The average result if you played forever — not what happens on any one go.
The problem
A game costs ₱50 to play. You win ₱200 with probability 0.2, otherwise nothing. Is it worth playing?
Expected winnings are 0.2 × ₱200 = ₱40, against a ₱50 ticket — so you lose ₱10 per play on average. B quotes the prize itself, but you only collect it one time in five. Compare the average return with the cost, never the prize with the cost.
E = Σ (value × probability)
| Outcome | Value | Probability | Product |
|---|---|---|---|
| win | 200 | 0.2 | 40 |
| lose | 0 | 0.8 | 0 |
| total | — | 1.0 | 40 |
The winnings are worth ₱40 a game on average — before you pay to play.
The probabilities must total 1. If they do not, an outcome has been missed.
expected winnings = 40
from Step 1
cost = 50
paid every time
net = 40 − 50 = −10
a loss
₱50 leaves your pocket on every single game, so it is not weighted by anything. Only the uncertain amounts get multiplied by probabilities.
A raffle. 400 tickets are sold at ₱15 each, with one ₱2,000 prize and four ₱250 prizes. Each ticket has the same chance at every prize.
prize money = 2000 + 4(250) = 3000
add every prize
expected winnings = 3000 ÷ 400 = 7.50
shared over all the tickets
net = 7.50 − 15 = −₱7.50
minus the ticket price, once
The seller’s side is the same sum turned over: an insurer who charges ₱500 and pays ₱20,000 with probability 0.02 expects to gain 500 − 0.02(20 000) = ₱100 per policy.
you will lose ₱10 this game
you win 200 or win 0 — never −10
₱1,000 down over 100 games
the long-run average
100 games, 20 wins expected
20% of 100
won 20 × 200 = 4,000
total winnings
paid 100 × 50 = 5,000
total cost
down 1,000
₱10 a game ✓
| Expected value | Verdict |
|---|---|
| positive | favours the player |
| zero | fair |
| negative | favours the house |
what price would be fair?
set net to 0
40 − cost = 0
solve
cost = ₱40
anything above is a losing bet
Real lotteries and casinos are always negative for the player. That gap is exactly how they make money.
The average roll of a fair die.
each face has probability ⅙
fair die
E = (1+2+3+4+5+6) ÷ 6
equal weights
= 21 ÷ 6 = 3.5
a value the die cannot show
An expected value need not be a possible outcome. It is an average, not a prediction.
With unequal probabilities the same method still works — build the table, multiply each value by its own chance, and add the column.
A missing probability. X takes the values 1, 2 and 4 with probabilities 0.5, k and 0.2. Find the expected value of X.
0.5 + k + 0.2 = 1, so k = 0.3
the probabilities must total 1
E = 1(0.5) + 2(0.3) + 4(0.2)
value × probability
= 0.5 + 0.6 + 0.8 = 1.9
between 1 and 4 ✓
What is different: one probability is hidden. Find it from “they add to 1” before building the table.
A penalty for wrong answers. A test has 20 items with 4 choices each. A right answer scores 1 point and a wrong one loses 13 of a point. What is the expected score from guessing every item?
one item: 1 × 14 + (−13) × 34
the loss goes in as a negative value
= 14 − 14 = 0
per item
20 items: 20 × 0 = 0
expected values add
What is different: a loss is a negative value in the table, and the expected value of many identical items is n times the value of one. The penalty is set so that blind guessing is a fair game.
20 × 14 = 5 is the expected number of right answers, not the score. The 15 expected wrong answers cost 15 × 13 = 5 points back.
expected winnings = 16x + 26(12)
weight each prize
x6 + 4 = 15
fair: expected winnings = the fee
x6 = 11
x = ₱66
Check: (66 + 12 + 12) ÷ 6 = 90 ÷ 6 = 15 ✓.
The fee is paid every game, so “fair” means the expected winnings equal the fee (a net value of 0), not that the winnings themselves are 0.
Expect raffles (total prize money ÷ tickets, minus the price), spinners and dice games, insurance, or a missing prize or fee that makes a game fair. Build the value × probability table, then subtract the certain cost once. Ten-second check: the expected value must lie between the worst and the best single outcome.
Netting the cost inside each outcome gives the same −10. Either route works.
The problem
A raffle ticket costs ₱100. There is one prize of ₱50,000 and 1,000 tickets sold. What is the expected value of a ticket?
The prize is worth 50,000 ÷ 1,000 = ₱50 on average, and the ticket costs ₱100, so the expected value is 50 − 100 = −₱50. C is the expected winnings before paying. A negative value is normal for a raffle — the gap is what funds the cause.
