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I was playing around with a ROI simulator today, and I ran a simulation of 100000 sets of 5 games. It gave the chance for a +10% ROI player making a loss over these 5 games as 52%.
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What does it mean to be a +10% ROI player? My take on it (correct me if I'm way off) is that there is a probability p1 that the player makes X1 above his buy-in, p2 that the player makes X2, ... pN probabliity that the player makes XN with the numbers chosen so that p1*X1+...+pN*XN = 1.1*buy-in. So there is a 1-p1-p2-...-pN chance that the player loses his buy-in. So here's a simple case with a 10% ROI:
* This is an example, not an offer!!!*
You give me a dollar. I pick a [real] number between 0 and 10. If the number is between 0 and 5.5, I'll give you $2, otherwise I'll keep your dollar. In the long run, you make $0.10 for every dollar you play with, for an ROI of 10%. However, if you only play 5 times, you need to win 3, 4, or 5 times to profit. (If you don't win at all, you lose $5. If you win once, you lose $3. If you win twice, you lose $1.)
So what is the probability that you lose money? Well, you have a 55% chance of winning each particular round. Using the
Binomial Distribution, we see that the probability of you not winning at all is 0.45^5. The probability of only winning once is 5*0.55*(0.45^4). The probability of only winning twice is 10*(0.55^2)*(0.45^3). Summing these up, you will lose money 40.6% of the time!
If you change the game so that 27.5% of the time you win $3 (i.e. I give you $4) and 72.5% of the time you lose $1, then your ROI is still 10%. In this case, you come out ahead provided you win twice. The probability of winning only once (or not at all) is 58%!!
As you can see, if you make it more difficult to win (but with much higher payouts) then the odds of not being profitable after just 5 attempts increases. This is variance at work. (The other way to look at it is that the times you _do_ come out ahead, you'll likely come out WAY ahead)