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I am trying to calculate the probability of rolling a 4 or more on six 6 sided dice and have managed to get the probability of getting 1 successful 4+ via the equation:

$\frac{6^6-3^6}{6^6}=0.9844$

the problem I'm having is calculating the probability for 2 or 3 successful 4+ results.

can any one help me out?

B.Rose
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    The problem is worded oddly. If I understand you correctly, you might as well take six fair coins and ask about throwing $3$ Heads, yes? After all, for any single die the probability of throwing $4$ or greater is $\frac 12$. Also, you should clarify whether throwing $1$ success means throwing "exactly one" or "at least one" (your formula suggests you mean the latter). In either case, it is a Bernoulli Distribution and standard formulas apply. – lulu Aug 08 '17 at 23:29

2 Answers2

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The probability of exactly two 4+'s is $p_2 = \binom{6}{2}\cdot (1/2)^6$, and the probability of exactly three 4+'s is $p_3 = \binom{6}{3}\cdot (1/2)^6$. If you want there to be at least two 4+'s, for instance, the probability is $p_2+p_3+p_4+p_5+p_6$.

primes.against.humanity
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Let $S(i)$ be the rolls where die $i$ has a $4+$. Then $$ \begin{align} N(j) &=\sum_{|A|=j}\left|\,\bigcap_{i\in A} S(i)\,\right|\\ &=\binom{6}{j}2^{-j} \end{align} $$ Then, according to the Generalized Inclusion-Exclusion Principle, we get the number of rolls in exactly $k$ of the $S(i)$ to be $$ \begin{align} \sum_{j=0}^6(-1)^{j-k}\binom{j}{k}N(j) &=\sum_{j=k}^6(-1)^{j-k}\binom{j}{k}\binom{6}{j}2^{-j}\\ &=\sum_{j=k}^6(-1)^{j-k}\binom{6}{k}2^{-k}\binom{6-k}{j-k}2^{k-j}\\ &=\frac1{64}\binom{6}{k} \end{align} $$ which upon reflection is obvious without inclusion-exclusion. Inclusion-Exclusion would be more useful in the case of rolling $5+$ or $2+$.

robjohn
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