I know I must be missing something very rudimentary but I can't find it.
On a roulette there are:
- 18 black values
- 18 red values
- 2 values
Betting on either red or black yields a winning probability of 18/38 (obvious advantage to the house).
The odds on this kind of bet are 1:1 (You win a matching sum to what you bet, or you lose your bet).
Another type of bet, is betting on a specific number. This yields a winning probability of 1/38 (even more obvious advantage to the house). The odds on this kind of bet are 35:1 (if you win you get your dollar plus 35 if you lose, the dollar is gone.).
So far so good.
If I bet $1 on 9 black numbers (35:1 odds) my chance of wining is 9/38 (9 outcomes will be a win). I will At the worst case I will lose all my bets; best case, I will lose 8 dollars, and win 35 (if one of my numbers won). So I have a 9/38 chance of being profitable.
Now if in addition to betting on 9 black numbers, I also bet $10 on red to win, to me it seems like I would have a 18/38 chance of having my color win and an additional 9/38 chance of having my number win (of opposite color).
Therefore, I have a 27/38 (or just over 70%) chance to win (come out profitable) every turn?
- I have a 30% chance of losing 19 dollars
- 23% chance of having one of my numbers win: if I win a number I lose 18 dollars and get an additional 35 dollars
- 47% chance of having my color win: If I win the color, I lose 9 dollars and win an additional 10 dollars
Could this be? Surely not! What am I missing?