Kurt Von Haller Roulette Lexikon

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Kurt Von Haller Roulette Lexikon Pdf

I bet Kurt Von Haller made more money selling his book than he did playing roulette

Dec 19, 2018 Hello, I am new in this forum and would like to ask a question regarding European Roulette. (Single Zero Roulette) In the Roulette Lexikon from Kurt von Haller he writes, that the last open number should show up on average after 132 spins. Von Kurt von Haller (Autor) Alle Formate und Ausgaben anzeigen Andere Formate und Ausgaben ausblenden. Preis Neu ab. Zugleich Lehrbuch und Tabellenwerk der Wahrscheinlichkeitsmathematik des Rouletts 5,0 von 5 Sternen 2. Gebundene Ausgabe.

If he made $1 selling books, then this statement is true. Come to think of it, even if he lost money on his book it is probably still true.

Beware, I work for the dark side.... We have cookies

Wizard
Administrator

If the question is how long will it take for every number to appear in double-zero roulette, the answer is 160.6602765, on average. This is the sum of the inverse of every integer from 1 to 38.

It's not whether you win or lose; it's whether or not you had a good bet.

mustangsally

If the question is how long will it take for every number to appear in double-zero roulette, the answer is 160.6602765, on average. This is the sum of the inverse of every integer from 1 to 38.

the OP questions 37. Where it came from
that is just 1/p where p=1/37
handy tables
0 Roulette (155.4586903)

# of numbers	average # of spins	cumulative sum
1	1	1
2	1.027777778	2.027777778
3	1.057142857	3.084920635
4	1.088235294	4.173155929
5	1.121212121	5.29436805
6	1.15625	6.45061805
7	1.193548387	7.644166437
8	1.233333333	8.877499771
9	1.275862069	10.15336184
10	1.321428571	11.47479041
11	1.37037037	12.84516078
12	1.423076923	14.2682377
13	1.48	15.7482377
14	1.541666667	17.28990437
15	1.608695652	18.89860002
16	1.681818182	20.58041821
17	1.761904762	22.34232297
18	1.85	24.19232297
19	1.947368421	26.13969139
20	2.055555556	28.19524694
21	2.176470588	30.37171753
22	2.3125	32.68421753
23	2.466666667	35.1508842
24	2.642857143	37.79374134
25	2.846153846	40.63989519
26	3.083333333	43.72322852
27	3.363636364	47.08686488
28	3.7	50.78686488
29	4.111111111	54.897976
30	4.625	59.522976
31	5.285714286	64.80869028
32	6.166666667	70.97535695
33	7.4	78.37535695
34	9.25	87.62535695
35	12.33333333	99.95869028
36	18.5	118.4586903
37	37	155.4586903

The average is not the mode (The 'mode' is the value that occurs most often)
or median (The 'median' is the 'middle' value or close to 50%)
median = spin 147 @ 0.501522154
mode = 133 @ 0.0106293156
00 Roulette (160.6602765)

# of numbers	average # of spins	cumulative sum
1	1	1
2	1.027027027	2.027027027
3	1.055555556	3.082582583
4	1.085714286	4.168296868
5	1.117647059	5.285943927
6	1.151515152	6.437459079
7	1.1875	7.624959079
8	1.225806452	8.85076553
9	1.266666667	10.1174322
10	1.310344828	11.42777702
11	1.357142857	12.78491988
12	1.407407407	14.19232729
13	1.461538462	15.65386575
14	1.52	17.17386575
15	1.583333333	18.75719908
16	1.652173913	20.409373
17	1.727272727	22.13664572
18	1.80952381	23.94616953
19	1.9	25.84616953
20	2	27.84616953
21	2.111111111	29.95728064
22	2.235294118	32.19257476
23	2.375	34.56757476
24	2.533333333	37.1009081
25	2.714285714	39.81519381
26	2.923076923	42.73827073
27	3.166666667	45.9049374
28	3.454545455	49.35948285
29	3.8	53.15948285
30	4.222222222	57.38170508
31	4.75	62.13170508
32	5.428571429	67.56027651
33	6.333333333	73.89360984
34	7.6	81.49360984
35	9.5	90.99360984
36	12.66666667	103.6602765
37	19	122.6602765
38	38	160.6602765

Kurt Von Haller Roulette Lexikon Pdf

median = spin 152 @ 0.501599171
mode = 138 @ 0.010333952
still interesting one brings up this question
Sally

I Heart Vi Hart

TomG

If the question is how long will it take for every number to appear in double-zero roulette, the answer is 160.6602765, on average. This is the sum of the inverse of every integer from 1 to 38.

The sum of the inverse of every integer from 1 to 38 is 4.2279020133. Using the inverse of every integer, it would take like 10^70 of them to get to 160. (160.66 is the harmonic series up to 38 times 38; using 37 for single zero roulette it is is 155)
(I learned this one when calculating the expected longest drought for a Super Bowl or World Series)

mustangsally

(160.66 is the harmonic series up to 38 times 38; using 37 for single zero roulette it is is 155)

being specific using pari/gp calculator found here
https://pari.math.u-bordeaux.fr/gp.html
a=sum(k=1,37,37/(37-(k-1)))
0 Roulette
00 Roulette
this is the short way to an answer
Sally

I Heart Vi Hart

Wizard
Administrator

You're right. I forgot to say to multiply by 38.

