I agree Misfire is incorrect, and I don't see an answer in sqroot's calculations, aside from how to receive one box twice.

The calculation needs to be how to get 8 different pairs of identical boxes (b1,b1) (b2,b2) (b3,b3) (b4,b4) (b5,b5) (b6,b6) (b7,b7) (b8,b8) in 16 drawings from 8 items.

I guess I'll spend some time on it in the next few days.

Again, this is not difficult, and the answer is in my post.

To do *any* calculations with probability, you need to assume a distribution. The stuff you (miss)learned in school is for Bernoulli trials with a uniform distribution.

A uniform distribution is quite literally that you have the same chance to receive each item. To make any statements about probability, we need to assume the probability to get each item and a corresponding distribution, and that is statistics, not probability!

Disregarding confidence, Noobles' post suggests exactly a uniform distribution. I'm gonna spare you with calculating the confidence (which is only dependent on the 16 trials, here), but by the law of big numbers, the expected values of a uniform distribution actually converge towards exactly all boxes being dropped the same amount of times, so Noobles' result is optimal for the given hypothesis!

This is very intuitive, and even people that have never heard of probability will find this intuitive.

What you're doing is assuming a uniform distribution and then trying to disprove that it is a uniform distribution.

Every attempt at this kind of "proof" will be wrong, because it would prove that there are no uniform distributions, which is trivially wrong itself.

For your question, ask yourself this: Which result to your problem would confirm that it is a uniform distribution, and which result would disprove it? (Hint: There's no correct answer to this question, because you're trying to solve the wrong problem!)

I suggest you go pick up a book and relearn the stuff about probability you misslearned in school, instead of investing time into a pointless endeavor.

Anyways, since I enjoy pointless endeavors, the answer to your question is ~0.029%.

Let B(n, k, p) := (n over k) * 1/p^k * (1-1/p)^(n-k).

If "Getting exactly 2 boxes of a kind i when drawing n times with probability p to get the box" is our event E_i(n, p) for every box i, then we're looking for the probability P(E_1(16, 1/8) ∩ E_2(16, 1/8) ∩ ... ∩ E_8(16, 1/8)).

Applying the theorem of multiplication, we get

P(E_1(16, 1/8) ∩ E_2(16, 1/8) ∩ ... ∩ E_8(16, 1/8))

= P(E_1(16, 1/8)) * P(E_2(16, 1/8) | E_1(16, 1/8)) * P(E_3(16, 1/8) | E_2(16, 1/8) ∩ E_1(16, 1/8)) * ... * P(E_8(16, 1/8) | E_7(16, 1/8) ∩ ... ∩ E_1(16, 1/8))

= P(E_1(16, 1/8)) * P(E_2(14, 1/7)) * ... * P(E_8(2, 1/1))

= B(16, 2, 1/8) * B(14, 2, 1/7) * ... * B(2, 2, 1/1)

~= 0.029%.

A quick experiment confirms this:

- def f(m):
- A = [1, 2, 3, 4, 5, 6, 7, 8]
- n = 0
- for i in range(0, m):
- amounts = {a : 0 for a in A}
- for j in range(0, 16):
- amounts[random.choice(A)] += 1
- if all(v == 2 for _, v in amounts.items()):
- n += 1
- print(n/m)
- >>> f(10000000)
- 0.0002914

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