3 nonrepeating roulette spins. All even money bets win once.

3 random numbers, all different, are picked between 0-36.

Scarlett wins if at least one is red
Jet wins if at least one is black
Oddette wins if at least one is odd
Evan wins if at least one is even
Magnus wins if at least one is 19-36
Minos wins if at least one is 1-18

If one of the numbers is 0 it helps nobody.

What are the odds all 6 players win?

EDIT: I completely missed the ‘different numbers’ part.
My math below allows duplicates.

——

I assume you mean for the purposes of this calculation, the ball always lands in 1-36.

Let's think about a single condition, and work backwards.

What's the odds that all three spins will be red?

18/36 * 18/36 * 18/36, or 1/2 * 1/2 * 1/2 = 1/8.

Since we don't care if it's all black or all red, the odds of all black OR all red is 1/8 * 2 or 1/4.

Therefore, it's 3/4 that there's at least one red and one black.

In other words, there's a 3/4 chance that both of your black/red people will win.

Similarly, there's a 3/4 chance that both of your even/odd people will win, and 3/4 that both high and low will win.

Therefore, it's ( 3/4 ) ^ 3 that all all six people will win.

( 3/4 ) ^ 3 = 42.1875%

Mind you, I'm no math expert, but I think that's right. Anybody got a different answer?

Assuming my logic is right, here's how to calculated it for a double zero wheel:

Start with failure: Any color, then red or green, then red or green:

38/38 * 20/38 * 20/38 = 0.277008

So one success is 1 - 0.277008 = 0.722992

So all three successes is 0.722992 ^ 3 = 0.377920 = 37.792%

FYI: Single zero: 39.9182%
Triple zero: 35.8000%

Quote:

I assume you mean for the purposes of this calculation, the ball always lands in 1-36

Nope, single zero table, 3 random unique numbers. One of them can be 0, in that case there are only 2 "useful" numbers for the players.

For example the chance of Scarlett winning as an isolated case is 1-((19/37)*(18/36)*(17/35)) or 87.529%

I've just realised that the odds of getting a green and 2 alike colours is 0.03828 (3/37)*(17/36)
The probability of getting 3 alike non greens is (34/37)*(17/36)*(16/35)=0.19836.


Adding them comes to 0.23664

(1-0.23664)^3 (not quite as colour, size and parity are not completely independent, but good enough) 0.44482 for a single zero table I think.

Actually 3 alike non greens is (34/37)*(17/35)*(16/34) or 0.21003

Adding is 0.24831

1-(0.24831)^3=0.42473


I think I've solved my own question and got it right now, but I'd be happy if any mathemagicians were to check.

OK here we go again

a green and 2 alike colours is actually 0.03938 (3/37)*(17/35)

total 0.24941

(1-0.24941)^3= 0.42287 final chance

I did a brute-force search on all 7770 sets of 3 different numbers drawn from the 37 possible (I assume it's a single-zero wheel), and found 3294 that won, which is about 42.394% of the time.

Quote: ThatDonGuy

I did a brute-force search on all 7770 sets of 3 different numbers drawn from the 37 possible (I assume it's a single-zero wheel), and found 3294 that won, which is about 42.394% of the time.
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Nice one, thanks.

Quote: ThatDonGuy

I did a brute-force search on all 7770 sets of 3 different numbers…
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First of all, 3 different numbers?
Oh. I missed that part. That changes my math a bit.

Secondly, how’d you get 7770?
37 * 36 * 35 = 46,620

Quote: DJTeddyBear

Secondly, how’d you get 7770?
37 * 36 * 35 = 46,620

6 combinations

Quote: rumba434

Quote: DJTeddyBear

Secondly, how’d you get 7770?
37 * 36 * 35 = 46,620

6 combinations
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6 combinations of WHAT?
Show the math.

I tried to figure out what you meant, but 6 ^ 5 = 7,776
So I have no clue.

123, 132, 213, 231, 312 and 321 are all the same. There are 6 ways to get the same 3 numbers, so 7770 unique sets of 3 numbers out of 46620 ways