Q:

Simulate a double chance bet with two single bets

If you bet on the result of a soccer match, you usually have three possibilities. You bet

1 - if you think the home team will win
X - if you think the match ends in a draw
2 - if you think the away team will win

Lets say we have the following soccer match with the following betting quotes:
Kansas City vs. Portland - 1 = 1.92, X = 3.57, 2 = 5.00
This means: If you think Portland (In the opinion of the bookie, the underdog) will win the match, you bet on 2
Example (I bet \$100): In case Portland wins I win \$400
$$100*5.00-100 = 400$$
$$stake*quote-stake = net win$$
(When Portland loses the match, or it ends in a draw, I'll lose my stake)
Now, some bookies offer a so-called double chance bet. This kind of bet takes one possibility out. That leaves you to following bets. You bet

1/X - if you think the home team will win or the match ends in a draw
X/2 - if you think the away team will win or the match ends in a draw

This variant is perfect if you think Portland will win the match, or at least it will end up in a draw. To calculate this quotes I use the following formula: (Q1 = 1st quote 1, Q2 = 2nd quote)
$$ 1/(1/Q1+1/Q2) $$
For the 1/X bet
$$ 1/(1/1.92+1/3.57) = 1.25 $$
For the X/2 bet
$$ 1/(1/3.57+1/5) = 2.08 $$
Now comes my math problem: When the bookie does not offer a double chance bet, I want to create it my self: With two single bets. For the Kansas City vs. Portland bet I'd like to place a X/2 bet. The quote for the bet is as I showed before 2.08. I want to place \$100 on it. When I win the bet, I'll get \$108 net win:
$$100*2.08-100 = 108$$
How do I have to split the money on two (X and 2) single bets, to win \$108, when Portland wins or the match ends in a draw?
I got to the solution for this case by trying out. But with the result in my hand, I still don't get the formula to calculate it.
I bet \$58.35 on X and \$41.65 on 2
$$ 58.35*3.57-58.53-41.65 ≈ 108$$
and
$$ 41.65*5.00-41.65-58.53 ≈ 108$$
Notice the last subtraction. You have to subtract the stake of the other bet. Because when Portland wins, I win only the 2 bet and lose the stake for the X bet.

A:

Let's examine the X/2 case. Denote by $Q_X=3.57$ the bookie's quote for X and $Q_2=5.00$ for 2. Denote by $Q=1/(1/Q_X+1/Q_2)=(Q_X Q_2)/(Q_X+Q_2)\approx 2.0828$ the quote you have calculated for the X/2 bet. You want to split the total bet $B=\$100$ into two bets $B_X$ (for X) and $B_2$ (for 2) so that $B_X Q_X=BQ=B_2 Q_2$. From the first equation you get $B_X=B(Q/Q_X)$. Similarly from the second equation you get $B_2=B(Q/Q_2)$, or alternatively $B_2=B-B_X$ (as the value of $B_1$ is already known). Let's substitute the values:
$$B_X=100(2.0828/3.57)\approx 58.34\quad\text{and}\quad B_2=100-58.34=41.66.$$
In fact, you can do this more easily without calculating $Q$ at all, since
$$B_X=B\frac{Q}{Q_X}=B\frac{Q_2}{Q_X+Q_2}.$$

