Baccarat Probability Calculator

 

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Baccarat probability calculator

Description of Baccarat Probability Calculator / 百家乐计算器 / 바카라 계산기 Free install.ApkMod file 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.

  1. This free probability calculator can calculate the probability of two events, as well as that of a normal distribution. Learn more about different types of probabilities, or explore hundreds of other calculators covering the topics of math, finance, fitness, and health, among others.
  2. 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.

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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.
Baccarat Probability Calculator
Doc
Calculator

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

Baccarat Probability Calculator


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.


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

Baccarat Probability Calculator Chart


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

Baccarat Probability Calculator Estimate

“Man Babes” #AxelFabulous