The player plays the following game in a casino. He bets c (some amount in dollars), says a number from 1 to 6, after what croupier rolls three dice. If the number player has said hasn't shown up on all dices (no dice with player's number at), then the player receives nothing, that is, loses 100 dollars (his bet); we believe that in this case his gain is −100. If the number has shown up on any dice, the player gets his bet back and receives a win - for each dice with specified number (which player has said), the casino pays the player the bet that he placed (100 dollars in this example). So, if a player bets one hundred dollars and his number falls twice (shows up on two out of three dice), the player will receive a win of 200 dollars, and if he doesn’t fall even once, then his win is −100 dollars. Find the expected value of the player win at a bet of 100 dollars.
I've already tried the following solution, however not sure whether it is correct: Expected value is equal to the sum of possible winnings multiplied by the their chance to happen, so the answer is: -100 * A + 100 * B + 200 * C + 300 * D, where $A=\frac{5^3}{6^3}, B=\frac{3\cdot5^2}{6^3}, C=\frac{3\cdot5}{6^3}, D=\frac{1}{6^3}$
Your approach in the second paragraph is correct, but one of your probabilities is wrong. You can tell from the fact that they don’t add up to $1$. Why do you have $5$ in $B$ but $6$ in $C$? If you fix that, you should get the right result.