Blackjack math question
May 26th, 2016 at 7:39:55 AM
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I read somewhere before that players bust 21% of the time and dealer busts 28%. What is the % where neither one busts? Any approximation is appreciated.
May 26th, 2016 at 7:43:15 AM
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S17, dealer bust 29.1% of the time.
H17, dealer bust 29.6% of the time.
I would expect the player to bust less because they can choose when to hit/stand, and quite often a player stands to the dealers "bust card."
H17, dealer bust 29.6% of the time.
I would expect the player to bust less because they can choose when to hit/stand, and quite often a player stands to the dealers "bust card."
Playing it correctly means you've already won.
May 26th, 2016 at 7:47:27 AM
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Thanks for the answers. What % would U say that both player and dealer do not bust? That is the question.
May 26th, 2016 at 7:52:27 AM
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Well, if the dealer busts ~29% of the time, and I don't have the player figures but sure let's say your 21% figure is correct... then the odds of either one not busting are:
P(Player No Bust) = 79%
P(Dealer No Bust) = 71%
thus,
P(Neither Bust) = P(Player No Bust) * P(Dealer No Bust) = .79*.71 = .5609, or 56.09%
P(Player No Bust) = 79%
P(Dealer No Bust) = 71%
thus,
P(Neither Bust) = P(Player No Bust) * P(Dealer No Bust) = .79*.71 = .5609, or 56.09%
Playing it correctly means you've already won.
May 26th, 2016 at 7:57:12 AM
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Thanks.
