January 3rd, 2013 at 7:20:34 PM
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Can anyone figure out the strategy of how to play with roughly 66% probability of winning on each bet?  It can be done in two games of my choice. One bet has a 16% chance of losing the other has roughly 35% chance of losing. Risk, however, is not 1 to 1.
                    January 3rd, 2013 at 7:27:39 PM
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Last edited by: sodawater on Oct 1, 2018
 
                    January 3rd, 2013 at 7:29:08 PM
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Never tried that... but interesting!
                    January 3rd, 2013 at 7:31:46 PM
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Quote: thezoneCan anyone figure out the strategy of how to play with roughly 66% probability of winning on each bet?
Hell, that's easy. Just fade my picks in the WoV picks game. You must have seen my W/L record.
"I am a man devoured by the passion for gambling." --Dostoevsky,  1871
 
                    January 3rd, 2013 at 7:38:58 PM
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Last edited by: sodawater on Oct 1, 2018
 
                    January 3rd, 2013 at 7:40:02 PM
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Last edited by: sodawater on Oct 1, 2018
 
                    January 3rd, 2013 at 7:49:30 PM
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Correct with option 3...   play any 2 of the 3 blocks (1st 12, 2nd 12, 3rd 12) .   As far as option 2 goes, you dont have a 66% chance of winning on every roll..  laying 2 to 1 with probability of 50% over a series of rolls until either hits.
                    January 3rd, 2013 at 7:50:44 PM
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not to mention the house juice.. which is 5%
                    January 3rd, 2013 at 8:02:06 PM
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Last edited by: sodawater on Oct 1, 2018
 

 
                         
                                                             
                                                             
                                                             
  
  
  
 