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Ask The Wizard #397

On average, how many spins would it take in double-zero roulette for every number to occur at least twice?

heatmap

Approximately 234.832663 spins

 

Here is my solution (PDF).

 

The following button shows additional answers for single-zero, double-zero and triple-zero roulette for requiring at least one, two and three occurrences of each number.

 

Single-Zero Roulette:

At least once: 155.458690
At least twice: 227.513340
At least thrice: 290.543597

Double-Zero Roulette:

At least once: 160.660277
At least twice: 234.832663
At least thrice: 298.396127

Triple-Zero Roulette:

At least once: 165.888179
At least twice: 242.181868
At least thrice: 308.880287

 

The next button shows integrals for the nine situations mentioned above.

Once 0: 1-(1-exp(-x/37))^37

00: 1-(1-exp(-x/38))^38

000: 1-(1-exp(-x/39))^39

Twice

0: 1-(1-exp(-x/37)*(1+x/37))^37

00: 1-(1-exp(-x/38)*(1+x/38))^38

000: 1-(1-exp(-x/39)*(1+x/39))^39

Thrice

0: 1-(1-exp(-x/37)*(1+x/37+x^2/2738))^37

00: 1-(1-exp(-x/38)*(1+x/38+x^2/2888))^38

000: 1-(1-exp(-x/39)*(1+x/39+x^2/3042))^39

 

Here is my recommended integral calculator.

What is the "law of thirds" in roulette?

anonymous

The "law of thirds" says that if you spin a roulette wheel once for every number on the wheel, about 1/3 of the numbers will never occur.

1/3 is really a pretty poor estimate. A much better one would be 1/e =~ 36.79%. The true percentage, in double-zero roulette, is 36.30%.

The following table shows the probability of 1 to 38 distinct numbers being observed in 38 spins of double-zero roulette.

Law of Thirds -- Double-Zero Roulette

Distinct
Numbers		 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 table shows the most likely outcome is 24 distinct numbers at 20.49%. The average is 24.20656478.

Some quacks make the argument that the player should observe the first nine distinct outcomes and then bet on them, under the incorrect belief that they are more likely to occur than other numbers. This is strictly not true! The wheel and ball do not have a memory. On a fair wheel, every number is equally likely and the past does not matter.

Suppose you are playing a board game with three to five players. Is it possible to construct a set of dice to determine the order of play with every order being equally likely and no chance of a tie?

anonymous

Here are dice for the three-player case:

  • Die #1: 3,4,9,10,13,18
  • Die #2: 2,5,7,12,15,16
  • Die #3: 1,6,8,11,14,17

For four players, I had to go up to 12-sided dice, as follows:

  • Die #1: 5,6,11,12,15,20,31,32,37,38,41,46
  • Die #2: 4,7,9,14,17,18,30,33,35,40,43,44
  • Die #3: 3,8,10,13,16,19,29,34,36,39,42,45
  • Die #4: 1,2,21,22,23,24,25,26,27,28,47,48

For five players, the best I could do is 840-sided dice. I indicate their faces in this post in my forum at Wizard of Vegas.

I go through how I arrived at the dice in my March 21, 2024 newsletter.