Last edited by: Wizard on Dec 20, 2018

It's not whether you win or lose; it's whether or not you had a good bet.

masterj

Thanks guys!

Wizard
Administrator

the OP questions 37. Where it came from
that is just 1/p where p=1/37
The average is not the mode (The 'mode' is the value that occurs most often)
or median (The 'median' is the 'middle' value or close to 50%)
median = spin 147 @ 0.501522154
mode = 133 @ 0.0106293156

I agree on the median. Here is my transition matrix.

0.027	0.973	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0
0	0.0541	0.9459	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0
0	0	0.0811	0.9189	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0
0	0	0	0.1081	0.8919	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0
0	0	0	0	0.1351	0.8649	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0
0	0	0	0	0	0.1622	0.8378	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0
0	0	0	0	0	0	0.1892	0.8108	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0
0	0	0	0	0	0	0	0.2162	0.7838	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0
0	0	0	0	0	0	0	0	0.2432	0.7568	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0
0	0	0	0	0	0	0	0	0	0.2703	0.7297	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0
0	0	0	0	0	0	0	0	0	0	0.2973	0.7027	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0
0	0	0	0	0	0	0	0	0	0	0	0.3243	0.6757	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0
0	0	0	0	0	0	0	0	0	0	0	0	0.3514	0.6486	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0
0	0	0	0	0	0	0	0	0	0	0	0	0	0.3784	0.6216	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0
0	0	0	0	0	0	0	0	0	0	0	0	0	0	0.4054	0.5946	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0
0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0.4324	0.5676	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0
0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0.4595	0.5405	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0
0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0.4865	0.5135	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0
0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0.5135	0.4865	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0
0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0.5405	0.4595	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0
0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0.5676	0.4324	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0
0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0.5946	0.4054	0	0	0	0	0	0	0	0	0	0	0	0	0	0
0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0.6216	0.3784	0	0	0	0	0	0	0	0	0	0	0	0	0
0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0.6486	0.3514	0	0	0	0	0	0	0	0	0	0	0	0
0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0.6757	0.3243	0	0	0	0	0	0	0	0	0	0	0
0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0.7027	0.2973	0	0	0	0	0	0	0	0	0	0
0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0.7297	0.2703	0	0	0	0	0	0	0	0	0
0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0.7568	0.2432	0	0	0	0	0	0	0	0
0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0.7838	0.2162	0	0	0	0	0	0	0
0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0.8108	0.1892	0	0	0	0	0	0
0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0.8378	0.1622	0	0	0	0	0
0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0.8649	0.1351	0	0	0	0
0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0.8919	0.1081	0	0	0
0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0.9189	0.0811	0	0
0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0.9459	0.0541	0
0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0.973	0.027
0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	1

If the table is too big, cell (x,x) = x/37 and cell (x,x+1) = (37-x)/37), and every other cell is zero.

It's not whether you win or lose; it's whether or not you had a good bet.

7craps

I agree on the median. Here is my transition matrix.

to point out little differences in building a transition matrix (both are correct, one needs some special attention after calculations)
The Wizard's transition matrix (where rows sum to 1) starts with the 1st spin already completed.
I was taught to always start at 0 to make sure one does not forget to add 1 to calculations of the matrix.
(both methods are perfectly fine to use)
My TM is A, the Wizards is B (in the photo below - Wizard's values have been rounded down)
the column 1,2,3,4 is the row number and not the 'state name' for Matrix A but is correct for Matrix B
(Matrix A row names is just the row value - 1)
after raising the Wizards TM to the 146th power
(in the photo below)
we find the median (0.50141 - values have been rounded)
we must add 1 to 146 = 147 for the median (for the example 37 number Roulette)
distribution to only 160 spins
using R code section 3r.
https://sites.google.com/view/krapstuff/coupon-collecting

remember, we can only raise a square matrix to a power (as in B^146)
Enjoy

winsome johnny (not Win some johnny)

ThatDonGuy

How about a math proof?

Assume there are N numbers, and K of them have already come up at least once.
This is equivalent to, 'If you have N balls, K of which are red and the other N-K are white, how many draws with replacement (i.e. when you draw a ball, you put it back) should it take before you draw a white ball?'
The probability of doing it in exactly D draws is (K / N)^D-1 x (N - K) / N
= (K^D-1 (N - K)) / N^D
The expected number is
1 x (N - K) / N
+ 2 x K (N - K) / N²
+ 3 x K² (N - K) / N³
+ 4 x K³ (N - K) / N⁴
+ ...
= (N - K) / N x (1 + 2 (K / N) + 3 (K / N)² + 4 (K / N)³ + ...)
= (N - K) / N x (1 + (K / N) + (K / N)² + (K / N)³ + ...)²
= (N - K) / N x (1 / (1 - (K / N))², since K < N
= (N - K) / N x (1 / ((N - K) / N))²
= (N - K) / N x (N / (N - K))²
= (N - K) / N x N² / (N - K)²
= N / (N - K)
At the start, K = 0; after each number is drawn for the first time, K increases by 1.
The total number is the number needed to get the first number + the number to get the
second different number once you have already drawn one + the number needed to get the
third different number once you have already drawn two different numbers + ... + the number
needed to get the Nth different number once you have already drawn N-1 different numbers
This is is N / N + N / (N-1) + N / (N-2) + ... + N / 2 + N
= N x (1 / N + 1 / (N-1) + ... + 1 / 3 + 1 / 2 + 1)