"$n$-Yahtzee" question
I believe the question divides into two parts:
- How many rolls to make a "point" of 2 or more dice with the same number.
- How many rolls of the remaining dice to achieve that "point".
Given that the rolls of dice are independent geometrically distributed variables we can use:
$P(Y)$ is the geometric probability distribution of the number $Y=X−1$ of failures before the first success, supported on the set $\{0,1,2,3,...\}$ in setting the "point". The expected value of this is $E_1=\frac{1−p}{p}$, where p is the probability of setting the point (i.e. getting a pair or better).
$R(Z)$ is the distribution for the remaining dice ($r$) to converge on the point. This the maximum of $r$ i.i.d. geometric random variables with $q=\frac{1}{6}$. The answer to Expectation of the maximum of IID geometric random variables gives an expected value of $$ \mathrm E_2=\sum_{k\geqslant0}(1-\mathrm P(Z\leqslant k))=\sum_{k\geqslant0}(1-\mathrm P(Y\leqslant k)^r)=\sum_{k\geqslant0}(1-(1-(1-q)^{k+1})^r). $$
The answers below have been checked against an empirical calculation of 999,999 results using a visual basic program. Empirical results are shown italicized.
$n=1$
Any number will do
$$E_1+E_2=0$$
$n=2$
$$\begin{align}E_1+E_2&=\frac{1-\left(1-\left(\frac{5}{6}\right)\right)}{1-\left(\frac{5}{6}\right)}+\sum_{k\geqslant0}(1-(1-(1-\frac{1}{6})^{k+1})^{2-2})\\ & =\frac{\left(\frac{5}{6}\right)}{\left(\frac{1}{6}\right)}\\ &=5\\ \end{align}$$
5.00 SD 5.48
I'm confident that this answer is correct since it only relies on setting the point, the term relating to the rolls of 0 dice to match it is 0 - which makes complete sense. Let's see how we go with $n=3$
$$\begin{align}E_1&=\frac{1-\left(1-\left(\frac{5}{6}\right)\left(\frac{4}{6}\right)\right)}{1-\left(\frac{5}{6}\right)\left(\frac{4}{6}\right)}\\ &=\frac{20}{16}\\ &=1.25 \end{align}$$
1.25 SD 1.68
$$\begin{align}E_2&=\sum_{k\geqslant0}(1-(1-(1-\frac{1}{6})^{k+1})^{3-2})\\ &=\sum_{k\geqslant0}\left(\frac{5}{6}\right)^{k+1}\\ &=\frac{1}{1-\frac{5}{6}}-1\\ &=5\ \end{align}$$
5.63 SD 5.49 - this is a little high but only by 0.11 SD
$$E_1+E_2=6.25$$
6.88 SD 5.74
Moving on to $n=4$
$$\begin{align}E_1&=\frac{1-\left(1-\left(\frac{5}{6}\right)\left(\frac{4}{6}\right)\left(\frac{3}{6}\right)\right)}{1-\left(\frac{5}{6}\right)\left(\frac{4}{6}\right)\left(\frac{3}{6}\right)}\\ &=\frac{60}{156}\\ &=0.38 \end{align}$$
0.39 SD .73
This is going down as expected, with more dice in play pairs are easier to come by.
$$\begin{align}E_2&=\sum_{k\geqslant0}(1-(1-(1-\frac{1}{6})^{k+1})^{4-2})\\ &=\sum_{k\geqslant0}(1-(1-(\frac{5}{6})^{k+1})^{2})\\ &=\sum_{k\geqslant0}(1-(1-2(\frac{5}{6})^{k+1}+(\frac{5}{6})^{2k+2}))\\ &=2\sum_{k\geqslant0}(\frac{5}{6})^{k+1}-\sum_{k\geqslant0}\left(\left(\frac{5}{6}\right)^2\right)^{k+1}\\ &=2\left(\frac{1}{1-\frac{5}{6}}-1\right)-\left(\frac{1}{1-\frac{25}{36}}-1\right)\\ &=10-\left(\frac{36}{11}-1\right)\\ &=10-\frac{25}{11}\\ &\approx7.73\\ \end{align}$$
8.31 SD 6.14
$$E_1+E_2\approx8.11$$
8.7 SD 6.19
The real Yahtzee to $n=5$
$$\begin{align}E_1&=\frac{1-\left(1-\left(\frac{5}{6}\right)\left(\frac{4}{6}\right)\left(\frac{3}{6}\right)\left(\frac{2}{6}\right)\right)}{1-\left(\frac{5}{6}\right)\left(\frac{4}{6}\right)\left(\frac{3}{6}\right)\left(\frac{2}{6}\right)}\\ &=\frac{120}{1176}\\ &=0.102 \end{align}$$
0.102 SD 0.336
$$\begin{align}E_2&=\sum_{k\geqslant0}(1-(1-(1-\frac{1}{6})^{k+1})^{5-2})\\ &=\sum_{k\geqslant0}(1-(1-3(\frac{5}{6})^{k+1}+3(\frac{5}{6})^{2(k+1)}-(\frac{5}{6})^{3(k+1)}))\\ &=3\sum_{k\geqslant0}(\frac{5}{6})^{k+1}-3\sum_{k\geqslant0}(\frac{5}{6})^{2(k+1)}+\sum_{k\geqslant0}(\frac{5}{6})^{3(k+1)}\\ &=3\left(\frac{1}{1-\frac{5}{6}}-1\right)-3\left(\frac{1}{1-\frac{25}{36}}-1\right)+\left(\frac{1}{1-\frac{125}{216}}-1\right)\\ &=15-\frac{75}{11}+\frac{125}{91}\\ &\approx9.56\\ \end{align}$$
10.0 SD 6.40
$$E_1+E_2\approx10.1$$
I think these are the correct answer even though the empirical results are a bit high for all of them (but well within 1 standard deviation).
The program I used to generate it is a 9kB file which I would post if I knew how. The spreadsheet it generates, however, is 165MB.
For interest, here are the results of the 5-Yahtzee using 3 slightly different algorithms. The D (for dumb) algorithm sets the point at the highest mode on the first roll (i.e. it will match a single die), the normal algorithm sets the point on the first pair or better, the S (for smart) sets the point on the first pair or better but will switch if 3 of a kind come up on the other 3 dice.
The means of the three algorithms are 10.24, 10.16 and 10.10 respectively.
With the 3 rolls you get in a game of Yahtzee you have a 0.0456 chance of getting a Yahtzee on any given turn. You get 13 turns in a game so if you go for nothing else (which is probably not a winning strategy), you have a 0.455 chance of at least one per game. In a 4 person game there is a 0.911 chance of at least one.