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 such simple excercise you can imagine to throw the die a big number of times, obtaining perfect statistics.
In your case let's throw the die 600 times, obtaining 100 ones, 100 twos and exactly 100 polls for all the figures.
Our earnings would have been $$100*(-1)+100*2+100*(3)+100*(-4)+100*(5)+100*(-6)=-100$$ so, for 600 throws we are "expected" to lose 100, dividing these values we have the game payoff of $-1/6$.
If you do the division before the sum you have exactly the solution of your book: $${100*(-1)+100*2+100*(3)+100*(-4)+100*(5)+100*(-6) \over 600}={-100 \over 600}$$.
The game have a negative payoff because, although you have equal probabilities to win or lose one match, not all matches have the same earnings so you can't assume the game is fair, but you have to evaluate your earnings as I did before.
