December 3rd, 2009 at 12:44:03 PM
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Assuming typical Las Vegas table rules for Blackjack (dealer hits soft 17, re-splitting allowed to 4 hands, except aces which can only be split once with a one card draw to each, no surrender, and doubling allowed after splitting) - With 6 or 8 decks, what would be the expected percentage of hands won, hands lost, and hands pushed?
December 3rd, 2009 at 11:30:20 PM
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Standard deviation in blackjack
The second table has basic win/loss/push stats. Those stats probably will differ for other rules, but shouldn't be too far off.
The second table has basic win/loss/push stats. Those stats probably will differ for other rules, but shouldn't be too far off.
December 5th, 2009 at 8:36:34 AM
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Thanks for the response. The table you referenced seems to answer my question, but I have one more that isn't apparent to me from the table. Not counting pushes, what would be the probability of winning two single hands (not doubled or split) in a row or losing two single hands in a row?
December 6th, 2009 at 10:55:54 PM
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Well, that might be taken care of by the next table, although not perfectly (because of surrender, which is often not an option). But rolling wins and BJ together, and surrenders and losses together (surrenders aren't a big part of basic strategy, and when they are, they're hands you're likely to lose anyway), you get about 40.98% win and 50.50% loss.
Once you find out the probability of something happening, you can find out the probability of streaks by (probability)^(streak), so 2 wins in a row are (.4098)^2 = 16.8% and 2 losses in a row are (.5050)^2 = 25.5%.
Once you find out the probability of something happening, you can find out the probability of streaks by (probability)^(streak), so 2 wins in a row are (.4098)^2 = 16.8% and 2 losses in a row are (.5050)^2 = 25.5%.