Loading...
Loading...
The Problem with Direct Counting:
"At least one 6 in 3 rolls" means: exactly 1, exactly 2, OR exactly 3 sixes. That's many cases to count! There's an easier way.
The Clever Shortcut:
Instead of counting all the ways to get "at least one," count the ONE way to get NONE. Then subtract from 1.
Why does this work?
There are only two possibilities: either you get at least one 6, or you get zero 6s. These are complements! So:
If you roll a die twice, what's the probability of getting at least one 6?
Think about what "no 6s at all" means
For zero 6s, BOTH rolls must be NOT a 6. Each roll independently must avoid 6.
First roll: P(NOT 6) = 5/6 (could be 1, 2, 3, 4, or 5)
Second roll: P(NOT 6) = 5/6 (same options)
Multiply because both must happen
For "no 6s," the first roll must avoid 6 AND the second roll must avoid 6. When we need BOTH independent events to happen, we multiply their probabilities.
Subtract from 1 to get "at least one"
Since "at least one 6" and "no 6s" are the only two possibilities:
Final Answer:
About a 31% chance of getting at least one 6 when rolling twice.
Intuition:
Think of it as "shrinking" the probability. If you need TWO lucky things to happen, it's harder than needing just one. Multiplying fractions (less than 1) gives a smaller number.
Example:
If there's a 5/6 chance of avoiding 6 on the first roll, only 5/6 of that 5/6 will also avoid 6 on the second roll. So: (5/6) × (5/6) = 25/36.
Step 1: Find P(no 6s in three rolls)
Each roll must avoid 6:
Step 2: Subtract from 1
Notice how your chances improve with more rolls, but never reach 100%!
| Rolls | P(no 6s) = (5/6)ⁿ | P(at least one 6) |
|---|---|---|
| 1 | ||
| 2 | ||
| 3 | ||
| 4 | (first time > 50%!) | |
| 6 |
Insight: You need about 4 rolls to have better than 50-50 odds of getting at least one 6.
For "at least one" problems, always use the complement!
The formula for n independent trials:
Where P(failure) is the probability of NOT getting what you want on a single try.