Roulette Law of Thirds
March 25th, 2024 at 8:48:26 AM
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Somebody 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.
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.
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.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
March 25th, 2024 at 8:56:52 AM
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Which numbers won't hit?
May the cards fall in your favor.
March 25th, 2024 at 10:16:11 AM
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It was not difficult to solve this recursively in Excel. Here is the probability of 1 to 38 distinct numbers in 38 spins.
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.
| 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.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
March 25th, 2024 at 10:19:34 AM
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I might be wrong, but isn't it simply about the binomial distribution?
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:
The simulated numbers of occurrence match the theoretical values quite well.
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.
"When it comes to probability and statistics, intuition is a bad advisor. Don't speculate. Calculate." - a math textbook author (name not recalled)
March 25th, 2024 at 12:02:24 PM
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If 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.
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 tell you it’s wonderful to be here, man. I don’t give a damn who wins or loses. It’s just wonderful to be here with you people.
https://wizardofvegas.com/forum/gambling/betting-systems/33908-the-adventures-of-mdawg/
March 25th, 2024 at 12:09:59 PM
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I was betting black and there were only 4 blacks out of 26 spins on this tote board. This wheel is about erasing my bankroll.
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.
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.
Last edited by: ChumpChange on Mar 25, 2024
March 25th, 2024 at 12:27:08 PM
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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.
link to original post
(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.
Gambling is a math contest where the score is tracked in dollars. Try not to get a negative score.
March 25th, 2024 at 12:39:57 PM
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You do have a tendency to say the same thing someone else just said, but in a very elongated roundabout way.
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.
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.
I tell you it’s wonderful to be here, man. I don’t give a damn who wins or loses. It’s just wonderful to be here with you people.
https://wizardofvegas.com/forum/gambling/betting-systems/33908-the-adventures-of-mdawg/
March 25th, 2024 at 1:08:05 PM
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Building on the result that on average with a probability of 0.362985 13.793435 numbers will not be rolled, one could assume a normal distribution for numbers not rolled.
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
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
"When it comes to probability and statistics, intuition is a bad advisor. Don't speculate. Calculate." - a math textbook author (name not recalled)
March 25th, 2024 at 1:58:32 PM
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.
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/
.
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/
.
Last edited by: lilredrooster on Mar 25, 2024
the foolish sayings of a rich man often pass for words of wisdom by the fools around him
March 25th, 2024 at 2:15:55 PM
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Well, 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/
https://wizardofodds.com/play/roulette/
March 25th, 2024 at 2:17:49 PM
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Rooster if you want to test things out on games from actual makers
Casino.org has free versions from many makers slots and table games.
Casino.org has free versions from many makers slots and table games.
March 25th, 2024 at 2:27:03 PM
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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/
link to original post
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 foolish sayings of a rich man often pass for words of wisdom by the fools around him
March 25th, 2024 at 3:03:10 PM
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my very first "tracking/predicting" algorithm (lol) was built to mimic the ballys digital roulette games
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
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
March 25th, 2024 at 3:07:00 PM
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I could bet the last 9 numbers that came up and hope for a repeat, idk.
March 25th, 2024 at 4:09:30 PM
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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.
link to original post
March 25th, 2024 at 4:39:13 PM
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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.
For Whom the bus tolls; The bus tolls for thee
March 25th, 2024 at 6:07:56 PM
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It's simple. You can't add up negative numbers correctly and get positive numbers.
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?
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?
♪♪Now you swear and kick and beg us That you're not a gamblin' man Then you find you're back in Vegas With a handle in your hand♪♪ Your black cards can make you money So you hide them when you're able In the land of casinos and money You must put them on the table♪♪ You go back Jack do it again roulette wheels turinin' 'round and 'round♪♪ You go back Jack do it again♪♪
March 25th, 2024 at 7:55:53 PM
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Wouldn'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?
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?
For Whom the bus tolls; The bus tolls for thee
March 25th, 2024 at 8:15:42 PM
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First post here, and I am not good at math by any means. But this is doable by PIE (principle inclusion exclusion). I prefer this to partitions.
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.
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.
March 25th, 2024 at 10:52:20 PM
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Played a 10 spin Roulette Tourney last night with a $25K buy-in on the home game. The virtual players are much more aggressive than me and I can't see what they bet, only what they win if I care to check the field for winning bets. One player turned $25K into $158K after 8 spins and another turned it into $211K. I was just $450 ahead hoping they'd go broke. The 3rd player was down to $11K. But the two leaders lost $27K on the 9th spin and the loser won $10K, I won $300. The loser went all-in on odd for the last spin and doubled up to $44K. The other 2 just lost around $5K. It doesn't show final tallies after the 10th spin, it just awards $25K to the the leader. That's basically how I get ahead on this software, the tourneys. Regular play is just so damn difficult and a downward spiral I have to rely on beating 3 other virtual players at the tourney for a grand prize equal to the buy-in. Whatever I lose or win during the tourney adds or subtracts from my bankroll, so I must be careful.
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.
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.
March 26th, 2024 at 4:32:21 AM
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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.
♪♪Now you swear and kick and beg us That you're not a gamblin' man Then you find you're back in Vegas With a handle in your hand♪♪ Your black cards can make you money So you hide them when you're able In the land of casinos and money You must put them on the table♪♪ You go back Jack do it again roulette wheels turinin' 'round and 'round♪♪ You go back Jack do it again♪♪
March 26th, 2024 at 8:33:56 AM
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In order to verify my conjecture of the normal distribution of numbers not rolled, I set up a simulation in Excel:
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:
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:
"When it comes to probability and statistics, intuition is a bad advisor. Don't speculate. Calculate." - a math textbook author (name not recalled)
March 26th, 2024 at 11:03:59 AM
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If 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.
March 26th, 2024 at 3:46:19 PM
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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
link to original post
I answer this in my latest Ask the Wizard column.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
March 26th, 2024 at 5:11:17 PM
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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.
link to original post
May 26th, 2024 at 6:58:06 AM
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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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the foolish sayings of a rich man often pass for words of wisdom by the fools around him
May 26th, 2024 at 8:01:20 AM
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I want to say that, as the size of the wheel approaches infinity, the fraction of numbers that are hit exactly N times, for all N in {0, 1, 2, ...}, approaches 1 / (e N!).
May 26th, 2024 at 12:23:03 PM
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aka Poisson
It’s all about making that GTA
