Baccarat Accumulation Tournament Strategy and Calculations

Discussion in 'Baccarat Tournament Strategy' started by KenSmith, Oct 10, 2013.

  1. KenSmith

    KenSmith Administrator Staff Member

    Note: This discussion began as part of the thread about a specific event at the Venetian in Las Vegas. You can see that thread here:

    A reasonable target goal in that event was to turn a starting bankroll of $50,000 into $450,000 in 10 hands, with a max bet of $50,000 on player or banker....

    Assumption 1: The tie bet is limited to $6250 or less making it not useful for getting big results.
    Assumption 2: There is no commission on banker bets.

    We need a net win of 8 max bets in 10 hands to get to $450,000 ending bankroll.
    I did some quick spreadsheet work and came up with an estimate of success 1 in 138.1 tries.
    At a $50 buyin, that's a cost of $6905.
    Assuming $450k will get you in the top four, your avg daily win is $6250.
    Then I guess you get one free entry into the Ruby session worth another $250.
    Still short.

    I took a lot of shortcuts with my math to get to 138.1, so if anyone works it out exactly I would be curious to see how close I am to accurate.
    Last edited: Oct 20, 2013
  2. gronbog

    gronbog Top Member

    Both assumptions are correct. Turns out the tie bet was limited to 5k at 8 to 1.
  3. S. Yama

    S. Yama Active Member

    The fast way to calculate chances of turning $50K to $450K with 1K to 50K bets, assuming that we always bet max bet on Bank is to use help of some internet binomial calculators (for example
    We know that if we win 9 Banks we can lose one Player, so probability of success on a single trial (Bank) is 0.4586, number of trials is 10, number of success is >=9, cumulative probability is 0.00527.
    We also have to add chances for 8 wins and two Ties. This can be done by calculating a single chance for this to occur, (chances of Bank to the eight power times chances of Tie squared), and since there are 28 possible combinations of this ( TTBB.., TBTB.., B...BTBT, B...BTT) times 28. It adds up to 0.00236
    Total chances are 0.0076, about once in 130 trials.

    However, since $450K is such a high score ending up with 448K to 452K has similar (unknown) chance to advance. The better strategy would be to stop betting max once we had eight wins in a row (1.96%) and bet minimum, so we end up plus/minus two minimum bets around the goal of $450K.
    To calculating chance for this strategy I would back up to situations with two bets left.
    To have chances of reaching the goal, after playing eight rounds we have to be in one of the four states:
    Winning 8 Banks, winning 7 Banks and 1 Tie or 1 Player, winning 6 Banks and 2 Ties.
    With 8 Banks we reached the goal. With 7 Banks and 1 Tie we need to have in the next two rounds 1 Bank and 1 Tie. Chances of Bank to the seventh power time chances of Tie times 8 (combinations) times the desired outcome of the last two hands (Bank times Tie times 2 [combinations]). This adds up to 0.00325
    In a similar manner we can calculate success for 7B1P and 6B2T.
    The total chance of reaching $450K plus/minus two minimum bets is 2.56%. This happens about once every 39 trials.
    I think it was a nice shot at a positive EV.
    Haven’t seen the daily scores but I think there was interesting play if you happened to win eight first rounds.

    S. Yama
  4. KenSmith

    KenSmith Administrator Staff Member

    S. Yama, I think you were in more of a hurry than me with your calculations. :p

    Your first paragraph describes one approach, but has two issues...
    Problem 1 is that you can't afford for the first hand to be Player. So you actually need the chance of 1 Banker, followed by any combination of 8 Banker and 1 Player hands.
    Problem 2 is the number of combinations of 8B2T. It's not 28 combinations, but 36 instead.

    Your second paragraph starts over with a different approach, with a different issue:
    For the 7B1T start, your desired last two hands are actually TB and Bx. You don't care about the final hand if the ninth hand is Banker.
    However, the 0.00325 you mention was just 7B1T * 8 regardless, without the effect of the last two hands.

