Probability when rolling 6 true dice

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The discussion centers on calculating the probability of rolling exactly two aces (1s) with six dice. The user initially applies the binomial theorem correctly but arrives at an incorrect probability of 0.2009 instead of the correct 0.040. A participant confirms that the user's method is accurate, indicating that the mistake lies elsewhere. The conversation highlights the importance of careful calculation in probability problems involving multiple outcomes.
doublemint
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Hello,

So the question is: You have 6 true dice and you want to find the probability of rolling exactly two aces (im assumming ace means 1).
I tried using the binomial theroem:
P = (6!/(2!4!)) (1/6)^2 (5/6)^4 = 0.2009
But the answer is 0.040..So where did I go wrong?

Thanks
DoubleMint
 
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Your answer is correct; you did nothing wrong.

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