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Thread: Question for Don Please

  1. #1


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    Question for Don Please

    Hello everyone. It's been a few months since I have visited this Forum.

    And Hi Don,

    Just wanted to clarify something. Please see the following thread that I started in July of last year:
    https://www.blackjacktheforum.com/showthread.php?44466-Questions-about-Player-Advantage-and-Next-Card-is-An-Ace-Math&p=298884&highlight=overkill#post298884

    In Post #17 of that thread, you stated the following (I highlighted the portion of the post of interest with red font):

    So, there are a couple of points to be made. Obviously, we all understand that, if an ace were certain, the edge on that round would be 51%. That's
    always the case: if your first card is an ace, your edge is 51%. But, as Overkill mentioned, usually, that probability is only 1/13 (7.69%), and it balances out with all the other first-card possibilities to give whatever the BS house edge is, say, 0.5%. So now, we have to rejigger the edges and the probabilities. Without any special knowledge, and to get to -0.5%, we would have 1/13(51%) + (12/13)x = -0.5%. Solving for x, we get -4.79%. So far, so good.

    But now, the +51% enjoys an inflated 15.1% probability. So the new calculation for the overall edge becomes: .151(51%) + .849(-4.79%) = edge. Player edge is 3.63%.

    Of course, this is the edge only when the 6 (or a ten, for that matter) appears. Naturally, it isn't your overall edge for the game, because, well, sixes appear only 1/13 of the time. So, that 3.63% edge would then be divided by 13 to give your flat-bet advantage of 0.28%. But, it goes without saying that, when the 6 (or ten) would appear, you'd bet a great deal more than your minimum bet. And, it's here that the frequency of the tens (4/13) would come in. That 0.28% would be multiplied by four (1.12%), for your flat-bet edge if you knew aces followed tens, instead of sixes, with 15.1% accuracy.

    Think I've covered it all.

    Don

    Don, after doing the arithmetic, rather than the player edge being 3.63%, isn't the player edge instead 7.291%, or did I miss a step? Your answer of 3.63% is roughly half of my answer of 7.291%.

    Thanks,
    Overkill






  2. #2


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    Are you questioning the math of the red portion? That equals 3.63%.

    Don

  3. #3


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    Yes, the math of the red portion.

    I believe you may have forgotten to move the decimal of the negative term two places to the left because it is actually a percentage.

    My edge would be about 7.3%

    Thank you again for showing how to calculate edge in these kind of scenarios.

  4. #4


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    Oh, geez - my apologies. I moved the decimal 3 places rather than 2!

  5. #5


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    Quote Originally Posted by Overkill View Post
    Oh, geez - my apologies. I moved the decimal 3 places rather than 2!


    Don

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