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xEnOnn

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

  2. xEnOnn
    • 4 years ago
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    If I want to consider, what should I do?

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

  5. KingGeorge
    • 4 years ago
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    long story short, no. I'll explain in a minute.

  6. xEnOnn
    • 4 years ago
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    ok sure. I will wait for your explanation. Thanks! :)

  7. KingGeorge
    • 4 years ago
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    Hold on, for your first question, are you throwing all three dice at the same time? Or one a a time?

  8. xEnOnn
    • 4 years ago
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    Wouldn't it be the same? They are all iid.

  9. KingGeorge
    • 4 years ago
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    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
    • 4 years ago
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    I think it is all thrown together.

  11. xEnOnn
    • 4 years ago
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    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
    • 4 years ago
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    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
    • 4 years ago
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    hmm... with your second equation, does it count the combinations of it?

  14. KingGeorge
    • 4 years ago
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    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
    • 4 years ago
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    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
    • 4 years ago
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    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
    • 4 years ago
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    use \( instead of \[

  18. xEnOnn
    • 4 years ago
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    ohh...haha...thanks for your help and explanation! :)

  19. KingGeorge
    • 4 years ago
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    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
    • 4 years ago
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    Thanks! :)

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