Probability when rolling 6 true dice

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SUMMARY

The forum discussion centers on calculating the probability of rolling exactly two aces (1s) with six true dice. The user DoubleMint initially applied the binomial theorem correctly, using the formula P = (6!/(2!4!)) (1/6)^2 (5/6)^4, which yields a probability of 0.2009. However, the expected probability is actually 0.040, indicating a misunderstanding in the application of the binomial theorem or the interpretation of the results. RGV confirms that DoubleMint's calculations were correct, suggesting that the discrepancy lies elsewhere.

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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.

RGV
 

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