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Subject : Of course it would posted by ETFan on Nov-29-2006 03:16pm |
| You just take the new bets and plug them into the equations I laid out. The pertinent part is: << Therefore, on each round in positive counts, the counter will not average $1.03, but $1.03 x 146% = $1.504. In negative counts he still bets the minimum, so he loses 2 x $0.11 = $0.22. The percentage of positive vs. negative rounds stays the same here, but the number of rounds per hour changes. Each round takes 5 x 12 secs + 2 x 8 secs = 76 seconds. Rounds per hour = 3600/76 = 47.368, and total EV per hour = ($1.5038 x 27.3% - $0.22 x 72.7%) x 47.368 = $11.87. >> If you have to bet $20 x 2 on negs, this becomes: << Therefore, on each round in positive counts, the counter will not average $1.03, but $1.03 x 146% = $1.504. In negative counts he bets twice the minimum, so he loses 2 x $0.22 = $0.44. The percentage of positive vs. negative rounds stays the same here, but the number of rounds per hour changes. Each round takes 5 x 12 secs + 2 x 8 secs = 76 seconds. Rounds per hour = 3600/76 = 47.368, and total EV per hour = ($1.5038 x 27.3% - $0.44 x 72.7%) x 47.368 = $4.29. >> Ie. less than half of what you can make just playing one hand. And in fact, it's even worse. By overbetting on negs, you've increased your risk of ruin, and you should actually bet less on high counts to compensate -- which drags down the EV even more. I was not putting up my post as "the final answer" to whether one should play two hands. As I tried to explain in previous posts, it very much depends on the specific details. There is no "one size fits all" answer. ETF |
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