How do you compute the expected value of a $1 bet on a single roulette number?
answer
- weighted average, not a plain average
- count every pocket, greens included
- net gain, not gross return
- the two greens are the entire edge
basics
~20 sMultiply each outcome by its probability and add the pieces up. On a 38-pocket wheel a $1 straight-up bet nets +$35 with probability 1/38 and -$1 with probability 37/38, giving about -$0.053 per dollar staked.
solid answer
~40 sExpected value is the probability-weighted average of outcomes: `E[X] = sum over x of x * P(X = x)`. An American wheel has 38 pockets (1 to 36 plus 0 and 00) and a straight-up number pays 35 to 1, so the net gain is +35 with probability 1/38 and -1 with probability 37/38. That gives `E[X] = 35*(1/38) - 1*(37/38) = -2/38 = -0.0526`, a loss of about 5.3 cents per dollar staked. That number is the house edge per bet. Note that -$0.053 is not an outcome anyone ever experiences on a spin: every spin returns +$35 or -$1. Expectation is a long-run average, the number the per-bet mean settles toward over many independent repetitions, not a prediction of any single spin.
go deeper
Be ready to write the sum of value times probability on the spot and to state the result in cents per dollar staked. Say out loud that no single spin returns the expected value.
Explain why the answer is identical for a red bet and a straight-up bet, and locate the edge in the mismatch between the 35-to-1 payout and the 38 pockets.
Show what expectation does and does not promise over a finite session, and use variance rather than mean to compare the risk of the two bets when someone asks which is safer.
Own the framing question: when a decision is repeated many times, expectation is the right summary; when it is a one-shot bet with ruinous downside, expected value alone is the wrong basis for the call.
## What expectation is The expected value of a random variable X, written `E[X]` and often called the mean of X, is the probability-weighted average of the values X can take. - Discrete case: `E[X] = sum over x of x * P(X = x)`. Every possible value is counted once, weighted by how likely it is. - Continuous case with density f: `E[X] = integral of x * f(x) dx` over the support. Same idea, with the sum replaced by an integral. The weights are probabilities, so they are non-negative and add to 1. That is what makes the result an average rather than just a sum, and it is why expectation always lands somewhere between the smallest and largest attainable value (inclusive). ## The roulette computation, step by step An American roulette wheel has 38 pockets: the numbers 1 through 36, plus 0 and 00. A straight-up bet on one number pays 35 to 1, meaning you keep your $1 stake and receive $35 on top of it. Define X as the net change to your bankroll from one $1 bet: - You win with probability 1/38, and the net change is +35. - You lose with probability 37/38, and the net change is -1. So `E[X] = 35*(1/38) + (-1)*(37/38) = (35 - 37)/38 = -2/38 = -0.05263...` The expected loss is about 5.26 cents per dollar staked, usually quoted as a house edge of 5.26 percent. The two green pockets are the entire source of that edge: if the wheel had only 36 pockets and still paid 35 to 1, the expectation would be `35/36 - 35/36 = 0` and the game would be fair. The most common arithmetic slip is treating the payout as 36 rather than 35. A 35-to-1 payout is quoted as net odds, so the win term is +35, not +36. The other frequent slip is dividing by the number of outcomes instead of weighting by probability: the outcomes here are not equally likely, so a plain average of +35 and -1 is meaningless. ## Why the answer is not an attainable value No spin ever moves your bankroll by -$0.053. Expectation is not the most likely outcome (that is the mode, here a $1 loss), and it is not the middle outcome (that is the median, also a $1 loss). It is the balance point of the distribution. Its operational meaning comes from the law of large numbers: as you repeat independent bets, the average net result per bet converges to -0.0526. Over 1,000 one-dollar bets your expected total loss is about $53, though the actual result has a wide spread because a single win moves the total by $36. A useful reframing for interviews: expectation is what you would pay to enter the game if you were exactly indifferent. Since the expectation of this bet is negative, no betting system can rescue it. Progressive staking (doubling after a loss, and similar schemes) changes the shape of the payoff distribution but not the sign of the expectation, because linearity of expectation makes the expected total of any finite sequence of negative-expectation bets negative. ## Doing it for other bets The same recipe handles any wager. An even-money bet on red covers 18 pockets, so `E[X] = 1*(18/38) - 1*(20/38) = -2/38`, the identical -5.26 percent. On this wheel almost every bet carries the same edge, because the payouts are all set as if there were 36 pockets while 38 exist. What differs enormously across bets is the variance, not the mean: the straight-up bet has a huge spread of outcomes, the red bet a small one. A European wheel has 37 pockets (a single zero) and still pays 35 to 1 straight up, so `E[X] = 35/37 - 36/37 = -1/37`, about -2.70 percent. Halving the number of green pockets halves the edge. ## The general recipe 1. Define the random variable precisely, including whether it is net gain or gross return. Ambiguity here causes most wrong answers. 2. List every outcome with its probability, and check the probabilities sum to 1. 3. Multiply and add. 4. Sanity-check the sign and the magnitude against the structure of the game. That recipe is the whole of elementary expected value, and interviewers use this question mainly to see whether you weight by probability, get the payout convention right, and can explain what a long-run average does and does not promise about one trial.
- What does a house edge of 5.26 percent actually promise a player who makes 1,000 one-dollar bets?It promises an expected total loss of about $53, because expectations add over bets whether or not the bets are independent. It promises nothing about a particular session: the spread around that figure is large, since a single straight-up win swings the total by $36. The law of large numbers only says the average per bet tightens around -0.0526 as the number of bets grows.
- Does betting on red instead of a single number change the expected value per dollar staked?No. Red covers 18 of the 38 pockets and pays even money, so the expectation is `18/38 - 20/38 = -2/38`, the same -5.26 percent. What changes is the variance: the red bet pays plus or minus one dollar, while the straight-up bet pays +35 or -1. Same mean, wildly different risk.
- How does the calculation change on a European wheel with a single zero?There are 37 pockets and the straight-up payout is still 35 to 1, so `E[X] = 35*(1/37) - 1*(36/37) = -1/37`, about -2.70 percent. Removing one green pocket roughly halves the house edge, which shows that the edge lives entirely in the mismatch between the 35-to-1 payout and the pocket count.
It is the balance point of a seesaw loaded with weights: the weights are probabilities, and the balance point need not sit under any one of them.
saying these in an interview costs you the question
- Calls the expected value the most likely outcome of a spin
- Averages the outcomes without weighting by probability
- Treats a 35-to-1 payout as a net gain of 36
- Forgets the green pockets and concludes the game is fair
- Claims a doubling betting system turns a negative expectation positive