TITLE: Simulate a double chance bet with two single bets
QUESTION [5 upvotes]: 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.

REPLY [2 votes]: The defining feature of a double chance bet is that, if either of the events you bet on happens, you win the same amount regardless of which event it was.
To simulate a double chance bet with single bets, you need to divide the stake so that the same will happen.
So, let $Q_1$ and $Q_2$ be the quotes offered for the two events.  We seek a value $0 \le \alpha \le 1$ such that, if we bet a fraction $\alpha$ of our total stake on event 1 and the rest on event 2, the payout in either case will be the same, i.e.
$$\alpha Q_1 = (1 - \alpha) Q_2.$$
To solve this, expand the right hand side, collect the $\alpha$ terms together on one side and divide to get
$$\alpha = \frac{Q_2}{Q_1 + Q_2}.$$
Then, to ensure equal payout in either case, you should bet $\alpha$ times you total stake on event 1, and the rest on event 2.

REPLY [2 votes]: Planning on betting that either team wins?
$S_1*Q_1-S_1-S_2=S_2*Q_2-S_1-S_2$
$S_1=S_2*\frac{Q_2}{Q_1}, S_2=S_1*\frac{Q_1}{Q_2}$
$S_1+S_2=1$ (to find $S_1$ and $S_2$ as percentages)
$S_1+S_1\frac{Q_1}{Q_2}=1, S_2+S_2\frac{Q_2}{Q_1}$
$S_1=\frac1{Q_2/Q_1+1}, S_2=\frac1{Q_1/Q_2+1}$
For X/2, S_1 = 58.3% just as above.