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Consider the following problem, from Understanding Probability by Henk Tijms:

What is the fewest number of dice one can roll such that, when they are rolled simultaneously, there will be at least 50% probability of rolling two or more sixes?

Call $r$ the required number.

My first idea: count the number of sequences of length $r$ with at least two sixes, and then impose that the relative probability is at least $1/2$. This leads to the equation: $$ \binom{r}{2}\frac{6^{r-2}}{6^r}=\frac12, $$ which clearly has no sense since the left hand side tends to $\infty$ with $r$.

Second idea: $$ P(\text{no sixes})=\left(\frac56\right)^r $$ $$ P(\text{exactly one six})= \binom{r}{1}\frac16\left(\frac56\right)^{r-1} $$ and $$ P(\text{at least two sixes})=1-P(\text{no sixes})-P(\text{exactly one six})=\frac12 $$ should solve the problem. Summing the series with Mathematica leads to: $$ r=-\frac{W_{c_1}\left(-\frac{15625 (\log (2)+\log (3)-\log (5))}{15552}\right)}{\log (2)+\log (3)-\log (5)}-5 $$ here $W_{c_1}$ denotes the branch $c_1$ of the Lambert W function. After trial and error, I found that the branch $c_1=-1$ gives a real solution, which is roughly $9.727$, and I know that the correct solution to this problem is $r=10$, but this approach seems way too complicated to me, so I expect that there is a simpler solution. This same question was asked here, but it does not solve my problem.

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Let $P_r=1-\left(\frac{5}{6}\right)^r-\frac{r}{6}\left(\frac{5}{6}\right)^{r-1}$, then $P_r\geq 1/2$ iff $$a_r:=\left(\frac{5}{6}\right)^r+\frac{r}{6}\left(\frac{5}{6}\right)^{r-1}\leq \frac{1}{2}.$$ Now note that the sequence $(a_r)_{r\geq 1}$ is strictly decreasing: $a_r> a_{r+1}$ is equivalent to $$\left(\frac{5}{6}\right)+\frac{r}{6}> \left(\frac{5}{6}\right)^2+\frac{r+1}{6}\left(\frac{5}{6}\right)\Leftrightarrow r>0.$$ Finally it is easy to verify by using a pocket calculator that $a_9>1/2>a_{10}$.

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  • $\begingroup$ Thanks. It is not always simple for me to understand when it is better to stop and use some numerical approximation. In this case seeing the term $r p^r = \text{something}$ should probably have warned against the presence of the $W$ function and hint at a numerical solution. I like the analysis on the monotonicity of $a_r$. $\endgroup$
    – J. D.
    Jan 8, 2018 at 18:55
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Your second approach is fine.

So, if you roll $n$ dice, the probability of at least two $6$s is given by:

$1-\left(\frac56\right)^n-n\cdot \left(\frac56\right)^{n-1}\cdot\frac16$.

You can then verify numerically (trial and error) that $9$ dice gives a probability somewhat less than $\frac12$ and $10$ dice gives a probability somewhat greater than $\frac12$. So you need at least $10$ dice to have the desired probability of at least two $6$s.

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  • $\begingroup$ I see, indeed the resulting equation in $n$ would not be too hard to check for a few values of $n$. $\endgroup$
    – J. D.
    Jan 8, 2018 at 18:52

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