    I'll post my revised work next.
    Last edited: Oct 15, 2013
  5. KenSmith

    KenSmith Administrator Staff Member

    I wish I had saved my original scratchpad spreadsheet, because now I don't remember what shortcuts I took to arrive at 1 in 138.1.
    That number appears close, but not accurate. I think the accurate number is 1 in 130.1.

    I believe my improved work is correct, but feel free to critique.

    I looked at 4 ways to get there:
    p(B) = 0.458597
    p(P) = 0.446247
    p(T) = 0.095156
                                 Ways   Total
    (8B)xx      0.0019563641     1     0.0019563641
    (7B1T)Bx    0.0001861598     8     0.0014892782
    (7B2T)B     0.0000177142     36    0.0006377119
    B(8B1P)     0.0004003651     9     0.0036032857
    Sum 0.0076866399  or 1 in 130.1 tries
    The parentheses indicate the group can be in any order.
    For example (7B1T) means 7 bankers and 1 tie in any order.
  6. S. Yama

    S. Yama Active Member

    Thanks for looking at it again, Ken.
    I am often in a hurry (which may have affected my calculations, and now am traveling. Will be back to check it again in a couple of weeks.
    S. Yama
  7. gronbog

    gronbog Top Member

    This comment caught my eye. Would this play be to bet minimum (100) on tie for the final two hands, thus making the lowest possible finishing score 449,800 (vs 448,000 for banker/player)? The possible high score would be reduced to 451,600 (vs 452,000 for banker/player)
  8. hopinglarry

    hopinglarry Top Member

    I look at this slightly differently. Using the binomial calculator there are 3 different scenarios that have at least 8 net wins assuming always betting bank.
    P(10) wins - .000411453
    P(9) wins - .004857374, however the other hand is going to be a player or tie, there are 20 of these combinations one of which is a player win on first hand. So I think we can only use 95% of that number which = .0046145053
    The only other possibility is 8 banks and 2 ties.
    P(8) wins - .02580498 times the probability of 2 ties - .095156 squared = .000233655 . There are 45 ways to arrange this so times 45 which = .0105145.
    Adding.000411453 + .0046145053 + .0105145 = .01554 probability which translate into 1 in 64.55.

    Open to comments.
  9. KenSmith

    KenSmith Administrator Staff Member

    This isn't quite right because the tie and player outcomes aren't equally likely in the 20 combinations.
    The actual 9-win success probability is a little lower at 0.0044570089.
    Here are two different ways to correct that part of your calculation:

    1) Correct your 95% constant to be the real percentage of the 20 cases we are OK.
    Of the 20 cases, B/(P+B) divided by 10 is the percentage chance of the one miss being a first-hand Player. (c)
    The correct percentage is therefore 1 - 0.0044570089 = 91.75758169%
    So, 0.004857374 * 91.75758169% = 0.0044570089

    2) The easier method: Just use the raw binomial p(9 out of 10 are banker) number, and subtract out the one problematic case:
    0.004857374 - p(Player followed by 9 bankers) = 0.004857374 - 0.0004003651 = 0.0044570089

    OK, that fixes the problem with the p(9) part of your process. The p(8) part will have to wait until after lunch. :)
  10. hopinglarry

    hopinglarry Top Member

    I actually had thought of this after I made the post, but being a good person with a working spouse, I was diligently cleaning house and just finished and was going to compute this and edit the post. I was going to use your method 1 above. Thanks for computing it for me so I didn't have to.
  11. hopinglarry

    hopinglarry Top Member

    Just FYI, the chain of events that made me post above shows how our lives get molded.
    A mechanic didn't finish working on my wife's car yesterday and had to keep it overnight. She took my car to work today.
    If I had my car, I would have gone to the casino to probably lose 2K on 6K-10K hands of VP. Could have been a good thing that I didn't go:)
    Since I was home, I decided to think about the problem before cleaning house.
    So the mechanic put you and me to work on this issue. I am not sure I would have attempted the computation without this scenario.

