Baccarat Probability Calculator

  

Get in depth info on probability and odds of baccarat to calculate your chances of winning. Gather useful information on probability and baccarat odds. A Basic Discussion On Calculating Odds Of Baccarat. Learn how to calculate odds of baccarat and how the house edges can vary. Calculating odds of baccarat helps you in dealing out. Friends, This application computes the probability, change in probability, house edge and Kelly criteria of the Player, Banker, Tie and 2 side wagers from a choice of 7, in Baccarat on a coup-by-coup and card-by-card basis. Considering that many baccarat players make big bets any extra edge you can get should always be used. The probabilities after cards are already dealt can also be worked out. The Wizard of Odds has an excellent Baccarat Probability article about the first two, three, four or five cards have been dealt. Of course you cannot change your bet once.

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Magical
Hello all,
I have a newbie question which I hope the more experienced of you can help me with (I've forgotten most of my high school math!)
Some baccarat tables offer a player and/or banker pair bet. The payouts vary. (ranging from 1 to 6, to 1 to 12?)
My question is as such: How do I calculate the probability that the very first hand of an 8-deck baccarat game will consist of a pair? And subsequently, as more and more cards are used, how do we calculate the changing probabilities?
My thoughts will be to simply assume the first card is, lets say a nine. Then the probability that the next card will be a nine will be (32-1)/(416-1)= 0.74699. Since one nine and one card out of the 416 cards have come out. This is not what the wizard says. Where am I going wrong? And how do I calculate the probabilities that a pair will appear once a number of cards (with their values recorded) have been used?
Thank you very very much in advance to anyone who replies.
Doc

I've forgotten most of my high school math!
....
(32-1)/(416-1)= 0.74699


Proved your point. ;-)
Sorry. Couldn't pass up the opportunity. I'll let someone else do the math correctly.
miplet

Hello all,
I have a newbie question which I hope the more experienced of you can help me with (I've forgotten most of my high school math!)
Some baccarat tables offer a player and/or banker pair bet. The payouts vary. (ranging from 1 to 6, to 1 to 12?)
My question is as such: How do I calculate the probability that the very first hand of an 8-deck baccarat game will consist of a pair? And subsequently, as more and more cards are used, how do we calculate the changing probabilities?
My thoughts will be to simply assume the first card is, lets say a nine. Then the probability that the next card will be a nine will be (32-1)/(416-1)= 0.74699. Since one nine and one card out of the 416 cards have come out. This is not what the wizard (http://wizardofodds.com/baccarat/baccaratapx5.html) says. Where am I going wrong? And how do I calculate the probabilities that a pair will appear once a number of cards (with their values recorded) have been used?
Thank you very very much in advance to anyone who replies.


Your decimal is in the wrong spot. 31/415 = 0.074699 . That's the the probability that the players first 2 cards are a pair. Sometimes the player gets a third card, which may result in three of a kind, or may pair up one of the first 2 cards. (I think that the side bet here only counts pairs in the first 2 cards, and three of a kind, but not pairs made by the third card.)
“Man Babes” #AxelFabulous
Wizard
Administrator

I'll let someone else do the math correctly.


Baccarat probability calculator equation(32-1)/(416-1) = 31/415 = 0.074699
It's not whether you win or lose; it's whether or not you had a good bet.
Baccarat Probability Calculator
Doc
Yes, Wizard, I had done that part of the math, which was why I made my snide remark. It was such things as the twists and turns that miplet mentioned that I didn't want to get into myself. I don't even know the game itself.
Magical
Yes, that was a mistake on my part, i put the decimal at the wrong place. My bad.
However the question is still the same, the answer i come up with is 0.074699. However the numbers on the WOO site shows 0.071663 and 0.071864 for player and banker respectively. And yes, I am talking about the first 2 cards of the player and banker, the third card is not involved in determining if the hand has a Pair. I realize my calculations ignore the fact that the cards are dealt player, banker, player, banker. How do I include the situation where the cards are dealt this way? And how are the numbers 0.071663 and 0.071864 calculated?
Thank you to those who replied so far :)
miplet

Yes, that was a mistake on my part, i put the decimal at the wrong place. My bad.
However the question is still the same, the answer i come up with is 0.074699. However the numbers on the WOO site shows 0.071663 and 0.071864 for player and banker respectively. And yes, I am talking about the first 2 cards of the player and banker, the third card is not involved in determining if the hand has a Pair. I realize my calculations ignore the fact that the cards are dealt player, banker, player, banker. How do I include the situation where the cards are dealt this way? And how are the numbers 0.071663 and 0.071864 calculated?
Thank you to those who replied so far :)


You did the calculations just fine. The 0.071663 and 0.071864 aren't counting pairs the become 3 of a kind when a third card is delt.
“Man Babes” #AxelFabulous
Magical
Hmm... I'm not counting the pairs that become 3 of a kind either. I'm not including the third card into the calculations. What am I missing? And how do i calculate the probability of a pair after lets say all the aces and 8 kings are used up already?
Doc
Partial answer: I think he is saying that to get the right figure for pairs, you have to consider the third card. If you don't, some of the pairs that you find will end up not being pairs after the third card is dealt -- they will be part of a three of a kind and should be excluded. Excluding those gives the slightly lower probability of a plain pair. Or did I misunderstand your post?
Edit for clarification (hopefully): When the first two cards are a pair, you can't know whether they will remain a 'pair' until you determine whether there will be a third card and what the chance is that it will change the 'pair' into trips.
ProbabilitymipletCalculator
Thanks for this post from:

Hmm... I'm not counting the pairs that become 3 of a kind either. I'm not including the third card into the calculations. What am I missing? And how do i calculate the probability of a pair after lets say all the aces and 8 kings are used up already?


If you are only calculating if the 1st 2 cards make a pair use (d*4-1)/(d*52-1) where d is the number of decks. This is what you already did. The side bet in the baccarat appendex pays more when the pair becomes a three of a kind. I think that is why you are confused. Here are the numbers for player pair:

Baccarat Probability Calculator Formula

pairs that become trips (0.003036) + pairs that don't become trips (0.071663) = original pairs (.074699)

Baccarat Probability Calculator Value

The main baccarat page also lists pairs just on the 1st 2 cards.
I'll answere your other question when I get my brain into thinking mode. I know how, its putting it into words.

Baccarat Probability Calculator Equation

“Man Babes” #AxelFabulous

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