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this is my 3rd question today ^^ Thanks a lot for any help :-)

I need to calculate how probable it is if i roll a die 12 rounds and i get each side twice.

Idea:

Because every side needs to be present twice and we roll the die 12 times:

$$ \left ( \frac{2}{6*2} \right )^{6} = \left ( \frac{2}{12} \right )^{6} = \left ( \frac{1}{6} \right )^{6} $$

Is this correct?

hukachaka
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2 Answers2

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We want $12$ letter words over the alphabet $\{1,2,3,\dotsc,6\}$ with exactly two of each letter. There are $$ \binom{12}{2,2,2,2,2,2} $$ such words out of $$ 6^{12} $$ words without any restrictions. Hence the probability is $$ \frac{\binom{12}{2,2,2,2,2,2}}{6^{12}}. $$

Sri-Amirthan Theivendran
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  • It might be worth pointing out that this is more that $160$ times the $\frac1{6^6}$ given in the question. – robjohn Oct 22 '17 at 16:05
  • since both of you gave me the same answer and i see something similar to that in my script i gave a +1 to you. Because you were first i gave you the right answer. I will try to understand that first and maybe later write back again here :-) – hukachaka Oct 22 '17 at 16:06
  • i just found (n choose k) in my script. I think this is what my teacher wanted to tell us. But what does those commas mean in (12 choose 2,2,2,2,2,2) ? – hukachaka Oct 22 '17 at 16:14
  • These are called "multinomial coefficients" and is used to e.g. give the number of words that can be formed with the letters M, i, s, s, i, s, s, i, p, p, i. – Qi Zhu Oct 22 '17 at 16:21
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    okay, this is new for me. so for (12 choose 2,2,2,2,2,2) this should be 12!/2!2!2!2!2!*2! is this right? i am sorry but i haven't seen something like this before – hukachaka Oct 22 '17 at 16:28
  • Yes, correct :) – Qi Zhu Oct 22 '17 at 16:56
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I'm not quite sure what you are trying to do with your idea but unfortunately, the answer is wrong.

There are probably better approaches but this is a possible one:
There are exactly $\binom{12}{2,2,2,2,2,2}$ possible results to satisfy the given conditions. (Example: One possible result is 112233445566. That is, we roll 1, then 1, then 2, then 2, ...)

That's the key idea! Those are all possibilities to hit each side exactly twice. We have to compute the probability of each of those results and add them.

Luckily, this is easy. The result of each single roll is already fixed. Hence, the probability for any of the aforementioned results is $\left(\frac16 \right)^{12}$, so our answer is $ \left(\frac16 \right)^{12} \cdot \binom{12}{2,2,2,2,2,2}$.

EDIT: I guess I was sniped, heh.

Qi Zhu
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  • Snipe means to shoot at individuals from a concealed place, to shoot snipe (long-billed wading birds), or to make malicious, underhanded remarks or attacks (American Heritage Dictionary). I do not see any evidence that any of these events occurred. – N. F. Taussig Oct 22 '17 at 16:11
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    I believe you do not see any evidence of any of these events not occurring, do you? ;-) – Qi Zhu Oct 22 '17 at 16:16
  • You could make this answer even better by explaining what ({12\n2,2,2,2,2,2}) means (i.e. 12!/(2!^6)) and why, giving the final answer. Remember $$\text{$$ line syntax $$}$$. – wizzwizz4 Oct 22 '17 at 16:40
  • Thanks for the constructive feedback! I want to explain why I've done what I've chosen to do: The answer I've given, explains better why we ended up with that answer which is why I decided not to still do the trivial computation. About the $$ syntax, I'm familiar with that but I usually don't like using it for one short term. – Qi Zhu Oct 22 '17 at 16:59
  • @N.F.Taussig This is closer to the use of snipe above, if not exact. – Yakk Oct 22 '17 at 22:01