In regards to a game scenario I am designing

Question

In the game there is a 6 sided die which has 5 different faces (i.e. two of the faces are the same.) Considering that the die is fair and has 1/6 chance to roll any one side on each individual roll. What is the average number of rolls until I see the 5 different faces of the 6 sided die.

(I've seen a lot of different answers in regards to if each face of the die is different but not that pertains to if two sides have the same face. I'm sure it would change the formula significantly but I may be wrong.)

The above scenario is in the actual game. I'm trying to make my own scenario that is comparable. Thank you very much with your help.

Answer 1

When you break the symmetry of a problem you make it (often much) harder. With all the faces equiprobable you have the coupon collector's problem which you say you have reviewed. Now you need to condition on whether you have gotten the common face or not. In the usual coupon collector's problem, if you have seen $k$ coupons out of $n$ possible the chance that a new coupon is different from all the previous is $\frac {n-k}n$ Now if you have seen $k$ faces including the duplicated one the chance that a new roll is different is $\frac {5-k}6$ while if you have not seen the duplicated one it is $\frac {6-k}6$. This shows that it will be less than the chance to get all of six equiprobably distributed numbers. The expected time for that is $6H_6\approx 14.7$ The expected time for your experiment is greater than $5H_5 \approx 11.4$ so there is not much of a range.

Answer 2

In general, if your probability of "success" for an event is $p$, then the expected number of trials until success is $1/p$. If there are $k$ faces of a die you haven't seen, then the probability of seeing a face you haven't seen yet is $k/6$ and you'll have to throw the dice an expected amount $6/k$ before you see one of them.

Consider the situation where you roll the die repeatedly and you see every face. It happens that the last face to come up is the doubled one. In this case, there are six faces you haven't seen at first, so you expect to wait wait 6/6=1 round (i.e. you see a new face immediately). Then there are five faces you haven't seen. You expect to wait 6/5 rounds. Then there are four faces you haven't seen: 6/4 rounds. Then three: 6/3 rounds. Then two: 6/2. The last two faces are identical, so we're finished once we've seen one of them. The expected wait time in this special case is $6/6 + 6/5 + 6/4 + 6/3 + 6/2$. If you had seen the doubled side earlier, you can see that the subsequent wait times would have been different.

We can compute the expected wait time overall. We compute the expected wait time depending on if you see the doubled face first, second, third, and so on. We weight each of those times by the likelihood of seeing the doubled face first, second, third, and so on. Then we add up the weighted expectations.

$$\frac{2}{6}\left(\frac{6}{6}+\frac{6}{4} + \frac{6}{3} + \frac{6}{2} + \frac{6}{1}\right) + \frac{4}{6}\cdot\frac{2}{5}\left(\frac{6}{6} + \frac{6}{5} + \frac{6}{3} + \frac{6}{2} + \frac{6}{1}\right) +\\ \frac{4}{6}\cdot\frac{3}{5}\cdot\frac{2}{4}\left(\frac{6}{6} + \frac{6}{5} + \frac{6}{4} + \frac{6}{2} + \frac{6}{1}\right) + \frac{4}{6}\cdot\frac{3}{5}\cdot\frac{2}{4}\cdot \frac{2}{3}\left(\frac{6}{6} + \frac{6}{5} + \frac{6}{4} + \frac{6}{3} + \frac{6}{1}\right) + \\\frac{4}{6}\cdot\frac{3}{5}\cdot\frac{2}{4}\cdot \frac{1}{3}\cdot \frac{2}{2}\left(\frac{6}{6} + \frac{6}{5} + \frac{6}{4} + \frac{6}{3} + \frac{6}{2}\right)$$

$$= \frac{9}{2} + \frac{88}{25} + \frac{127}{50} + \frac{39}{25} + \frac{29}{50}\\= \frac{225 + 176 + 127 + 78 + 29}{50} = \frac{635}{50}$$ $$\frac{635}{50} = 12.7 \text{ rounds}$$

as opposed to 14.7 if you have six different faces.