I need a formula, if one exists.

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SUMMARY

This discussion focuses on calculating optimal betting amounts for two different bets to achieve a desired return using specific odds. The user seeks a formula to determine the amounts to bet based on the odds and the desired return. The derived formulas are X = G*(1+Ny)/(Nx*Ny-1) and Y = G*(1+Nx)/(Nx*Ny-1), where G is the desired return, Nx and Ny are the net winnings from each bet. The example provided confirms that betting $40 on odds of 5.50 and $80 on odds of 2.75 results in a net gain of $100 if either bet wins.

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  • Understanding of basic algebra and equations
  • Familiarity with betting odds and payouts
  • Knowledge of net winnings calculation
  • Ability to manipulate formulas for practical applications
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  • Research advanced betting strategies and risk management techniques
  • Learn about probability theory as it applies to betting scenarios
  • Explore tools for calculating expected value in betting
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This discussion is beneficial for gamblers, betting strategists, and anyone interested in optimizing their betting returns through mathematical formulas and calculations.

Tups
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Hello, I am not a mathematician and therefore unsure about what or how i should be asking to make it clear about what i am after.
So i think just showing a practical example is best.

If you have two different options for a bet and both are paying over $2 then it is possible to place a certain amount of money on both bets that will not only pay you for your desired return but also pays for the money invested on the losing bet.

E.g

Desired Return = $100
Bet 1 Odds = $5.50
Bet 2 odds = $2.75

If i put on $40 @ $5.50 it’ll return $220,
And if i put $80 @ $2.75 it’ll return $220.

$220 – $120 (both bet amounts) = $100 my desired return.
Just as long as one of these two bets wins.

So what i'd like to know is if there is a formula that could be used to find out the amount to put on the two bets when only the odds and desired return are known?

The trick obviously is that both unknown figures are reliant on knowing what the other one is.

Many thanks.
 
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When you speak of "odds" you are giving the winning payout (presumably on a $1 bet)? I ask because there is also an Expected payout, i.e. winnings times chance of winning - cost of the bet. I'll assume you are giving the winning amounts which seems to fit your reasoning.

Try a little Algebra: Bet X dollars on the first and Y dollars on the 2nd. Call Px the payoff for the first and Py the payoff for the 2nd.
Outcomes:

Lose both: Net = -(X+Y).
Win 1st Lose 2nd: Net= X*Px-X-Y = (Px-1)X - Y
Win 2nd Lose 1st: Net= Y*Py-X-Y = X + (Py-1)Y
Win both: Net = X(Px-1)+Y(Py-1).

It is easier to work with the Net payoffs for each bet. Nx = Px - 1, Ny = Py -1.

Winning 1 bet yields your Goal amount (G=$100 in your example):
G = Nx *X - Y
G = Ny* Y - X
Solve this system for X and Y:
X = Ny*Y - G --> G = Nx * ( Ny * Y - G) - Y
(Nx Ny - 1)Y = (1+Nx)G
Y = G * (1+Nx)/(Nx*Ny-1)

X = G [Ny*(1+Nx)/(Nx*Ny-1) - 1] = G(Ny + Nx Ny - Nx Ny + 1)/(Nx Ny - 1)
Check!

Given your goal amount G, and the Net winnings on each bet Nx and Ny you would bet the amounts:
X = G*(1+Ny)/(Nx*Ny-1)
and
Y = G * (1+Nx)/(Nx*Ny-1)

And If you win one of the two you will net G dollars.

With your example: Nx = $5.50 - $1 = 4.5, Ny = $2.75 - $1 = 1.75, and G = $100.
X = 100(1+1.75)/(4.50 * 1.75 - 1) = 275/6.875 = $40.
Y = 100(1+4.5)/(4.5*1.75-1) = 550/6.875 = $80.

Don't loose too much money!
 
Thanks Jambaugh. That is very good of you.
And the formula work a treat also.

I'll look forward to going through your workings tomorrow (one too many drinks tonight) and see where i went wrong.

Thanks again and wish you all the best.
 

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