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The normal probability of a number in a regular die (6 faces) is $\dfrac{1}{6}$. Let in an addicted [that is, "loaded"] die, the probability of a even number (2, 4 and 6) be twice the normal probability;

I've got such outcome: $regular\space probability \space on \space evens \space is \space \dfrac{3}{6} $, doubling it, it would become $\dfrac{6}{6}$, in other words, a certain event, it sounds strange to me, is that right ?

Thanks in advance;

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    If the probability of rolling an even number on the loaded die really is twice what it would be on a fair die, then yes: with the loaded die you are certain to roll an even number.2012-04-12
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    I'm not sure if others use "right" in the context you imply when you say "right event". You should say "certain event".2012-04-12
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    yes, I meant "certain". Thanks2012-04-12
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    The singular of "dice" is "die".2012-04-17
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    I have done some light editing to make the question read a little better.2012-04-17

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You might check the wording of the question. If the probability of each even number is twice that of a normal die, you are correct. If the probability of each even number is twice that of each odd number, the result is different-then the evens come up $2/9$ each for a total probability of $6/9=2/3$