Example 4.4.5: Suppose that there is a 6-sided die that is weighted in such a way that each time the die is rolled, the probabilities of rolling any of the numbers from 1 to 5 are all equal, but the probability of rolling a 6 is twice the probability of roll- ing a 1. When you roll the die once, the 6 outcomes are not equally likely. What are the probabilities of the 6 outcomes?

On the basis of the above question a question has been asked to evaluate the mean. enter image description here

My problem is that, when they evaluate the mean, they multiplied with the probability by 1, 2 , 3 , 4 , 5 and 6 why? In die , all are they equal probable.

  • $\begingroup$ mean is $\sum_i Y_i P(Y_i)$ $\endgroup$ – Memming Jul 10 '14 at 19:37

The mean of a discrete random variable is defined as

$$\mathbb{E}(X)=\sum\limits_{x\in X} xp(x)$$

In this case $X=\{1,2,3,4,5,6\}$, so that's where you are getting the multiplication. You can think of the probabilities as weighting the importance of each of these $6$ numbers.

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  • $\begingroup$ The question is about 6 sided die. So the probability of 6 can be 2/7 and this is not saying that there are 6 possibilities that we can have 6 from the die or 5 possibilities that we can have 5. We only have one 6, one 5 , one 4 etc on the die. $\endgroup$ – user52950 Jul 10 '14 at 19:43
  • $\begingroup$ @ComplexGuy What is your definition of 'mean'? We are not saying that $\dfrac{2}{7}$ happens six times - it doesn't even happen one time. We are saying that $6$ happens two-sevenths of the time. $\endgroup$ – Peter Woolfitt Jul 10 '14 at 19:51
  • $\begingroup$ 6 happens two-sevenths of the time. But why do you multiply the 6 with the 2/7? Let an apple happens two-sevenths of the time then will we multiply the apple with the probability? Sorry i have misunderstood in it this is why i am clearing my concept $\endgroup$ – user52950 Jul 10 '14 at 19:55
  • $\begingroup$ @ComplexGuy Well, if an apple happened two-sevenths of the time, and nothing happened the other five-sevenths of the time, your the mean over the time would be two-sevenths of the apple. If an orange happened the other five-sevenths, then your (fruit) mean would be two-sevenths of an apple and five-sevenths of an orange. 'The mean' intuitively means what you are expecting to roll. Perhaps you are getting it confused with the arithmetic mean. Here we are talking about a weighted arithmetic mean. $\endgroup$ – Peter Woolfitt Jul 10 '14 at 20:15
  • $\begingroup$ ^where the sum of the weights is equal to $1$. Also take a look at the expected value page - this is what you have been asked to calculate. $\endgroup$ – Peter Woolfitt Jul 10 '14 at 20:17

As Peter stated, to calculate the mean, you multiply each outcome (the number that you roll) by the probability of getting that roll, and add those products up.

On a regular or fair die, yes, the probabilities of each outcome are the same. But this isn't a fair die; $6$ is twice as probable as each of the other numbers.

For a fair die, $$E(X) = \frac{1}{6}(1) + \frac{1}{6}(2) + \frac{1}{6}(3) + \frac{1}{6}(4) + \frac{1}{6}(5) + \frac{1}{6}(6) = 3.5.$$

For the unfair die in your problem, $E(X)$ is as calculated.

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  • $\begingroup$ $E(X) = \frac{1}{6}(6) + \frac{1}{6}(3) + \frac{1}{4}(4) + \frac{1}{6}(2) + \frac{1}{6}(5) + \frac{1}{6}(1) $Can we write like this? What does the X represent? $\endgroup$ – user52950 Jul 10 '14 at 19:48
  • $\begingroup$ $X$ is your random variable. As for what you wrote in the comment, you have a typo with the $1/4$. The sum of the probabilities $p(x)$ must be $1$. $\endgroup$ – John Jul 10 '14 at 20:36

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