question about baccarat strategy

Discussion in 'Probability and Statistics' started by poopick, Feb 19, 2022.

  1. poopick


    Feb 19, 2022
    Likes Received:
    if you google baccarat odds you will get
    a player = 0.446 banker = 0.458 and tie = 0.096
    a bet on the player is 1:1 odds on the banker is 95:100
    my question is my strategy is good or am i missing something it goas like this:
    you always bet on the player the basic bet is x and if you lose once you bet 2x lose twice bet 4x lose 3 times reset to 1x
    you also reset when winning and if there is a tie keep the bet size the same
    the formula is 0.446 * x + 0.458 * 0.446 * x + 0.458 * 0.458 * 0.446 * x - 0.458 * 0.458 * 0.458 * 7x = 0.07131936
    also little cleaner 0.446 * x + 0.458 * 0.446 * x + 0.458 ^ 2 * 0.446 * x - 0.458 ^ 3 * 7x = 0.07131936
    each time i am winning i win 1x (2x - x or 4x - 3) and if i lose all 3 times i lose 7x (1 + 2 +4)
    i am ignoring tie because i will multiply them by 0
    i really feel that the casino wont let a strategy like this exist can someone confirm or deny that logic before i loose al my money
    Last edited: Feb 19, 2022
    poopick, Feb 19, 2022
  2. poopick

    Country Boy

    Dec 15, 2021
    Likes Received:
    That is a variation of what is called the "gamblers ruin" strategy! It works UNTIL you hit a string of losses and cannot keep going.

    Here if you lose three times in a row (which has probability about 0.17) you will lose -(x+ 2x+ 3x)= -6x. If you win three time in a row (which has probability about 0.09) you will win x+ x+ x= 3x. If you win two and lose one out of three (which has probability about 0.10) you will win x+ x- x= x. If you win one and lose two out of three (which has probability about 0.13) you will lose x- x- 2x= -2x.

    So, on average, every three games you can expect to lose 3x(0.09)+ x(.10)- 2x(.13)- 6x(.17)= (.27+ .10- .26- 1.02)x= -0.91 x.
    Country Boy, Feb 19, 2022
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