Quote:
Originally Posted by Andy Asaro
http://blog.horseplayersassociation....ch?q=roulette+
Excerpt:
Q. How many 'zeroes' do you have to add to a roulette wheel to turn roulette into the equivalent of what a newbie horse bettor faces? (Where random WPS selections produce long term net losses bordering on minus 25 percent?)
A. Believe it or not you have to add TEN ADDITIONAL ZEROES - until the layout itself has TWELVE ZEROES ON IT! – in order to turn roulette into the equivalent gamble (net losses of minus 25 percent) faced by a horse bettor making random WPS selections at 16 percent takeout!
TWELVE zeroes! (Mull that over for a few minutes until it sinks in.)
No one in their right mind is going to spend any serious amount of time immersed in study trying to develop a system to beat a roulette wheel with TWELVE freaking zeroes on it. (Can we at least agree on that much?)
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This suffers from a pretty basic math error.
Imagine the following two games:
Game 1 involves incredibly hard to hit big payoffs with a negative takeout-- i.e., you actually make money on the game in the long term if you are an average player. However, you only hit those payoffs once in 250 times or more that you play it.
Game 2 involves a positive takeout, i.e., the average player loses. However, there are lots of moderate payoffs and a ton of variance, so something like 35 percent of people who play it on a given day come out as winners.
Which game is going to deliver a better experience to the AVERAGE fan?
(By the way, it's worth noting that the vast majority of successful slot machines nowadays follow the paradigm of game 2.)
