Probabilty of guessing correctly with an uneven dice Suppose there is an uneven dice. The probability of getting $1$ to $6$ is not exactly $\frac16$, instead, it is some other values of $P(1)$, $P(2)$ to $P(6)$. Apparently $P(1) + P(2) + \cdots + P(6) = 1$.
A person knows this dice is uneven and knows the exact value of $P(1)$ to $P(6)$. He throws dice for $n$ times and tries to use his psychic power to guess the outcome. He either guesses it correctly or incorrectly. So, his guesses will be a series of right and wrong. How to calculate how much better his psychic performance is compared with merely guessing?
Thanks for all the answers :) Please let me clarify my question. As we know, people with psychic power or who believe they have psychic power are typically not good at math or game theory. So, by "merely guessing", I mean this person's guesses will exactly match the probability of p1 to p6. That is the case that this person does not have any psychic power, but merely guessing according to his knowledge of the dice. However, if he or she does have psychic power, the hit rate will be somewhat higher. So, how to calculate how much psychic power this person has?
 A: Just guessing would imply that he's equally likely to say any one of the six faces, independent of what they actually are.  So $pr(\text{Guess Correctly})=pr(\text{guess 1, roll 1}) + pr(2,2) +...+pr(6,6)$.  Since these are independent events, this probability is 
$$p(\text{guess 1})\cdot p(\text{roll 1})+\ldots+p(\text{guess 6})\cdot p(\text{roll 6})$$
$$=\frac{1}{6}p_1+\frac{1}{6}p_2+\frac{1}{6}p_3+\frac{1}{6}p_4+\frac{1}{6}p_5+\frac{1}{6}p_6 $$
$$=\frac{1}{6}(p_1+p_2+p_3+p_4+p_5+p_6)$$
$$=\frac{1}{6}\cdot1=\frac{1}{6}$$
So regardless of how the die is weighted, we still expect him to guess correctly 1 out of 6 times.  From there, we could do a standard t-test on proportions.
A: Assume the question is to guess the exact outcome of the die ($1$ to $6$) every time.
For the case of pure guessing (implying that we assume the die to be fair), the number of times $1$ to $6$ appeared in sufficiently large number of guesses should be equal (i.e. the frequency of every number $1$ - $6$ appeared in the guess will be $\frac{1}{6}$ in the long run). So, the expected value for the probability of correct guess is:
$\frac{1}{6}\times p_1 + \frac{1}{6}\times p_2 + \frac{1}{6}\times p_3 + \frac{1}{6}\times p_4 + \frac{1}{6}\times p_5 + \frac{1}{6}\times p_6 = \frac{1}{6} $ 
(i.e. $1$ out of $6$ guesses is correct in the long run)
If the person knows the exact probability of $p_i (i = 1, 2, 3, 4, 5, 6) $, then in his/her guess in the long run, $i$ will appear in the guess with a frequency $p_i$ out of the total number of guesses. Hence the expected value for the probability of correct guess is:
$\ p_1\times p_1 + p_2 \times p_2 + p_3 \times p_3 + p_4 \times p_4 + p_5 \times p_5 + p_6 \times p_6$ 
This will always be greater than or equal to $\frac{1}{6}$ (the equal situation happens only when $p_1 = p_2 = p_3 = p_4 = p_5 = p_6 = \frac{1}{6}$).
Added: As a related topic to the question, in fact if a person knows the die is uneven or unfair, and he/she knows the exact probability of $p_i (i = 1, 2, 3, 4, 5, 6) $, the best way to maximize the expected value for the probability of correct guess is to always guess the number $i$ with the maximum $p_i$. The expected value will be $p_i$ in this case. Proof:
Without loss of generality, assume $p_1 > p_i$ for $i = 2, 3, 4, 5, 6$. Then,
$\ p_1\times p_1 + p_2 \times p_2 + p_3 \times p_3 + p_4 \times p_4 + p_5 \times p_5 + p_6 \times p_6$ 
$< p_1\times p_1 + p_1 \times p_2 + p_1 \times p_3 + p_1 \times p_4 + p_1 \times p_5 + p_1 \times p_6$ 
$= p_1$
A: Probability of his guess to be correct:
For first time, let him guess 3, then probability that he will be correct is $p_3$. If 4, then $p_4$ and if $i$ then $p_i$. Suppose the dice is thrown for n times.For an example take $n=3$ and let him guess $3,1,2$ then probability of him being correct is $p_3p_1p_2$. So for any order the probability of him being correct can be expressed as: $$\prod_{i\in\{1,2,..6\}}^np_i$$
Knowing the values of all $p_i$ his psychic powers will select a particular value all the time, i.e. $(\max p_i)$. So we are comparing these two:
$$(\max p_i)^n\quad\text{and}\quad \left\langle\prod_{i\in\{1,2,..6\}}^np_i\right\rangle$$
and I took the average value since the outcome is undetermined to us and it will be random ("merely guessing"). Therefore the ratio of these would be:
$$\frac{(\max p_i)^n}{\displaystyle \frac1{6^n}\sum\left(\prod_{i\in\{1,2,..6\}}^np_i\right)}$$
I am working on simplification of the last term but this thing can be said that:
$$\frac1{6^n}\sum\left(\prod_{i\in\{1,2,..6\}}^np_i\right)\le\frac1{6^n}\sum(\max p_i)^n=(\max p_i)^n$$
So the "ratio of guessing powers" is always greater than one (that when $p_i=1/6\forall i\in\{1,2,..6\}$).
A: Suppose that the person guesses side $i$ with probability $q_i$, and assume for generality that there are $n$ sides. Her success probability is
$$ f(q) = \sum_{i=1}^n p_i q_i. $$
The question arises – what is the person's best strategy? Suppose that $p_i \geq p_j$. Given a vector $q$, construct a new vector $q'$ by defining $q'_i = q_i + q_j$, $q'_j = 0$, and $q'_k = q_k$ for $k \neq i,j$. Then
$$ f(q') - f(q) = (q_i+q_j)p_i - q_ip_i - q_jp_j = q_j(p_i - p_j) \geq 0. $$
Therefore the vector $q'$ is always at least as good as $q$. Now let $p_i = \max_{j=1}^n p_j$. By repeating the previous argument $n-1$ times, we find that an optimal strategy is to always guess $i$. This strategy succeeds with probability $p_i$, compared to $1/n$ which is a random guess.

Here is a more sophisticated way to solve this problem. The success probability of the optimal strategy is the solution of the linear program
$$
\begin{align*}
&\max \sum_i p_i q_i \\
s.t. & \sum_i q_i = 1 \\
& q_1,\ldots,q_n \geq 0
\end{align*}
$$
If we eliminate one of the variables using the equation, we obtain a linear program with $n-1$ variables and $n$ constraints. We know that if there is any optimal solution at all, there must be one which is a vertex, i.e., it satisfies $n-1$ constraints tightly. Thus at most one of the $q_i$ can be non-zero, and we quickly conclude that an optimal strategy bets on a single side.
