Probability with Loaded Dice.
By James Williams
The Question
Suppose that a die is biased (or loaded) so that 3 appears twice as often as each other number but that the other five outcomes are equally likely. What is the probability that an odd number appears when we roll this die?
My Question
The solution from my book is as follows:
We want to find the probability of the event $E = {1, 3, 5}$. By Exercise 2, we have $p(1) = p(2) = p(4) = p(5) = p(6) = 1/7; p(3) = 2/7$. It follows that $p(E) = p(1) + p(3) + p(5) = 1/7 + 2/7 + 1/7 = 4/7$.
I'm wondering why is it that $p(1) = p(2) .. = 1/7$. Where do we get this number?
$\endgroup$ 21 Answer
$\begingroup$On a normal die, the probability is 1/6 for each side, but since the 3 is twice as likely, then the probabilities are split up like this:
1: 1/7
2: 1/7
3: 2/7
4: 1/7
5: 1/7
6: 1/7
which equals 1 when summed.
$\endgroup$ 8
