Why is the chance of at least one six in four dice rolls not 4/6?

## Why the additive answer is wrong Probabilities add only across **mutually exclusive** events - events that cannot both occur. "Six on roll 1" and "six on roll 2" can both occur, so `P(A or B) = P(A) + P(B)` does not apply. Adding `1/6` four times counts an outcome with two sixes twice, one with three sixes three times, and so on. The fastest way to see the error is to push it: seven rolls would give `7/6`, a probability greater than 1. Any method that can exceed 1 is not a probability method. (What `n x p` *is* is a valid upper bound - the union bound - which is why it is close to right when `n x p` is small and badly wrong when it is not.) ## The complement trick The event "at least one six" has a messy structure: exactly one six, exactly two, exactly three, exactly four. Its negation has a single clean structure: **every** roll is a non-six. Because the rolls are independent, the probability that all four miss is a product: ``` P(no six) = (5/6) x (5/6) x (5/6) x (5/6) = (5/6)^4 = 625/1296 = 0.4823 ``` And then ``` P(at least one six) = 1 - 625/1296 = 671/1296 = 0.5177 ``` So the bet is slightly favourable - just over 51.7 percent, not the 66.7 percent the additive error suggests. ## The general form For `n` independent trials each with success probability `p`: ``` P(at least one success) = 1 - (1 - p)^n ``` This single formula covers a large share of "at least one" interview questions: at least one head in five flips (`1 - (1/2)^5 = 31/32`), at least one defective in a batch, at least one of `n` independent components failing during a mission. Two properties are worth stating: - It is **increasing in `n`** and approaches 1, but never reaches it. More trials always help; no finite number guarantees a success. - For small `p`, `1 - (1-p)^n` is slightly **less than** `n x p`. The additive estimate is always an over-estimate, and the gap is what the double counting represents. ## Solving for n "How many rolls do I need for a 90 percent chance of at least one six?" Set `1 - (5/6)^n >= 0.9`, so `(5/6)^n <= 0.1`, so `n >= ln(0.1)/ln(5/6) = 12.63`. Round **up**: 13 rolls. (Twelve rolls give about 0.888; thirteen give about 0.906.) The rounding direction matters - rounding down would leave you short of the target. ## The classic pair of bets A seventeenth-century gambler's puzzle contrasts two bets that look equivalent under the additive intuition: - At least one six in **4 rolls of one die**: `1 - (5/6)^4 = 0.5177` - favourable. - At least one double-six in **24 rolls of two dice**: `1 - (35/36)^24 = 0.4914` - unfavourable. The naive scaling argument says the second bet should match the first, because `24/36` equals `4/6`. It does not, and the gap is small enough (about 0.026) that it can only be found by computing rather than by intuiting. This is the canonical demonstration that "at least one" probabilities do not scale linearly with the number of trials. ## The independence assumption The complement is *always* valid - `P(at least one) = 1 - P(none)` is just the complement rule and needs no assumptions. What needs independence is **factorising** `P(none)` into a product of per-trial probabilities. If the trials are dependent, that step fails. Drawing cards without replacement is the standard trap: `P(no ace in 5 cards)` is not `(48/52)^5` but `(48/52) x (47/51) x (46/50) x (45/49) x (44/48)`, a chain of conditional probabilities. Say which assumption you are using and why it holds; an interviewer who slips "without replacement" into the prompt is testing exactly this. ## Where candidates go wrong 1. **Adding overlapping probabilities** and not noticing the result can exceed 1. 2. **Computing `P(none)` correctly and then forgetting to subtract from 1** - reporting 0.482 instead of 0.518. 3. **Confusing "at least one" with "exactly one".** Exactly one six in four rolls is `4 x (1/6) x (5/6)^3 = 0.386`, a different and smaller number. 4. **Factorising under dependence** without flagging it. ## The compact answer Name the overlap, show that the additive method breaks at seven rolls, then compute `1 - (5/6)^4 = 0.5177`. Three sentences, and it demonstrates both the error and the repair.