The sum of all nummers in the American Roulette (0.00)

Hi There;
It´s my first thread, hope to be correct.

Quote:

https://wizardofodds.com/roulette

Number Placement

To the casual observer, it would appear that the numbers on the wheel are not organized and seem to be distributed randomly. The only obvious patterns are that red and black numbers alternate and that usually two odd numbers alternate with two even numbers. However the distribution of numbers was carefully arranged so that the sum of the numbers for any given section of the wheel would be roughly equal to any other section of equal size. Most numbers are part of a pair, with one number between them. These pairs add to either 37 or 39.

For what it is worth, the sum of all the numbers in roulette is 666.

I´m adding the numbers by sectors; these are the result:

Sector 0 (left)
2+14+35+23+4+16 = 94

Sector 0 (Right)
28+9+26+30+11+7 = 111

Orphan (Left)
33+21+6+18+31+19 = 128

Orphan (Right)
20+32+17+5+22+34 = 130

Sector 00 Left
8+12+29+25+10+27 = 111

Sector 00 (Right)
1+13+36+24+3+15 = 92

Overall = 94 + 111 + 128 + 130 + 111 + 92 = 666

Would you give us any inside about these discrepancies betweeen some sectors?
The right side of the roulette sums 333; the left side 333; (= 666)

Thanks
reytheking
Caracas, Venezuela.

Another interesting thing about Roulette. All the number in the "streets" on the layout all total 6 when added together, or when the number of the total is added together.

For example

1+2+3 = 6

10+11+12 = 33
3+3 = 6

28+29+30 = 87
8+7 = 15
1+5 = 6

It works diagonally as well,

eg 1.5.9 add up to 15, 1+5 =6.

That will work for any three consecutive numbers where the largest one is divisible by 3. If the largest can be expressed as 3n, then the sum will be (3n-2) + (3n-1) + 3n = 9n-3 = 3*(3n-1), which is divisible be 3.

You can tell if any number is divisible by 3 if the sum of the digits is divisible by 3. You can take the sum of the sum of the digits, for any number of sums and it still works. For example, let's look at 340987897290834. The sum of the digits is 81, and 8+1=9. 9 is divisible by 3, so 340987897290834 is divisible by 3.

I can prove this if anyone is interested. The roulette numbers is an extension of the same concept.

Quote: Wizard

That will work for any three consecutive numbers where the largest one is divisible by 3. If the largest can be expressed as 3n, then the sum will be (3n-2) + (3n-1) + 3n = 9n-3 = 3*(3n-1), which is divisible be 3.

You can tell if any number is divisible by 3 if the sum of the digits is divisible by 3. You can take the sum of the sum of the digits, for any number of sums and it still works. For example, let's look at 340987897290834. The sum of the digits is 81, and 8+1=9. 9 is divisible by 3, so 340987897290834 is divisible by 3.

I can prove this if anyone is interested. The roulette numbers is an extension of the same concept.

There is another "shortcut" I learned on this one years ago. Instead of adding all the numbers you can "cast out" those divisible by 3 then add the rest or remove more divisions 3.

For example:

340987897290834 and we remove the numbers divisible by 3 (3,6,9,0)

Gives us 48787284

4+8=12 which is divisible by 3 so we will remive those sets and get

7872

7+8=15, divisible by 3, so remove and get

72

7+2=9, divisible by 3.

Thus the number is divisible by 3.

This can be faster if you are crossing off on paper or if you want to do "smaller math" in your head instead of adding a long string.

In the lesson we were taught "shortcuts" to find out if any number is divisible by 1-10. I forget most of them (did remember this one) and I think it was only "7" that had no shortcut, though I could be wrong on that.

I think what is interesting is why it works. Although nobody asked, I'll try to explain it. Let's look at the phone number of the Mandalay Bay, as an example (7026327777).

7026327777 =
700000000 + 20000000 + 300000 + 20000 + 7000 + 700 + 70 + 7 =
7*(99999999 + 1) + 2*(999999 + 1) + 3*(9999+1) + 2*(9999+1) + 7*(999+1) + 7*(99+1) + 7*(9+1) + 7 =
[7*(99999999) + 2*(999999) + 3*(9999) + 2*(9999) + 7*(999) + 7*(99) + 7*(9)] + [ 7+2+3+2+7+7+7+7 ]

If both parts in brackets are divisiable by 3, then the sum must be divisible by 3. The first expression in brackets must be divisible by 3, because each term has a factor of a number consisting of only nines, which must be divisible by 3. So if the second expression is divisible by 3, then the whole number is. So...

7+2+3+2+7+7+7+7 = 42.
4+2=6, which is divisible by 3, thus 7026327777 is divisible by 3.

Quote: AZDuffman

In the lesson we were taught "shortcuts" to find out if any number is divisible by 1-10. I forget most of them (did remember this one) and I think it was only "7" that had no shortcut, though I could be wrong on that.

7 is a fun one. From here:

Quote:

Dividing by 7

To find out if a number is divisible by seven, take the last digit, double it, and subtract it from the rest of the number.
Example: If you had 203, you would double the last digit to get six, and subtract that from 20 to get 14. If you get an answer divisible by 7 (including zero), then the original number is divisible by seven. If you don't know the new number's divisibility, you can apply the rule again.

Not on the same "intelectual wave lenght" but still a fun one for kids is using your fingers for the 9 times table.
Extend your hands palm towards youso they are fully extended. You should be looking at 10 digits (your thumbs and fingers).
For the sum 1 x 9 simply fold your left thumb into the cleft of your hand, then count the other fingers, and hey presto the answer is 9
For the sun 2 x 9, fold the 2nd digit (your finger) into your palm then look at what you have left 1 thumb, a space and then 8 fingers, so the answer is 18
repeat as necessary.

I've always remembered this since I was a kid and I use it to impress other kids at parties. Works a treat.

madmike: i love it!

Quote: madmike

Not on the same "intelectual wave lenght" but still a fun one for kids is using your fingers for the 9 times table.
Extend your hands palm towards youso they are fully extended. You should be looking at 10 digits (your thumbs and fingers).
For the sum 1 x 9 simply fold your left thumb into the cleft of your hand, then count the other fingers, and hey presto the answer is 9
For the sun 2 x 9, fold the 2nd digit (your finger) into your palm then look at what you have left 1 thumb, a space and then 8 fingers, so the answer is 18
repeat as necessary.

I've always remembered this since I was a kid and I use it to impress other kids at parties. Works a treat.

Too bad my niece and nephew are too young for this one. Sometime I would like to know who comes up with this stuff.