# How many dice needed to make rolling at least $3$ sixes probable ($p>0.5$)

I'm wondering how to solve a questions of the example form:

How many dice are needed to make rolling at least 3 sixes in a single throw probable (p>0.5)

I know how to solve the question by graphing out all of the binomial probabilities for successive numbers of dice (for $n\geq3$ dice, compute $P_n(\geq 3$ sixes$) = 1 - P_n(0$ sixes$) - P_n(1$ six$) - P_n(2$ sixes$)$ each time and find the least $n$ such that $P_n(\geq 3$ sixes$) \geq 0.5$).

Enumerating out the probabilities for different numbers of dice, the answer I get is 16.

But one could of course consider generalising the question to arbitrary-sided dice, different number of successes, different target probability, etc. I'm wondering if there a more direct method to solve the general form of such a problem?

(There are already lots of similar questions on here, but the ones I found tackle the more classical question of computing the probability of a specific number of successes for a specific number of trials, not the number of trials required to make [at least] a fixed number of successes probable.)

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Are your binomial coefficients correct? $\binom{13}{1} = 13$, $\binom{13}{2} = 78$. Hence, $(5/6)^{13} + 13(1/6)(5/6)^{12} + 78(1/6)^2(5/6)^{11} = 0.628...$. – badroit Jul 8 '13 at 17:58
For the case of 16, $\binom{16}{1} = 16$, $\binom{16}{2} = 120$. Now, $(5/6)^{16} + 16(1/6)(5/6)^{15} + 120(1/6)^2(5/6)^{14} = 0.487...$. – badroit Jul 8 '13 at 18:01

Consider rolling a six to be a "failure", and model the chance of getting the third failure on the $n$th throw using a negative binomial distribution, shifted right by 3 (since the negative binomial distribution counts only successes). The CDF of the negative binomial distribution then represents the probability of having thrown at least three sixes by the time you've thrown $n+3$ dice. You can then solve for a CDF of 0.5, and round up.