Dice Probability Explained: Odds for 1 to 6 Dice

Every dice game is a probability problem wearing a costume. Whether you are deciding which dice to hold in Yahtzee, whether to bank 300 points in Farkle, or whether the Pass Line in Craps is a fair bet, the answer comes from counting outcomes. This guide walks through the odds for one to six dice, shows the arithmetic behind each figure, and ends with the tools you need to compute any dice probability yourself.

How Dice Probability Works

A fair six-sided die has six faces that are equally likely, so each face has a probability of 1/6, about 16.67%. When you roll more than one die, the dice are independent: what one shows has no effect on the others. That single fact does almost all the work in dice mathematics, because it means you can count outcomes by multiplication.

• One die: 6 outcomes.
• Two dice: 6 × 6 = 36 outcomes.
• Three dice: 6 × 6 × 6 = 216 outcomes.
• Four dice: 1,296 outcomes.
• Five dice: 7,776 outcomes.
• Six dice: 46,656 outcomes.

Every probability in this article is a fraction with one of those numbers on the bottom. On top is the count of outcomes that satisfy whatever you are asking about. Rolling a 7 with two dice is 6/36 because six of the 36 ordered outcomes add to 7. Rolling a Yahtzee in one throw is 6/7,776 because only six of the 7,776 ordered outcomes show the same face on all five dice.

One trap to avoid: treat the dice as distinguishable even when they look identical. “A 3 and a 4” is two outcomes (3-4 and 4-3), while “a 3 and a 3” is one. Forgetting this is the most common way to get dice odds wrong, and it is why the Craps table pays hardways (a 4 made as 2-2) far better than easy ways (1-3 or 3-1).

One Die and Expected Value

The expected value of a roll is the average you would see over a very large number of rolls. For a single die it is the sum of every face times its probability:

(1 + 2 + 3 + 4 + 5 + 6) × 1/6 = 21/6 = 3.5

No face shows 3.5, but that is the long-run average. Expected values add, so the expected total of two dice is 7, of three dice is 10.5, and of five dice is 17.5. In Ship, Captain and Crew the cargo is the sum of two dice, so a typical cargo is 7 and anything from 10 up is above average. In Qwixx, the two white dice sum to 7 more often than any other number, which is why the middle of each row fills up fastest.

Single-die odds also settle the basic question in Pig: each roll has a 1/6 chance of showing the 1 that wipes out your turn, and a 5/6 chance of adding an average of (2 + 3 + 4 + 5 + 6) ÷ 5 = 4 points. We will use that to find Pig’s famous break-even point later.

Two Dice: The 36-Outcome Table

With two dice there are 36 equally likely ordered outcomes. The table lists every total and how many ways it can be made.

The shape is a triangle peaking at 7. A few consequences worth memorizing:

• Doubles happen 6 times in 36, so 1 roll in 6.
• 7 or 11 (a natural on the come-out roll in Craps) is 8/36, or 22.2%.
• 2, 3 or 12 (craps on the come-out) is 4/36, or 11.1%.
• A total of 10 or more is 6/36, or 16.7%.

Craps is the two-dice table turned into a betting game, and the Pass Line odds fall straight out of it. You win at once with 7 or 11 (8/36). You lose at once with 2, 3 or 12 (4/36). Otherwise a point is set and you need that number before a 7. For a point of 4, three outcomes make 4 and six make 7, so the chance of winning is 3/(3 + 6) = 1/3. Adding up every branch:

8/36 + (3/36 × 3/9) × 2 + (4/36 × 4/10) × 2 + (5/36 × 5/11) × 2 = 244/495 = 49.29%

The house keeps the other 50.71%, and the gap between the two, 1.41%, is the house edge on the Pass Line. It is one of the lowest of any casino bet, and you can see it in action on our Craps page.

At Least One Six: One to Six Dice

“What is the chance of at least one six?” is the most useful question in dice probability, because the trick that answers it works for almost everything. Do not count the ways to get a six. Count the ways to get none, and subtract from 1.

Each die fails to show a six with probability 5/6. Because the dice are independent, n dice all fail with probability (5/6)^n. So:

P(at least one six with n dice) = 1 − (5/6)^n

Notice that six dice do not give you a 100% chance of a six, or even six times the single-die chance. Each extra die adds less than the one before, because it can only help on the rolls where every earlier die already missed.

This formula turns up everywhere. In Ship, Captain and Crew you need a 6 before anything else counts, and your first throw of five dice finds one 59.8% of the time. In Farkle and Ten Thousand the question is whether six dice contain at least one 1, which is the same 66.5%, or at least one 1 or 5, which is 1 − (4/6)^6 = 91.2%. In Bunco, three dice show at least one copy of the target number 42.1% of the time.

Five Dice: Poker Combinations Out of 7,776

Five dice have 7,776 ordered outcomes. Every roll falls into exactly one of the seven poker-style patterns below, and the counts add up to 7,776, which is a handy check that nothing has been missed.

Straights are counted separately because they overlap with the patterns above (a large straight is also a “no pair” roll):

The arithmetic behind each count

• Yahtzee: 6 faces, one way each. 6.
• Four of a kind: 6 choices for the quad face, 5 for the odd die, and the odd die can sit in any of 5 positions. 6 × 5 × 5 = 150.
• Full house: 6 choices for the triple, 5 for the pair, and 5!/(3! × 2!) = 10 arrangements. 6 × 5 × 10 = 300.
• Three of a kind: 6 choices for the triple, 10 ways to pick which three dice show it, then the other two dice must be different from the triple and from each other: 5 × 4 = 20 ordered choices. 6 × 10 × 20 = 1,200.
• Two pair: 15 ways to choose two pair faces from six, 4 choices for the kicker, and 5!/(2! × 2!) = 30 arrangements. 15 × 4 × 30 = 1,800.
• One pair: 6 choices for the pair, 10 ways to choose the three other faces from the remaining five, and 5!/2! = 60 arrangements. 6 × 10 × 60 = 3,600.
• No pair: 6 × 5 × 4 × 3 × 2 = 720.
• Large straight: 2 possible sequences, each in 5! = 120 orders. 240.

