# 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. - ## 4 Answers 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 $$B_X=B\frac{Q}{Q_X}=B\frac{Q_2}{Q_X+Q_2}.$$

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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.

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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.

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If you think it isn't a tie with these odds, put 72.2% on 1 and 27.7% on 2 for +38.7% on a win. Remember that you never win a double-chance if $\frac1{1/Q_1+1/Q_2}<=1$. –  user474632 Aug 17 '11 at 23:50

You bet $\$100Q2 / (Q1 + Q2)$on X and$\$100$ $Q1 / (Q1 + Q2)$ on 2. To the nearest cent, these work out at $\$58.34$and$\$41.66$.

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