# Expected Loss for American Roulette

• Feb 5th 2013, 05:33 PM
Shapeshift
Expected Loss for American Roulette
Hey everyone! So I'm getting a bit stuck when it comes to calculating expected loss for American Roulette (38 spots: 18 are red (odd numbers between 1 and 35), 18 are black (even numbers between 1 and 36), and 2 spots are green (one is 0 and the other is 00)).

For example, I understand that If I bet on red, my expected loss is (18/38 - 20/38) = -1/19 = -0.0526

However, I am confused when it comes to betting on a colour as well as numbers. For example, if I bet on red and the numbers 1, 2, 3, and 4, how would I go about calculating the expected loss then?

I figure it should be (18/38 - 20/38 + 20/38 - 18/38), but this would give 0, so this seems wrong.

Any help would be very much appreciated!
• Feb 5th 2013, 11:49 PM
chiro
Re: Expected Loss for American Roulette
Hey Shapeshift.

Can you outline how roulette pays out (for people like me not familiar with the game)?
• Feb 8th 2013, 11:36 AM
Shapeshift
Re: Expected Loss for American Roulette
Hey sorry I forgot to include that!
The payoff for red is 1 to 1, so a \$1 bet would have a net gain of \$1.
The payoff for the block of four numbers is 8 to 1, so a \$1 bet would have a net gain of \$8.

So if I place a \$1 bet on the red, and a \$1 bet on the block of four numbers, I end up getting

18/38 + 8(4/38) - 20/38 - 34/38 = -0.105263 which I believe is right?
• Feb 8th 2013, 05:49 PM
chiro
Re: Expected Loss for American Roulette
That looks right.