What are the odds of rolling a 3 number straight throwing 6d6 If you throw six fair dice, what are the odds that at least three dice make a straight (i.e. 123, 234, 345, or 456)
I am certain that I am making a mistake in calculating it?
 A: In Mathematica, here's a brute force solution.  There are 27720 rolls of the 6 dice that will give that straight.
Length[Select[Tuples[Range[6], {6}], Max[Table[Length[Intersection[#,
{{1, 2, 3}, {2, 3, 4}, {3, 4, 5}, {4, 5, 6}}[[n]]]], {n, 1, 4}]] == 3 &]]/6^6
$385/648 = 0.5941358024691358024691\dots$
A: Here's another way. Let ${\bf k}$ be the number of distinct die numbers, let $S$ ("success") be the event that at least one of the four required straights occurred.
Then $$P(S) = \sum_{k=1}^6 P(S | {\bf k}=k) \; P({\bf k}=k)$$
It's clear that, for a given fixed $k$, all configurations are equiprobable.
It's trivial that $P(S | {\bf k}=0)=P(S | {\bf k}=1)=P(S | {\bf k}=2)=0$. The other are not difficult:
$P(S | {\bf k}=3)$ : There are 4 possible success configurations, out of ${6 \choose 3}$, hence $P(S | {\bf k}=3) = 1/5$
$P(S | {\bf k}=4)$ : This is the most difficult one, let's count the unsuccessful configurations: These are : [o x x o x x] [x o x o x x] [x o x x o x] [x x o o x x] plus the mirroring of the first two: 6 unsuccessful configurations out of  ${6 \choose 4}$
hence $P(S | {\bf k}=4) = 1- 2/5 = 3/5$   [*]
Further, $P(S | {\bf k}=5)= P(S | {\bf k}=6) = 1$
Now, to compute $P({\bf k}=k)$, we must count the number of ways of filling $k$ positions with 6 throws: for example, $$P({\bf k}=4)=\frac{{6 \choose 4} \;  4! \; S_2(6,4)}{6^6}$$
where $S_2(n,m)$ are the Stirling numbers of the second kind. So finally
$$P(S) = \sum_{k=3}^6 a_k \; \frac{{6 \choose k} \;  k! \; S_2(6,k)}{6^6} \approx 0.59413
$$ 
with $a_3=1/5$,  $a_4=3/5$,  $a_5=1$, $a_6=1$. 
[*] Added: to generalize this  for arbitrary number of dice or runlengths, notice that this counting of unsuccessful configurations is equivalent to count the ways of expressing the number $k=4$ as a sum of $n-k+1=3$ non-negative terms less that $m=3$ (runlength), with order (in the example, 0+2+2 , 1+1+2, 1+2+1, 2+0+2, 2+2+0,2+1+1). Hence $P(S | {\bf k}) = 1 - b_{k,n-k+1,m}/{n \choose k}$ where $b_{r,s,t}$ counts the $s$-terms weak-compositions of the integer $r$, with the restriction that each term is less than $t$. An expression is given in Enumerative combinatorics (Stanley)  page 120, problem 28.
Added 2: Because the exact formula for arbitrary dice is quite formidable, here's some quick asymptotics. Let $n$ be the number of throws, $c$ the number of dice faces, $m$ the straight length. For large $c,n$, the probability that a particular face number does not appear is $q=(1-1/c)^n \approx \exp(- n/c)$. Asymptotically, we can assume that these events are independent, and disregard border effects, and regard each outcome as a sequence of runlengths $(a_1 b_1 a_2 b_2 ... a_r b_r)$ where $a_i$ is the length of consecutive (run) die faces that appeared, $b_i$ the faces that didn't (and $a_1 + b_1 + ... = c$). Each run can then be approximated by independent geometric variables (starting at 1) with stopping probabilities $q$ and $1-q$. Their expectations are respectively $1/q$ and $1/(1-q)$, so that equating the expected sum 
we get $r=c \; q \;(1-q)$. The event of no-success, correspond to $a_i<m$, and its probability is given, in this approximation, by $(1-(1-q)^{m-1})^r$. Finally:
$$P(S) \approx 1- \left[ 1- (1-q)^{m-1} \right]^r$$
with 
$$ r = c \; q \; (1-q) \hspace{20px} q = (1-1/c)^n \approx e^{-n/c}$$
In our original question... we have $c=n=6$,$m=3$, the numbers are too small to apply this asymptotics, but anyway, I get: $P(S) \approx 0.50922$ ($0.54184$ if using the "exact" $q$)
