Ask The Wizard #397
What is the estimated number of spins required in double-zero roulette for every number to appear at least twice?
The button below provides further insights regarding single-zero, double-zero, and triple-zero roulette, focusing on the occurrence of numbers once, twice, or three times each.
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 will display the integral calculations for the nine scenarios outlined previously.
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 .
Can you explain the 'law of thirds' in relation to roulette?
The 'law of thirds' suggests that when spinning a roulette wheel for every number available, around one-third of those numbers will not show up.
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 table below illustrates the chances of seeing between 1 to 38 unique numbers during a session of 38 spins of double-zero roulette.
Understanding the Law of Thirds in 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 |
According to the table, the most common result is that 24 distinct numbers appear, with a probability of 20.49%. The average is approximately 24.20656478.
Some misguided individuals suggest that players should track the first nine unique results and then place bets on those numbers, believing they are more likely to repeat. This notion is fundamentally flawed! The roulette wheel and ball have no memory, ensuring that each number has an equal chance of coming up, regardless of prior results.
Imagine you're engaged in a board game with anywhere from three to five participants. Is it feasible to design a set of dice that ensures every possible order of play is equally probable, without any ties?
Here are the dice configurations suitable for a game with three players:
- Die #1: 3,4,9,10,13,18
- Die #2: 2,5,7,12,15,16
- Die #3: 1,6,8,11,14,17
To accommodate four players, I had to utilize 12-sided dice, organized 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 a scenario involving five players, the optimal solution I found involved using 840-sided dice, whose faces I will specify in the following. this post in my forum at Wizard of Vegas.
I explain how I created the dice in my previous statement. March 21, 2024 newsletter .