- #1
elusiveshame
- 170
- 35
Okay, so the title is a bit clickbaity as most people will say baccarat isn't a beatable game, and I'm on the fence myself, but only because I have tons of questions that I can't seem to find reasonable answers to. I think doing the math will help me, but it's been quite a while since I did anything requiring more than algebra 1 (maybe 2) in recent years.
So the first thing I was trying to determine was if counting cards in baccarat was a worthwhile pursuit. It's not, at least in any meaningful way, since you don't get to choose to hit or stay on a given hand, and cards are dealt based on set rules. Maybe there's something there, but nobody has been able to discover anything that will help your position in the game.
Board pattern matching would be a way, but there's n = 416! possible arrangements for an 8 deck shoe, that even if you had let the shoe play out until the last hand before pattern matching a shoe, it would still take more time to determine the next play than you have to place your bet (and even then, you don't know what the burn cards are or what cards remain after the end cut card, so at best, there's 15! possible outcomes (if you draw an A for the burn card, total burn cards are 2, and the end range of the 2nd cut card is typically 14 cards or more).
That leaves me with probability as a potentially last effort.
The probabilities are:
Player: 44.6%
Banker: 45.8%
Tie: 9.6%
(honestly, I'm not sure how they get these values - any place I've stumbled upon baccarat probabilities doesn't show any of the math behind these numbers, so I'm a bit lost on how to get these). I understand why banker has a higher probability (because of how the rules govern when banker draws a third card as opposed to anything lower than a 6, like in the players case), otherwise it *should* be an even game.
Now, people parrot that the previous event has no effect on the next event (i.e., a player win won't guarantee a banker win), but I don't understand this. Wouldn't the removal of cards from the shoe alter the probability of a player/banker/tie win?
Maybe it would be better to determine the probability of a certain hand to come out? If there's 8 decks/416 cards, where 10, J, Q, K all = 0, would the probability of drawing a hand of 0 points (3 value 0 cards) be something like number of cards valued 0 (a) divided by the number of cards remaining (b), multiplied each time a card is drawn? So something like (128/416) * (127/415) * (126/414) - assuming a full shoe to start and no burn cards.
I know I'm a bit all over the place with this, but I hope some of this makes sense with what I'm asking, and if not, I'll try to clear up what I was asking.
So the first thing I was trying to determine was if counting cards in baccarat was a worthwhile pursuit. It's not, at least in any meaningful way, since you don't get to choose to hit or stay on a given hand, and cards are dealt based on set rules. Maybe there's something there, but nobody has been able to discover anything that will help your position in the game.
Board pattern matching would be a way, but there's n = 416! possible arrangements for an 8 deck shoe, that even if you had let the shoe play out until the last hand before pattern matching a shoe, it would still take more time to determine the next play than you have to place your bet (and even then, you don't know what the burn cards are or what cards remain after the end cut card, so at best, there's 15! possible outcomes (if you draw an A for the burn card, total burn cards are 2, and the end range of the 2nd cut card is typically 14 cards or more).
That leaves me with probability as a potentially last effort.
The probabilities are:
Player: 44.6%
Banker: 45.8%
Tie: 9.6%
(honestly, I'm not sure how they get these values - any place I've stumbled upon baccarat probabilities doesn't show any of the math behind these numbers, so I'm a bit lost on how to get these). I understand why banker has a higher probability (because of how the rules govern when banker draws a third card as opposed to anything lower than a 6, like in the players case), otherwise it *should* be an even game.
Now, people parrot that the previous event has no effect on the next event (i.e., a player win won't guarantee a banker win), but I don't understand this. Wouldn't the removal of cards from the shoe alter the probability of a player/banker/tie win?
Maybe it would be better to determine the probability of a certain hand to come out? If there's 8 decks/416 cards, where 10, J, Q, K all = 0, would the probability of drawing a hand of 0 points (3 value 0 cards) be something like number of cards valued 0 (a) divided by the number of cards remaining (b), multiplied each time a card is drawn? So something like (128/416) * (127/415) * (126/414) - assuming a full shoe to start and no burn cards.
I know I'm a bit all over the place with this, but I hope some of this makes sense with what I'm asking, and if not, I'll try to clear up what I was asking.
