2. $-1.00
  3. $0.00 (correct answer)
  4. $0.40
Explanation: For this prize wheel, expected value is the long-run average net gain or loss per spin. Calculate net gains by subtracting $1 cost: $6 (1/10), 0(3/10),−0 (3/10), -1 (6/10), resulting in (1/10)×6 + (3/10)×0 + (6/10)×(-1) = 0.00. Zero EV indicates a fair game with no long-term advantage or disadvantage. It relates to long-run outcomes as the average settles near zero over many spins. A common misconception is ignoring cost in nets, leading to expected prize of $1.00, mistakenly implying profit. Understanding this prevents overoptimism in game evaluations.

Question 14

A prize wheel costs $3 to spin once. The wheel has 8 equal sections: 1 section pays $9, 2 sections pay $5, and 5 sections pay $0. From the player's perspective, what is the expected value (long-run average net gain/loss) of 1 spin?

  1. $0.75
  2. $1.50
  3. $-1.50
  4. $-0.75 (correct answer)
Explanation: The expected value represents the average net gain or loss per spin if you played many times. Since the player pays $3 to spin, we must subtract this cost from any winnings to find net gains. With 8 equal sections, the probability of landing on the $9 section is 1/81/8 (net gain of $6), the probability of landing on a $5 section is 2/82/8 (net gain of $2), and the probability of landing on a $0 section is 5/85/8 (net loss of $3). The expected value is $(18\frac{1}{8} ×\times 6) + (28\frac{2}{8} ×\times 2) + (\frac{5}{8} \times (-3)) = 0.75 + 0.50 - 1.875 = -0.75$. This negative expected value indicates that players will lose an average of 75 cents per spin in the long run. Students often mistakenly calculate the expected winnings without accounting for the cost to play.

Question 15

A school club sells a lottery-style ticket for $2. One ticket is drawn at random from a box containing 20 tickets: 1 ticket pays $10, 3 tickets pay $4, and the other 16 tickets pay $0. From the player's perspective, what is the expected value (long-run average net gain/loss) of buying 1 ticket?

  1. $0.00
  2. $-0.90 (correct answer)
  3. $-1.10
  4. $1.10
Explanation: To find the expected value from the player's perspective, we calculate the average net gain/loss by considering all possible outcomes and their probabilities. The player pays $2 to play and has a 1/201/20 chance of winning $10 (net gain of $8), a 3/203/20 chance of winning $4 (net gain of $2), and a 16/2016/20 chance of winning $0 (net loss of $2). The expected value is $120×8+320×2+1620×(−2)=0.40+0.30−1.60=−0.90\frac{1}{20} \times 8 + \frac{3}{20} \times 2 + \frac{16}{20} \times (-2) = 0.40 + 0.30 - 1.60 = -0.90 $. This negative expected value means that over many plays, a player would lose an average of 90 cents per ticket. A common misconception is forgetting to subtract the cost of playing when calculating net gains, which would incorrectly give a positive expected value.

Question 16

A school booth runs a simple wager: you pay $5 to play. With probability 0.100.10 you win $20; with probability 0.900.90 you win $0. From the player's perspective, what is the expected value (long-run average net gain/loss) of 1 play?

  1. $-3.00 (correct answer)
  2. $-5.00
  3. $3.00
  4. $15.00
Explanation: The expected value tells us the average net gain or loss per play over many repetitions. Since the player pays $5 to play, we must subtract this from any winnings to find the net result. With probability 0.10, the player wins $20 (net gain of $15), and with probability 0.90, the player wins $0 (net loss of 5).Theexpectedvalueis:0.10×(5). The expected value is: 0.10×(20-5)+0.90×(5) + 0.90×(0-5)=0.10×5) = 0.10×15 + 0.90×(-$5) = $1.50 - 4.50=−4.50 = -3.00. This negative expected value means that on average, players lose 3perplayinthelongrun.Acommonmisconceptioniscalculatingonlytheexpectedwinnings(3 per play in the long run. A common misconception is calculating only the expected winnings (2) without accounting for the cost to play, which would miss the fact that this is a losing game.

