This example illustrates the difference between expected

**value** and expected

**outcome**. The expected value is +X/30, where X is the amount you put into the machine. The expected outcome of method 2 is something quite different though. Over the course of six games, you can expect one of each result, netting you 0.7*0.8*0.9*1.1*1.2*1.5*X - X = -0.00208X. Hence, the expected outcome is negative, but the expected value is positive. This might seem counterintuitive but it is not surprising, considering that you can never lose even one full dollar, while your potential winnings are unlimited.

Suppose you play using method 2 for a very long time, starting with one dollar. What can you expect? Say you play N times, and let E>0 (small number). The probability that your bankroll is less than E approaches 1 as N approaches infinity. In other words, in the long run you will almost certainly lose almost the whole dollar.

Suppose you try to play using method 2 until you have won 1 million dollars, and then you stop. Once again you're out of luck (unless you're in a whole lot of luck). The probability that your bankroll will ever exceed M approaches 0 as M approaches infinity. For M = $1,000,000, it is very unlikely that you will ever win that much.

On the other hand, suppose we play using method 1 for a very long time, one dollar every game. What can you expect? Say you play N times, and let W>0 (big number). The probability that your winnings exceed W approaches 1 as N approaches infinity. In other words, in the long run you will almost certainly win as much as you want.

So **why** will you get rich by method 1 but not by method 2? Because in method 1 your bankroll is infinite, while in method 2 your bankroll is a measly $1. In method 1, no matter how much you might lose in the short run, you will always be able to bet another dollar for the next game. This means you were already rich from the start, you have an infinite bankroll, and winning a finite amount of money is easy. In method 2, no matter how much you might lose in the short run, you will never bring in any outside money for the next game. This means you have a $1 bankroll, and winning a large amount of money is hard.

The more you win using method 2, the more you stand to lose (as your bet goes up). But the more you lose, the less you stand to win (as your bet goes down).

What of Prime's argument about your money growing exponentially with the factor of 1.0333... each game? If we were to actually increase our bet by that factor with each consecutive game, regardless of how the game went, then his argument would hold true. But then again we would have an infinite bankroll, as we would be able to compensate for any losses using outside money. In method 2, whenever we lose our next bet will be smaller, and it will be harder for us to make up for that loss. But whenever we win our next bet will be larger, and it will be easier for us to lose those winnings.

So after 422 games, the expected value is over +1,000,000 dollars. But we can **not** expect to have nearly that much. We can expect to have (0.7*0.8*0.9*1.1*1.2*1.5)^{422/6} dollars, which is about 86 cents, because that is the expected outcome. This is an important concept for any gambler to be aware of. **You can never expect to gain your full expected value.** The expected outcome is always smaller than the expected value, and the reason for this is your limited bankroll. If you are only looking to maximize expected value, you will bet your entire bankroll on every favorable game you come across (higher bet equals higher EV), and inevitably lose all your money at some point.

Finally, the big question, can you win money using this machine? We must disqualify method 1 because it assumes an infinite bankroll. We can't use method 2 because it expects to lose money. Let's say your bankroll is $1 to start with, and you want it to grow to $1,000,000. Can you do it? Can you make a poor man rich? The answer is yes. Rather than throwing in your whole bankroll each time, suppose you throw in a fraction of it. Let that fraction be k. We have 0<k<1. You have 1-k of your bankroll left over on the side. The possible returns are 0.7k, 0.8k, 0.9k, 1.1k, 1.2k, and 1.5k. Your new bankroll will be either 1-0.3k, 1-0.2k, 1-0.1k, 1+0.1k, 1+0.2k, or 1+0.5k. The expected outcome over six games is (1-0.3k)(1-0.2k)(1-0.1k)(1+0.1k)(1+0.2k)(1+0.5k), which has a local maximum of ~1.0492607 at k~0.491298 (courtesy of Wolfram|Alpha). A good approximation for the optimal strategy is to bet half your bankroll each game, for an expected bankroll growth factor of a little over 1.049 per six games. After no more than 1728 games you can truly expect to be a millionnaire.