Check: 6 + 150 + 300 + 1,200 + 1,800 + 3,600 + 720 = 7,776.

These are single-roll figures. In Yahtzee and Dice Poker you get up to three rolls with holds in between, which changes everything. Keeping your most common face and re-rolling the rest lifts the chance of a Yahtzee over a turn to roughly 4.6%, about 1 in 22, and over a 13-turn game you will see at least one Yahtzee about 46% of the time. The Yahtzee score sheet guide uses these numbers to explain why the upper-section bonus is worth chasing.

Six Dice: Farkle Odds

Farkle and Ten Thousand roll six dice and score 1s, 5s, three or more of a kind, three pairs and a 1-to-6 straight. A roll with none of those is a farkle, and it costs you everything set aside that turn. The chance of that disaster depends on how many dice you are throwing.

Where the 1,080 comes from

A six-dice farkle can only use the faces 2, 3, 4 and 6, and no face may appear three times (that would be a triple) or in three matched pairs. Distributing six dice across four faces with no face above two copies leaves two patterns: 2-2-2-0, which is three pairs and scores, and 2-2-1-1, which does not. For 2-2-1-1 there are C(4, 2) = 6 ways to choose the paired faces, the other two faces are the singletons, and the six dice can be arranged in 6!/(2! × 2!) = 180 orders. 6 × 180 = 1,080. Divide by 46,656 and you get 2.31%.

The smaller cases work the same way. With three dice, 4 × 4 × 4 = 64 rolls avoid 1s and 5s, and 4 of them are triples that score, leaving 60 farkles out of 216. With two dice, 4 × 4 = 16 of 36 rolls contain neither a 1 nor a 5. With one die, 4 of the 6 faces are blanks.

Some other six-dice figures that matter at the table: three pairs shows up 1,800 times in 46,656 (3.86%), a 1-to-6 straight 720 times (1.54%), and six of a kind 6 times (0.013%, or 1 in 7,776). The full breakdown is on the Farkle page, and the Farkle vs Ten Thousand comparison shows how different scoring tables shift these values.

Expected Value and When to Stop Rolling

Push-your-luck games ask a single question over and over: is one more roll worth it? Expected value answers it. Compare the expected gain from the roll against the expected loss, which is the chance of busting multiplied by what you would lose.

Pig. With a turn total of T, a roll gains an average of 4 points on the 5/6 of rolls that are not a 1, and loses T on the 1/6 that are. Expected change = (5/6 × 4) − (1/6 × T) = 3.33 − T/6. That is zero when T = 20, which is why “hold at 20” is the standard rule of thumb in Pig. Above 20, the average roll costs you points.

Farkle. With three dice, the average score from a roll (counting farkles as zero) is about 84 points, and the farkle chance is 27.78%. Expected change = 84 − 0.2778 × T, which hits zero around T = 302. That is the arithmetic behind the common advice to bank at 300 with three dice. With four dice the break-even is roughly 133 ÷ 0.1574 = 845 points, and with two dice it is 50 ÷ 0.4444 = 113 points.

Zombie Dice. The same logic drives Zombie Dice, except green, yellow and red dice carry different shotgun counts, so the stop decision depends on which colors are left in the cup as well as how many shotguns you have taken.

Liar’s Dice. Liar’s Dice hides most of the dice, so you estimate instead of count. With 1s wild, each hidden die has a 2/6 = 1/3 chance of matching any bid face, so n hidden dice hold about n/3 matches on average. With 10 hidden dice, the binomial formula gives a 44.1% chance that at least 4 match and 21.3% that at least 5 do.

Expected value is a long-run guide, not a law. If you are far behind in Ten Thousand on the last round, a negative-expectation roll is still correct because banking a small total cannot win. Probability tells you the price of the gamble; the score tells you whether you can afford not to take it.

How to Compute Dice Odds Yourself

Four techniques cover almost every dice question.

1. List the outcomes. For one or two dice, write them all down. The 36-outcome grid above is the whole of two-dice probability.

2. Use the complement. For “at least one” questions, compute the chance of “none” and subtract from 1. P(at least one 1 or 5 on six dice) = 1 − (4/6)^6 = 91.2%.

3. Use the binomial formula when you want exactly k successes in n independent tries with success chance p:

P(exactly k) = C(n, k) × p^k × (1 − p)^(n − k)

Example: the chance of exactly two 6s in five dice is C(5, 2) × (1/6)^2 × (5/6)^3 = 10 × 1/36 × 125/216 = 1,250/7,776 = 16.1%.

4. Count arrangements for pattern questions (full house, three pairs). Choose the faces, then multiply by the number of ways to arrange them, using n!/(a! × b! × …) for repeated faces.

When the pattern gets awkward, enumerate by computer. Every table in this article can be regenerated with a dozen lines of Python:

```python from itertools import product from collections import Counter

total, full_house = 0, 0 for roll in product(range(1, 7), repeat=5): total += 1 counts = sorted(Counter(roll).values()) if counts == [2, 3]: full_house += 1 print(full_house, total, full_house / total) # 300 7776 0.0386 ```

Change repeat=5 to repeat=6 and swap the condition to test any Farkle rule, or to check a house rule before you argue about it.

For the terms used above, see the dice game glossary. To find out which games use one, two, three, five or six dice, see dice games by number of dice, and for where these games came from, read the history of dice.