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Probability in Dice Games
Discrete Mathematics · Chapter 7: Probability in Games, Gambling, and Strategic DecisionSection 7.1 · lesson 40 of 8711 min · free with a free account
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In this lesson
From 36 equally likely outcomes to the house edge in craps.
- Build correct sample spaces for 1 and 2 dice
- Derive the distribution of a 2-dice sum
- Compute craps probabilities, including the point phase
- Derive a house edge from expected value
Transcript
Everything said in the video, section by section (1,656 words). Read along, or search it.
Probability in Games, Gambling, and Strategic Decision
Probability in Dice Games
Chapter seven applies discrete probability to games, gambling, and strategic decisions. It starts with dice. This section builds the sample space for one and two dice, derives the distribution of a two-dice sum, and uses it to work out the game of craps. That includes the point phase, and finally the house edge that keeps a casino profitable.
The Opening Problem
Odds Set Before the Table Opens
Casinos don't merely observe that some dice outcomes happen more often than others. They calculate the exact probability of every outcome, before a table ever opens. The house edge that keeps a casino profitable is built entirely from these numbers, not from trial and error. The same tools are used by tabletop game designers, calibrating a two-dice resource mechanic. And by software engineers, testing whether a dice app's random number generator behaves like a pair of fair physical dice. In every case, the underlying question is the same. Given a well-defined set of equally likely outcomes, what's the probability of a particular event?
7.1.1 Sample Spaces
Why 2 Dice Must Be Told Apart
Every calculation starts with a sample space, the set of all possible outcomes. Getting it wrong, even subtly, is the most common source of error in introductory probability. When every outcome is equally likely, the probability of an event is its size over the size of the sample space. That's the discrete uniform model. One fair die has six equally likely outcomes, each with probability one sixth. Two dice give the Cartesian product, every ordered pair i comma j, for thirty-six outcomes in all. Even if the two dice look identical, they must be treated as distinguishable. The die thrown first and the die thrown second, or mentally colored red and blue. If you use unordered pairs instead, there are only twenty-one of them, and they aren't equally likely. The pair one, two is two physical outcomes. The pair one, one is just one. Applying the counting formula to outcomes that aren't equally likely gives a meaningless number. So the ordered-pair sample space is the right starting point.
Examples 7.1.1 and 7.1.2
Doubles, and a Six on Either Die
First, rolling doubles, both dice showing the same value. There are six such outcomes, so the probability is six thirty-sixths, which is one sixth. That matches a single die. It's a useful check, since a common misconception is that doubles should be rarer, just because two dice are involved. Next, the probability of at least one six. Counting directly is possible, but the complement is much faster. Neither die shows a six with probability five sixths times five sixths, or twenty-five thirty-sixths. So at least one six has probability eleven thirty-sixths, about zero point three one. As a check, count directly. Six pairs have a six on the first die, and six on the second. Subtract the pair six, six, which was counted twice. That's eleven.
Theorem 7.1.1
Counting the Ways to Roll Each Sum
Most dice games only care about the sum, i plus j. So the key modeling step is turning thirty-six outcomes into a distribution over sums. The theorem says the number of ways to roll sum s is s minus one, for sums up to seven. And thirteen minus s, for sums from seven up. Here's why. For a target sum s, an outcome works exactly when j equals s minus i. So we just count valid values of i. Two constraints apply. i must be between one and six, and so must s minus i. That forces i to run from the larger of one and s minus six, up to the smaller of six and s minus one. For sums up to seven, that range is one through s minus one, which has s minus one values. For sums from seven up, the range is s minus six through six, which has thirteen minus s values. That proves both branches.
Table 7.1.1
A Triangle Peaking at 7
Here's the full distribution. The counts rise from one way at a sum of two, to a peak of six ways at seven, then fall back to one way at twelve. Each probability is the count over thirty-six. It's the same symmetric, triangular shape plotted in section six point one. Reading from the table, the probability that the sum is at most four is one plus two plus three, over thirty-six. That's six thirty-sixths, or one sixth.
Example 7.1.4
The Expected Sum Is 7
What's the expected sum? Computing it straight from the table means eleven weighted terms. Linearity of expectation shortens the work. The expected value of a sum is the sum of the expected values, whether or not the variables are independent. For one fair die, the expected value is twenty-one over six, which is three and a half. The second die is the same by symmetry. So the expected sum is three and a half plus three and a half, which is seven. And a direct check from the table gives two hundred fifty-two over thirty-six, also seven. That's why seven is the mode, the median, and the mean of the two-dice sum. And the rules of craps rely on it directly.
