The question
Roll a fair six-sided die repeatedly and note each face as it appears. How many rolls, on average, until you have seen all six at least once? The answer is about 14.7. Many people guess closer to 6 or 8, because the first few faces come quickly. The slow part is the end, when you are waiting for the last missing face.
This is an example of the coupon collector problem, named after the puzzle of collecting a full set of prizes found in packets. The same result holds for any set of equally likely items.
Waiting for each new face
Break the process into stages. In each stage you wait for a face you have not yet seen.
- Stage 1: any face is new. The chance is 6/6 = 1, so you wait 1 roll.
- Stage 2: five faces are still unseen. The chance of a new one on each roll is 5/6.
- Stage 3: four unseen, chance 4/6.
- Stage 4: three unseen, chance 3/6.
- Stage 5: two unseen, chance 2/6.
- Stage 6: one unseen, chance 1/6.
When a success has probability p on each independent try, the average number of tries until the first success is 1/p. This is the standard result for waiting times of this kind. So the average waiting times for the stages are:
| Stage | Success chance | Average rolls |
|---|---|---|
| 1 | 6/6 | 1 |
| 2 | 5/6 | 1.2 |
| 3 | 4/6 | 1.5 |
| 4 | 3/6 | 2 |
| 5 | 2/6 | 3 |
| 6 | 1/6 | 6 |
Add them: 1 + 1.2 + 1.5 + 2 + 3 + 6 = 14.7.
The formula
Notice the pattern. Each average is 6 divided by the number of unseen faces. So the total is
6 x (1/6 + 1/5 + 1/4 + 1/3 + 1/2 + 1) = 6 x 2.45 = 14.7
The bracket is a harmonic sum. For n equally likely faces or items, the expected number of rolls is n multiplied by the sum 1 + 1/2 + … + 1/n. For a coin (n = 2) it is 2 x 1.5 = 3 tosses to see both sides. For a twenty-sided die it is about 72, which is much higher, though it requires a longer sum I will not reproduce here.
Why the last face dominates
Of the 14.7 expected rolls, the final face accounts for 6 and the final two for 9. In other words, around 61 percent of the average waiting time (9 out of 14.7) goes on the last two faces. The first face is free, and by the time five are seen the chance of any given roll helping is only 1 in 6.
An average is not a guarantee
An expected value is a long-run average over many repeated experiments. It does not say you will need 14 or 15 rolls this time. The actual number varies a lot. The minimum is 6 rolls, when all six faces appear in the first six rolls, which has probability 6!/6^6 = 720/46,656, about 1.5 percent. The number can be much more than 15 as well, since there is no upper limit. Some runs of rolls will still be missing a face after 25 or 30 rolls, though this is uncommon.
That is why the number is best read as typical rather than certain. Simulating the process many times, with a spreadsheet or Python dice simulation, gives an average close to 14.7 but each trial differs. The match gets closer as the number of trials grows, in line with the law of large numbers.
Where this shows up
The idea is not limited to dice. In a game with a random reward from a fixed set of equally likely items, the same calculation estimates how many draws are needed to obtain a complete set, if the drops really are uniform. Real games often have unequal drop rates, which usually makes the wait longer, and they may add features such as pity timers. Reading a published rate and understanding the odds is part of making an informed decision about time or money.
A caution
This is a maths explanation for a fair die. A complete set of faces does not have any special value, and the calculation does not help with predicting rolls. In a game of chance with money at stake, understanding expected waiting times does not alter the house edge. Gambling can be harmful, and anyone concerned about their play can seek help from a local support service.