# Which probability calculation is "more correct" in Baccarat games?

• B
• bernd
In summary: B threw a total of 1 or more so...then A and B get to shake hands and go home.In summary, In this analogy, the player who threw more dice is the winner.
bernd
TL;DR Summary
baccarat odds calculations, weighing results on their likeliness or not?
Hello, this question isnt really much about calculation but rather which view point is more correct.
See, in a gambling game called Baccarat a game is played where a player A ("player") and a player B ("bank") draw cards according to a fixed ruleset from a given card pot.
First A and B both draw 2 card.
depending on A's cards, he might draw a third card or not.
Depending on that, B might also draw a third card or not.

The ruels are complicated.
But in the end, the final game situation is 4-6 cards.
Now I went ahead (in a java Program) and first calculated all possible end situations, by first building all possible sixtuples (where the slot for the respectively third card is set to -1 if no such third card was drawn. )
Then I use the super complicated rule set and eleminate sixtuples that are impossible to achieve.

so I got a list of legit endgame scenarios.
now for each such scenario e, I can first calculate the winner (just add A's cards togehter modulo 10, do same for Bs cards and compare.)
And, more important, I can calculate the odds for that scenario happening and by using all thos odds, calculate who wins more often.

1. How I initially did it:
for each scenario, I calculated the odds of it with happening by simple stochastic rules with putting bakc and such.
Like if you first draw a card from a pool of 52 cards, then draw another card from a pool of 51 cards, etc.
you get the drill. obviously keping in mind that some cards jsut appear more often in the pool and other details.

so I this way get odds for each scenario.
and I go through the list, add up all the odds of scenarios where A wins, all odds of scenarios where B wins.
and all odds for scenarios where its a draw.

this way I get 3 odds for 3 different outcomes.

however this result doesnt fit in what is plastered all over the internet.
internet says, player and bank win in about 45% of cases.
however I get odds of more lik 35%.
so big descrepancy.2. now how some other person I spoke to did it:
list of legit endgames as usual.
he goes also through the list, looks for each one whos winner and jsut counts them up depending on winner.
so, like you would do on votes "1 vote for A, 1 vote for B, another vote for A, etc."
so, basically he weights all outcomes the same, ignoring how likely each one of them is to happen in reality.
so he jsut counts hopw many games end in A winning, how many in B, how many in a draw.
dvidie by total naumber of games.
bam, got the 3 outcomes that are all over the internet.

I tried to simulate this too, by jsut using "odds" of fixed 1 for each scenario and in the end dividing by total number of games.
And yep, I get otucomes of 46% for A, 44,5 for B and 10% for draw.
still off by like .5% each from the ideal result,(which might be due to programming issues, rounding and such)
but basically, confirmed his methods does get the desired result.
now unintendedly long story short:
I base my calculations on how likely each end result is to actually happening, therefor obviously weighing super unlikely events less than others.
friends jsut counts how many games gives each winner, basically weighing all endresults equally likely, in a way.

friends endresults are wha<t the internet says is right.
I think mine is actually more right as it incorporates how less or more likely an event is to happen.

Who is "in the right" here?
I shpould mention that this whole calculation is important to me as I want to do some expectancy value calculations for baccarat
where i then basically want to go
exp value for A winning=odds for A winning * payout for A winning - odds of A not winning * payout in case A doesnt win

same for B and draw.
Then just compare which of the 3 outcomes I would be best of betting on (yeah, internet tells me that already too but I want to derive it on my own!)
So I kind of want my calculations to be realsitic in the way that obviously the unlikeliness of an event should be reflected in the final determiantion on what to bet on! :-)

I'm not familiar with baccarat, but in any case I'd need to see more precisely what is being calculated to venture an opinion on it.

Well, I could copy in the source code of those 3 java files but not sure if it is of much help.

If I want to describe an analog issue instead:
Say we have play A and B and a 6 sided dice.

first A throws the dice 2 times.
then B throws the dice 2 times.
if A threw a total of 8 or more so far, he now throws the dice a third time.
now B may throw also a third dice depending on some things:
if(a threw third dice +Bs first 2 dice add up to >=11) then B throws a third dice.

