Expected Value Worksheet - mathdotcom.org - Free Printable

Let’s go through each problem one by one. We’ll calculate the Expected Value (EV) first — that’s how much you’d expect to win on average per game if you played many times. Then we’ll suggest a fair charge for playing — enough so the house makes a small profit, but not so high that no one wants to play.

Remember:
→ EV = (Probability of winning) × (Prize money)
→ Prize money is $15 per game (as given).
→ Charge should be a bit more than EV, so the house profits slightly.

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Problem 1: Roll a 2 and then roll another 2 on a die.

- Probability of rolling a 2 on one die = 1/6
- Since two rolls are independent: P(both 2s) = (1/6) × (1/6) = 1/36
- EV = (1/36) × $15 = $15 ÷ 36 ≈ $0.4167 → about 42¢

Now, what should you charge?
You want to make a profit, so charge more than 42¢. But not too much — maybe round up to something reasonable like $1.00. That gives the house about 58¢ profit per game on average, which is fair since the chance of winning is very low.

✔ Final for #1:
EV ≈ $0.42
Charge = $1.00

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Problem 2: Roll a sum of 7 or 11 with 2 dice.

First, find probability of getting 7 or 11.

Total possible outcomes with 2 dice = 6 × 6 = 36

Ways to get 7:
(1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6 ways

Ways to get 11:
(5,6), (6,5) → 2 ways

Total favorable = 6 + 2 = 8
P(win) = 8/36 = 2/9

EV = (2/9) × $15 = $30 ÷ 9 ≈ $3.33

What to charge?
Since EV is $3.33, charging $4.00 gives the house about 67¢ profit per game — still attractive because players have a decent chance (about 22%) to win $15.

✔ Final for #2:
EV ≈ $3.33
Charge = $4.00

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Problem 3: Toss a coin 3 times and get the same thing every time.

Same thing means all heads OR all tails.

P(all heads) = (1/2)^3 = 1/8
P(all tails) = (1/2)^3 = 1/8
Total P(win) = 1/8 + 1/8 = 2/8 = 1/4

EV = (1/4) × $15 = $15 ÷ 4 = $3.75

Charge? Round up to $4.00 — house makes 25¢ profit per game. Fair deal.

✔ Final for #3:
EV = $3.75
Charge = $4.00

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Problem 4: Choose 2 kings out of a regular deck without replacement.

A standard deck has 52 cards, 4 kings.

We pick 2 cards without replacement.

P(first card is king) = 4/52
P(second card is king | first was king) = 3/51

So P(both kings) = (4/52) × (3/51)

Simplify:
4/52 = 1/13
3/51 = 1/17
So P = (1/13) × (1/17) = 1/221

EV = (1/221) × $15 ≈ $15 ÷ 221 ≈ $0.0679 → about 7¢

That’s super low! So even though the prize is $15, it’s almost impossible to win.

What to charge? You could charge $1.00 — house makes about 93¢ profit per game. Still, some people might play for fun hoping to hit the jackpot.

✔ Final for #4:
EV ≈ $0.07
Charge = $1.00

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Problem 5: Toss a head and then roll a 5.

These are independent events.

P(head) = 1/2
P(roll 5) = 1/6
P(both) = (1/2) × (1/6) = 1/12

EV = (1/12) × $15 = $15 ÷ 12 = $1.25

Charge? Round up to $2.00 — house makes 75¢ profit. Reasonable since chance is 1 in 12 (~8.3%).

✔ Final for #5:
EV = $1.25
Charge = $2.00

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Final Answer:

1) EV ≈ $0.42, Charge = $1.00
2) EV ≈ $3.33, Charge = $4.00
3) EV = $3.75, Charge = $4.00
4) EV ≈ $0.07, Charge = $1.00
5) EV = $1.25, Charge = $2.00