7.1.3 Craps
The Come-Out Roll
Craps is played in rounds, built directly on this distribution. On the first roll of a round, the come-out roll, a pass-line bettor wins immediately on a seven or eleven. That's called a natural. Its probability is six plus two, over thirty-six, which is two ninths. The bettor loses immediately on two, three, or twelve, called craps. That's one plus two plus one, over thirty-six, or one ninth. Any other sum becomes the point. That happens with probability one minus two ninths minus one ninth, which is two thirds. And counting the ways directly gives twenty-four thirty-sixths, the same. Once a point is set, the shooter keeps rolling until that point reappears, a win, or a seven appears first, a loss called sevening out. Every other sum is simply ignored.
The Structure of a Round
A Craps Round, Mapped Out
Here's the structure of a round. The come-out roll either resolves the bet at once, as a natural or craps, or sets a point. With a point set, the point phase repeats, ignoring other sums, until either the point or a seven appears. The chance the point comes first is the ratio we'll derive next.
Example 7.1.6
Making a Point of 4
That derivation needs independence. Two events are independent if the probability of both equals the product of their probabilities. Successive rolls are independent. One roll has no effect on the probabilities of the next. Suppose the point is four. The probability of a four is three thirty-sixths, one twelfth. Call it p. A seven has probability one sixth. Call it q. Everything else, with probability three quarters, is ignored. Call that r. The point can be made on the first roll, or after one ignored roll, or after two, and so on. Those possibilities are disjoint, and independence makes each one a power of r times p. They form a geometric series, from chapter three. It sums to p over one minus r. And one minus r is just p plus q. So the probability of making the point is p over p plus q. Here, one twelfth over three twelfths, which is one third.
The General Rule
Points Near 7 Are Easier
The same argument works for any point. The probability of winning is the point's probability, over the point's probability plus the probability of a seven. For a point of four or ten, that's one third. For five or nine, it's four over four plus six, which is two fifths. And for six or eight, it's five over five plus six, which is five elevenths, about zero point four five five. Points closer to seven have more ways to be rolled, against the same fixed chance of a seven. So they're mathematically easier to make.
7.1.4 The House Edge
Deriving 1.41% on the Pass Line
Now the house edge. For a bet of one unit, the house edge is minus the expected value, as a percentage. It's the long-run fraction of each unit wagered that the house expects to keep. A pass-line bettor wins on the come-out roll with probability two ninths. Or by making a point, with probability of that point times the chance of making it, for each point value. Adding all of those gives a total win probability of two hundred forty-four over four hundred ninety-five, about zero point four nine three. The bet pays one to one. So the expected value is twice the win probability, minus one. That's minus seven over four hundred ninety-five, about minus zero point zero one four one. So the house edge is about one point four one percent. That matches the widely published figure for the pass line, and here it's derived, not just asserted.
Table 7.1.2
House Edges Across Craps Bets
This table compares house edges across common craps bets. The pass line, derived here, is about one point four one percent. Don't pass, its mirror image with a tie on twelve, is about one point three six. The field bet, a one-roll bet on several sums, runs from about two point seven to five point five percent, depending on the payout. Any craps, a one-roll bet on two, three, or twelve, is about eleven point one percent. And big six or big eight, betting that six or eight comes before a seven, is about nine point zero nine. The figures other than the pass line are standard published values, and they shift slightly between casinos whose payouts differ.
Takeaways
Equally Likely Outcomes, All the Way Up
To summarize. Two dice must be treated as distinguishable, ordered outcomes, or the sample space stops being equally likely. The sum distribution is triangular and symmetric about seven. The number of ways is s minus one up to seven, and thirteen minus s from seven up. Independence of rolls gives the point-phase rule, the point's probability over itself plus a seven's. And expected value reproduces the one point four one percent pass-line edge. Here's something to think about. A shooter rolls three sevens in a row, on three come-out rolls. A player argues the dice are now due for something else. Using independence, explain exactly what's wrong with that for fair dice. Then explain why the argument changes if you were told, beforehand, that the dice might be weighted. Why would several sevens in a row count as evidence then?
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