Roulette odds question

Factual Questions
1

In American style roulette, what configuration of bets will give you the best odds of making a profit on a single spin given that:

  • You cannot bet more than $60.
  • The table have 0 and 00.
  • $10 minimum on inside (the numbers) and outside bets.

For example, if you put $10 on 1st 12 and $10 on 2nd 12, you have a ~63% chance of hitting a good number and making $10 on a single spin. Is it possible to get odds higher than that? I’m only considering one spin, I’m not interested in expected value which of course is always negetive.

2

Oops, I forgot one important piece - you can’t bet on the numbers.

3

I actually don’t think it makes any difference. The expected value is always the same on the outside bets.

How about $10 on 1-12 and $10 on Column 1?

Expected profit = (4/38 x $40) + (16/38 x $10) + (18/38 x -$20) = -$1.053

Compare that to your suggestion, which gives (24/38 x $10) + (14/38 x -$20) = -$1.053

Or $20 on red = (18/38 x $20) + (20/38 x -$20) = -$1.053.

Look here. The expected value for any bet (except the 0-3 five number bet which is marginally worse) is the same. Any combination will lose the same amount.

4

OK so on second glance that’s not what you’re asking. :smack:
If you’re interested in maximising the chance of making any profit, then yes, covering two thirds or two columns would be the best bet.

My first example gives you a 52.6% chance of making any profit, but there is a 10.5% chance of trebling your money.

Your example, as you say, gives a 63.2% chance of making 50% profit. I can’t see how you can beat that.

How about combining a column and a colour?
e.g. £10 on 1-12 and $10 on red.
You now have a 63.2% chance of not losing, with a 15.8% chance of making 150% profit, 15.8% chance of making 50% profit, and 31.6% chance of breaking even.

5

This is the kind of thing I’m looking for. Any other interesting combos?

6

OK, how about two columns and one 12?

You’re now covering 73.7% of the numbers (28/38)
But the two column bets can’t both win, so you have:

20/38 (52.6%) = break even (either of the two columns you covered but outside your 12, or the third column in the 12 you covered)
10/38 (26.3%) = lose 100%
8/38 (21.1%) = win 100%

Or you could cover all except one ninth of the board, with two columns and two 12s. That is pretty stupid though, as although you have little chance of losing the lot, you are still more likely to lose than win:

6/38 (15.8%) = lose 100%
16/38 (42.1%) = win 50% (stake = $40, return = $60 from two winning bets)
16/38 (42.1%) = lose 25% (stake $40, return = $30 from one winning bet)

So, cover two columns or two 12s. Or stick the lot on Chelsea to win the Premier league for a 70% profit come May…