Question

Practical Probability Implementation Question:

So you know those roulette wheels at casinos where there's 18 red spaces, and a total of 38 spaces. By the law of large numbers, you should be losing money in the long-run. But......

What if you multiplied your lost bet by a value of a power of 2:

I bet 'x' dollars on a red space, if I win, I win. If I lose, I lose 'x' dollars, and so my next bet would be 'x*(2^1)' dollars. If I win, I win 'x' dollars in total, If I lose, I lose 3x dollars. My next bet would then be x*(2^2) or 4x dollars. etc...

Guarantee win of 'x' dollars for each round? :)

Answer 1

Yes, your scheme works.....but
but not practical for two reasons.

1. Say a player bets for $100 at the first bet.
Using the strategy, and he loses 10 times in a row, he would be betting 100*2^10=$102,400 hoping to win back $100. Not everyone has that much ready cash.
2. All casinos have limits for bets, varying between 1000 and 10000. Some smaller casinos have \(payout\) limits of 1750, which really limits your winnings considerably. For inside bets, (those on particular numbers) are limited to between 25 to 1000.

Ref. http://www.casinonewsdaily.com/roulette-guide/roulette-betting-limitations

Answer 2

Erghhhh, Just when I thought I didn't have to get a job.

Answer 3

@mhchen
Next time you have a new scheme, pm me first. If you don't see me at OS anymore, your scheme works! lol

Answer 4

Your scheme does not work. Such things are designed to increase THEIR cash flow, not yours. You are probably also overlooking the Green spaces and you have not defined the amount you would win.