Question

A die is tampered in such a way that the probability of observing an even number is twice as likely to observe an odd number. Find the expected value of the upper most face obtained after rolling the die.

Answer 1

11/3

Answer 2

The problem states that a die is tampered such that the probability of observing an even number is twice as likely as observing an odd number. Let $P(i)$ be the probability of observing the face with number $i$. The possible outcomes are {1, 2, 3, 4, 5, 6}. The odd numbers are {1, 3, 5} and the even numbers are {2, 4, 6}.

Let $P(\text{Odd})$ be the probability of observing an odd number, and $P(\text{Even})$ be the probability of observing an even number. $P(\text{Odd}) = P(1) + P(3) + P(5)$ $P(\text{Even}) = P(2) + P(4) + P(6)$

The problem states that $P(\text{Even}) = 2 \times P(\text{Odd})$. Also, the sum of probabilities of all possible outcomes is 1: $P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 1$ This means $P(\text{Odd}) + P(\text{Even}) = 1$.

Substitute the first equation into the second one: $P(\text{Odd}) + 2 \times P(\text{Odd}) = 1$ $3 \times P(\text{Odd}) = 1$ $P(\text{Odd}) = 1/3$.

Then, $P(\text{Even}) = 2 \times P(\text{Odd}) = 2 \times (1/3) = 2/3$.

Now we need to determine the probabilities of individual faces. In the absence of further information, we assume that all odd outcomes are equally likely, and all even outcomes are equally likely. Let $P(1) = P(3) = P(5) = p$. Let $P(2) = P(4) = P(6) = q$.

The sum of probabilities of odd outcomes is $P(1) + P(3) + P(5) = p + p + p = 3p$. We know $P(\text{Odd}) = 1/3$, so $3p = 1/3$, which gives $p = 1/9$.

The sum of probabilities of even outcomes is $P(2) + P(4) + P(6) = q + q + q = 3q$. We know $P(\text{Even}) = 2/3$, so $3q = 2/3$, which gives $q = 2/9$.

Thus, the probabilities for each face are: $P(1) = 1/9$ $P(2) = 2/9$ $P(3) = 1/9$ $P(4) = 2/9$ $P(5) = 1/9$ $P(6) = 2/9$

The expected value of the upper most face obtained after rolling the die is given by the formula $E[X] = \sum_{i=1}^{6} i \cdot P(i)$. $E[X] = 1 \cdot P(1) + 2 \cdot P(2) + 3 \cdot P(3) + 4 \cdot P(4) + 5 \cdot P(5) + 6 \cdot P(6)$ $E[X] = 1 \cdot (1/9) + 2 \cdot (2/9) + 3 \cdot (1/9) + 4 \cdot (2/9) + 5 \cdot (1/9) + 6 \cdot (2/9)$ $E[X] = (1 \times 1 + 2 \times 2 + 3 \times 1 + 4 \times 2 + 5 \times 1 + 6 \times 2) / 9$ $E[X] = (1 + 4 + 3 + 8 + 5 + 12) / 9$ $E[X] = (5 + 3 + 8 + 5 + 12) / 9$ $E[X] = (8 + 8 + 5 + 12) / 9$ $E[X] = (16 + 5 + 12) / 9$ $E[X] = (21 + 12) / 9$ $E[X] = 33 / 9$ $E[X] = 11 / 3$

The expected value of the upper most face is $11/3$.