I tried to calculate the probability of exactly any specific total to distinct numbers hitting and the math got messy quickly. The extreme cases close to 1 and 38 are simple, but it gets messier the closer you get to half of them. The way I approached it became a problem of partitions, which I think are known to be mathematically tedious.

To make an example, what would be a formula for the probability exactly 10 different numbers hit in 38 spins in double-zero roulette. I think you would have to look at the number of ways to partition 10 items.

Any thoughts from the math wizards of the forum?

Helpful links: List of first 49 partitions.

Which numbers won't hit?

Distinct Number | Probability |
---|---|

1 | 0.000000000 |

2 | 0.000000000 |

3 | 0.000000000 |

4 | 0.000000000 |

5 | 0.000000000 |

6 | 0.000000000 |

7 | 0.000000000 |

8 | 0.000000000 |

9 | 0.000000000 |

10 | 0.000000000 |

11 | 0.000000000 |

12 | 0.000000000 |

13 | 0.000000005 |

14 | 0.000000124 |

15 | 0.000001991 |

16 | 0.000022848 |

17 | 0.000191281 |

18 | 0.001186530 |

19 | 0.005519547 |

20 | 0.019434593 |

21 | 0.052152293 |

22 | 0.107159339 |

23 | 0.169042497 |

24 | 0.204864337 |

25 | 0.190490321 |

26 | 0.135436876 |

27 | 0.073211471 |

28 | 0.029838199 |

29 | 0.009063960 |

30 | 0.002020713 |

31 | 0.000323888 |

32 | 0.000036309 |

33 | 0.000002742 |

34 | 0.000000132 |

35 | 0.000000004 |

36 | 0.000000000 |

37 | 0.000000000 |

38 | 0.000000000 |

Total | 1.000000000 |

The expected number of distinct numbers hit is 24.20656478, which is 63.70149% of all 38 numbers. The percentage of numbers not hit is 36.29851%.

I think a better term than the "law of thirds" would be the "law of 1/e." In this case the 1/e estimate gives us 36.7879%, which is a lot closer to the actual probability than 1/3.

That is what I played around with quite a bit in the past:

The probability of any number not occuring within 38 rolls is 0.362985.

Multiplying by the 38 rolls this also means that on average a little less than 14 numbers will not be rolled.

The same reads for more occurrences:

Any chosen number will on average occur only once with a probability of 0.372796.

Within 38 rolls on average a little more than 14 numbers will be rolled only once.

Same for 2, 3, etc.

To support this result I also did a simulation in Excel:

1) 3800 random numbers -1 (represents "00") through 36, grouped into 100 trials of 38 numbers.

2) Pivot table counting the number of occurrences of the randomly chosen numbers within each trial.

3) Another Pivot table counting the counts of occurrences within each trial.

4) Adding the count of zero occurrances to that table for this result:

