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Norway

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- TL;DR Summary
- You throw 14 (fair, six-sided) dice. What is the probability of getting exactly 7 of a kind, when the number 1 is a "joker"?

We're playing a game with regular, fair, six-sided dice, and we're throwing 14 of them.

The "1" is a "joker", which means it can count as any other number. So, for example, if we roll three "1"s and four "2"s, we have seven "2"s.

My latest attempt was simply considering the probability of success with a specific number, for example "2".

For each dice, the probability of getting a "2" is 1/3 (both "1" and "2" is a success). Therefore, the probability of getting seven "2"s out of 14 is: 14C7*(1/3)

Then I thought I could just multiply this by 5, because what goes for "2" obviously goes for "3", "4", "5" and "6" as well. ...Right?

Alas, no.

I completely forgot that we can get seven of more than just one value. In fact we can get seven of four values simultaneously. Consider, for example, the following roll: {1, 1, 1, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6}. Here we have seven of both "2"s, "3"s, "4"s and "5"s.

At this point I actually gave up on all calculations, and just wrote a Python script to simulate it. It seems to converge around 35.7%, so assuming I coded correctly, I guess that is the answer.

But

If you read through all this, and want to have a go – thank you so much!

The "1" is a "joker", which means it can count as any other number. So, for example, if we roll three "1"s and four "2"s, we have seven "2"s.

**What is that probability that we get exactly seven of any face value?**At first glance, this seemed like simple, high school probability stuff, but I kept crashing into more and more walls trying to calculate it, so now I'm not so sure. There was always something I missed, which made the calculation much, much harder once I corrected my mistake. It might still be easy, though, I don't know. I've been banging my head to the point I've become blind.After a few tries with horrible logical flaws, I eventually started thinking about probability trees. For example, for the first dice, there is a 5/6 probability that it comes out {2,3,4,5,6}, in which case I can do a binomial distribution for the 13 remaining dice, where I have 13C6, and p=1/3. However, there is a 1/6 chance that the first dice is a "1", in which case the problem is just transposed by one notch, giving a 5/6 probability for {2,3,4,5,6} on the second dice and binomial 12C5 for the rest, and then 1/6 for getting a "1" on the third aswell, and ... ... ... ... This is getting deep. I had to rethink.My latest attempt was simply considering the probability of success with a specific number, for example "2".

For each dice, the probability of getting a "2" is 1/3 (both "1" and "2" is a success). Therefore, the probability of getting seven "2"s out of 14 is: 14C7*(1/3)

^{7}*(2/3)^{7}= 0,0918...Then I thought I could just multiply this by 5, because what goes for "2" obviously goes for "3", "4", "5" and "6" as well. ...Right?

Alas, no.

I completely forgot that we can get seven of more than just one value. In fact we can get seven of four values simultaneously. Consider, for example, the following roll: {1, 1, 1, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6}. Here we have seven of both "2"s, "3"s, "4"s and "5"s.

At this point I actually gave up on all calculations, and just wrote a Python script to simulate it. It seems to converge around 35.7%, so assuming I coded correctly, I guess that is the answer.

But

**why**is it 35.7%? How would you calculate this in a clean and simple manner by hand?If you read through all this, and want to have a go – thank you so much!

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