Outside bet

OK, here's a question I should know but somebody is challenging me on this.....

Roulette: Inside bet minimum = $5. Outside bet minimum = $5.

Fully understand the inside bet where the money can placed on any of the available combinations, straight up, squares, splits, etc. Just so long as at least $5 is wagered "inside".

Now, here's the argument....do the same rules apply for the "outside", specifically, a player can put any amount on any outside spot so long as the total wager by the player is at least $5? For example, $2 on "odd" and $2 on "red" and $1 on "1st 12".

Or.....does it mean that "any" wager on the outside must be for the minimum of $5? For example, if you want to bet "red", it must be for at least $5. And, if you wanted to bet both the red and the odd, the player would have $10 in play?

Thanks

Normally inside bets are aggregate, outside bets are individual. You can't bet red and odd for $2 and $3 respectively at a $5 table.

Quote: Riva

Or.....does it mean that "any" wager on the outside must be for the minimum of $5?

Usually that is it. Outside bets must actually be for at least the minimum, rather than aggregate to the minimum.

Quote: MathExtremist

Normally inside bets are aggregate, outside bets are individual. You can't bet red and odd for $2 and $3 respectively at a $5 table.

Emphasis on, "Normally," Wheeling Island (for at least one) used to let you do it as long as the total bet was at least $5 on the outside. I don't know if they do anymore, or not, obviously, I don't play Roulette anymore and haven't for some time...with exception to Binion's non-negotiable on a single number when I was in Vegas. Lost, of course!