trial 0 1 2 3 4 5 6 plausibility

----------------------------------------------------------------------------

1 15 14 5 2 2 0 0 38

2 16 11 6 5 0 0 0 38

3 17 10 7 2 2 0 0 38

4 11 20 4 2 1 0 0 38

5 13 16 6 2 1 0 0 38

6 15 12 7 4 0 0 0 38

7 15 15 4 3 0 0 1 38

8 16 11 9 0 1 1 0 38

9 15 13 5 5 0 0 0 38

10 15 13 6 3 1 0 0 38

11 13 16 6 2 1 0 0 38

12 13 13 11 1 0 0 0 38

13 13 13 11 1 0 0 0 38

14 12 15 10 1 0 0 0 38

15 14 13 8 3 0 0 0 38

16 15 11 9 3 0 0 0 38

17 13 17 4 3 1 0 0 38

18 12 18 6 1 0 1 0 38

19 16 13 4 3 2 0 0 38

20 13 16 6 2 1 0 0 38

21 14 12 10 2 0 0 0 38

22 14 12 10 2 0 0 0 38

23 13 13 11 1 0 0 0 38

24 12 15 10 1 0 0 0 38

25 14 12 10 2 0 0 0 38

26 17 10 6 4 1 0 0 38

27 13 15 7 3 0 0 0 38

28 18 10 5 3 1 1 0 38

29 13 15 7 3 0 0 0 38

30 14 13 9 1 1 0 0 38

31 14 16 4 2 2 0 0 38

32 16 12 5 4 1 0 0 38

33 13 15 7 3 0 0 0 38

34 15 14 5 2 2 0 0 38

35 15 12 8 2 1 0 0 38

36 13 17 3 5 0 0 0 38

37 11 20 4 2 1 0 0 38

38 15 13 5 5 0 0 0 38

39 13 16 5 4 0 0 0 38

40 13 15 7 3 0 0 0 38

41 13 15 7 3 0 0 0 38

42 16 14 2 5 0 1 0 38

43 18 8 7 4 1 0 0 38

44 15 11 10 1 1 0 0 38

45 10 20 7 0 1 0 0 38

46 12 15 10 1 0 0 0 38

47 15 13 7 2 0 1 0 38

48 14 13 8 3 0 0 0 38

49 12 18 4 4 0 0 0 38

50 13 13 11 1 0 0 0 38

51 16 10 8 4 0 0 0 38

52 13 14 9 2 0 0 0 38

53 16 12 6 3 0 1 0 38

54 13 14 9 2 0 0 0 38

55 14 14 6 4 0 0 0 38

56 18 7 9 3 1 0 0 38

57 12 16 8 2 0 0 0 38

58 17 13 3 3 1 0 1 38

59 12 15 10 1 0 0 0 38

60 12 17 6 3 0 0 0 38

61 15 12 8 2 1 0 0 38

62 12 17 6 3 0 0 0 38

63 13 15 8 1 1 0 0 38

64 13 16 5 4 0 0 0 38

65 13 16 6 2 1 0 0 38

66 13 16 5 4 0 0 0 38

67 15 11 10 1 1 0 0 38

68 12 16 8 2 0 0 0 38

69 11 17 9 1 0 0 0 38

70 13 16 7 1 0 1 0 38

71 15 11 9 3 0 0 0 38

72 15 10 11 2 0 0 0 38

73 14 12 10 2 0 0 0 38

74 9 21 7 1 0 0 0 38

75 15 16 2 3 1 1 0 38

76 15 14 5 3 0 1 0 38

77 13 16 5 4 0 0 0 38

78 13 17 4 3 1 0 0 38

79 14 12 10 2 0 0 0 38

80 14 15 4 5 0 0 0 38

81 11 18 7 2 0 0 0 38

82 18 5 12 3 0 0 0 38

83 15 12 7 4 0 0 0 38

84 17 10 6 4 1 0 0 38

85 17 11 5 3 2 0 0 38

86 17 8 9 4 0 0 0 38

87 17 11 5 3 2 0 0 38

88 14 14 8 1 0 1 0 38

89 11 19 6 1 1 0 0 38

90 13 16 7 0 2 0 0 38

91 12 15 10 1 0 0 0 38

92 13 15 8 1 1 0 0 38

93 14 14 6 4 0 0 0 38

94 14 13 8 3 0 0 0 38

95 16 15 3 1 1 2 0 38

96 13 17 5 2 0 1 0 38

97 13 16 5 4 0 0 0 38

98 12 18 6 0 2 0 0 38

99 16 12 6 2 2 0 0 38

100 11 18 7 2 0 0 0 38

The simulated numbers of occurrence match the theoretical values quite well.

But that isn't the same as waiting for a streak of 5, and then placing the same bet on that there will not be a 6th.

I did win a 9 in a row later on and it had me debating whether I should stay at my last max bet or reduce to a new minimum bet. If there were 5 more winners, I could have won 5 more max level bets and made up for 2 or 3 lost sessions. Instead there were a dozen losers in a row and I had reduced to a new minimum bet and I wound up below where I started the 9 in a row at. I'm thinking I should switch from red to black or vice versa after winning 4 or more in a row of them after losing one. Then I get charts like the one below where switching sides will not pay off. The green 0's just punctuate that the winning streak on whichever side, ends here.

