<s>[INST] What is the probability that the sum of the die rolls is odd? 
Two fair coins are to be tossed once. For each head that results, one fair die is to be rolled. What is the probability that the sum of the die rolls is odd? (Note that if no die is rolled, the sum is $0$.)

I thought that since odd+even =odd, we then have to get (odd,even) and (even,odd) pairs from $\{0,1,2,3,4,5,6\}$. Thus, since there are a total of $7*7$ possibilities for pairs and $4*3*2$ pairs that add to an odd number I get $\dfrac{4*3*2}{7*7} = \dfrac{24}{49}$ but the right answer is $\dfrac{3}{8}$. What did I do wrong?
 [/INST] Notice that not your sample set cannot be one of $\{0, 1, 2, 3, 4, 5, 6\}$ because they do not all occur with equal probabilities. Specifically, the set $\{1, 2, 3, 4, 5, 6\}$ only occurs with probability $\frac{1}{2}$ for each die (you might roll $T$ and not have those outcomes at all). Similarly, $\{0\}$ occurs with probability $\frac{1}{2}.$
Adjusting your calculation to account for this, we find that there is a $(\frac{1}{2} + \frac{1}{2}(\frac{1}{2})) \cdot (\frac{1}{2}(\frac{1}{2})) = \frac{3}{16}$ chance to get (even, odd) and another $\frac{3}{16}$ chance to get (odd, even) by symmetry. Our final answer is $2 \cdot \frac{3}{16} = \boxed{\frac{3}{8}}.$
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