Question 17

A student plays a simple wager: pay $2 to play. With probability $\tfrac{1}{4}youwin$10;withprobability$34 you win $10; with probability $\tfrac{3}{4} you win $0. From the player's perspective, what is the expected value (long-run average net gain/loss) of 1 play?

  1. $-0.50
  2. $-2.50
  3. $2.50
  4. $0.50 (correct answer)
Explanation: The expected value tells us the average net gain or loss per play if this game is repeated many times. The player pays $2 to play and has a 1/4 probability of winning $10 (net gain of $8) and a 3/4 probability of winning $0 (net loss of 2).Theexpectedvalueis:(1/4)×(2). The expected value is: (1/4)×(10-2)+(3/4)×(2) + (3/4)×(0-2)=(1/4)×2) = (1/4)×8 + (3/4)×(-$2) = $2.00 - $1.50 = 0.50.Thispositiveexpectedvaluemeansthatonaverage,aplayergains50centsperplayinthelongrun.Studentsoftenmistakenlycalculatetheexpectedwinnings(0.50. This positive expected value means that on average, a player gains 50 cents per play in the long run. Students often mistakenly calculate the expected winnings (2.50) instead of the expected net gain, forgetting to subtract the cost of playing from all outcomes.

Question 18

In a classroom wager, you pay $3 to draw one marble from a bag of 10 marbles. If you draw a red marble (4 marbles), you receive $7. If you draw a blue marble (6 marbles), you receive $0. From the player's perspective, what is the expected value (long-run average net gain/loss) of one draw?

  1. $4.00
  2. -\0.20$ (correct answer)
  3. $1.60
  4. -\1.60$
Explanation: The expected value (EV) for this marble draw measures the average net payoff over many repeated plays. Net payoffs are 4forred(probability4/10)and−4 for red (probability 4/10) and -3 for blue (probability 6/10), resulting in EV = (4)(4/10) + (-3)(6/10) = -0.20. In the long run, you'd anticipate losing $0.20 per draw on average, linking EV to overall trends rather than single results. This negative EV indicates the game benefits the house over time. A common misconception is averaging payoffs without probabilities, such as simply averaging $7 and $0, ignoring costs and chances. Grasping EV correctly reveals the game's inherent bias.

Question 19

A prize wheel costs $1 to spin. The wheel has 6 equal sections: 1 section pays $7, 2 sections pay $2, and 3 sections pay $0. From the player's perspective, what is the expected value (long-run average net gain/loss) of 1 spin?

  1. $1.83
  2. $0.83 (correct answer)
  3. $-1.83
  4. $-0.83
Explanation: Expected value represents the average net gain or loss per spin if you played this wheel many times. The player pays $1 to spin and has a 1/6 chance of winning $7 (net gain of $6), a 2/6 chance of winning $2 (net gain of $1), and a 3/6 chance of winning $0 (net loss of $1). The expected value calculation is: $(1/6)×6+(2/6)×1+(3/6)×(−1)=1.00+0.33−0.50=0.83(1/6) \times 6 + (2/6) \times 1 + (3/6) \times (-1) = 1.00 + 0.33 - 0.50 = 0.83 $. This positive expected value of 83 cents means that over many spins, a player would gain an average of 83 cents per spin. A common error students make is forgetting to account for the cost of playing when determining net gains and losses from each outcome.

Question 20

A simple wager: you pay $1 to play. With probability 0.400.40 you receive $4; with probability 0.600.60 you receive $0. From the player's perspective, what is the expected value (long-run average net gain/loss) of 1 play?

  1. $0.60 (correct answer)
  2. $-0.60
  3. $-1.60
  4. $1.60
Explanation: The expected value for this wager reflects the average net outcome per play in the long run. Find net gains by subtracting the $1 cost from receipts: 3(probability0.40)and−3 (probability 0.40) and -1 (probability 0.60), then compute 0.40×3 + 0.60×(-1) = 0.60. This positive EV suggests a long-term gain of 60 cents per play, favoring the player. It connects to long-run outcomes by averaging results over many wagers, showing overall profitability. A common misconception is overlooking the cost deduction, giving an expected receipt of $1.60, which overstates the benefit. Appreciating this helps evaluate if a game offers a genuine advantage.