Those are not remotely the same bets.Quote:MDawgIf you could make a single even money bet that a given Baccarat shoe will not have a streak of 6 or more, you should do fine over time, making that same bet before each shuffled and ready to go shoe.

But that isn't the same as waiting for a streak of 5, and then placing the same bet on that there will not be a 6th.

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(A) an even money bet that a given Baccarat shoe will not have a streak of 6 or more

(B) an even money bet that a given Baccarat shoe will not have a streak of 7 or more

(B) an even money bet that a given Baccarat shoe will not have a streak of 8 or more

All very different probabilities.

(A) waiting for a streak of 5, and then placing the same bet on that there will not be a 6th.

(B) waiting for a streak of 6, and then placing the same bet on that there will not be a 7th.

(C) waiting for a streak of 7, and then placing the same bet on that there will not be a 8th.

Essentially the same probabilities except for some tiny EOR correlations.

Of course the bet becomes better if betting against that there will be a streak of 7, 8, 9, etc. in a given shoe, versus of course it doesn't matter if you're betting against the occurrence of the next in a sequence of streaks 7, 8 or 9.

For the binomial distribution the following parameters are known:

mean: n*p

variance: n*p*(1-p)

For the normal distribution the following parameters are needed:

mean

standard deviation (=sqrt(variance))

Putting all together and substituting

distinct numbers rolled = 38 - numbers not rolled

results in

hows about this for a system -

you flat bet either hi/lo or even/odd or red/black

you track the results and when it's tied such as 0-0 or 3-3 you either make a very small bet or just watch

when one side or the other goes up by 1 unit you bet on the side that's winning

your goal is for one of the even chances to go ahead by enough that you have won 2 bets (units) or down by 3 bets (units)

and then you start over doing the same thing, maybe with a different even chance to break up the monotony

it might not be a long run winner but it's kinna fun - I've been winning with it on the Wizard's free roulette game

perfect for a guy like me who doesn't wanna make real bets anymore for various reasons I won't bother to bore anyone with -

.

https://wizardofodds.com/play/roulette/

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https://wizardofodds.com/play/roulette/

Casino.org has free versions from many makers slots and table games.

Quote:ChumpChangeWell, it's a single 0 game, but the limits are $3 to $500 (same as the local casino roulette machine). Double 00 games are so bad and it's so hard to find single 0 games except in select high limit rooms. I may try the double 00 casino roulette machine for 1 to 8 hours and see if I can win some sessions. I think the plunger on it is rigged, but the ball still has to bounce around. If I find a single 0 machine, I'd expand the time trial out to 27 hours at 50 spins/hour.

https://wizardofodds.com/play/roulette/

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you could do the same thing on the free bacc games

(try to go up by 2 bets or down by 3 bets and then start over as in my post 2 up at 1:58)

lower house edge

Bank will go ahead more often but Player will go ahead often enough to make it interesting

.

https://wizardofodds.com/play/baccarat/

.

the premise was simple... how long ON AVERAGE does it take to repeat EACH number and most of them would go hundreds of numbers before repeating a second time... some would repeat more often than not

yup i have no clue what im doing when it comes to math or if that even did anything in general but i was satisfied i did anything like that in general

i knew some of the casino staff personally so they allowed me (probably laughing while i did) to use the phone while i was playing the game and to track the numbers

nothing ever came of it though

That's exactly what a lot of "roulette masters" recommend, lol. Or watch for 18 rolls and bet any number that came up twice.Quote:ChumpChangeI could bet the last 9 numbers that came up and hope for a repeat, idk.

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Quote:ChumpChangeI could bet the last 9 numbers that came up and hope for a repeat, idk.

link to original post

My experience with similar systems.

