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December 10

Roulette

The numbers on a roulette wheel are not sequential. They run (on the French wheel):

0-32-15-19-4-21-2-25-17-34-6-27-13-36-11-30-8-23-10-5-24-16-33-1-20-14-31-9-22-18-29-7-28-12-35-3-26

and (on the American wheel):

0-28-9-26-30-11-7-20-32-17-5-22-34-15-3-24-36-13-1-00-27-10-25-29-12-8-19-31-18-6-21-33-16-4-23-35-14-2.

These are not random sequences - black and red numbers alternate according to the rule that even numbers between 2 and 10 and 20 and 28 (inclusive) are black. The French sequence was not standardised originally, so how did these sequences originate? Is there some underlying pattern and did the arrangement of the dominant manufacturer prevail as happened with the typewriter? How many different sequences can be generated within the constraint of the red/black rule? 92.27.49.50 (talk) 13:29, 10 December 2017 (UTC)[reply]

One of your questions is easy to answer. If you already have a rule that assigns red to half of the numbers and black to the other half, and have already constructed and painted your wheel, then there are 18! ways of numbering the black pockets and 18! ways of numbering the red pockets. See factorial and permutation. Thus the total number of different sequences which can be generated within the constraint of the red/black rule is the product of those two numbers, or (18!)² ≈ 4.1×10³¹ (a very big number). Without following a predetermined red/black rule (but still keeping the zeros where they are), then the total number of different sequences is 36! ≈ 3.7×10⁴¹ (an even bigger big number). -- ToE 16:26, 10 December 2017 (UTC)[reply]

The relevant section of our article, Roulette#Roulette wheel number sequence, gives the sequences but not their history. -- ToE 16:33, 10 December 2017 (UTC)[reply]

There is a pattern in the placement of high (19-36) and low (1-18). The French wheel alternates high, low, high, low, ... in either direction from the zero, meeting half-way across the wheel at two adjacent lows, 10 & 5. This, combined with the alternating red and black pockets, ensures that half of either color is high and half is low. The American wheel alternates once, then alternates in pairs, that is, H,L,H,H,L,L,H,H,L,L,..., clockwise around the wheel, starting from either zero. Since the pockets adjacent to a zero (either zero) are the same color on an American wheel, this pattern also ensures that half of either color is high and half is low. Since 18 is not divisible by 4, each half of the American wheel could not alternate in pairs while keeping the same number of high and lows on each half, hence the initial singleton alternation. -- ToE 17:03, 10 December 2017 (UTC)[reply]

Additionally, for the half of the American wheel clockwise from 0 to 00, the highs are all even and the lows are all odd, with the opposite the case for the other half of the wheel, and with the consecutive numbers 1&2, 3&4, etc. diametrically opposed on the wheel. -- ToE 17:47, 10 December 2017 (UTC)[reply]

Looking at the sequences, there appears to be an effort to make it so that the bets listed at Roulette#Outside bets don't cluster on one half of the wheel. I presume that this is to avoid the possibility of a skilled player being able to hit one side more often than the other. I wonder if either of the traditional sequences are actually optimized for this. Sounds like an interesting math puzzle. --Guy Macon (talk) 19:58, 10 December 2017 (UTC)[reply]

Detail correction: it's not the people betting who launch the ball, it's the croupier. But it doesn't matter, since no honest casino wants their croupiers to be able to influence who wins. --69.159.60.147 (talk) 05:34, 11 December 2017 (UTC)[reply]

I see what is almost one more pattern in the American wheel, but I don't quite know what to make of it. Since one side of the wheel determines the other, it is sufficient to look for patterns within and between the high/even and the low/odd sequences on the first half of the wheel, separating them as 28-26-30-20-32-22-34-24-36 and 9-11-7-17-5-15-3-13-1. For relative ordering of the first sequence, subtract 18 and divide by two, and for the second sequence, add one and divide by two, giving 5-4-6-1-7-2-8-3-9 and 5-6-4-9-3-8-2-7-1. If you cast out the 5 (which is from the singleton high/low alternation before they start alternating in pairs) then you get two quite similar sequences: 4-6-1-7-2-8-3-9 and 6-4-9-3-8-2-7-1. Rotate the first sequence left by two (i.e., take the 4-6 from the start and put it on the end), and you have the the reverse of the second sequence. This seems too significant to be mere coincidence, but not significant enough to have much meaning. The individual sequence 1-7-2-8-3-9-4-6 isn't quite broken at the end, as it comes in pairs of n, (n mod 4) + 6. This reflects the pattern in the alternating high, high and low, low pairs where within each pair there is one number from the upper half of the range and one from the lower, with the difference between them 12 except for when that take the higher number beyond its range in which case the difference is 4. For example, break the even highs into the lower highs, 20, 22, 24, & 26, the middle high 28, and the upper highs, 30, 32, 34, & 36. 28, the middle high comes first, then the pair 26 & 30, namely the fourth of the lower highs and the first of the upper highs, then 20 & 32, namely the first of the lower highs and the second of the upper highs, then 22 & 34, namely the second of the lower highs and the third of the upper highs, and so on. Using the symbols Even/Odd, High/Low, Upper/Middle/Lower, & numbers 1..4, we have EH:(M--L4-U1--L1-U2--L2-U3--L3-U4). The odd lows run OL:(M--U1-L4--U4-L3--U3-L2--U2-L1). (And the two are appropriately interleaved.) So there is a pattern there, with the two running in the opposite direction, but jumbled up just a bit at the start. -- ToE 20:03, 10 December 2017 (UTC)[reply]