The question at hand:
On average in every 7th chocoloate egg there is a figure to be drawn from a known list of special figures (e.g 15 distinct pieces) the other draws are on average fails.
With what probability in a given sample size (n) can you at least get 1/2/3/4/5 distinct specific figures out of these 15 figures (duplicates do not matter, though I could ask for it as well)?
The case with at least one distinct specific figure is: $$\small 1-\left(\frac{104}{105}\right)^n$$
For two I get:
$$\small -2^{3 n+1} \left(\frac{13}{105}\right)^n+\left(\frac{103}{105}\right)^n+1$$
For three I get: $$\small 105^{-n} \left(-102^n+3\dot\ 103^n-3\dot\ 104^n+105^n\right)$$
Which looks kind of nice.
For four I used again the modified code:
p1[a_, b_, n_] :=
n - Sum[Subscript[k, a - i + 1], {i, 1, b}] - ((a - b))
bubu[a_, n_] :=
Flatten[Append[
Reverse[Table[{Subscript[k, i - 1], 1, p1[a - 1, a - i, n]}, {i, 1,
a}]], {l, 0, p1[a - 1, a - 1, n] - Subscript[k, 0]}], 0]
(*Get at least 4 distinct special figures in your sample n, add up \
probability of all applicable possible series*)
calc22[n_] :=
Sum[(n!/((k0)!*(k1)!*(k2)!*(k3)!*
l!*(n - l - k0 - k1 - k2 -
k3)!))*(1/(105))^(k0)*(1/(105))^(k1)*(1/(105))^(k2)*(1/(105)\
)^(k3)*((11/105)^l)*((90/105)^(n - l - k0 - k1 - k2 - k3)), {k3, 1,
n - 3}, {k2, 1, n - k3 - 2}, {k1, 1, n - k2 - k3 - 1}, {k0, 1,
n - k1 - k2 - k3}, {l, 0, n - k0 - k1 - k2 - k3}]
For more I have written the general code:
p1[a_, b_, n_] :=
n - Sum[Subscript[k, a - i + 1], {i, 1, b}] - ((a - b))
bubu[a_, n_] :=
Flatten[Append[
Reverse[Table[{Subscript[k, i - 1], 1, p1[a - 1, a - i, n]}, {i, 1,
a}]], {l, 0, p1[a - 1, a - 1, n] - Subscript[k, 0]}], 0]
(*Get at least #(A+1) distinct special figures in your sample n, add \
up probability of all applicable possible series*)
(*Choose low sample size <30*)
calc24[n_, A_] :=
Sum[(n!/(Product[(Subscript[k, A - i])!, {i, 0, A}]*
l!*((n - l -
Sum[Subscript[k, A - i], {i, 0, A}])!)))*(1/(105))^(Sum[
Subscript[k, A - i], {i, 0, A}])*(((15 - (A + 1))/105)^
l)*((90/105)^(n - l - (Sum[Subscript[k, A - i], {i, 0, A}]))),
Evaluate[Sequence @@ bubu[A + 1, n]]]
Using the Multinomial function you can substitute the last block beneath the comment with these two:
buba[A_, n_] :=
Flatten[Append[
Table[(Subscript[k, A - i]), {i, 0, A}], {l,
n - l - Sum[Subscript[k, A - i], {i, 0, A}]}], 1]
calc24[n_, A_] :=
Sum[(Multinomial[
Evaluate[Sequence @@ buba[A, n]]])*(1/(105))^(Sum[Subscript[k,
A - i], {i, 0, A}])*(((15 - (A + 1))/105)^
l)*((90/105)^(n - l - (Sum[Subscript[k, A - i], {i, 0, A}]))),
Evaluate[Sequence @@ bubu[A + 1, n]]]
Somehow it takes too long to solve for larger values of A (number of distinct figures -1) or n (Sample size).
On this site [crossreference] the probability can be calculated but without stating the mechanics.
My number is slightly off after about 10 digits, I am still astonished that the code actually works to some precision.
Am I doing it overly complicated or is there a way to work with probability in complements for this kind of question?
Thank you.
Multinomial[]
is built-in? $\endgroup$ – J. M. will be back soon♦ Sep 2 '15 at 13:54