Fair game calculation

tart22 said:

Here are the rules of a dice gamble game:
1. The player wins if the die is rolled 1 or 2
2. The casino wins if the die is rolled 3, 4, 5, or 6.
3. The casino pays two dollars for every dollar you bet if you win.
That is if you place a $10 bet, when you win, you would triple your money. You get your original $10 plus a $20 for your win.
Would you say it is a “fair” gamble although the odds of winning for the casino is twice as those of you? Why? Why not? Support your position mathematically if you can.

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HallsofIvy said:

There are 6 numbers on a die from "1" to "6", all equally likely. That is, the probability of either "1" or "2" is 1/6 so the probability of " 1 or 2" is 2(1/6)= 1/3. The probability of any one of "3", "4", "5", or "6" is also 1/6 so the probability of "3, 4, 6, of 6" is 4(1/6)= 2/3 (and 1/3+ 2/3= 1, of course). The "expected value" is (1/3)(2)-(2/3)(1). The game is "fair" if and only if the "expected value" is 0.

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