Baccarat Probability Calculator
2021年5月29日Register here: http://gg.gg/urwhw
*Baccarat Probability Calculator
*Baccarat Probability Calculator Rules
*Baccarat Probability Calculator Value
*Baccarat Probability Calculator Percentage
*Baccarat Probability Calculator Equation
Baccarat is a simple game to play and is especially popular with high rollers and players of an Asian demographic. General Rules of Play. Baccarat in its standard form is a card game normally played with 8 decks though the number of decks in play does vary when playing at online casinos (usually 6). 1: Baccarat - The Number of Combinations of Each Possible Hand; 2: Baccarat card counting - Effects of removing a card; 3: Probability of a Win at Various Mid-game Stages. In most baccarat games you rarely bet on the pair and it will depend on the table. In some instances a pair will result when you deal the cards. If the table allows you to bet on this pair you have an 11 to 1 payout. There are only 6400 combinations to get pairs with a probability of.075. At Baccarat Andrew MacDonald Abstract. Multiplied by the probability of winning multiplied by the probability of losing. Probability (Heads) 0.5 X 90 45. For example, it is useful if full play details are known to calculate the probability of the result. To do this the formula is solved to determine the value C. Probability theory for baccarat players Assuming you play the ‘Player’ or ‘Banker’ in Baccarat it has the distinction of being one of the best bets in the casino. The player is paid at even money when he wins and he wins just under 50% of the time.
Thread Rating: MagicalHello 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
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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” #AxelFabulousWizard
Administrator
I’ll let someone else do the math correctly.
(32-1)/(416-1) = 31/415 = 0.Baccarat Probability Calculator074699It’s not whether you win or lose; it’s whether or not you had a good bet.DocYes, 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.MagicalYes, 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” #AxelFabulousMagicalHmm.. 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?DocPartial 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.mipletThanks 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:
pairs that become trips (0.003036) + pairs that don’t become trips (0.071663) = original pairs (.074699)
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.“Man Babes” #AxelFabulous
We first present the probabilities attached to card dealing and initial predictions. In making this calculus, circumstantial information such as fraudulent dealing is not taken into account (as in all situations corresponding to card games). All probabilities are calculated for cases using one or two decks of cards. Let us look at the probabilities for a favorable initial hand (the first two cards dealt) to be achieved. The total number of possible combinations for each of the two cards is C(52, 2) = 1326, for the 1-deck game and C(104, 2)=5356for the 2-deck game.
Probability of obtaining a natural blackjack isP= 8/663 = 1.20663% in the case of a 1-deck game andP = 16/1339 = 1.19492%in the case of a 2-deck game.
Probability of obtaining a blackjack from the first two cards isP= 32/663 = 4.82654%in the case of a 1-deck game andP = 64/1339= 4.77968%in the case of a 2-deck game.
Similarly, we can calculate the following probabilities:
Probability of obtaining 20 points from the first two cards isP = 68/663 = 10.25641% in the case of a 1-deck game andP= 140/1339 = 10.45556%in the case of a 2-deck game.
Probability of obtaining 19 points from the first two cards isP = 40/663 = 6.03318%in the case of a 1-deck game andP = 80/1339 = 5.97460%in the case of a 2-deck game.
Probability of obtaining 18 points from the first two cards isP= 43/663 = 6.48567%in the case of a 1-deck game andP= 87/1339 = 6.4973% in the case of a 2-deck game.
Probability of getting 17 points from the first two cards isP= 16/221 = 7.23981%in the case of a 1-deck game andP= 96/1339 = 7.16952%in the case of a 2-deck game.
A good initial hand (which you can stay with) could be a blackjack or a hand of 20, 19 or 18 points. The probability of obtaining such a hand is calculated by totaling the corresponding probabilities calculated above: P = 32/663 + 68/663 + 40/663 + 43/663 = 183/663, in the case of a 1-deck game and P = 64/1339 + 140/1339 + 80/1339 + 87/1339 = 371/1339, in the case of a 2-deck game.
Probability of obtaining a good initial hand isP= 183/663 = 27.60180%in the case of a 1-deck game andP= 371/1339 = 27.70724%in the case of a 2-deck game.
The probabilities of events predicted during the game are calculated on the basis of the played cards (the cards showing) from a certain moment. This requires counting certain favorable cards showing for the dealer and for the other players, as well as in your own hand. Any blackjack strategy is based on counting the cards played. Unlike a baccarat game, where a maximum of three cards are played for each player, at blackjack many cards could be played at a certain moment, especially when many players are at the table. Thus, both following and memorizing certain cards require some ability and prior training on the player’s part. Card counting techniques cannot however be applied in online blackjack.