    (add this FYI, since I had my car back on Thursday I went to the casino and hit a 5778 royal so I guess it was a good thing the mechanic kept my wife's car:p
    Last edited: Oct 18, 2013
  12. KenSmith

    KenSmith Administrator Staff Member

    Or, today could have been the day you hit more than one royal. ;)

    Back to your calculations in the P(8) section:
    The main problem is that this statement isn't true: "The only other possibility is 8 banks and 2 ties."
    There are actually 3 ways to get there with 8 bank outcomes:
    (a) 8 banks and 2 ties in any order
    (b) 8 banks, 1 tie, and 1 player, as long as all 8 banks occur before the player outcome. (You'll be betting the minimum.)
    (c) 8 banks and 2 player outcomes, as long as we get our 8 bank outcomes in the first 8 hands.

    Your attempt to calculate (a) was also a problem.
    The binomal stat for 8 wins also includes those times that one or both of the two misses were a player outcome.
    Multiplying it by the probability of two tie outcomes is an apples and oranges problem.

    Instead, we can figure the (a) 8B2T trials without the binomial stat this way:
    (a) = B^8 * T^2 * 45 ways = 0.0007971399

    Now for (b), where we have 1 player outcome:
    Let's break it into two parts... If the sole player outcome happens on hand 9 (after 8 banks), hand 10 must be the tie.
    So, (b1) = B^8 * P * T = 0.0000830732425

    If the sole player outcome is the final hand, we need any combination of 8 banks and 1 tie, followed by player:
    (b2) = B^8 * T * 9 ways * P = 0.0007476592

    Hmm, I see that (b1) and (b2) are the same formula with a factor of 1 in (b1) and a factor of 9 in (b2).
    So we can simplify to
    (b) = (b1) + (b2) = B^8 * P * T * 10 = 0.0008307324
    (c) is an easy one: (c) = B^8 * P^2 = 0.0003895833
    Add it all up: P(8) = (a) + (b) + (c) = 0.0007971399 + 0.0008307324 + 0.0003895833 = 0.0020174556

    However, when I add up P(8) + P(9) + P(10) I don't get the expected match to my method's answer.
    There has to be another problem here somewhere.
    I'll be thinking about that while I'm running a few errands.
  13. KenSmith

    KenSmith Administrator Staff Member

    Urgh. Whose idea was this anyway? :)

    Here is the bottom line from the post 18 method, which I do not see any fault with (at least yet!):
    P(8) + P(9) + P(10) = 0.0068859098 which is 1 in 145.2 trials.

    That doesn't match my result in post 11 above, but I see a problem with post 11.
    In that result, I inadvertently counted some combinations more than once.
    The B(8B1P) line overlaps some of the 8Bxx line. Namely, 8BPB and 9BP.
    But subtracting those two combinations from post 11's result does not make the methods match, so I'm still missing something.
  14. KenSmith

    KenSmith Administrator Staff Member

    Ding, ding, ding. We have a winner.
    Subtracting the two mentioned overlaps in the post 11 method DOES match exactly the P(8,9,10) method from post 18.
    (My previous claim that they didn't match was due to a spreadsheet mistake.)

    Since these two methods both arrive at the same answer, I have some degree of confidence that it's finally correct.
    And that answer is 1 in 145.2 trials.
  15. hopinglarry

    hopinglarry Top Member

    Thanks for correcting my thinking I might even remember the process the next time I look at something.
    Well I hope you are proud of me for doing something the wrong way and leading you down the path to the right answer:)
    I guess we can blame the mechanic:)
    KenSmith likes this.
  16. gronbog

    gronbog Top Member

    Seeing the complexities that can arise when approaching probabilistic calculations reminds me of why I prefer to use simulation :)
    The following sim, confirms Ken's result:

