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Carnival Game - Answer

The player would have an expected loss of 7.87 cents per game.

There are 216 (6³) possible dice throw permutations. If the player always bets on the same number, let's say 1, then there are four possible distributions of 1 and "not 1." Their probabilities are like this, where x stands for "not 1."

That leads to:

E = ((0)125 + (1 + 1)75 + (1 + 2)15 + (1 + 3)1) / 216 - 1 = -17/216 = -$0.0787.