The formula of probability for obtaining a certain favorable value is similar to that for baccarat and depends on the number of decks of cards used. If we denote by x a favorable value, by nx the number of cards showing with the value x (from your hand, the hands of the other players and the face up card in the dealer’s hand) and by nv the total number of cards showing, then the probability of the next card from the deck (the one you receive if you ask for an additional card) having the value x is:
This formula holds for the case of a 1-deck game. In the case of a 2-deck game, the probability is:
Generally speaking, if playing with m decks, the probability of obtaining a card with the value x is:
Example of application of the formula: Assume play with one deck, you are the only player at table, you hold Q, 2, 4, A (total value 17) and the face up card of the dealer is a 4. Let us calculate the probability of achieving 21 points (receiving a 4).
We havenx = 2, nv = 5, so:
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.
For the probability of achieving 20 points (receiving a 3), we havenx = 0, nv = 5, so:Baccarat Probability Calculator Rules
.
For the probability of achieving 19 points (receiving a 2), we havenx = 1, nv = 5, so:Baccarat Probability Calculator Value
.
If we want to calculate the probability of achieving 19, 20 or 21 points, all we must do is total the three probabilities just calculated. We obtainP = 9/47 = 19.14893%. Baccarat Probability Calculator Percentage
Unlike in baccarat, where fewer cards are played, the number of players is constant (two), and the number of gaming situations is very limited, in blackjack, the number of possible playing configurations is in the thousands and, as a practical matter, cannot be entirely covered by tables of values.Baccarat Probability Calculator EquationSourcesA big part of the gaming situations that require a decision, where the total value held is 15, 16, 17, 18, 19 or 20 points, is comprised in tables in the section titled Blackjack of the book PROBABILITY GUIDE TO GAMBLING: The Mathematics of Dice, Slots, Roulette, Baccarat, Blackjack, Poker, Lottery and Sport Bets.You will also find there other issues of probability-based blackjack strategy . See the Books section for details.
Register here: http://gg.gg/urwhw
https://diarynote.indered.space
*Baccarat Probability Calculator
*Baccarat Probability Calculator Rules
*Baccarat Probability Calculator Value
*Baccarat Probability Calculator Percentage
*Baccarat Probability Calculator Equation
Baccarat is a simple game to play and is especially popular with high rollers and players of an Asian demographic. General Rules of Play. Baccarat in its standard form is a card game normally played with 8 decks though the number of decks in play does vary when playing at online casinos (usually 6). 1: Baccarat - The Number of Combinations of Each Possible Hand; 2: Baccarat card counting - Effects of removing a card; 3: Probability of a Win at Various Mid-game Stages. In most baccarat games you rarely bet on the pair and it will depend on the table. In some instances a pair will result when you deal the cards. If the table allows you to bet on this pair you have an 11 to 1 payout. There are only 6400 combinations to get pairs with a probability of.075. At Baccarat Andrew MacDonald Abstract. Multiplied by the probability of winning multiplied by the probability of losing. Probability (Heads) 0.5 X 90 45. For example, it is useful if full play details are known to calculate the probability of the result. To do this the formula is solved to determine the value C. Probability theory for baccarat players Assuming you play the ‘Player’ or ‘Banker’ in Baccarat it has the distinction of being one of the best bets in the casino. The player is paid at even money when he wins and he wins just under 50% of the time.
Thread Rating: MagicalHello 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
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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” #AxelFabulousWizard
Administrator
I’ll let someone else do the math correctly.
(32-1)/(416-1) = 31/415 = 0.Baccarat Probability Calculator074699It’s not whether you win or lose; it’s whether or not you had a good bet.DocYes, 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.MagicalYes, 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” #AxelFabulousMagicalHmm.. 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?DocPartial 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.mipletThanks 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:
pairs that become trips (0.003036) + pairs that don’t become trips (0.071663) = original pairs (.074699)
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.“Man Babes” #AxelFabulous
We first present the probabilities attached to card dealing and initial predictions. In making this calculus, circumstantial information such as fraudulent dealing is not taken into account (as in all situations corresponding to card games). All probabilities are calculated for cases using one or two decks of cards. Let us look at the probabilities for a favorable initial hand (the first two cards dealt) to be achieved. The total number of possible combinations for each of the two cards is C(52, 2) = 1326, for the 1-deck game and C(104, 2)=5356for the 2-deck game.
