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anonymous
 4 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?
anonymous
 4 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?

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KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.2You have not considered those other cases. You're assuming you get a 1, 2, and 5 in that order.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If I want to consider, what should I do?

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.2You 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Suppose if there were repeated values in side it: {2, 2, 5}. Then in this case, would I still multiply by 3!?

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.2long story short, no. I'll explain in a minute.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok sure. I will wait for your explanation. Thanks! :)

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.2Hold on, for your first question, are you throwing all three dice at the same time? Or one a a time?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Wouldn't it be the same? They are all iid.

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.2It'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
 4 years ago
Best ResponseYou've already chosen the best response.0I think it is all thrown together.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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
 4 years ago
Best ResponseYou've already chosen the best response.2There's a difference between thrown together and thrown at the same time. You get different information about which dice need to which numbers.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hmm... with your second equation, does it count the combinations of it?

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.2If 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
 4 years ago
Best ResponseYou've already chosen the best response.2Also, you were correct in thinking I was wrong with my other explanation. It doesn't actually matter. I was merely getting confused.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh 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
 4 years ago
Best ResponseYou've already chosen the best response.2use \( instead of \[

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ohh...haha...thanks for your help and explanation! :)

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.2I 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.
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