What "at least one" covers
Roll a fair dice twice. "At least one six" means a six on the first roll, a six on the second roll, or a six on both. The possibility diagram has 36 equally likely outcomes, the first roll down the side and the second across the top; every cell with a six in it is marked.
The first roll down the side and the second across the top. A cell shows 6 when either roll is a six, and a dot when neither is.
Count the outcomes with no six
Counting the sixes means sorting out one six from two. The other side is simpler: no six at all. The first roll misses a six in 5 ways and so does the second, so the cells with no six form a block of 5 rows by 5 columns, 5 × 5 = 25 cells.
The rolls are independent, so the same number comes from multiplying: the chance of no six is .
The 25 cells with no six fill a block of 5 rows by 5 columns.
Everything else has a six
Every outcome either has no six or has at least one six, and never both. So at least one six is the rest of the grid: 36 − 25 = 11 outcomes, and the chance of at least one six is .
Counting the sixes directly gives the same 11: 6 cells in the last row and 6 in the last column, with the double six in both, so 6 + 6 − 1 = 11.
The 11 cells with at least one six: the last row and the last column, which share the double six.
One minus the chance of none
"At least one" and "none" are complements, so P(at least one) = 1 − P(none). This works for any number of tries.
With three rolls there are 6 × 6 × 6 = 216 outcomes, too many for a grid. The chance of no six is , so the chance of at least one six is , a little less than a half.
Counting the sixes directly would need three cases. Exactly one six can be on any of the 3 rolls, with 5 × 5 ways for the other two: 3 × 25 = 75. Exactly two sixes: 3 places for the roll that is not a six, with 5 ways each, so 15. Three sixes: 1. And 75 + 15 + 1 = 91. The complement needs one case instead of three.
The chance of at least one six in n rolls, , for n = 1 to 6. It rises with each roll, passes 0.5 at 4 rolls, and never reaches 1.
Two tries at one chance in four
Two independent tries each win with probability . Each misses with probability , so the chance of no win is , and the chance of at least one win is .
Picture each try as 4 equally likely results, 1 of them a win. The grid has 16 cells, the 9 with no win form a 3 by 3 block, and the other 7 are the first row and the first column.
Result 1 is the win on each try. The 7 ticked cells have at least one win, and the 9 colored cells have none.
The usual mistakes
Adding the chances: instead of . The cell where both tries win is inside both chances, so adding counts it twice. With dice, adding for each of seven rolls would give , more than 1.
Taking 1 minus the chance of exactly one win. Exactly one win is , and is wrong, because exactly one win is part of "at least one". The complement of at least one is none.
Multiplying the chances of a win, . That is the chance that both tries win, the smallest part of "at least one".
A fairground game
In the application below, a stall pays a prize for at least one six in three rolls, and the stall-holder claims the chance is . The complement gives the true chance, and then the number of rolls needed for a better than even chance.
Worked example: A Fairground Game That Pays on at Least One Six in Three Rolls, Worked by the Complement
Question A fairground stall charges a player to roll a fair dice three times, and pays a prize if at least one six comes up. The stall-holder says the chance of a prize is 36, because each roll has a 16 chance of a six. (a) Find the true probability of at least one six in three rolls. (b) What is the smallest number of rolls for which the chance of at least one six is more than 12?
1.The complement of at least one six is no six in any of the three rolls. Each roll misses a six with probability 56.
On one roll the chance of no six is 56. The complement of at least one six is no six on any roll. 2.The rolls are independent, so P(no six) = 56 × 56 × 56 = 125216.
The rolls are independent, so the chance of no six shrinks by 56 with each roll: 125216 after three. 3.(a) P(at least one six) = 1 − 125216 = 91216, which is about 0.421 — the game is worse than even, not the stall-holder's even chance. Adding 16 three times counts a roll with two or three sixes more than once, because those events can happen together; with seven rolls the same reasoning would give 76, which is more than 1.
(a) 1 − 125216 = 91216, about 0.421. Adding 16 three times would count some throws more than once. 4.Three rolls give less than a half, since 91216 is less than 108216. Try a fourth roll: P(no six) = 6251296, so P(at least one six) = 6711296.
Three rolls fall short of a half, so try a fourth: 6711296. 5.(b) The smallest number is 4 rolls: 6711296 is more than 12 = 6481296, and three rolls give less.
(b) Four rolls is the fewest: 6711296 is more than a half.
Answer: (a) 91216, about 0.421; (b) 4 rolls
Common mistakes
- Adding 16 three times to get 36. A six on the first roll and a six on the second roll can both happen, so adding counts those throws more than once. The complement has no overlap to worry about.
- Working out 16 × 16 × 16. That is the chance of a six on every one of the three rolls, which is far smaller than the chance of at least one six.
More probability with several events problems, worked step by step →