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Last edited by garnabby; 06-01-2011 at 02:58 AM.
Garnabby. If I'm reading this right, basically you are saying that given a stop-win of 1 and no stop-loss you will succeed around 9/10 times.
Say I stop at 1 for 9 shoes in a row.
I am now +9 all up (not including commission)
On the 10th shoe I lose a huge amount (for example: more than 9) and thus I lose overall. Is that what you are saying?
Plasia,
What Garn is attempting to explain, in his own simple words (and a tad of mathematical obsfucation), is why some people from BTC, who "claim" to have won 15 or 19 or 20 shoes in a row, most probably did it by winning 1 unit and calling it a winning shoe.
Of course his premis is all wrong, as is his motive, but what he appears to be trying is to tell everyone that only by making 1 unit could you possibly run 20 shoes in a row as winners.
I'd would beg to differ, but then I'd be on the losing side of any argument with "The Appointed One" as anything I would say on the matter would be total lies.
The best I could come up with was 19 out of 20. I can assure the G man that it was more than 1 unit per shoe.
As I now tend to just ignore G's message posts, he'll have to talk to you or himself about it.
AD
The operative words here are "statistical basis." Of course G is absolutely correct. Particularly he is correct about the compounded house edge making the edge "seem" higher. In terms of a single finite bankroll one might see that the house edge is 100%!
The above is not to say that a player can go an awful long time picking spots and winning. An awful long time.
It also assumes random trials but that is another thread.
A
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