Super simplified since there are like tons of rules when A or B can and cannot draw a third card or where the game ends isntantly if A or B gets a sum of 11 with the first 2 dice throws, etc.

Anyways this way, we have a result of 4-6 dice numbers.
(1 6 3, 3 4 -1) for example means that A threw 3 and 6 and threw a 3 on the third dice.
and that B first threw 3 and 4 and didnt throw a third time cause he didnt fulfill the condition to do so.
in such a way I have (legit) configurations with 6 numbers in them
(2 of which might be -1 depending on the whole thing)

Now the winner is the guy with bigger total dice number (ignoring the -1s of course).
so A got 1+6+3=10, B got 3+4=7.
B won.

And this given play can happen about
1/6*1/6*1/6*1/6*1/6=1/(6^5)
Cause, well, we have 5 numbers and dice sites have probability 1/6.
in my baccarat I obviously kept in mind that we don't put cards back, so its a bit more complicated to calculate the odds.

now what I would have done:
went through the lsit of all (legially possible) (maximum 6) dice configurations,
check whos the winner and how likely the odds are for it to happen (as I calculated above, same way)
and jsut add that odd to whoever won.

like, you know, if you go through all the paths in a stochastic tree diagram where the end result is A winning.

hile my bestie goes ahead and jsut counts thorugh the games
"that is +1 win for A, this game is +1 win for B, etc."
obviously dividing by the total of games to get a ratio.
and my friend ends up with the result that is written all over the internet.

I just cant comprehend why this should be correct?
it doesnt take the likeliness of different event happening less likely into account at all.
Like it is more likely to throw two 6s and a 1 in a row than two 9s alone (later has odds of 1/(6^2), first one has 1/(6^3)!)

I didn't really understand the rules. But, you should be able to get an accurate estimate simply by simulating a large number of games.

Is that what your friend did?

You don't need to go through each possible game and calculate its probability. Is that what you tried to do?

FactChecker
bernd said:
If I want to describe an analog issue instead:
Bad idea. Now you have mixed two problems, one with cards and one with dice.

Better to decide what problem you want help with rather than leading us all over the place.

Well It's still about the baccarat game.
Just though that using a similar game in comparison would make me question a bit easier to understand.
Here's the rules if anyone cares:
https://www.wikihow.com/Play-Baccarat
Just ignore anything that wont matter for the online version anyways (like where on the table cards are placed and such non-important stuff. all that matters how cards are drawn and according to what rules)

Last edited:
PeroK said:
I didn't really understand the rules. But, you should be able to get an accurate estimate simply by simulating a large number of games.

Is that what your friend did?

You don't need to go through each possible game and calculate its probability. Is that what you tried to do?
Well, yeah.
since if I add the odds of games where A wins, then that number should be As overall win likeliness.
same for B and draw result.

Sure, I can go and jsut do 1 million games or such, but I thought to be more precisely just evaluate every possible game and add up the winning odds for each of the 3 possible outcomes, since that should give up the msot precise answer (leaving some rounding and programming issues aside).
but what I got as a result can even be explained by rounding errors, as getting like 35% of the around 45% as expected was jsut too wrong.

so either there was still some big programming error in it.
or my calculations were right but what I calculated wasnt the odds of a certain event happening at all.
even though simplest stochastic school class material would suggest that my method should work.check all possible "paths" in the tree diagram, look how likely each one happens.
and add up the probabilities of all those path probabilities that end in the desired result.

