Gambling risk:reward and winning

[MathsFormula]

Say I play the same gambling game and set the risk:reward of each game at:
1:1 (Bet A)
1:2 (Bet B)
1:3 (Bet C)

Say I place £10 on each bet.

Examples of what can happen:

If Bet A hits -£10 then Bets B and C will also hit -£10 and I'll be -£30 (loss).

If Bet A hits £10 then games B and C will continue either to a win (B £20, C £30) or a loss (B -£10, C -£10).

If Bet C hits £30 then along the way the £10 (Bet A) and £20 (Bet C) will also have been hit and gains made.

If Bet B finishes with a win (£20) then Bet A will also have ended at £10 but Bet C will continue (to either £-10 or £30)

Will always win money this way of betting ????

The possible outcomes are: BetA BetB BetC
£10 -£10 -£10 = -£10
£10 £20 -£10 = £20
£10 £20 £30 = £60
£-10 -£10 -£10 = -£30

Total = £40

Will I take the casino down if I bet 1:1, 1:2, 1:3 at the same time ? Do I have the edge?

 

[MathsFormula]

I just wanted to add the the description of the game above:

Think of a number line.

The beginning is point 0 (zero).

Point X (-£10) is one unit to the left of point 0

Point A (£10) is one unit to the right of point 0

Point B (£20) is two unit to the right of point 0

Point C (£30) is three units to the right of point 0


Assume it takes equal energy to travel between the each marker on the number line.

Thank you

 

[JeffM]

This is virtually incohereht. If I understand it, there are 8 possible outcomes if you wager 10 pounds each on A, B, and C. So in every case, you first plunk down 30 pounds.

A loses, B loses, and C loses: you are down 30.
A loses, B loses, C wins: you break even.
A loses, B wins, C loses: you are down 10.
A loses, B wins, C wins: you are up 20.
A wins, B loses, C loses: you are down 20.
A wins, B loses, C wins: you are up 10.
A wins, B wins, C loses: you break even.
A wins, B wins, C wins: you are up 30.

If the probability of winning at bets A, B, and C all equal 1/2, you will break even playing this repeatedly.

Of course, no sane casino would offer this game because a sane gambler would take only bet C.

 

[MathsFormula]

A loses, B loses, and C loses: you are down 30.

A loses, B loses, C wins: you break even.

A loses, B wins, C loses: you are down 10.

A loses, B wins, C wins: you are up 20.

A wins, B loses, C loses: you are down 20.

A wins, B loses, C wins: you are up 10.

A wins, B wins, C loses: you break even.

A wins, B wins, C wins: you are up 30.

Hello, thank you for helping me to organise my thoughts. My question probably is incoherent .... sorry. But I think you managed to decipher my ramblings.

I'll go through you list of possibilities now.

Each bet A, B, C are placed at the same time.
Maximum that can be lost is £10 per bet and winnings are £10, £20 and £30 for bets A, B, C respectively.

Please IGNORE my wager in the calculation.

Think of number line. Its just as easy to lose £10 as it is win £10 because they are equidistant from £0.
£20 is more difficult to reach on the number line and £30 is more difficult.


A loses, B loses, and C loses:
£-10, -£10, -£10 = -£30

A loses, B loses, C wins:
£-10, -£10, £30 = -£10

A loses, B wins, C loses:
£-10, £20, -£10 = £0

A loses, B wins, C wins:
£-10, £20, £30 = £40

A wins, B loses, C loses
£10, -£10, -£10 = -£20

A wins, B loses, C wins
£10, -£10, £30 = £30

A wins, B wins, C loses: you break even.
£10, £20, -£10= £20


A wins, B wins, C wins:
£10, £20, £30 = £60

£90 win???


Thanks for your help. I'm not a mathematician so just asking for help. Not a good gambler either but just trying to find an edge

 

[JeffM]

Again, I am not sure what you are trying to say, but it looks as though you are talking about some implied set of probabilities.

In a fair game, the probability of winning A would be 1/2; the probability of winning B would 1/3, and the probability of winning C would 1/4.

Why is this? Assume n is a large number.

If the probability of winning A is 1/2, then if you bet n times on A, you would win about
n / 2 times and lose about n / 2 and so break about even.

If the probability of winning B is 1/3, then you would lose about n * 2/ 3 times and win about n * 1/3 times. So you would lose twice as often as you would win. But your monetary gain on a win is twice your monetary loss when you lose so again you will approximately break even.

If the probability of winning C is 1/4, then you would lose about n * 3/ 4 times and win about n * 1/4 times. So you would lose thrice as often as you would win. But your monetary gain on a win is thrice your monetary loss when you lose so again you will approximately break even.

In fact, we have just defined a fair game. If the odds were different, one person or the other is virtually guaranteed to bankrupt the other in the end.