Theoretical Probability: Calculate a Bet’s Chance of Winning

Theoretical probability is the chance of an event calculated from a mathematical model, rather than from observed results. In gambling, that model starts with the game’s possible outcomes and rules. It tells you how likely a result is—not whether the payout makes the wager fair.

When all individual outcomes are equally likely, the calculation is:

P(event) = number of favorable outcomes ÷ total number of possible outcomes

Here, “favorable” simply means outcomes that satisfy the event you are counting. The collection of all possible outcomes is the sample space. The equal-likelihood condition matters: counting outcomes alone works only when each has the same chance. OpenStax explains this counting method.

Worked example: red on single-zero roulette

Assume a fair single-zero wheel with 37 pockets: 18 red, 18 black and one green zero. Also assume a red bet pays 1:1 profit and loses its entire stake on zero—no half-back or imprisonment rule. These are the standard single-zero rules described in Wizard of Odds’ roulette guide; Plaza’s published tutorial also lists red/black bets as paying even money.

To calculate the probability of winning a red bet:

1. Define the event: the ball lands in any red pocket.
2. Count favorable outcomes: 18 red pockets.
3. Count all equally likely outcomes: 37 pockets, including zero.
4. Divide: P(red) = 18 ÷ 37 ≈ 48.65%.

The losing probability is 19 ÷ 37 ≈ 51.35%, because both black and zero lose.

Red is therefore not a 50% bet. Nor can you count “red, black, green” as three equally likely outcomes: those categories contain different numbers of pockets.

Theoretical probability versus experimental probability

Theoretical probability comes from the model. Experimental probability, also called empirical probability, comes from observations:

Experimental probability = observed occurrences ÷ total trials

Suppose a hypothetical record of 100 spins contains 55 red results. Its experimental probability of red is 55 ÷ 100 = 55%. That does not replace the theoretical 48.65% under the fair-wheel model. It is one sample’s observed frequency. OpenStax distinguishes these two methods.

Assuming successive spins are independent, those 55 reds do not make red less likely next time. Equally, a run of black does not make red “due.” Independence means that knowing a previous result does not change the next event’s probability, as explained in OpenStax’s treatment of independent events.

Add the payout to find the bet’s price

Probability alone does not tell you the long-run cost. For that, calculate expected value using net profit or loss—not the total money returned.

For a $10 red bet under the stated rules:

Result	Probability	Net result
Red	18/37	+$10
Black or zero	19/37	−$10

EV = (18/37 × $10) + (19/37 × −$10) EV = −$10/37 ≈ −$0.27 per bet

The house edge is therefore approximately 2.70% of the stake. Across 100 fixed $10 bets, the expected net result is about −$27.03.

That is an average, not a session forecast. The hypothetical 55-win record above would produce a $100 profit: 55 × $10 − 45 × $10. Random variation around the expected result—measured mathematically by variance—allows winning sessions even when a wager has negative expected value. It also allows losses much larger than the average.

Our expected-value guide explains how to apply the same calculation to wagers with more than two payout outcomes.

Before using any probability figure, check the exact rules, whether individual outcomes are equally likely, and whether earlier results change the remaining possibilities. Cards dealt without replacement, for example, are not independent draws. Treat the calculated cost as information for setting limits, not as a promise that actual losses will stay near the average.