however i jsut don't get, why the internet says I cant calculate this way but have to "assume" that each event (displayed beind differently likely) has equal chance of happening.
and it's only jsut about how many of the total game end with A winning, B winning or draw. :-/

bernd said:
Well, yeah.
since if I add the odds of games where A wins, then that number should be As overall win likeliness.
same for B and draw result.
Yes, but you have to calculate the odds.
bernd said:
Sure, I can go and jsut do 1 million games or such, but I thought to be more precisely just evaluate every possible game and add up the winning odds for each of the 3 possible outcomes, since that should give up the msot precise answer (leaving some rounding and programming issues aside).
Yes, but it might be a lot of extra work for little benefit.
bernd said:
so either there was still some big programming error in it.
or my calculations were right but what I calculated wasnt the odds of a certain event happening at all.
even though simplest stochastic school class material would suggest that my method should work.
Looking at the rules, it can't be a trivial program. Is it such a surprise if there is a bug in your code?
bernd said:
however i jsut don't get, why the internet says I cant calculate this way but have to "assume" that each event (displayed beind differently likely) has equal chance of happening.
The different outcomes won't be equally likely.

It sounds like your friend assumed that all outcomes are equally likely and just counted them up. Based on a brief look at the rules, that seems wrong because some outcomes have different numbers of cards drawn. On the other hand, your approach of calculating the probabilities of all the possible outcomes could be complicated. If you had outcomes with different probabilities, you should check that and determine if it is a valid example where your friend's assumption is wrong.
Although the rules are complicated, the implementation of a game in a computer simulation should be very straightforward (if tedious). I would tend to think that @PeroK 's idea of simulating a large number of games is the more reliable method. Getting the program correct requires careful attention to following the rules correctly.

bernd and PeroK
FactChecker said:
It sounds like your friend assumed that all outcomes are equally likely and just counted them up. Based on a brief look at the rules, that seems wrong because some outcomes have different numbers of cards drawn. On the other hand, your approach of calculating the probabilities of all the possible outcomes could be complicated. If you had outcomes with different probabilities, you should check that and determine if it is a valid example where your friend's assumption is wrong.
Although the rules are complicated, the implementation of a game in a computer simulation should be very straightforward (if tedious). I would tend to think that @PeroK 's idea of simulating a large number of games is the more reliable method. Getting the program correct requires careful attention to following the rules correctly.
Well, while in PeroK's case in order to play a game, I basically have to randomly draws cards following the rules,
in my case I have to eliminate not-possible scenarios by finding rules broken by a given end table.
one way or another I wont get around checking every single rules for a game.
it's jsut either checking if all rules are followed. or if at least one rule is not followed.

@PeroK: Oh, I aint surprised by bugs at all.
you don't even want to know how often I had to go about all parts of my code (especially the one for checking if a rule was broken) and change it again and again fpr it to be in a state where I would call it "pretty likely to be right".

the baccarat rules by themselves feel like a made up random mess anyways, who comes up with soemthing so inconsistent? :O

well, either way, the "playing tons of games" idea is likely the easier one I guess.
I should probably make my life easier and jsut do some millions of games to just estimate the odds for such games.

## 1. What is the difference between the probability calculation for the Player and Banker in Baccarat?

The probability calculation for the Player and Banker in Baccarat is based on the number of possible outcomes and the likelihood of each outcome occurring. The main difference between the two is that the Banker has a slightly higher chance of winning due to the rules of the game.

## 2. Is there a "correct" probability calculation for Baccarat games?

There is no one "correct" probability calculation for Baccarat games. The probability of winning for the Player and Banker will always be slightly different due to the rules of the game. However, both calculations are based on statistical analysis and can be considered accurate.

## 3. How do you calculate the probability of a tie in Baccarat?

The probability of a tie in Baccarat is calculated by taking the total number of possible outcomes and dividing it by the number of possible ties. For example, if there are 100 possible outcomes and 10 possible ties, the probability of a tie would be 10/100 or 10%.

## 4. Can the probability of winning change throughout the course of a Baccarat game?

Yes, the probability of winning can change throughout a Baccarat game. This is because the probability is based on the current state of the game, including the cards that have been dealt and the remaining cards in the shoe. As the game progresses, the probability of winning may increase or decrease.

## 5. How important is understanding probability in Baccarat games?

Understanding probability is crucial in Baccarat games as it can help players make more informed decisions. By knowing the probability of certain outcomes, players can adjust their betting strategies and make more calculated risks. However, it is important to remember that probability is not a guarantee and there is always an element of chance in Baccarat games.

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