Gambling risk:reward and winning

MathsFormula

Junior Member
Joined
Jul 13, 2014
Messages
90
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

Junior Member
Joined
Jul 13, 2014
Messages
90
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

Elite Member
Joined
Sep 14, 2012
Messages
3,935
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.
 
Last edited:

MathsFormula

Junior Member
Joined
Jul 13, 2014
Messages
90
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

Elite Member
Joined
Sep 14, 2012
Messages
3,935
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.
 
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