So you're saying you wouldn't play a lottery with a 1/1,000,000 chance of winning $5,000,000 with a $1 ticket?

+1

Precisely. For $1 you'd be buying $5 worth of chances (provided the pot can't be split between several winners.) That's how I play lottery. In gambling EV is what counts.

.....

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<f_{i}<0.1667, for all i) approaches 1 as N approaches infinity. But if 0.1666<f_{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 think, you've misunderstood what I was asking to prove.

With **N** rolls there are **6**^{N} total variations. There are **T**_{W} variations ending in a payoff **P > 1** and **T**_{L} variations ending with the final payoff **P < 1**. **T**_{W}+T_{L}=6^{N}.

Find the limit of **T**_{W}/6^{N} as **N** tends to infinity.

I agree that "Expected Outcome" tends to zero as the number of rolls N tends to infinity. I just don't see it all that relevant to winning in this game. If this game was offered in casinos, I would have most certainly won millions in a matter of days starting with a bankroll of $100 or so. And I wouldn't let the entire $100 ride in just one sitting. Money management still counts. However, exponential accumulation of winnings should be the predominant scheme to win big.

Another sure thing is -- the Casino offering that game would go bust very quickly.

I proved that the probability of a randomly selected variation ending in a payoff P>1 approaches 0 as N approaches infinity. T_{W}/6^{N} is the probability of a random variation ending in a payoff P>1. Hence I have proven that T_{W}/6^{N} approaches 0 as N approaches infinity. Simultaneously, I also proved that this is true not just for payoffs P>1, but for payoffs P>E, as long as E>0. For example, if we let T_{p} be the number of variations that give at least a penny back, then T_{p}/6^{N} also approaches 0 as N approaches infinity.

I will try to explain my proof step by step:

Given a randomly created variation of N rolls, we can count the relative frequency of each result. Let f_{1} be the relative frequency of 0.7-rolls, which is defined as the number of 0.7-rolls divided by the total number of rolls N. It follows that the number of 0.7-rolls is f_{1}*N. Similarly, let f_{2} be the relative frequency of 0.8-rolls, f_{3} the relative frequency of 0.9-rolls... and so on until f_{6}, being the relative frequency of 1.5-rolls.

Now, consider the interval 0.1666<x<0.1667. Clearly 1/6 is an internal point in this interval. Because the expected values of f_{1} through f_{6} all equal 1/6, the law of large numbers guarantees the following as N approaches infinity: **the probability approaches 1 that f**_{1} through f_{6} will all be within this interval. A direct implication* of f_{1} through f_{6} all being within this interval is that the payoff is less than 1. Replacing the clause "f_{1} through f_{6} will all be within this interval" with its direct implication "the payoff is less than 1" in the bold-face text above, we get that **the probability approaches 1 that the payoff is less than 1****. If the probability of A approaches 1, then the probability of (not A) approaches 0. So the probability approaches 0 that the payoff is **not** less than 1. Consequently the probability that you get a payoff P>1 approaches 0.

Since all variations are equally likely, the probability that you get a payoff P>1 is equal to the number of such variations divided by the total number of variations, i.e. T_{W}/6^{N}. So replacing "the probability that you get a payoff P>1" with its equal "T_{W}/6^{N}", we get the sought result: T_{W}/6^{N} approaches 0.

*I proved this implication in my original proof (the inequality part), but I will try to explain that proof as well. So what I'm trying to prove is the implication "if f_{1} through f_{6} are all within the interval 0.1666<x<0.1667, then the payoff is less than 1".

We can write the payoff as P = 0.7^{A}*0.8^{B}*0.9^{C}*1.1^{D}*1.2^{E}*1.5^{F}, where A is the number of 0.7-rolls, B is the number of 0.8-rolls, etc. But as already stated, the number of 0.7-rolls is f_{1}*N, the number of 0.8-rolls is f_{2}*N, etc. So replacing A through F with f_{1}*N through f_{6}*N, and extracting the common exponent N, we get the payoff as P = (0.7^{f1}*0.8^{f2}*0.9^{f3}*1.1^{f4}*1.2^{f5}*1.5^{f6})^{N}. This function will increase as f_{1},f_{2}, and f_{3} decrease (because they are exponents for base numbers less than 1), and it will increase as f_{4},f_{5}, and f_{6} increase (because they are exponents for base numbers greater than 1). So to maximize the payoff, we should minimize f_{1} through f_{3}, and maximize f_{4} through f_{6}. Setting f_{1} through f_{3} as 0.1666 and f_{4} through f_{6} as 0.1667, we get an upper bound for the payoff: P < (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}, which can be verified by calculator to imply P < 0.9998^{N}, which in turn implies P < 1.

**If B is a direct implication of A, then P(B) is greater than or equal to P(A). Since P(A) approaches 1, and P(A) ^{<}_{=} P(B) ^{<}_{=} 1, it follows that P(B) approaches 1 as well. So we can replace the original clause with its direct implication.