Payoff of a dice game I came across this question today: 

"A fair die is tossed. If 2,3 or 5 occurs, the player wins that number of rupees, but if 1, 4 or 6 occurs, the player loses that number if rupees. Then find the possible payoffs for the player". 

My textbook has then proceeded to solve it like this: 

What is the logic behind directly adding all the values? Secondly, how is it that the game is unfavorable to the player? There is an equal chance that the player will get 2, 3 or 5 and 1,4 or 6! 
Please help. (Also note that this doubt is not specific to this problem alone, since this concept is crucial to understanding this part of the chapter "Probability".) Much thanks in advance :) Regards.
Edit: Thanks ever so much for these answers :) I read up on "Expected values" but encountered another important doubt. I have posted it here and hope that you will clear my doubt :) 

For your second question, let's exaggerate the values: suppose that with probability 1/2, you lose 10000 and with probability 1/2 you win 1. Then even though the probabilities are equal, you'd admit that the game is unfavorable.
The expected value in the above example is $1/2 \times 1 - 1/2 \times 100000$.
The conclusion is however not appropriate. Consider another game/bet where you win $10^{12}$ with probability $10^{-6}$ and lose 10^5 with the remaining probability. The expected value is positive and large (close to $10^6-10^5$), but if you play this bet just once, it would be foolish to take it on (depending on how valuable/large $10^5$ is).
In the absence of complete information about the probability distribution, the expected value alone does not tell you whether the outcome is likely to be close to that value. For this, you will need to look into concepts like variance and standard deviation.
