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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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ACCEPT SOME ANSWERS! – Gerry Myerson Apr 12 '12 at 0:45
I've just figured it out now, I'm starting doing it. – aajjbb Apr 12 '12 at 0:48
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. – Brian M. Scott Apr 12 '12 at 0:53
I'm not sure if others use "right" in the context you imply when you say "right event". You should say "certain event". – David Mitra Apr 12 '12 at 0:57
yes, I meant "certain". Thanks – aajjbb Apr 12 '12 at 1:52
up vote 2 down vote accepted

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 $6/9=2/3$

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