## xEnOnn 2 years ago 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?

1. KingGeorge

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

2. xEnOnn

If I want to consider, what should I do?

3. 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.

4. xEnOnn

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

5. KingGeorge

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

6. xEnOnn

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

7. KingGeorge

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

8. xEnOnn

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

9. 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.

10. xEnOnn

I think it is all thrown together.

11. xEnOnn

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.

12. 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.

13. xEnOnn

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

14. 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.

15. KingGeorge

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

16. xEnOnn

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. 17. KingGeorge use \( instead of \[ 18. xEnOnn ohh...haha...thanks for your help and explanation! :) 19. 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.

20. xEnOnn

Thanks! :)