callipygian
Well-Known Member
Depending on the level of accuracy you need, this approximation may be fine: add +0.5% advantage for every positive true count. A normal 6D game is around -0.6%, so +1 = -0.1%, +2 = +0.4%, etc.itakeyourmoney said:I was just wondering if someone could direct me to somewhere where I could calculate the player's advantage depending on the true count of 6 decks.
(1) Expected Value, EV, is a ratio of what you expect to win in the long run divided by your bet. If you expect to win $1 on a $5 bet, your EV is 0.2; if you expect to win $1 on a $10 bet, your EV is 0.1. Obviously, for complex bets (such as blackjack, where your bet can double halfway through a hand) this is not trivial to calculate. Generally people will calculate expected winnings and then divide through by an average bet.itakeyourmoney said:I might need some help from a better mathimatician than myself here: to calculate my expected winnings I took .50408^9 (That is, the odds that I do not win the hand -- either I lose or I push -- 9 times in a row, starting with a min bet of $5 up to a max bet of $1,280.
(2) EV is calculated from the general formula:
ExpectedValue = ProbabilityWin * WinAmount - ProbabilityLose * LossAmount
In your example, your EV is not $1,280, because it doesn't take into account the probability that you lose it all - even if the probability that you lose it all is small, the fact that you will lose a large amount will change the EV.
(3) In general, EV for a Martingale progression is no different from EV for a single hand. That's because you divide the winnings by the amount bet - you end up expecting to win a large percentage of the time, but only a small amount, and expect to lose a small percentage of the time, but a large amount. Martingale won't change your EV (or not by much).