    Simulating 10000000 sessions of 10 hands of Baccarat
    Bankroll: 50000, Goal: 450000, Max: 50000, Min 1000
    Best session result: 450000
    Worst session result: 0
    Distribution of Session Results
    Result 0 achieved 7365069 times = 73.65%
    Result 50000 achieved 258121 times = 2.58%
    Result 100000 achieved 379706 times = 3.80%
    Result 150000 achieved 551534 times = 5.52%
    Result 200000 achieved 438431 times = 4.38%
    Result 250000 achieved 443123 times = 4.43%
    Result 300000 achieved 244151 times = 2.44%
    Result 350000 achieved 188413 times = 1.88%
    Result 400000 achieved 62704 times = 0.63%
    Result 450000 achieved 68748 times = 0.69%

    Goal of 450000 reached 68748 times = 0.69%
    Max number of hands to reach goal = 8032
    Min number of hands to reach goal = 8
    Average number of hands to reach goal = 641.48
    Max number of sessions to reach goal = 1861
    Min number of sessions to reach goal = 1
    Average number of sessions to reach goal = 145.46
    KenSmith and hopinglarry like this.
  17. tirle_bj

    tirle_bj Member

    Here's the method to use for exact probability:
    Since there is no commission on banker bets, our best bet will be max ($50,000) on "Banker", and then if we reached the goal (+8 bets) to lower bet to the min. So our goal is $450,000+$2,000.
    Lets consider all "good" tipe of sequences:
    10 Bankers 1 sequence, probability is (.4586)^10(tenth power)= .00041
    9 Bankers 1Player 9 sequences(1 hand cannot be Player), probability 9 x (.4586)^9 x (.4462) = .0036
    8 Bankers 2Ties 45 sequences, probability 45 x (.4586)^8 x (.0952)^2 = .0008
    9 Bankers 1Tie 10 sequences, probability 10 x (.0952) x (.4586)^9 = .00085
    First 8 Bankers 2 Players 1 sequence, probability (.4586)^8 x (.4462)^2 = .00039
    8 Bankers 1 Player 1 Tie 10 sequences (in this case we should lose Min bet on Player on next to last hand after 8
    wins on Banker, or on last hand after 8 Bankers and 1 Tie), probability
    10 x (.4586)^8 x (.0952) x (.4462) = .00083
    Total probability to obtain the Goal of $450,000+ $2,000 in 10 hands is (.00688), which is 688 out of 100,000 ~ 145.3
    So, Ken, it's even worse than your result, but not to far from it (within 5%)
    KenSmith likes this.
  18. KenSmith

    KenSmith Administrator Staff Member

    As usual, tirle's work is excellent and he arrives at the answer in one post when I required a dozen! Thanks.

    And, I agree with gronbog that this is easier to do with a sim! You don't even need to program it with cards at all; just use the P,B,T probabilities directly.
    Thanks g
    gronbog likes this.
  19. gronbog

    gronbog Top Member

    Thanks for the thanks Ken ;)

    Regarding the simulation of cards, I do go the extra mile and simulate the handling of the cards. That way, if there are any additional nuances associated with that (e.g. the cut card effect in black jack), they are automatically accounted for.
  20. PlayHunter

    PlayHunter Active Member

    Still in this area, how long of a shot is to catch two ties in a row ? (is it about 1 in 83 shots ?)

    What I meant to ask in fact, is to find out what is the best way to achieve a 81000 BR, starting with 1000 for a tournament where each player can play up to 50 hands ? Max bet 10000 on either banker, player or tie. (banker 5% commission taken, and tie bet payout 9 to 1)

    - Is it best to bet all-in on tie until 2 ties in a row occur or there can be something better ? (like betting on banker until 4 (or 3?) wins in a row occur, and after that betting 9000 on tie ? But if there would not be the 10000 max limit, but 100000 (or no limit instead) ? (having the same 81000 goal)
    Last edited: Feb 1, 2014

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