Probability Basics: From Win Chance to a Bet’s Cost

Probability tells you how likely an outcome is. It does not tell you whether a bet is good value: that also depends on what the bet pays. A two-dice example connects the basic rules to the long-run price of a wager.

Count outcomes, not just possible totals

A probability runs from 0 to 1, or from 0% to 100%. Zero means impossible; one means certain. An outcome is one possible result, an event is the result or group of results you care about, and the sample space contains every possible outcome. OpenStax explains these definitions and the counting rule:

Probability of an event = favorable outcomes ÷ all possible outcomes

That shortcut requires the outcomes being counted to be equally likely.

Assume two fair, six-sided dice whose results are independent. Label them die A and die B. There are 6 × 6 = 36 equally likely pairs. The pairs (1, 6) and (6, 1) count separately because different dice show the six.

Six pairs total seven:

• (1, 6), (2, 5), (3, 4)
• (4, 3), (5, 2), (6, 1)

So:

P(seven) = 6/36 = 1/6 ≈ 16.67%

The notation P(seven) simply means “the probability of rolling a total of seven.” Although there are 11 possible totals, from 2 through 12, they are not equally likely. Dividing by 11 would give the wrong answer. For another example of this counting method, see theoretical probability.

Three rules for combining chances

“Not” means subtract from one. The complement of rolling seven is rolling anything else:

P(not seven) = 1 − 1/6 = 5/6 ≈ 83.33%.

“Or” means add, provided the events cannot both happen. Such events are called mutually exclusive. Seven and eleven cannot occur on the same roll. Eleven has two pairs, (5, 6) and (6, 5), so:

P(seven or eleven) = 6/36 + 2/36 = 8/36 ≈ 22.22%.

If events overlap, subtract the probability of their overlap so it is not counted twice.

“And” means multiply for independent events. Assuming successive rolls are independent:

P(seven on both of the next two rolls) = 1/6 × 1/6 = 1/36 ≈ 2.78%.

Independence means knowing one result does not change the next result’s probability. A run without a seven does not make seven “due”: the next-roll chance remains 1/6. Do not apply this assumption automatically to cards dealt without replacement, where earlier cards change what remains. OpenStax distinguishes independent events from mutually exclusive events and explains replacement.

Win probability is not payout odds

With six winning pairs and 30 losing pairs, the odds against rolling seven are 30:6, or 5:1. That compares losing outcomes with winning outcomes—not the casino’s payout or a guaranteed sequence of results.

For the wager example, assume a $10 Any Seven craps bet that wins only when the next roll totals seven and pays 4 to 1, as listed in Casino Del Sol’s published craps rules.

Result	Probability	Net change
Seven	1/6	+$40
Anything else	5/6	−$10

A win returns the $10 stake plus $40 profit. Use profit—not the $50 total return—when calculating net expected value:

EV = (1/6 × $40) + (5/6 × −$10) = −$1.666… per bet

The expected loss is about $1.67 per $10 wager. Dividing by the stake gives a 16.67% house edge, matching the Wizard of Odds calculation for Any Seven paying 4:1. A break-even payout under these assumptions would be 5:1.

An average is not a schedule

“One chance in six” does not promise one win every six rolls. For six independent $10 Any Seven bets, the expected net result is −$10, but the chance of losing all six is:

(5/6)⁶ ≈ 33.49%

That outcome costs $60, not $10. Variance measures the spread around the expected result; expected value describes the average, not your session’s outcome.

For the next step, use expected value to calculate a wager’s long-run price. Keep any gambling spending within a limit set before play—not a prediction that the average will arrive before your money runs out.