Ask The Wizard #397
On average, how many spins would it take in double-zero roulette for every number to occur at least twice?
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: 155.458690At 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?
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?
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.