Is there a sure way to win? (Bet on given odds)

gambling

You have 25 numbers on a wheel.

  • 12 are number 1 : Doubles your bet. (x2)
  • 6 are number 3 : Triples your bet. (x3)
  • 4 are number 5 : Quintuples your bet. (x5)
  • 2 are number 10 : Decuples your bet. (x10)
  • 1 is number 20 : (x20) what is the name for x20?

You can bet on multiple numbers at a time and the amount you want.

For example :
You could bet 5 on 1, 5 on 3, 10 on 5, 0 on 10 and 6 on 20.

How can you maximise the amount of money you make ? Or minimise the amount you lose?

Thanks !

EDIT : One number only will win.

EDIT2 : I am aware of the martingale betting strategy, where I could put an amount of money on a number and double everytime I lose until I win, but I'm looking for a better strategy.

EDIT 3 : Ok so the answer I received made me realise some things,
which I tried.

Lets say I bet like this :

- 200 on 1 : (Win) 400, (Lose) -200, (Total Cost) 400, (Balance on win) : 0.
- 80 on 3 : (Win) 240, (Lose) -80, (Total Cost) 400, (Balance on win) : 240.
- 60 on 5 : (Win) 300, (Lose) -60, (Total Cost) 400, (Balance on win) : 300.
- 40 on 10 : (Win) 400, (Lose) 40, (Total Cost) 400, (Balance on win) : 400.
- 20 on 20 : (Win) 400, (Lose) 20, (Total Cost) 400, (Balance on win) : 400.

If this is correct, I would never lose. But is this correct? If it
is, is there a way to maximise this?

Best Answer

Compute the expected value of betting $1$ on each number. If you bet on $1$ you have $\frac {12}{25}$ chance of getting $2$, for an expected return of $2\cdot \frac {12}{25}=\frac {24}{25} \lt 1$. Similarly for each of the other numbers, the expected return is $\frac {18}{25}$ or $\frac {20}{25}$. It is a losing bet, so the way to minimize your losses is not to play.

In your system, you bet $400$. That should be the cost you deduct from the winnings for any number. You break even if $1,10$ or $20$ come up and lose if $3$ or $5$ comes up. Your nets should be $0,-160,-100,0,0$. You can never win this way and may lose.

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