Percentages of events in baccarat
Thank you all!
-SilverCoin
Quote: SilverCoinHello, Wizard and all. I have a question about percentages of certain events occurring within 5 rounds of baccarat. I have a maximum of 5 bets with 6 different outcomes, and I stop after banker wins one time or player wins 5 times (not a progressive betting system - something completely different). What are the percentages that banker will appear on the first round, second round, third round, fourth round, and fifth round or that player will appear five times in a row? Can someone calculate these ignoring tie and also regarding tie? Also, since the fifth bet is the final bet, what are the percentages that the fifth outcome will be player or banker ignoring tie and regarding tie?
Thank you all!
-SilverCoin
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SilverCoin,
For 8D baccarat, the WoO gives these probabilities:
Banker: 0.458597
Player: 0.446246
Tie: 0.095156
IF WE NEGLECT TIES, these are the win probabilities:
Player: 0.446246/(1-0.095156) = 0.4931756192 = p
Banker: 0.5068243808 = (1-p)
You asked for the probabilities of six outcomes:
1. Banker wins #1: prob = (1-p) = 0.5068...
2. Player wins #1; Banker wins #2: prob = p*(1-p) = 0.2500...
3. Player wins #1&2; Banker wins #3: prob = p^2*(1-p) = 0.1233...
4. Player wins #1&2&3; Banker wins #4: prob = p^3*(1-p) = 0.0608...
5. Player wins #1&2&3&4; Banker wins #5: prob = p^4*(1-p) = 0.0300...
6. Player wins all 5 rounds: prob = p^5 = 0.0292...
I'll leave the calculations with ties included to someone else.
Hope this helps!
Dog Hand
First, we need to consider the basic probabilities in baccarat: the banker wins about 45.86% of the time, the player wins around 44.62% of the time, and ties occur approximately 9.52% of the time. Since you mentioned ignoring ties and considering ties separately, I'll break it down for both cases.
If we ignore ties, the combined probability of banker or player winning is 90.48%. The probability of the banker appearing on a specific round, say the first, would then be approximately 45.86% (since it's just the probability of a banker win). The probability of the player appearing five times in a row would be (44.62%)^5, which is about 2.52%.
Now, if we consider ties, then in each round, the chance of banker winning is 45.86%, the player is 44.62%, and the tie is 9.52%. To find the specific probabilities you're interested in, we would have to calculate all possible combinations of outcomes for each round, considering the effect of ties.
For the fifth and final bet, the probability that the outcome will be a banker or player win ignoring ties is straightforward: it's just the probability of either banker (45.86%) or player (44.62%) winning.
If you want to consider ties for this final bet, then the probabilities are still 45.86% for banker, 44.62% for player, and 9.52% for a tie.
These are just the basic probabilities. If you're looking for a detailed breakdown with all potential outcomes across the rounds, that would require a more complex calculation considering all possible paths through the rounds, which could be done step by step or using a probabilistic model. If you need help with those detailed calculations, let me know!
Quote: SilverCoinOver the course of enough time, will I expect to see each result occur 1/6th of the time +/- the house edge?
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SilverCoin,
NO!
If you try 10,000 times, you should expect the following approximate results:
1. Banker wins #1: 5,068
2. Player wins #1; Banker wins #2: 2,500
3. Player wins #1&2; Banker wins #3: 1,233
4. Player wins #1&2&3; Banker wins #4: 608
5. Player wins #1&2&3&4; Banker wins #5: 300
6. Player wins all 5 rounds: 292
These total 10,001 because of round-off error.
At any rate, the probable outcomes are nowhere near 1,667 each.
Perhaps I misunderstood your question?
Dog Hand
-SilverCoin
Quote: SilverCoinNo. You definitely understood my question, and I thank you very much. I only use the 4 basic operations of math for all of my plays. I'm not good when it comes to more advanced math. I was just thinking that over the course of my career, BXXXX, PBXXX, PPBXX, PPPBX, PPPPB, and PPPPP would show up at roughly even amounts +/- the house edge. But I can do the calculations with what you've provided. I apologize if I'm bad at explaining it. I know what I'm trying to understand, but I don't know how to explain it efficiently, and I also don't want to expose my plays until I'm finished with my paper.
-SilverCoin
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Without taking the house edge into account, think of baccarat as a coin toss - heads, the player wins; tails, the banker wins.
There are 32 different possible sets of five results: PPPPP, PPPPB, PPPBP, PPPBB, PPBPP, and so on through BBBBB.
16 of them have a banker win on the first hand.
Eight have a player win followed by a banker win.
Four (PPBBB, PPBBP, PPBPB, PPBBB) have two player wins followed by a banker win.
Two (PPPBB, PPPBP) have three player wins followed by a banker win.
One (PPPPB) has four player wins followed by a banker win.
One (PPPPP) has five player wins.
Since, in this case, each set of results is equally likely, you are twice as likely to have BXXXX as PBXXX, twice as likely to have PBXXX as PPBXX, twice as likely to have PPBXX as PPPBX, twice as likely to have PPPBX as PPPPB, and just as likely to have PPPPB as PPPPP.
Quote: SilverCoinI'm just going to ignore all of the possible combinations of 5 because my play will not always have a combination of 5. I was just thinking that if I count all of the possibilities as combinations of 5, then I should see them all in roughly equal amounts over the course of my career, hence the math would be easier, but I want to be as accurate as possible.
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That’s not right because there are not 5 combinations.
BXXXX is twice as likely as PBXXX. They are not equally likely so cannot be compared in the way you are.
Don explains it above. There are 32 combinations.
