Suppose I have 3 dices. I throw them. I want to get a {1,2,5}.
So I do this to get the probabilities:
\[\frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{6} \]
In this case, have I considered other combinations like {2, 1, 5}, {5,2,1}, etc?

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- KingGeorge

You have not considered those other cases. You're assuming you get a 1, 2, and 5 in that order.

- anonymous

If I want to consider, what should I do?

- KingGeorge

You just have to do one more thing. You can just consider all possible orderings. There happen to be \(3!=6\) orderings, and they all have equal probability. Thus, just multiply your original answer by 6.

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## More answers

- anonymous

Suppose if there were repeated values in side it: {2, 2, 5}. Then in this case, would I still multiply by 3!?

- KingGeorge

long story short, no. I'll explain in a minute.

- anonymous

ok sure. I will wait for your explanation. Thanks! :)

- KingGeorge

Hold on, for your first question, are you throwing all three dice at the same time? Or one a a time?

- anonymous

Wouldn't it be the same? They are all iid.

- KingGeorge

It's actually a little bit different. If we throw them one at a time, we do the process I explained above. Otherwise, you have to do something different.

- anonymous

I think it is all thrown together.

- anonymous

I don't think this is it. Even if they are all thrown together, the dices are i.i.d. random variables on their own.

- KingGeorge

There's a difference between thrown together and thrown at the same time. You get different information about which dice need to which numbers.

- anonymous

hmm... with your second equation, does it count the combinations of it?

- KingGeorge

If we were looking at \(\{2, 2, 5\}\) instead, we have to look at a couple different cases. Drawing a tree is a good idea here.
Like above, we have a \(2/6\) chance of getting a 2 or a 5. Then, within this probability, there's a \(2/3\) chance we get a 2, and \(1/3\) chance we get a 5. If we get a 5, we need to find the probability of getting 2 2's, and if we get a 2, we find the probability of getting a 2 and a 5.

- KingGeorge

Also, you were correct in thinking I was wrong with my other explanation. It doesn't actually matter. I was merely getting confused.

- anonymous

oh I see. Thanks!! :)
And just one thing, how do you type the math equations inline? I used the \[ tags but it goes to the next line.

- KingGeorge

use \( instead of \[

- anonymous

ohh...haha...thanks for your help and explanation! :)

- KingGeorge

I should note that doing it the other way I suggested is also a valid way of doing it, but you get \[{3 \over 6} \cdot{2 \over 6}\cdot{1 \over 6}={3! \over 6^3}\]Not what I originally got.

- anonymous

Thanks! :)

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