Probability of obtaining a natural blackjack isP= 8/663 = 1.20663% in the case of a 1-deck game andP = 16/1339 = 1.19492%in the case of a 2-deck game.
Probability of obtaining a blackjack from the first two cards isP= 32/663 = 4.82654%in the case of a 1-deck game andP = 64/1339= 4.77968%in the case of a 2-deck game.
Similarly, we can calculate the following probabilities:
Probability of obtaining 20 points from the first two cards isP = 68/663 = 10.25641% in the case of a 1-deck game andP= 140/1339 = 10.45556%in the case of a 2-deck game.
Probability of obtaining 19 points from the first two cards isP = 40/663 = 6.03318%in the case of a 1-deck game andP = 80/1339 = 5.97460%in the case of a 2-deck game.
Probability of obtaining 18 points from the first two cards isP= 43/663 = 6.48567%in the case of a 1-deck game andP= 87/1339 = 6.4973% in the case of a 2-deck game.
Probability of getting 17 points from the first two cards isP= 16/221 = 7.23981%in the case of a 1-deck game andP= 96/1339 = 7.16952%in the case of a 2-deck game.
A good initial hand (which you can stay with) could be a blackjack or a hand of 20, 19 or 18 points. The probability of obtaining such a hand is calculated by totaling the corresponding probabilities calculated above: P = 32/663 + 68/663 + 40/663 + 43/663 = 183/663, in the case of a 1-deck game and P = 64/1339 + 140/1339 + 80/1339 + 87/1339 = 371/1339, in the case of a 2-deck game.
Probability of obtaining a good initial hand isP= 183/663 = 27.60180%in the case of a 1-deck game andP= 371/1339 = 27.70724%in the case of a 2-deck game.
The probabilities of events predicted during the game are calculated on the basis of the played cards (the cards showing) from a certain moment. This requires counting certain favorable cards showing for the dealer and for the other players, as well as in your own hand. Any blackjack strategy is based on counting the cards played. Unlike a baccarat game, where a maximum of three cards are played for each player, at blackjack many cards could be played at a certain moment, especially when many players are at the table. Thus, both following and memorizing certain cards require some ability and prior training on the player’s part. Card counting techniques cannot however be applied in online blackjack.
The formula of probability for obtaining a certain favorable value is similar to that for baccarat and depends on the number of decks of cards used. If we denote by x a favorable value, by nx the number of cards showing with the value x (from your hand, the hands of the other players and the face up card in the dealer’s hand) and by nv the total number of cards showing, then the probability of the next card from the deck (the one you receive if you ask for an additional card) having the value x is:
This formula holds for the case of a 1-deck game. In the case of a 2-deck game, the probability is:
Generally speaking, if playing with m decks, the probability of obtaining a card with the value x is:
Example of application of the formula: Assume play with one deck, you are the only player at table, you hold Q, 2, 4, A (total value 17) and the face up card of the dealer is a 4. Let us calculate the probability of achieving 21 points (receiving a 4).
We havenx = 2, nv = 5, so:
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.
For the probability of achieving 20 points (receiving a 3), we havenx = 0, nv = 5, so:Baccarat Probability Calculator Rules
.
For the probability of achieving 19 points (receiving a 2), we havenx = 1, nv = 5, so:Baccarat Probability Calculator Value
.
If we want to calculate the probability of achieving 19, 20 or 21 points, all we must do is total the three probabilities just calculated. We obtainP = 9/47 = 19.14893%. Baccarat Probability Calculator Percentage
Unlike in baccarat, where fewer cards are played, the number of players is constant (two), and the number of gaming situations is very limited, in blackjack, the number of possible playing configurations is in the thousands and, as a practical matter, cannot be entirely covered by tables of values.Baccarat Probability Calculator EquationSourcesA big part of the gaming situations that require a decision, where the total value held is 15, 16, 17, 18, 19 or 20 points, is comprised in tables in the section titled Blackjack of the book PROBABILITY GUIDE TO GAMBLING: The Mathematics of Dice, Slots, Roulette, Baccarat, Blackjack, Poker, Lottery and Sport Bets.You will also find there other issues of probability-based blackjack strategy . See the Books section for details.
Register here: http://gg.gg/urwhw
https://diarynote.indered.space
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