Perhaps I should have phrased my question differently UK

Why do EC bettors get half their stake back when they didn't bet zero ?

Questions I have but answers there come none.

Is choosing 1 from each of the 3 EC truly a 7/1 chance - ignoring the zero ?

1. My advice would be to engage a medium and see if you can make contact with one of the Blanc brothers and ask if they're aware of when the rule was first introduced and by whom.

2. Don't know where you get 7-1 from? Let's take 1-18, red, odd as an example bet combination:

First covers 18 numbers (1-18)

Second covers an additional 9 numbers (19,21,23,25,27,30,32,34,36)

Third covers an additional 4 numbers (29,31,33,35)

Total numbers covered = 31

So (I think I've got this right, but I'm sure someone will correct me if not):

If you're lucky and either of 1,3,5,7,9 red pop up you've won on all three bets (prob=5/37)

If any numbers that are 19-36/red/odd (x5), black/1-18/odd (x4) or even/1-18/red (x4) show, you win two and lose one (13/37)

If any numbers that are 19-36/black/odd (x4), 19-36/red/even (x4) or 1-18/black/even (x5) number shows you win one and lose two.(13/37)

If an even black number greater than 18 shows you lose all three. (5/37)

If a zero pops up, you lose either all or half your wagers (prob 1/37)

At best you may win all three bets made which represents a 1:1 payoff on all wagers. There's a 18/37 chance of coming out in front (winning two or three out of three) and a 19/37 of losing money on each spin (winning only one, or none out of three or the zero being spun).

To begin with I struggled to understand what "EC" referred to, but I've now worked it out as "evens-chance" bets (although of course they're 18/37 if you want to be pedantic about it).