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Theorem:

the probability of winning (or even getting at least a fixed non-zero return) approaches 0 as the number of rolls approaches infinity.Proof: There are six equally likely outcomes of one roll, let's denote them a

_{1}through a_{6}(a_{1}being 0.7 and ascending to a_{6}being 1.5). Let f_{i}denote the relative frequency of the outcome a_{i}during a sequence of rolls. For example, if we roll 12 times and get 1*a_{1}, 3*a_{2}, 0*a_{3}, 1*a_{4}, 2*a_{5}, and 5*a_{6}, then f_{1}= 1/12, f_{2}= 3/12, f_{3}= 0/12, f_{4}= 1/12, f_{5}= 2/12, and f_{6}= 5/12. The exact number of outcomes a_{i}is the relative frequency f_{i}times N. The final outcome of our game will be G = 0.7^{f}^{1}^{N}*0.8^{f2}^{N}*0.9^{f3N}*1.1^{f4}^{N}*1.2^{f5}^{N}*1.5^{f6N}= (0.7^{f}^{1}*0.8^{f2}*0.9^{f3}*1.1^{f4}*1.2^{f5}*1.5^{f6})^{N}.The expected value of each f

_{i}is 1/6. By the law of large numbers,P(0.1666<fapproaches 1 as N approaches infinity. But if 0.1666<f_{i}<0.1667, for all i)_{i}<0.1667, then G = (0.7^{f}^{1}*0.8^{f2}*0.9^{f3}*1.1^{f4}*1.2^{f5}*1.5^{f6})^{N}< (0.7^{0.1666}*0.8^{0.1666}*0.9^{0.1666}*1.1^{0.1667}*1.2^{0.1667}*1.5^{0.1667})^{N}< 0.9998^{N}. So the probability approaches 1 that your outcome is smaller than a value which approaches 0 as N approaches infinity. Conversely, for any given E>0, the probability approaches 0 that your outcome is at least E.

I like the reasoning, but the proof by example uses a slanted interval. The true Median of 1/6 is not in the middle of the interval 0.1666 to 0.16667. What happens if you use a truly median interval, e.g., from 9/60 to 11/60? It seems to be leading to the opposite conclusion.