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Thread: Counting 10s only

  1. #66


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    Quote Originally Posted by 21forme View Post

    Don, what is the significance of ln2 in solving this? How was that derived?
    Let's start with something simple: A coin-toss game.

    We want to know how many tosses we need to get an edge. We know that to have an edge, we need an EV of > 0; therefore our losses should be less than our wins, or our wins should be the majority of our outcomes.

    For simplicity, let's assume we want to know our break-even value. That would be 0.5-0.5 = 0. So, we start off with the following formulae:

    1-(1/2)^x = 0.5

    We then solve for x:

    -(1/2)^x = 0.5-1 = -0.5

    (1/2)^x = 0.5

    Here, we then solve for x using this law of Logarithms : Log(n^k)= k*Log(n); where Log(n) is the natural Log of n

    Log(0.5^x) = Log(0.5)

    x*Log(0.5) = -0.69314718

    x = -0.69314718/Log(0.5)

    x = 1

    Now we know we need 1 roll per game to average an EV of zero.

    Now, where does the Ln2 come from?

    Look back at our Law of Logs formulae : Log(n^k)= k*Log(n); where Log(n) is the natural Log of n

    Substitute n for 0.5

    Log(0.5^k)

    If you remember algebra, 2^-1 = 0.5; so sub that into the equation:

    Log(2^-1)

    and solve for the rest:

    Log(2^-1) = -1*Log(2) = -1*(-0.69314718) = ~0.69

    Therefore Log(2) ~ 0.69

  2. #67


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    Quote Originally Posted by DSchles View Post
    Where is it??
    What I meant was the answer to Freightman's, "what is ln2?", not the explanation of WHY it's ln2.

  3. #68


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    Ln2 is a good approximation when the denominator n>>1.

  4. #69


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    Thanks Dogman. I'm sure I knew the law of logarithms at one time, but it was overwritten in my RAM a long time ago.

  5. #70


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    Quote Originally Posted by Freightman View Post
    No he mirado las matemáticas, francamente, ni siquiera estoy seguro de cuál es la pregunta. Sin embargo, la penúltima publicación aclaró la siguiente publicación de OP.

    Por el bien del argumento, digamos que la ventaja es en realidad un enorme 15%. Luego, aclaró que la apuesta DEBE ser de $1000. ¿Puede su bankroll manejar la excavación profunda para apostar múltiples rondas para obtener el retorno?

    Si tu respuesta es sí, entonces bien. Si la respuesta es no, aléjate. El hecho de que las matemáticas (pueden) estar allí, no significa que sea un buen negocio para todos. Además, las apuestas de $1000 están más allá de la zona de confort de la mayoría de las personas. Si está más allá de los tuyos, aléjate.

    Cambiaré el escenario a uno familiar para muchos. Ese pésimo juego de ruleta (no tengo las reglas claras, así que aquí hay algo de licencia) promueve un pago de 105-1 por un pago normal de 35-1 en 00. Sin mínimo, siga apostando. Digamos que la apuesta mínima para la promoción es de $100. Para algunos, está fuera de la zona de confort. Seguiré jugando, no hay problema. Digamos que el mínimo para la promoción es de $1000.

    1, 10, 30, 60, 100 giros: ¿cuándo te compensa? ¿Puedes cavar tan profundo si tuvieras que hacerlo? Personalmente, me gusta dormir por la noche.
    Greetings, I managed to get a 20 percent refund at the time of making my first purchase of chips, therefore if I buy 300 USD they would give me 300 USD in normal chips + 60 in promotional chips which I must play until I lose because they are not redeemable for real money.


    According to the strategy that you recommend, would I have any advantage if I buy 300 and play all 360 in one hand?


    In case of winning, I would take the 60 promotional cards and play them until I lose it. While I am winning, I am accumulating chips.

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