Answer: Choice B
Game A is <u>not</u> fair, and game B is fair.
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We'll start with game A.
The sides of the dice are {1,2,3,4,5,6}
There are 6*6 = 36 ways to roll two dice.
If we roll two odd numbers, say 3 and 5, the product is odd because 3*5 = 15.
If any of the numbers are even, then the product is even. Eg: 4*5 = 20 or 2*6 = 12.
To be an even number, 2 must be a factor.
So you can probably see that it's more likely to get an even number product considering that in order to get an odd product, both values must be odd (which narrows the selection pool). We can say that game A is not fair. It favors player 1 more. If you're sure about this, then I recommend looking at a multiplication table but only focus on the first 6 rows and 6 columns. Ignore the row and column that has 0s in it.
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Now onto game B
The rules are nearly the same, except now instead of multiplying the dice values, we add them up.
We still have 6*6 = 36 outcomes.
Here are some handy rules we'll take into account
- even+even = even
- even+odd = odd
- odd+odd = even
If the first dice rolls to '1', then there are 3 ways to get an even number (rolling either 1,3,or 5 on the second dice) and there are 3 ways to get an odd number (rolling either 2,4,or 6 on the second dice). We can see that there are the same number of ways to get an even sum or an odd sum. Apply this in general to any result that the first dice shows, and this basically leads to the conclusion that half of the 36 sums are even, and the other half are odd. We have a fair game now. Refer to a sum of dice chart if you need more convincing.