You watch, it works

You watch, it works

You watch, it works

You watch and actually wager, it doesn't work.

Baccarat has a built-in-house advantage, it's a negative number for the player. No betting system can change the HA and turn a -EV game into a +EV game.

Just show me the math.

Is baccarat countable? Perhaps there is some method where freely being able to use a computer to keep track can gain a slight advantage.

What's that advantage?

Is that advantage negated if the casino either cheats or doesn't pay?

My thinking is count tens. Deck is high in tens wager tie.

For every deck there are a multiplicity of tens (10, j, q, K) and when the deck is ten heavy ties will happen more often (esch side getting tens equals zero tie).

With ties paying 8:1 a few losses should be overcome.

This is just a theory on my part. Anyone ever looked into this?

The TLDR is that you want the ways to hit exactly 10 numbers, but you are overcounting the times you hit 9 numbers, which overcounts the times you hit 8 numbers, etc.

Unfortunately, I don't know how to embed images or latex so this will be extraordinarily messy. I apologize in advance.

The probability you hit exactly n numbers is:

(38 choose n) * (sum from (k = 1, to n) (k/38)^38 * (n choose k) * (-1)^(n-k))

You can double check that this gets you the exact same answers as your table.

It looks like one player bet $750 on the red 16 and another bet $2,750 when it hit on the 6th spin so they got $27K and $99K respectively on just that bet.

Dragon or side bets, I don't know if they have those and what protocols are in place to stop AP.Quote:darkozWouldn't the best counting system for baccarat be yhe tie bet?

My thinking is count tens. Deck is high in tens wager tie.

For every deck there are a multiplicity of tens (10, j, q, K) and when the deck is ten heavy ties will happen more often (esch side getting tens equals zero tie).

With ties paying 8:1 a few losses should be overcome.

This is just a theory on my part. Anyone ever looked into this?

link to original post

There is a video out there of a major offshore provider cheating, so who knows if you would have any advantage.

MyBookie posted a video of someone getting a 400k 777 of diamonds jackpot. Somthing seemed off to me, but who knows? I'm thinking...they could set that stuff up for a friend or the company to win easily if they wanted to.

Is there money to be made at online gambling? Absolutely, but I wouldn't want to be playing anything with a slim edge.

1) 1048572 random rolls = 27594 trials * 38 rolls per trial

2) Pivot table listing the indvidual numbers rolled per trial

3) Another Pivot table counting the individual numbers per trial

4) For each trial subtracting the count from 38 to determine the number of numbers not rolled

5) Yet another Pivot table counting the trials for each amount of numbers not rolled

6) Dividing the counts by 27594, the number of trials, to obtain the probabilities

The graph shows that the Wizard's figures (red) and the simulated random rolls (blue) match perfectly. My assumption of a normal distribution based on the parameters of mean and standard deviation of the binomial distribution (grey) obviously is not the correct answer to the question:

Quote:heatmapthe premise was simple... how long ON AVERAGE does it take to repeat EACH number and most of them would go hundreds of numbers before repeating a second time... some would repeat more often than not

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I answer this in my latest Ask the Wizard column.

Bubble Roulette here we come...Quote:ChumpChangeIf you can bet 30 numbers with 5 double streets and win them all 38 times you might be onto something. At $10 per street, you'd win $10 per spin, and win $380 in one round. Then you need to adjust your streets to figure out which new numbers won't hit in the next 38 spins.

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Quote:WizardSomebody asked me recently about the "Law of Thirds" in roulette. The premise in that 38 spins of roulette, about 1/3 of the numbers will never hit. My quick approximation is that in double-zero roulette about 36.3% will never hit.

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36.3% of 38 is 13.794 let's say 14

so, this is true if you were a wheel watcher and you watched 24 spins with no repeats then you could bet on all of those 24 numbers, expecting a repeat of one of them, in which case you would be paid 35/1 - a profit of almost 1/2

of course, this could not possibly be a winning system, but it sure "seems" like it should be

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